context
stringlengths 545
71.9k
| questionsrc
stringlengths 16
10.2k
| question
stringlengths 11
563
|
|---|---|---|
welcome to the presentation on averages . averages is probably a concept that you 've already used before , maybe not in a mathematical way . but people will talk in terms of , the average voter wants a politician to do this , or the average student in a class wants to get out early . so you 're probably already familiar with the concept of an average . and you probably already intuitively knew that an average is just a number that represents the different values that a group could have . but it can represent that as one number as opposed to giving all the different values . and let 's give a couple of examples of how to compute an average , and you might already know how to do this . so let 's say i had the numbers 1 , 3 , 5 , and 20 . and i asked you , what is the average of these four numbers ? well , what we do is , we literally just add up the numbers . and then divide by the number of numbers we have . so we say 1 plus 3 is 4 . so let me write that . 1 plus 3 plus 5 plus 20 equals , let 's see , 1 plus 3 is 4 . 4 plus 5 is 9 . 9 plus 20 is 29 . and we had 4 numbers ; one , two , three , four . so 4 goes into 29 . and it goes , 7 , 7 , 28 . and then we have 10 , i did n't have to do that decimal there , oh well . 2 , 8 , 25 . so 4 goes into 29 7.25 times . so the average of these four numbers is equal to 7.25 . and that might make sense to you because 7.25 is someplace in between these numbers . and we can kind of view this , 7.25 , as one way to represent these four numbers without having to list these four numbers . there are other representations you 'll learn later on . like the mode . you 'll also the mean , which we 'll talk about later , is actually the same thing as the average . but the average is just one number that you can use to represent a set of numbers . so let 's do some problems which i think are going to be close to your heart . let 's say on the first four tests of an exam , i got a -- let 's see , i got an 80 , an 81 . an 87 , and an 88 . what 's my average in the class so far ? well , all i have to do is add up these four numbers . so i say , 80 plus 81 plus 87 plus 88 . well , zero plus 1 is 1 . 1 plus 7 is 8 . 8 plus 8 is 16 . i just ran eight miles , so i 'm a bit tired . and , 4/8 , so that 's 32 . plus 1 is 33 . and now we divide this number by 4 . 4 goes into 336 . goes into 33 , 8 times . 8 times 4 is 32 . 33 minus 32 is 1 , 16 . 4 . so the average is equal to 84 . so depending on what school you go to that 's either a b or a c. so , so far my average after the first four exams is an 84 . now let 's make this a little bit more difficult . we know that the average after four exams , at four exams , is equal to 84 . if i were to ask you what do i have to get on the next test to average an 88 , to average an 88 in the class . so let 's say that x is what i get on the next test . so now what we can say is , is that the first four exams , i could either list out the first four exams that i took . or i already know what the average is . so i know the sum of the first four exams is going to 4 times 84 . and now i want to add the , what i get on the 5th exam , x . and i 'm going to divide that by all five exams . so in other words , this number is the average of my first five exams . we just figured out the average of the first four exams . but now , we sum up the first four exams here . we add what i got on the fifth exam , and then we divide it by 5 , because now we 're averaging five exams . and i said that i need to get in an 88 in the class . and now we solve for x . let me make some space here . so , 5 times 88 is , let 's see . 5 times 80 is 400 , so it 's 440 . 440 equals 4 times 84 , we just saw that , is 320 plus 16 is 336 . 336 plus x is equal to 440 . well , it turns out if you subtract 336 from both sides , you get x is equal to 104 . so unless you have a exam that has some bonus problems on it , it 's probably impossible for you to get ah an 88 average in the class after just the next exam . you 'd have to get 104 on that next exam . and let 's just look at what we just did . we said , after 4 exams we had an 84 . what do i have to get on that next exam to average an 88 in the class after 5 exams ? and that 's what we solved for when we got x . now , let 's ask another question . i said after four exams , after four exams , i had an 84 average . if i said that there are 6 exams in the class , and the highest score i could get on an exam is 100 , what is the highest average i can finish in the class if i were to really study hard and get 100 on the next 2 exams ? well , once again , what we 'll want to do is assume we get 100 on the next 2 exams and then take the average . so we 'll have to solve all 6 exams . so we 're going to have the average of 6 , so in the denominator we 're going to have 6 . the first four exams , the sum , as we already learned , is 4 exams times the 84 average . and this dot is just times . plus , and there 's going to be 2 more exams , right ? because there 's 6 exams in the class . and i 'm going to get 100 in each . so that 's 200 . and what 's this average ? well , 4 times 84 , we already said , is 336 . plus 200 over 6 . so that 's 536 over 6 . 6 goes into 5 36 . i do n't know if if i gave myself enough space . but 6 goes into 53 , 8 times . 48 . 56 . 9 times . 9 times 6 is 54 . 6 minus is 20 6 goes into -- so we 'll see it 's actually 89.333333 , goes on forever . so 89.3 repeating . so no matter how hard i try in this class , the best i can do . because i only have two exams left , even if i were to get 100 on the next two exams . i can finish the class with an 89.333 average . hopefully , i think some of this might have been a little bit of a review for you . you already had kind of a sense of what an average is . and hopefully these last two problems not only taught you how to do some algebra problems involving average , but they 'll also help you figure out how well you have to do on your exams to get an a in your math class . i think you 're now ready for the average module . have fun .
|
like the mode . you 'll also the mean , which we 'll talk about later , is actually the same thing as the average . but the average is just one number that you can use to represent a set of numbers .
|
how changing the value effects the mean and median ?
|
welcome to the presentation on averages . averages is probably a concept that you 've already used before , maybe not in a mathematical way . but people will talk in terms of , the average voter wants a politician to do this , or the average student in a class wants to get out early . so you 're probably already familiar with the concept of an average . and you probably already intuitively knew that an average is just a number that represents the different values that a group could have . but it can represent that as one number as opposed to giving all the different values . and let 's give a couple of examples of how to compute an average , and you might already know how to do this . so let 's say i had the numbers 1 , 3 , 5 , and 20 . and i asked you , what is the average of these four numbers ? well , what we do is , we literally just add up the numbers . and then divide by the number of numbers we have . so we say 1 plus 3 is 4 . so let me write that . 1 plus 3 plus 5 plus 20 equals , let 's see , 1 plus 3 is 4 . 4 plus 5 is 9 . 9 plus 20 is 29 . and we had 4 numbers ; one , two , three , four . so 4 goes into 29 . and it goes , 7 , 7 , 28 . and then we have 10 , i did n't have to do that decimal there , oh well . 2 , 8 , 25 . so 4 goes into 29 7.25 times . so the average of these four numbers is equal to 7.25 . and that might make sense to you because 7.25 is someplace in between these numbers . and we can kind of view this , 7.25 , as one way to represent these four numbers without having to list these four numbers . there are other representations you 'll learn later on . like the mode . you 'll also the mean , which we 'll talk about later , is actually the same thing as the average . but the average is just one number that you can use to represent a set of numbers . so let 's do some problems which i think are going to be close to your heart . let 's say on the first four tests of an exam , i got a -- let 's see , i got an 80 , an 81 . an 87 , and an 88 . what 's my average in the class so far ? well , all i have to do is add up these four numbers . so i say , 80 plus 81 plus 87 plus 88 . well , zero plus 1 is 1 . 1 plus 7 is 8 . 8 plus 8 is 16 . i just ran eight miles , so i 'm a bit tired . and , 4/8 , so that 's 32 . plus 1 is 33 . and now we divide this number by 4 . 4 goes into 336 . goes into 33 , 8 times . 8 times 4 is 32 . 33 minus 32 is 1 , 16 . 4 . so the average is equal to 84 . so depending on what school you go to that 's either a b or a c. so , so far my average after the first four exams is an 84 . now let 's make this a little bit more difficult . we know that the average after four exams , at four exams , is equal to 84 . if i were to ask you what do i have to get on the next test to average an 88 , to average an 88 in the class . so let 's say that x is what i get on the next test . so now what we can say is , is that the first four exams , i could either list out the first four exams that i took . or i already know what the average is . so i know the sum of the first four exams is going to 4 times 84 . and now i want to add the , what i get on the 5th exam , x . and i 'm going to divide that by all five exams . so in other words , this number is the average of my first five exams . we just figured out the average of the first four exams . but now , we sum up the first four exams here . we add what i got on the fifth exam , and then we divide it by 5 , because now we 're averaging five exams . and i said that i need to get in an 88 in the class . and now we solve for x . let me make some space here . so , 5 times 88 is , let 's see . 5 times 80 is 400 , so it 's 440 . 440 equals 4 times 84 , we just saw that , is 320 plus 16 is 336 . 336 plus x is equal to 440 . well , it turns out if you subtract 336 from both sides , you get x is equal to 104 . so unless you have a exam that has some bonus problems on it , it 's probably impossible for you to get ah an 88 average in the class after just the next exam . you 'd have to get 104 on that next exam . and let 's just look at what we just did . we said , after 4 exams we had an 84 . what do i have to get on that next exam to average an 88 in the class after 5 exams ? and that 's what we solved for when we got x . now , let 's ask another question . i said after four exams , after four exams , i had an 84 average . if i said that there are 6 exams in the class , and the highest score i could get on an exam is 100 , what is the highest average i can finish in the class if i were to really study hard and get 100 on the next 2 exams ? well , once again , what we 'll want to do is assume we get 100 on the next 2 exams and then take the average . so we 'll have to solve all 6 exams . so we 're going to have the average of 6 , so in the denominator we 're going to have 6 . the first four exams , the sum , as we already learned , is 4 exams times the 84 average . and this dot is just times . plus , and there 's going to be 2 more exams , right ? because there 's 6 exams in the class . and i 'm going to get 100 in each . so that 's 200 . and what 's this average ? well , 4 times 84 , we already said , is 336 . plus 200 over 6 . so that 's 536 over 6 . 6 goes into 5 36 . i do n't know if if i gave myself enough space . but 6 goes into 53 , 8 times . 48 . 56 . 9 times . 9 times 6 is 54 . 6 minus is 20 6 goes into -- so we 'll see it 's actually 89.333333 , goes on forever . so 89.3 repeating . so no matter how hard i try in this class , the best i can do . because i only have two exams left , even if i were to get 100 on the next two exams . i can finish the class with an 89.333 average . hopefully , i think some of this might have been a little bit of a review for you . you already had kind of a sense of what an average is . and hopefully these last two problems not only taught you how to do some algebra problems involving average , but they 'll also help you figure out how well you have to do on your exams to get an a in your math class . i think you 're now ready for the average module . have fun .
|
but 6 goes into 53 , 8 times . 48 . 56 .
|
32,44,45,49,56 , what will happen if 45 becomes 48 ?
|
welcome to the presentation on averages . averages is probably a concept that you 've already used before , maybe not in a mathematical way . but people will talk in terms of , the average voter wants a politician to do this , or the average student in a class wants to get out early . so you 're probably already familiar with the concept of an average . and you probably already intuitively knew that an average is just a number that represents the different values that a group could have . but it can represent that as one number as opposed to giving all the different values . and let 's give a couple of examples of how to compute an average , and you might already know how to do this . so let 's say i had the numbers 1 , 3 , 5 , and 20 . and i asked you , what is the average of these four numbers ? well , what we do is , we literally just add up the numbers . and then divide by the number of numbers we have . so we say 1 plus 3 is 4 . so let me write that . 1 plus 3 plus 5 plus 20 equals , let 's see , 1 plus 3 is 4 . 4 plus 5 is 9 . 9 plus 20 is 29 . and we had 4 numbers ; one , two , three , four . so 4 goes into 29 . and it goes , 7 , 7 , 28 . and then we have 10 , i did n't have to do that decimal there , oh well . 2 , 8 , 25 . so 4 goes into 29 7.25 times . so the average of these four numbers is equal to 7.25 . and that might make sense to you because 7.25 is someplace in between these numbers . and we can kind of view this , 7.25 , as one way to represent these four numbers without having to list these four numbers . there are other representations you 'll learn later on . like the mode . you 'll also the mean , which we 'll talk about later , is actually the same thing as the average . but the average is just one number that you can use to represent a set of numbers . so let 's do some problems which i think are going to be close to your heart . let 's say on the first four tests of an exam , i got a -- let 's see , i got an 80 , an 81 . an 87 , and an 88 . what 's my average in the class so far ? well , all i have to do is add up these four numbers . so i say , 80 plus 81 plus 87 plus 88 . well , zero plus 1 is 1 . 1 plus 7 is 8 . 8 plus 8 is 16 . i just ran eight miles , so i 'm a bit tired . and , 4/8 , so that 's 32 . plus 1 is 33 . and now we divide this number by 4 . 4 goes into 336 . goes into 33 , 8 times . 8 times 4 is 32 . 33 minus 32 is 1 , 16 . 4 . so the average is equal to 84 . so depending on what school you go to that 's either a b or a c. so , so far my average after the first four exams is an 84 . now let 's make this a little bit more difficult . we know that the average after four exams , at four exams , is equal to 84 . if i were to ask you what do i have to get on the next test to average an 88 , to average an 88 in the class . so let 's say that x is what i get on the next test . so now what we can say is , is that the first four exams , i could either list out the first four exams that i took . or i already know what the average is . so i know the sum of the first four exams is going to 4 times 84 . and now i want to add the , what i get on the 5th exam , x . and i 'm going to divide that by all five exams . so in other words , this number is the average of my first five exams . we just figured out the average of the first four exams . but now , we sum up the first four exams here . we add what i got on the fifth exam , and then we divide it by 5 , because now we 're averaging five exams . and i said that i need to get in an 88 in the class . and now we solve for x . let me make some space here . so , 5 times 88 is , let 's see . 5 times 80 is 400 , so it 's 440 . 440 equals 4 times 84 , we just saw that , is 320 plus 16 is 336 . 336 plus x is equal to 440 . well , it turns out if you subtract 336 from both sides , you get x is equal to 104 . so unless you have a exam that has some bonus problems on it , it 's probably impossible for you to get ah an 88 average in the class after just the next exam . you 'd have to get 104 on that next exam . and let 's just look at what we just did . we said , after 4 exams we had an 84 . what do i have to get on that next exam to average an 88 in the class after 5 exams ? and that 's what we solved for when we got x . now , let 's ask another question . i said after four exams , after four exams , i had an 84 average . if i said that there are 6 exams in the class , and the highest score i could get on an exam is 100 , what is the highest average i can finish in the class if i were to really study hard and get 100 on the next 2 exams ? well , once again , what we 'll want to do is assume we get 100 on the next 2 exams and then take the average . so we 'll have to solve all 6 exams . so we 're going to have the average of 6 , so in the denominator we 're going to have 6 . the first four exams , the sum , as we already learned , is 4 exams times the 84 average . and this dot is just times . plus , and there 's going to be 2 more exams , right ? because there 's 6 exams in the class . and i 'm going to get 100 in each . so that 's 200 . and what 's this average ? well , 4 times 84 , we already said , is 336 . plus 200 over 6 . so that 's 536 over 6 . 6 goes into 5 36 . i do n't know if if i gave myself enough space . but 6 goes into 53 , 8 times . 48 . 56 . 9 times . 9 times 6 is 54 . 6 minus is 20 6 goes into -- so we 'll see it 's actually 89.333333 , goes on forever . so 89.3 repeating . so no matter how hard i try in this class , the best i can do . because i only have two exams left , even if i were to get 100 on the next two exams . i can finish the class with an 89.333 average . hopefully , i think some of this might have been a little bit of a review for you . you already had kind of a sense of what an average is . and hopefully these last two problems not only taught you how to do some algebra problems involving average , but they 'll also help you figure out how well you have to do on your exams to get an a in your math class . i think you 're now ready for the average module . have fun .
|
so 4 goes into 29 . and it goes , 7 , 7 , 28 . and then we have 10 , i did n't have to do that decimal there , oh well .
|
if you have a average rating of 2.7 and you want to determine the total rating and there were 100 people who gave ratings that would just be a rate problem involving average so it would be 2.7 rating/ person *100 people= 270 total ratings then right ?
|
welcome to the presentation on averages . averages is probably a concept that you 've already used before , maybe not in a mathematical way . but people will talk in terms of , the average voter wants a politician to do this , or the average student in a class wants to get out early . so you 're probably already familiar with the concept of an average . and you probably already intuitively knew that an average is just a number that represents the different values that a group could have . but it can represent that as one number as opposed to giving all the different values . and let 's give a couple of examples of how to compute an average , and you might already know how to do this . so let 's say i had the numbers 1 , 3 , 5 , and 20 . and i asked you , what is the average of these four numbers ? well , what we do is , we literally just add up the numbers . and then divide by the number of numbers we have . so we say 1 plus 3 is 4 . so let me write that . 1 plus 3 plus 5 plus 20 equals , let 's see , 1 plus 3 is 4 . 4 plus 5 is 9 . 9 plus 20 is 29 . and we had 4 numbers ; one , two , three , four . so 4 goes into 29 . and it goes , 7 , 7 , 28 . and then we have 10 , i did n't have to do that decimal there , oh well . 2 , 8 , 25 . so 4 goes into 29 7.25 times . so the average of these four numbers is equal to 7.25 . and that might make sense to you because 7.25 is someplace in between these numbers . and we can kind of view this , 7.25 , as one way to represent these four numbers without having to list these four numbers . there are other representations you 'll learn later on . like the mode . you 'll also the mean , which we 'll talk about later , is actually the same thing as the average . but the average is just one number that you can use to represent a set of numbers . so let 's do some problems which i think are going to be close to your heart . let 's say on the first four tests of an exam , i got a -- let 's see , i got an 80 , an 81 . an 87 , and an 88 . what 's my average in the class so far ? well , all i have to do is add up these four numbers . so i say , 80 plus 81 plus 87 plus 88 . well , zero plus 1 is 1 . 1 plus 7 is 8 . 8 plus 8 is 16 . i just ran eight miles , so i 'm a bit tired . and , 4/8 , so that 's 32 . plus 1 is 33 . and now we divide this number by 4 . 4 goes into 336 . goes into 33 , 8 times . 8 times 4 is 32 . 33 minus 32 is 1 , 16 . 4 . so the average is equal to 84 . so depending on what school you go to that 's either a b or a c. so , so far my average after the first four exams is an 84 . now let 's make this a little bit more difficult . we know that the average after four exams , at four exams , is equal to 84 . if i were to ask you what do i have to get on the next test to average an 88 , to average an 88 in the class . so let 's say that x is what i get on the next test . so now what we can say is , is that the first four exams , i could either list out the first four exams that i took . or i already know what the average is . so i know the sum of the first four exams is going to 4 times 84 . and now i want to add the , what i get on the 5th exam , x . and i 'm going to divide that by all five exams . so in other words , this number is the average of my first five exams . we just figured out the average of the first four exams . but now , we sum up the first four exams here . we add what i got on the fifth exam , and then we divide it by 5 , because now we 're averaging five exams . and i said that i need to get in an 88 in the class . and now we solve for x . let me make some space here . so , 5 times 88 is , let 's see . 5 times 80 is 400 , so it 's 440 . 440 equals 4 times 84 , we just saw that , is 320 plus 16 is 336 . 336 plus x is equal to 440 . well , it turns out if you subtract 336 from both sides , you get x is equal to 104 . so unless you have a exam that has some bonus problems on it , it 's probably impossible for you to get ah an 88 average in the class after just the next exam . you 'd have to get 104 on that next exam . and let 's just look at what we just did . we said , after 4 exams we had an 84 . what do i have to get on that next exam to average an 88 in the class after 5 exams ? and that 's what we solved for when we got x . now , let 's ask another question . i said after four exams , after four exams , i had an 84 average . if i said that there are 6 exams in the class , and the highest score i could get on an exam is 100 , what is the highest average i can finish in the class if i were to really study hard and get 100 on the next 2 exams ? well , once again , what we 'll want to do is assume we get 100 on the next 2 exams and then take the average . so we 'll have to solve all 6 exams . so we 're going to have the average of 6 , so in the denominator we 're going to have 6 . the first four exams , the sum , as we already learned , is 4 exams times the 84 average . and this dot is just times . plus , and there 's going to be 2 more exams , right ? because there 's 6 exams in the class . and i 'm going to get 100 in each . so that 's 200 . and what 's this average ? well , 4 times 84 , we already said , is 336 . plus 200 over 6 . so that 's 536 over 6 . 6 goes into 5 36 . i do n't know if if i gave myself enough space . but 6 goes into 53 , 8 times . 48 . 56 . 9 times . 9 times 6 is 54 . 6 minus is 20 6 goes into -- so we 'll see it 's actually 89.333333 , goes on forever . so 89.3 repeating . so no matter how hard i try in this class , the best i can do . because i only have two exams left , even if i were to get 100 on the next two exams . i can finish the class with an 89.333 average . hopefully , i think some of this might have been a little bit of a review for you . you already had kind of a sense of what an average is . and hopefully these last two problems not only taught you how to do some algebra problems involving average , but they 'll also help you figure out how well you have to do on your exams to get an a in your math class . i think you 're now ready for the average module . have fun .
|
so that 's 200 . and what 's this average ? well , 4 times 84 , we already said , is 336 .
|
what is the difference between the average and the mean ?
|
welcome to the presentation on averages . averages is probably a concept that you 've already used before , maybe not in a mathematical way . but people will talk in terms of , the average voter wants a politician to do this , or the average student in a class wants to get out early . so you 're probably already familiar with the concept of an average . and you probably already intuitively knew that an average is just a number that represents the different values that a group could have . but it can represent that as one number as opposed to giving all the different values . and let 's give a couple of examples of how to compute an average , and you might already know how to do this . so let 's say i had the numbers 1 , 3 , 5 , and 20 . and i asked you , what is the average of these four numbers ? well , what we do is , we literally just add up the numbers . and then divide by the number of numbers we have . so we say 1 plus 3 is 4 . so let me write that . 1 plus 3 plus 5 plus 20 equals , let 's see , 1 plus 3 is 4 . 4 plus 5 is 9 . 9 plus 20 is 29 . and we had 4 numbers ; one , two , three , four . so 4 goes into 29 . and it goes , 7 , 7 , 28 . and then we have 10 , i did n't have to do that decimal there , oh well . 2 , 8 , 25 . so 4 goes into 29 7.25 times . so the average of these four numbers is equal to 7.25 . and that might make sense to you because 7.25 is someplace in between these numbers . and we can kind of view this , 7.25 , as one way to represent these four numbers without having to list these four numbers . there are other representations you 'll learn later on . like the mode . you 'll also the mean , which we 'll talk about later , is actually the same thing as the average . but the average is just one number that you can use to represent a set of numbers . so let 's do some problems which i think are going to be close to your heart . let 's say on the first four tests of an exam , i got a -- let 's see , i got an 80 , an 81 . an 87 , and an 88 . what 's my average in the class so far ? well , all i have to do is add up these four numbers . so i say , 80 plus 81 plus 87 plus 88 . well , zero plus 1 is 1 . 1 plus 7 is 8 . 8 plus 8 is 16 . i just ran eight miles , so i 'm a bit tired . and , 4/8 , so that 's 32 . plus 1 is 33 . and now we divide this number by 4 . 4 goes into 336 . goes into 33 , 8 times . 8 times 4 is 32 . 33 minus 32 is 1 , 16 . 4 . so the average is equal to 84 . so depending on what school you go to that 's either a b or a c. so , so far my average after the first four exams is an 84 . now let 's make this a little bit more difficult . we know that the average after four exams , at four exams , is equal to 84 . if i were to ask you what do i have to get on the next test to average an 88 , to average an 88 in the class . so let 's say that x is what i get on the next test . so now what we can say is , is that the first four exams , i could either list out the first four exams that i took . or i already know what the average is . so i know the sum of the first four exams is going to 4 times 84 . and now i want to add the , what i get on the 5th exam , x . and i 'm going to divide that by all five exams . so in other words , this number is the average of my first five exams . we just figured out the average of the first four exams . but now , we sum up the first four exams here . we add what i got on the fifth exam , and then we divide it by 5 , because now we 're averaging five exams . and i said that i need to get in an 88 in the class . and now we solve for x . let me make some space here . so , 5 times 88 is , let 's see . 5 times 80 is 400 , so it 's 440 . 440 equals 4 times 84 , we just saw that , is 320 plus 16 is 336 . 336 plus x is equal to 440 . well , it turns out if you subtract 336 from both sides , you get x is equal to 104 . so unless you have a exam that has some bonus problems on it , it 's probably impossible for you to get ah an 88 average in the class after just the next exam . you 'd have to get 104 on that next exam . and let 's just look at what we just did . we said , after 4 exams we had an 84 . what do i have to get on that next exam to average an 88 in the class after 5 exams ? and that 's what we solved for when we got x . now , let 's ask another question . i said after four exams , after four exams , i had an 84 average . if i said that there are 6 exams in the class , and the highest score i could get on an exam is 100 , what is the highest average i can finish in the class if i were to really study hard and get 100 on the next 2 exams ? well , once again , what we 'll want to do is assume we get 100 on the next 2 exams and then take the average . so we 'll have to solve all 6 exams . so we 're going to have the average of 6 , so in the denominator we 're going to have 6 . the first four exams , the sum , as we already learned , is 4 exams times the 84 average . and this dot is just times . plus , and there 's going to be 2 more exams , right ? because there 's 6 exams in the class . and i 'm going to get 100 in each . so that 's 200 . and what 's this average ? well , 4 times 84 , we already said , is 336 . plus 200 over 6 . so that 's 536 over 6 . 6 goes into 5 36 . i do n't know if if i gave myself enough space . but 6 goes into 53 , 8 times . 48 . 56 . 9 times . 9 times 6 is 54 . 6 minus is 20 6 goes into -- so we 'll see it 's actually 89.333333 , goes on forever . so 89.3 repeating . so no matter how hard i try in this class , the best i can do . because i only have two exams left , even if i were to get 100 on the next two exams . i can finish the class with an 89.333 average . hopefully , i think some of this might have been a little bit of a review for you . you already had kind of a sense of what an average is . and hopefully these last two problems not only taught you how to do some algebra problems involving average , but they 'll also help you figure out how well you have to do on your exams to get an a in your math class . i think you 're now ready for the average module . have fun .
|
and i said that i need to get in an 88 in the class . and now we solve for x . let me make some space here .
|
if you have two groups with respective averages x and y , and the average of the groups added together is something closer to x than to y , does that mean that the number of samples in the first group ( x ) is necessarily larger than the number of samples in the second group ( y ) ?
|
welcome to the presentation on averages . averages is probably a concept that you 've already used before , maybe not in a mathematical way . but people will talk in terms of , the average voter wants a politician to do this , or the average student in a class wants to get out early . so you 're probably already familiar with the concept of an average . and you probably already intuitively knew that an average is just a number that represents the different values that a group could have . but it can represent that as one number as opposed to giving all the different values . and let 's give a couple of examples of how to compute an average , and you might already know how to do this . so let 's say i had the numbers 1 , 3 , 5 , and 20 . and i asked you , what is the average of these four numbers ? well , what we do is , we literally just add up the numbers . and then divide by the number of numbers we have . so we say 1 plus 3 is 4 . so let me write that . 1 plus 3 plus 5 plus 20 equals , let 's see , 1 plus 3 is 4 . 4 plus 5 is 9 . 9 plus 20 is 29 . and we had 4 numbers ; one , two , three , four . so 4 goes into 29 . and it goes , 7 , 7 , 28 . and then we have 10 , i did n't have to do that decimal there , oh well . 2 , 8 , 25 . so 4 goes into 29 7.25 times . so the average of these four numbers is equal to 7.25 . and that might make sense to you because 7.25 is someplace in between these numbers . and we can kind of view this , 7.25 , as one way to represent these four numbers without having to list these four numbers . there are other representations you 'll learn later on . like the mode . you 'll also the mean , which we 'll talk about later , is actually the same thing as the average . but the average is just one number that you can use to represent a set of numbers . so let 's do some problems which i think are going to be close to your heart . let 's say on the first four tests of an exam , i got a -- let 's see , i got an 80 , an 81 . an 87 , and an 88 . what 's my average in the class so far ? well , all i have to do is add up these four numbers . so i say , 80 plus 81 plus 87 plus 88 . well , zero plus 1 is 1 . 1 plus 7 is 8 . 8 plus 8 is 16 . i just ran eight miles , so i 'm a bit tired . and , 4/8 , so that 's 32 . plus 1 is 33 . and now we divide this number by 4 . 4 goes into 336 . goes into 33 , 8 times . 8 times 4 is 32 . 33 minus 32 is 1 , 16 . 4 . so the average is equal to 84 . so depending on what school you go to that 's either a b or a c. so , so far my average after the first four exams is an 84 . now let 's make this a little bit more difficult . we know that the average after four exams , at four exams , is equal to 84 . if i were to ask you what do i have to get on the next test to average an 88 , to average an 88 in the class . so let 's say that x is what i get on the next test . so now what we can say is , is that the first four exams , i could either list out the first four exams that i took . or i already know what the average is . so i know the sum of the first four exams is going to 4 times 84 . and now i want to add the , what i get on the 5th exam , x . and i 'm going to divide that by all five exams . so in other words , this number is the average of my first five exams . we just figured out the average of the first four exams . but now , we sum up the first four exams here . we add what i got on the fifth exam , and then we divide it by 5 , because now we 're averaging five exams . and i said that i need to get in an 88 in the class . and now we solve for x . let me make some space here . so , 5 times 88 is , let 's see . 5 times 80 is 400 , so it 's 440 . 440 equals 4 times 84 , we just saw that , is 320 plus 16 is 336 . 336 plus x is equal to 440 . well , it turns out if you subtract 336 from both sides , you get x is equal to 104 . so unless you have a exam that has some bonus problems on it , it 's probably impossible for you to get ah an 88 average in the class after just the next exam . you 'd have to get 104 on that next exam . and let 's just look at what we just did . we said , after 4 exams we had an 84 . what do i have to get on that next exam to average an 88 in the class after 5 exams ? and that 's what we solved for when we got x . now , let 's ask another question . i said after four exams , after four exams , i had an 84 average . if i said that there are 6 exams in the class , and the highest score i could get on an exam is 100 , what is the highest average i can finish in the class if i were to really study hard and get 100 on the next 2 exams ? well , once again , what we 'll want to do is assume we get 100 on the next 2 exams and then take the average . so we 'll have to solve all 6 exams . so we 're going to have the average of 6 , so in the denominator we 're going to have 6 . the first four exams , the sum , as we already learned , is 4 exams times the 84 average . and this dot is just times . plus , and there 's going to be 2 more exams , right ? because there 's 6 exams in the class . and i 'm going to get 100 in each . so that 's 200 . and what 's this average ? well , 4 times 84 , we already said , is 336 . plus 200 over 6 . so that 's 536 over 6 . 6 goes into 5 36 . i do n't know if if i gave myself enough space . but 6 goes into 53 , 8 times . 48 . 56 . 9 times . 9 times 6 is 54 . 6 minus is 20 6 goes into -- so we 'll see it 's actually 89.333333 , goes on forever . so 89.3 repeating . so no matter how hard i try in this class , the best i can do . because i only have two exams left , even if i were to get 100 on the next two exams . i can finish the class with an 89.333 average . hopefully , i think some of this might have been a little bit of a review for you . you already had kind of a sense of what an average is . and hopefully these last two problems not only taught you how to do some algebra problems involving average , but they 'll also help you figure out how well you have to do on your exams to get an a in your math class . i think you 're now ready for the average module . have fun .
|
so 4 goes into 29 7.25 times . so the average of these four numbers is equal to 7.25 . and that might make sense to you because 7.25 is someplace in between these numbers .
|
if the average of 7 numbers is 12 , what is their sum ?
|
welcome to the presentation on averages . averages is probably a concept that you 've already used before , maybe not in a mathematical way . but people will talk in terms of , the average voter wants a politician to do this , or the average student in a class wants to get out early . so you 're probably already familiar with the concept of an average . and you probably already intuitively knew that an average is just a number that represents the different values that a group could have . but it can represent that as one number as opposed to giving all the different values . and let 's give a couple of examples of how to compute an average , and you might already know how to do this . so let 's say i had the numbers 1 , 3 , 5 , and 20 . and i asked you , what is the average of these four numbers ? well , what we do is , we literally just add up the numbers . and then divide by the number of numbers we have . so we say 1 plus 3 is 4 . so let me write that . 1 plus 3 plus 5 plus 20 equals , let 's see , 1 plus 3 is 4 . 4 plus 5 is 9 . 9 plus 20 is 29 . and we had 4 numbers ; one , two , three , four . so 4 goes into 29 . and it goes , 7 , 7 , 28 . and then we have 10 , i did n't have to do that decimal there , oh well . 2 , 8 , 25 . so 4 goes into 29 7.25 times . so the average of these four numbers is equal to 7.25 . and that might make sense to you because 7.25 is someplace in between these numbers . and we can kind of view this , 7.25 , as one way to represent these four numbers without having to list these four numbers . there are other representations you 'll learn later on . like the mode . you 'll also the mean , which we 'll talk about later , is actually the same thing as the average . but the average is just one number that you can use to represent a set of numbers . so let 's do some problems which i think are going to be close to your heart . let 's say on the first four tests of an exam , i got a -- let 's see , i got an 80 , an 81 . an 87 , and an 88 . what 's my average in the class so far ? well , all i have to do is add up these four numbers . so i say , 80 plus 81 plus 87 plus 88 . well , zero plus 1 is 1 . 1 plus 7 is 8 . 8 plus 8 is 16 . i just ran eight miles , so i 'm a bit tired . and , 4/8 , so that 's 32 . plus 1 is 33 . and now we divide this number by 4 . 4 goes into 336 . goes into 33 , 8 times . 8 times 4 is 32 . 33 minus 32 is 1 , 16 . 4 . so the average is equal to 84 . so depending on what school you go to that 's either a b or a c. so , so far my average after the first four exams is an 84 . now let 's make this a little bit more difficult . we know that the average after four exams , at four exams , is equal to 84 . if i were to ask you what do i have to get on the next test to average an 88 , to average an 88 in the class . so let 's say that x is what i get on the next test . so now what we can say is , is that the first four exams , i could either list out the first four exams that i took . or i already know what the average is . so i know the sum of the first four exams is going to 4 times 84 . and now i want to add the , what i get on the 5th exam , x . and i 'm going to divide that by all five exams . so in other words , this number is the average of my first five exams . we just figured out the average of the first four exams . but now , we sum up the first four exams here . we add what i got on the fifth exam , and then we divide it by 5 , because now we 're averaging five exams . and i said that i need to get in an 88 in the class . and now we solve for x . let me make some space here . so , 5 times 88 is , let 's see . 5 times 80 is 400 , so it 's 440 . 440 equals 4 times 84 , we just saw that , is 320 plus 16 is 336 . 336 plus x is equal to 440 . well , it turns out if you subtract 336 from both sides , you get x is equal to 104 . so unless you have a exam that has some bonus problems on it , it 's probably impossible for you to get ah an 88 average in the class after just the next exam . you 'd have to get 104 on that next exam . and let 's just look at what we just did . we said , after 4 exams we had an 84 . what do i have to get on that next exam to average an 88 in the class after 5 exams ? and that 's what we solved for when we got x . now , let 's ask another question . i said after four exams , after four exams , i had an 84 average . if i said that there are 6 exams in the class , and the highest score i could get on an exam is 100 , what is the highest average i can finish in the class if i were to really study hard and get 100 on the next 2 exams ? well , once again , what we 'll want to do is assume we get 100 on the next 2 exams and then take the average . so we 'll have to solve all 6 exams . so we 're going to have the average of 6 , so in the denominator we 're going to have 6 . the first four exams , the sum , as we already learned , is 4 exams times the 84 average . and this dot is just times . plus , and there 's going to be 2 more exams , right ? because there 's 6 exams in the class . and i 'm going to get 100 in each . so that 's 200 . and what 's this average ? well , 4 times 84 , we already said , is 336 . plus 200 over 6 . so that 's 536 over 6 . 6 goes into 5 36 . i do n't know if if i gave myself enough space . but 6 goes into 53 , 8 times . 48 . 56 . 9 times . 9 times 6 is 54 . 6 minus is 20 6 goes into -- so we 'll see it 's actually 89.333333 , goes on forever . so 89.3 repeating . so no matter how hard i try in this class , the best i can do . because i only have two exams left , even if i were to get 100 on the next two exams . i can finish the class with an 89.333 average . hopefully , i think some of this might have been a little bit of a review for you . you already had kind of a sense of what an average is . and hopefully these last two problems not only taught you how to do some algebra problems involving average , but they 'll also help you figure out how well you have to do on your exams to get an a in your math class . i think you 're now ready for the average module . have fun .
|
33 minus 32 is 1 , 16 . 4 . so the average is equal to 84 .
|
so you just add all the numbers together and then divide by how ever many numbers there are so like 2+4+6+8=20 then divide 20/4=5 so 5 would be the average of 2,4,6,8 right ?
|
welcome to the presentation on averages . averages is probably a concept that you 've already used before , maybe not in a mathematical way . but people will talk in terms of , the average voter wants a politician to do this , or the average student in a class wants to get out early . so you 're probably already familiar with the concept of an average . and you probably already intuitively knew that an average is just a number that represents the different values that a group could have . but it can represent that as one number as opposed to giving all the different values . and let 's give a couple of examples of how to compute an average , and you might already know how to do this . so let 's say i had the numbers 1 , 3 , 5 , and 20 . and i asked you , what is the average of these four numbers ? well , what we do is , we literally just add up the numbers . and then divide by the number of numbers we have . so we say 1 plus 3 is 4 . so let me write that . 1 plus 3 plus 5 plus 20 equals , let 's see , 1 plus 3 is 4 . 4 plus 5 is 9 . 9 plus 20 is 29 . and we had 4 numbers ; one , two , three , four . so 4 goes into 29 . and it goes , 7 , 7 , 28 . and then we have 10 , i did n't have to do that decimal there , oh well . 2 , 8 , 25 . so 4 goes into 29 7.25 times . so the average of these four numbers is equal to 7.25 . and that might make sense to you because 7.25 is someplace in between these numbers . and we can kind of view this , 7.25 , as one way to represent these four numbers without having to list these four numbers . there are other representations you 'll learn later on . like the mode . you 'll also the mean , which we 'll talk about later , is actually the same thing as the average . but the average is just one number that you can use to represent a set of numbers . so let 's do some problems which i think are going to be close to your heart . let 's say on the first four tests of an exam , i got a -- let 's see , i got an 80 , an 81 . an 87 , and an 88 . what 's my average in the class so far ? well , all i have to do is add up these four numbers . so i say , 80 plus 81 plus 87 plus 88 . well , zero plus 1 is 1 . 1 plus 7 is 8 . 8 plus 8 is 16 . i just ran eight miles , so i 'm a bit tired . and , 4/8 , so that 's 32 . plus 1 is 33 . and now we divide this number by 4 . 4 goes into 336 . goes into 33 , 8 times . 8 times 4 is 32 . 33 minus 32 is 1 , 16 . 4 . so the average is equal to 84 . so depending on what school you go to that 's either a b or a c. so , so far my average after the first four exams is an 84 . now let 's make this a little bit more difficult . we know that the average after four exams , at four exams , is equal to 84 . if i were to ask you what do i have to get on the next test to average an 88 , to average an 88 in the class . so let 's say that x is what i get on the next test . so now what we can say is , is that the first four exams , i could either list out the first four exams that i took . or i already know what the average is . so i know the sum of the first four exams is going to 4 times 84 . and now i want to add the , what i get on the 5th exam , x . and i 'm going to divide that by all five exams . so in other words , this number is the average of my first five exams . we just figured out the average of the first four exams . but now , we sum up the first four exams here . we add what i got on the fifth exam , and then we divide it by 5 , because now we 're averaging five exams . and i said that i need to get in an 88 in the class . and now we solve for x . let me make some space here . so , 5 times 88 is , let 's see . 5 times 80 is 400 , so it 's 440 . 440 equals 4 times 84 , we just saw that , is 320 plus 16 is 336 . 336 plus x is equal to 440 . well , it turns out if you subtract 336 from both sides , you get x is equal to 104 . so unless you have a exam that has some bonus problems on it , it 's probably impossible for you to get ah an 88 average in the class after just the next exam . you 'd have to get 104 on that next exam . and let 's just look at what we just did . we said , after 4 exams we had an 84 . what do i have to get on that next exam to average an 88 in the class after 5 exams ? and that 's what we solved for when we got x . now , let 's ask another question . i said after four exams , after four exams , i had an 84 average . if i said that there are 6 exams in the class , and the highest score i could get on an exam is 100 , what is the highest average i can finish in the class if i were to really study hard and get 100 on the next 2 exams ? well , once again , what we 'll want to do is assume we get 100 on the next 2 exams and then take the average . so we 'll have to solve all 6 exams . so we 're going to have the average of 6 , so in the denominator we 're going to have 6 . the first four exams , the sum , as we already learned , is 4 exams times the 84 average . and this dot is just times . plus , and there 's going to be 2 more exams , right ? because there 's 6 exams in the class . and i 'm going to get 100 in each . so that 's 200 . and what 's this average ? well , 4 times 84 , we already said , is 336 . plus 200 over 6 . so that 's 536 over 6 . 6 goes into 5 36 . i do n't know if if i gave myself enough space . but 6 goes into 53 , 8 times . 48 . 56 . 9 times . 9 times 6 is 54 . 6 minus is 20 6 goes into -- so we 'll see it 's actually 89.333333 , goes on forever . so 89.3 repeating . so no matter how hard i try in this class , the best i can do . because i only have two exams left , even if i were to get 100 on the next two exams . i can finish the class with an 89.333 average . hopefully , i think some of this might have been a little bit of a review for you . you already had kind of a sense of what an average is . and hopefully these last two problems not only taught you how to do some algebra problems involving average , but they 'll also help you figure out how well you have to do on your exams to get an a in your math class . i think you 're now ready for the average module . have fun .
|
we said , after 4 exams we had an 84 . what do i have to get on that next exam to average an 88 in the class after 5 exams ? and that 's what we solved for when we got x .
|
how did you get the 440 during the average exam score ?
|
welcome to the presentation on averages . averages is probably a concept that you 've already used before , maybe not in a mathematical way . but people will talk in terms of , the average voter wants a politician to do this , or the average student in a class wants to get out early . so you 're probably already familiar with the concept of an average . and you probably already intuitively knew that an average is just a number that represents the different values that a group could have . but it can represent that as one number as opposed to giving all the different values . and let 's give a couple of examples of how to compute an average , and you might already know how to do this . so let 's say i had the numbers 1 , 3 , 5 , and 20 . and i asked you , what is the average of these four numbers ? well , what we do is , we literally just add up the numbers . and then divide by the number of numbers we have . so we say 1 plus 3 is 4 . so let me write that . 1 plus 3 plus 5 plus 20 equals , let 's see , 1 plus 3 is 4 . 4 plus 5 is 9 . 9 plus 20 is 29 . and we had 4 numbers ; one , two , three , four . so 4 goes into 29 . and it goes , 7 , 7 , 28 . and then we have 10 , i did n't have to do that decimal there , oh well . 2 , 8 , 25 . so 4 goes into 29 7.25 times . so the average of these four numbers is equal to 7.25 . and that might make sense to you because 7.25 is someplace in between these numbers . and we can kind of view this , 7.25 , as one way to represent these four numbers without having to list these four numbers . there are other representations you 'll learn later on . like the mode . you 'll also the mean , which we 'll talk about later , is actually the same thing as the average . but the average is just one number that you can use to represent a set of numbers . so let 's do some problems which i think are going to be close to your heart . let 's say on the first four tests of an exam , i got a -- let 's see , i got an 80 , an 81 . an 87 , and an 88 . what 's my average in the class so far ? well , all i have to do is add up these four numbers . so i say , 80 plus 81 plus 87 plus 88 . well , zero plus 1 is 1 . 1 plus 7 is 8 . 8 plus 8 is 16 . i just ran eight miles , so i 'm a bit tired . and , 4/8 , so that 's 32 . plus 1 is 33 . and now we divide this number by 4 . 4 goes into 336 . goes into 33 , 8 times . 8 times 4 is 32 . 33 minus 32 is 1 , 16 . 4 . so the average is equal to 84 . so depending on what school you go to that 's either a b or a c. so , so far my average after the first four exams is an 84 . now let 's make this a little bit more difficult . we know that the average after four exams , at four exams , is equal to 84 . if i were to ask you what do i have to get on the next test to average an 88 , to average an 88 in the class . so let 's say that x is what i get on the next test . so now what we can say is , is that the first four exams , i could either list out the first four exams that i took . or i already know what the average is . so i know the sum of the first four exams is going to 4 times 84 . and now i want to add the , what i get on the 5th exam , x . and i 'm going to divide that by all five exams . so in other words , this number is the average of my first five exams . we just figured out the average of the first four exams . but now , we sum up the first four exams here . we add what i got on the fifth exam , and then we divide it by 5 , because now we 're averaging five exams . and i said that i need to get in an 88 in the class . and now we solve for x . let me make some space here . so , 5 times 88 is , let 's see . 5 times 80 is 400 , so it 's 440 . 440 equals 4 times 84 , we just saw that , is 320 plus 16 is 336 . 336 plus x is equal to 440 . well , it turns out if you subtract 336 from both sides , you get x is equal to 104 . so unless you have a exam that has some bonus problems on it , it 's probably impossible for you to get ah an 88 average in the class after just the next exam . you 'd have to get 104 on that next exam . and let 's just look at what we just did . we said , after 4 exams we had an 84 . what do i have to get on that next exam to average an 88 in the class after 5 exams ? and that 's what we solved for when we got x . now , let 's ask another question . i said after four exams , after four exams , i had an 84 average . if i said that there are 6 exams in the class , and the highest score i could get on an exam is 100 , what is the highest average i can finish in the class if i were to really study hard and get 100 on the next 2 exams ? well , once again , what we 'll want to do is assume we get 100 on the next 2 exams and then take the average . so we 'll have to solve all 6 exams . so we 're going to have the average of 6 , so in the denominator we 're going to have 6 . the first four exams , the sum , as we already learned , is 4 exams times the 84 average . and this dot is just times . plus , and there 's going to be 2 more exams , right ? because there 's 6 exams in the class . and i 'm going to get 100 in each . so that 's 200 . and what 's this average ? well , 4 times 84 , we already said , is 336 . plus 200 over 6 . so that 's 536 over 6 . 6 goes into 5 36 . i do n't know if if i gave myself enough space . but 6 goes into 53 , 8 times . 48 . 56 . 9 times . 9 times 6 is 54 . 6 minus is 20 6 goes into -- so we 'll see it 's actually 89.333333 , goes on forever . so 89.3 repeating . so no matter how hard i try in this class , the best i can do . because i only have two exams left , even if i were to get 100 on the next two exams . i can finish the class with an 89.333 average . hopefully , i think some of this might have been a little bit of a review for you . you already had kind of a sense of what an average is . and hopefully these last two problems not only taught you how to do some algebra problems involving average , but they 'll also help you figure out how well you have to do on your exams to get an a in your math class . i think you 're now ready for the average module . have fun .
|
so that 's 200 . and what 's this average ? well , 4 times 84 , we already said , is 336 .
|
what is the difference between mean and average ?
|
welcome to the presentation on averages . averages is probably a concept that you 've already used before , maybe not in a mathematical way . but people will talk in terms of , the average voter wants a politician to do this , or the average student in a class wants to get out early . so you 're probably already familiar with the concept of an average . and you probably already intuitively knew that an average is just a number that represents the different values that a group could have . but it can represent that as one number as opposed to giving all the different values . and let 's give a couple of examples of how to compute an average , and you might already know how to do this . so let 's say i had the numbers 1 , 3 , 5 , and 20 . and i asked you , what is the average of these four numbers ? well , what we do is , we literally just add up the numbers . and then divide by the number of numbers we have . so we say 1 plus 3 is 4 . so let me write that . 1 plus 3 plus 5 plus 20 equals , let 's see , 1 plus 3 is 4 . 4 plus 5 is 9 . 9 plus 20 is 29 . and we had 4 numbers ; one , two , three , four . so 4 goes into 29 . and it goes , 7 , 7 , 28 . and then we have 10 , i did n't have to do that decimal there , oh well . 2 , 8 , 25 . so 4 goes into 29 7.25 times . so the average of these four numbers is equal to 7.25 . and that might make sense to you because 7.25 is someplace in between these numbers . and we can kind of view this , 7.25 , as one way to represent these four numbers without having to list these four numbers . there are other representations you 'll learn later on . like the mode . you 'll also the mean , which we 'll talk about later , is actually the same thing as the average . but the average is just one number that you can use to represent a set of numbers . so let 's do some problems which i think are going to be close to your heart . let 's say on the first four tests of an exam , i got a -- let 's see , i got an 80 , an 81 . an 87 , and an 88 . what 's my average in the class so far ? well , all i have to do is add up these four numbers . so i say , 80 plus 81 plus 87 plus 88 . well , zero plus 1 is 1 . 1 plus 7 is 8 . 8 plus 8 is 16 . i just ran eight miles , so i 'm a bit tired . and , 4/8 , so that 's 32 . plus 1 is 33 . and now we divide this number by 4 . 4 goes into 336 . goes into 33 , 8 times . 8 times 4 is 32 . 33 minus 32 is 1 , 16 . 4 . so the average is equal to 84 . so depending on what school you go to that 's either a b or a c. so , so far my average after the first four exams is an 84 . now let 's make this a little bit more difficult . we know that the average after four exams , at four exams , is equal to 84 . if i were to ask you what do i have to get on the next test to average an 88 , to average an 88 in the class . so let 's say that x is what i get on the next test . so now what we can say is , is that the first four exams , i could either list out the first four exams that i took . or i already know what the average is . so i know the sum of the first four exams is going to 4 times 84 . and now i want to add the , what i get on the 5th exam , x . and i 'm going to divide that by all five exams . so in other words , this number is the average of my first five exams . we just figured out the average of the first four exams . but now , we sum up the first four exams here . we add what i got on the fifth exam , and then we divide it by 5 , because now we 're averaging five exams . and i said that i need to get in an 88 in the class . and now we solve for x . let me make some space here . so , 5 times 88 is , let 's see . 5 times 80 is 400 , so it 's 440 . 440 equals 4 times 84 , we just saw that , is 320 plus 16 is 336 . 336 plus x is equal to 440 . well , it turns out if you subtract 336 from both sides , you get x is equal to 104 . so unless you have a exam that has some bonus problems on it , it 's probably impossible for you to get ah an 88 average in the class after just the next exam . you 'd have to get 104 on that next exam . and let 's just look at what we just did . we said , after 4 exams we had an 84 . what do i have to get on that next exam to average an 88 in the class after 5 exams ? and that 's what we solved for when we got x . now , let 's ask another question . i said after four exams , after four exams , i had an 84 average . if i said that there are 6 exams in the class , and the highest score i could get on an exam is 100 , what is the highest average i can finish in the class if i were to really study hard and get 100 on the next 2 exams ? well , once again , what we 'll want to do is assume we get 100 on the next 2 exams and then take the average . so we 'll have to solve all 6 exams . so we 're going to have the average of 6 , so in the denominator we 're going to have 6 . the first four exams , the sum , as we already learned , is 4 exams times the 84 average . and this dot is just times . plus , and there 's going to be 2 more exams , right ? because there 's 6 exams in the class . and i 'm going to get 100 in each . so that 's 200 . and what 's this average ? well , 4 times 84 , we already said , is 336 . plus 200 over 6 . so that 's 536 over 6 . 6 goes into 5 36 . i do n't know if if i gave myself enough space . but 6 goes into 53 , 8 times . 48 . 56 . 9 times . 9 times 6 is 54 . 6 minus is 20 6 goes into -- so we 'll see it 's actually 89.333333 , goes on forever . so 89.3 repeating . so no matter how hard i try in this class , the best i can do . because i only have two exams left , even if i were to get 100 on the next two exams . i can finish the class with an 89.333 average . hopefully , i think some of this might have been a little bit of a review for you . you already had kind of a sense of what an average is . and hopefully these last two problems not only taught you how to do some algebra problems involving average , but they 'll also help you figure out how well you have to do on your exams to get an a in your math class . i think you 're now ready for the average module . have fun .
|
let me make some space here . so , 5 times 88 is , let 's see . 5 times 80 is 400 , so it 's 440 .
|
why sal multiples 5.88 ( min ) ?
|
welcome to the presentation on averages . averages is probably a concept that you 've already used before , maybe not in a mathematical way . but people will talk in terms of , the average voter wants a politician to do this , or the average student in a class wants to get out early . so you 're probably already familiar with the concept of an average . and you probably already intuitively knew that an average is just a number that represents the different values that a group could have . but it can represent that as one number as opposed to giving all the different values . and let 's give a couple of examples of how to compute an average , and you might already know how to do this . so let 's say i had the numbers 1 , 3 , 5 , and 20 . and i asked you , what is the average of these four numbers ? well , what we do is , we literally just add up the numbers . and then divide by the number of numbers we have . so we say 1 plus 3 is 4 . so let me write that . 1 plus 3 plus 5 plus 20 equals , let 's see , 1 plus 3 is 4 . 4 plus 5 is 9 . 9 plus 20 is 29 . and we had 4 numbers ; one , two , three , four . so 4 goes into 29 . and it goes , 7 , 7 , 28 . and then we have 10 , i did n't have to do that decimal there , oh well . 2 , 8 , 25 . so 4 goes into 29 7.25 times . so the average of these four numbers is equal to 7.25 . and that might make sense to you because 7.25 is someplace in between these numbers . and we can kind of view this , 7.25 , as one way to represent these four numbers without having to list these four numbers . there are other representations you 'll learn later on . like the mode . you 'll also the mean , which we 'll talk about later , is actually the same thing as the average . but the average is just one number that you can use to represent a set of numbers . so let 's do some problems which i think are going to be close to your heart . let 's say on the first four tests of an exam , i got a -- let 's see , i got an 80 , an 81 . an 87 , and an 88 . what 's my average in the class so far ? well , all i have to do is add up these four numbers . so i say , 80 plus 81 plus 87 plus 88 . well , zero plus 1 is 1 . 1 plus 7 is 8 . 8 plus 8 is 16 . i just ran eight miles , so i 'm a bit tired . and , 4/8 , so that 's 32 . plus 1 is 33 . and now we divide this number by 4 . 4 goes into 336 . goes into 33 , 8 times . 8 times 4 is 32 . 33 minus 32 is 1 , 16 . 4 . so the average is equal to 84 . so depending on what school you go to that 's either a b or a c. so , so far my average after the first four exams is an 84 . now let 's make this a little bit more difficult . we know that the average after four exams , at four exams , is equal to 84 . if i were to ask you what do i have to get on the next test to average an 88 , to average an 88 in the class . so let 's say that x is what i get on the next test . so now what we can say is , is that the first four exams , i could either list out the first four exams that i took . or i already know what the average is . so i know the sum of the first four exams is going to 4 times 84 . and now i want to add the , what i get on the 5th exam , x . and i 'm going to divide that by all five exams . so in other words , this number is the average of my first five exams . we just figured out the average of the first four exams . but now , we sum up the first four exams here . we add what i got on the fifth exam , and then we divide it by 5 , because now we 're averaging five exams . and i said that i need to get in an 88 in the class . and now we solve for x . let me make some space here . so , 5 times 88 is , let 's see . 5 times 80 is 400 , so it 's 440 . 440 equals 4 times 84 , we just saw that , is 320 plus 16 is 336 . 336 plus x is equal to 440 . well , it turns out if you subtract 336 from both sides , you get x is equal to 104 . so unless you have a exam that has some bonus problems on it , it 's probably impossible for you to get ah an 88 average in the class after just the next exam . you 'd have to get 104 on that next exam . and let 's just look at what we just did . we said , after 4 exams we had an 84 . what do i have to get on that next exam to average an 88 in the class after 5 exams ? and that 's what we solved for when we got x . now , let 's ask another question . i said after four exams , after four exams , i had an 84 average . if i said that there are 6 exams in the class , and the highest score i could get on an exam is 100 , what is the highest average i can finish in the class if i were to really study hard and get 100 on the next 2 exams ? well , once again , what we 'll want to do is assume we get 100 on the next 2 exams and then take the average . so we 'll have to solve all 6 exams . so we 're going to have the average of 6 , so in the denominator we 're going to have 6 . the first four exams , the sum , as we already learned , is 4 exams times the 84 average . and this dot is just times . plus , and there 's going to be 2 more exams , right ? because there 's 6 exams in the class . and i 'm going to get 100 in each . so that 's 200 . and what 's this average ? well , 4 times 84 , we already said , is 336 . plus 200 over 6 . so that 's 536 over 6 . 6 goes into 5 36 . i do n't know if if i gave myself enough space . but 6 goes into 53 , 8 times . 48 . 56 . 9 times . 9 times 6 is 54 . 6 minus is 20 6 goes into -- so we 'll see it 's actually 89.333333 , goes on forever . so 89.3 repeating . so no matter how hard i try in this class , the best i can do . because i only have two exams left , even if i were to get 100 on the next two exams . i can finish the class with an 89.333 average . hopefully , i think some of this might have been a little bit of a review for you . you already had kind of a sense of what an average is . and hopefully these last two problems not only taught you how to do some algebra problems involving average , but they 'll also help you figure out how well you have to do on your exams to get an a in your math class . i think you 're now ready for the average module . have fun .
|
because there 's 6 exams in the class . and i 'm going to get 100 in each . so that 's 200 .
|
how did you get to be so good at math ?
|
welcome to the presentation on averages . averages is probably a concept that you 've already used before , maybe not in a mathematical way . but people will talk in terms of , the average voter wants a politician to do this , or the average student in a class wants to get out early . so you 're probably already familiar with the concept of an average . and you probably already intuitively knew that an average is just a number that represents the different values that a group could have . but it can represent that as one number as opposed to giving all the different values . and let 's give a couple of examples of how to compute an average , and you might already know how to do this . so let 's say i had the numbers 1 , 3 , 5 , and 20 . and i asked you , what is the average of these four numbers ? well , what we do is , we literally just add up the numbers . and then divide by the number of numbers we have . so we say 1 plus 3 is 4 . so let me write that . 1 plus 3 plus 5 plus 20 equals , let 's see , 1 plus 3 is 4 . 4 plus 5 is 9 . 9 plus 20 is 29 . and we had 4 numbers ; one , two , three , four . so 4 goes into 29 . and it goes , 7 , 7 , 28 . and then we have 10 , i did n't have to do that decimal there , oh well . 2 , 8 , 25 . so 4 goes into 29 7.25 times . so the average of these four numbers is equal to 7.25 . and that might make sense to you because 7.25 is someplace in between these numbers . and we can kind of view this , 7.25 , as one way to represent these four numbers without having to list these four numbers . there are other representations you 'll learn later on . like the mode . you 'll also the mean , which we 'll talk about later , is actually the same thing as the average . but the average is just one number that you can use to represent a set of numbers . so let 's do some problems which i think are going to be close to your heart . let 's say on the first four tests of an exam , i got a -- let 's see , i got an 80 , an 81 . an 87 , and an 88 . what 's my average in the class so far ? well , all i have to do is add up these four numbers . so i say , 80 plus 81 plus 87 plus 88 . well , zero plus 1 is 1 . 1 plus 7 is 8 . 8 plus 8 is 16 . i just ran eight miles , so i 'm a bit tired . and , 4/8 , so that 's 32 . plus 1 is 33 . and now we divide this number by 4 . 4 goes into 336 . goes into 33 , 8 times . 8 times 4 is 32 . 33 minus 32 is 1 , 16 . 4 . so the average is equal to 84 . so depending on what school you go to that 's either a b or a c. so , so far my average after the first four exams is an 84 . now let 's make this a little bit more difficult . we know that the average after four exams , at four exams , is equal to 84 . if i were to ask you what do i have to get on the next test to average an 88 , to average an 88 in the class . so let 's say that x is what i get on the next test . so now what we can say is , is that the first four exams , i could either list out the first four exams that i took . or i already know what the average is . so i know the sum of the first four exams is going to 4 times 84 . and now i want to add the , what i get on the 5th exam , x . and i 'm going to divide that by all five exams . so in other words , this number is the average of my first five exams . we just figured out the average of the first four exams . but now , we sum up the first four exams here . we add what i got on the fifth exam , and then we divide it by 5 , because now we 're averaging five exams . and i said that i need to get in an 88 in the class . and now we solve for x . let me make some space here . so , 5 times 88 is , let 's see . 5 times 80 is 400 , so it 's 440 . 440 equals 4 times 84 , we just saw that , is 320 plus 16 is 336 . 336 plus x is equal to 440 . well , it turns out if you subtract 336 from both sides , you get x is equal to 104 . so unless you have a exam that has some bonus problems on it , it 's probably impossible for you to get ah an 88 average in the class after just the next exam . you 'd have to get 104 on that next exam . and let 's just look at what we just did . we said , after 4 exams we had an 84 . what do i have to get on that next exam to average an 88 in the class after 5 exams ? and that 's what we solved for when we got x . now , let 's ask another question . i said after four exams , after four exams , i had an 84 average . if i said that there are 6 exams in the class , and the highest score i could get on an exam is 100 , what is the highest average i can finish in the class if i were to really study hard and get 100 on the next 2 exams ? well , once again , what we 'll want to do is assume we get 100 on the next 2 exams and then take the average . so we 'll have to solve all 6 exams . so we 're going to have the average of 6 , so in the denominator we 're going to have 6 . the first four exams , the sum , as we already learned , is 4 exams times the 84 average . and this dot is just times . plus , and there 's going to be 2 more exams , right ? because there 's 6 exams in the class . and i 'm going to get 100 in each . so that 's 200 . and what 's this average ? well , 4 times 84 , we already said , is 336 . plus 200 over 6 . so that 's 536 over 6 . 6 goes into 5 36 . i do n't know if if i gave myself enough space . but 6 goes into 53 , 8 times . 48 . 56 . 9 times . 9 times 6 is 54 . 6 minus is 20 6 goes into -- so we 'll see it 's actually 89.333333 , goes on forever . so 89.3 repeating . so no matter how hard i try in this class , the best i can do . because i only have two exams left , even if i were to get 100 on the next two exams . i can finish the class with an 89.333 average . hopefully , i think some of this might have been a little bit of a review for you . you already had kind of a sense of what an average is . and hopefully these last two problems not only taught you how to do some algebra problems involving average , but they 'll also help you figure out how well you have to do on your exams to get an a in your math class . i think you 're now ready for the average module . have fun .
|
welcome to the presentation on averages . averages is probably a concept that you 've already used before , maybe not in a mathematical way .
|
is averages how they grade your tests and sports skills ?
|
welcome to the presentation on averages . averages is probably a concept that you 've already used before , maybe not in a mathematical way . but people will talk in terms of , the average voter wants a politician to do this , or the average student in a class wants to get out early . so you 're probably already familiar with the concept of an average . and you probably already intuitively knew that an average is just a number that represents the different values that a group could have . but it can represent that as one number as opposed to giving all the different values . and let 's give a couple of examples of how to compute an average , and you might already know how to do this . so let 's say i had the numbers 1 , 3 , 5 , and 20 . and i asked you , what is the average of these four numbers ? well , what we do is , we literally just add up the numbers . and then divide by the number of numbers we have . so we say 1 plus 3 is 4 . so let me write that . 1 plus 3 plus 5 plus 20 equals , let 's see , 1 plus 3 is 4 . 4 plus 5 is 9 . 9 plus 20 is 29 . and we had 4 numbers ; one , two , three , four . so 4 goes into 29 . and it goes , 7 , 7 , 28 . and then we have 10 , i did n't have to do that decimal there , oh well . 2 , 8 , 25 . so 4 goes into 29 7.25 times . so the average of these four numbers is equal to 7.25 . and that might make sense to you because 7.25 is someplace in between these numbers . and we can kind of view this , 7.25 , as one way to represent these four numbers without having to list these four numbers . there are other representations you 'll learn later on . like the mode . you 'll also the mean , which we 'll talk about later , is actually the same thing as the average . but the average is just one number that you can use to represent a set of numbers . so let 's do some problems which i think are going to be close to your heart . let 's say on the first four tests of an exam , i got a -- let 's see , i got an 80 , an 81 . an 87 , and an 88 . what 's my average in the class so far ? well , all i have to do is add up these four numbers . so i say , 80 plus 81 plus 87 plus 88 . well , zero plus 1 is 1 . 1 plus 7 is 8 . 8 plus 8 is 16 . i just ran eight miles , so i 'm a bit tired . and , 4/8 , so that 's 32 . plus 1 is 33 . and now we divide this number by 4 . 4 goes into 336 . goes into 33 , 8 times . 8 times 4 is 32 . 33 minus 32 is 1 , 16 . 4 . so the average is equal to 84 . so depending on what school you go to that 's either a b or a c. so , so far my average after the first four exams is an 84 . now let 's make this a little bit more difficult . we know that the average after four exams , at four exams , is equal to 84 . if i were to ask you what do i have to get on the next test to average an 88 , to average an 88 in the class . so let 's say that x is what i get on the next test . so now what we can say is , is that the first four exams , i could either list out the first four exams that i took . or i already know what the average is . so i know the sum of the first four exams is going to 4 times 84 . and now i want to add the , what i get on the 5th exam , x . and i 'm going to divide that by all five exams . so in other words , this number is the average of my first five exams . we just figured out the average of the first four exams . but now , we sum up the first four exams here . we add what i got on the fifth exam , and then we divide it by 5 , because now we 're averaging five exams . and i said that i need to get in an 88 in the class . and now we solve for x . let me make some space here . so , 5 times 88 is , let 's see . 5 times 80 is 400 , so it 's 440 . 440 equals 4 times 84 , we just saw that , is 320 plus 16 is 336 . 336 plus x is equal to 440 . well , it turns out if you subtract 336 from both sides , you get x is equal to 104 . so unless you have a exam that has some bonus problems on it , it 's probably impossible for you to get ah an 88 average in the class after just the next exam . you 'd have to get 104 on that next exam . and let 's just look at what we just did . we said , after 4 exams we had an 84 . what do i have to get on that next exam to average an 88 in the class after 5 exams ? and that 's what we solved for when we got x . now , let 's ask another question . i said after four exams , after four exams , i had an 84 average . if i said that there are 6 exams in the class , and the highest score i could get on an exam is 100 , what is the highest average i can finish in the class if i were to really study hard and get 100 on the next 2 exams ? well , once again , what we 'll want to do is assume we get 100 on the next 2 exams and then take the average . so we 'll have to solve all 6 exams . so we 're going to have the average of 6 , so in the denominator we 're going to have 6 . the first four exams , the sum , as we already learned , is 4 exams times the 84 average . and this dot is just times . plus , and there 's going to be 2 more exams , right ? because there 's 6 exams in the class . and i 'm going to get 100 in each . so that 's 200 . and what 's this average ? well , 4 times 84 , we already said , is 336 . plus 200 over 6 . so that 's 536 over 6 . 6 goes into 5 36 . i do n't know if if i gave myself enough space . but 6 goes into 53 , 8 times . 48 . 56 . 9 times . 9 times 6 is 54 . 6 minus is 20 6 goes into -- so we 'll see it 's actually 89.333333 , goes on forever . so 89.3 repeating . so no matter how hard i try in this class , the best i can do . because i only have two exams left , even if i were to get 100 on the next two exams . i can finish the class with an 89.333 average . hopefully , i think some of this might have been a little bit of a review for you . you already had kind of a sense of what an average is . and hopefully these last two problems not only taught you how to do some algebra problems involving average , but they 'll also help you figure out how well you have to do on your exams to get an a in your math class . i think you 're now ready for the average module . have fun .
|
33 minus 32 is 1 , 16 . 4 . so the average is equal to 84 .
|
why sal does n't make a shortcut , and use the average of the past 4 exams as is ?
|
welcome to the presentation on averages . averages is probably a concept that you 've already used before , maybe not in a mathematical way . but people will talk in terms of , the average voter wants a politician to do this , or the average student in a class wants to get out early . so you 're probably already familiar with the concept of an average . and you probably already intuitively knew that an average is just a number that represents the different values that a group could have . but it can represent that as one number as opposed to giving all the different values . and let 's give a couple of examples of how to compute an average , and you might already know how to do this . so let 's say i had the numbers 1 , 3 , 5 , and 20 . and i asked you , what is the average of these four numbers ? well , what we do is , we literally just add up the numbers . and then divide by the number of numbers we have . so we say 1 plus 3 is 4 . so let me write that . 1 plus 3 plus 5 plus 20 equals , let 's see , 1 plus 3 is 4 . 4 plus 5 is 9 . 9 plus 20 is 29 . and we had 4 numbers ; one , two , three , four . so 4 goes into 29 . and it goes , 7 , 7 , 28 . and then we have 10 , i did n't have to do that decimal there , oh well . 2 , 8 , 25 . so 4 goes into 29 7.25 times . so the average of these four numbers is equal to 7.25 . and that might make sense to you because 7.25 is someplace in between these numbers . and we can kind of view this , 7.25 , as one way to represent these four numbers without having to list these four numbers . there are other representations you 'll learn later on . like the mode . you 'll also the mean , which we 'll talk about later , is actually the same thing as the average . but the average is just one number that you can use to represent a set of numbers . so let 's do some problems which i think are going to be close to your heart . let 's say on the first four tests of an exam , i got a -- let 's see , i got an 80 , an 81 . an 87 , and an 88 . what 's my average in the class so far ? well , all i have to do is add up these four numbers . so i say , 80 plus 81 plus 87 plus 88 . well , zero plus 1 is 1 . 1 plus 7 is 8 . 8 plus 8 is 16 . i just ran eight miles , so i 'm a bit tired . and , 4/8 , so that 's 32 . plus 1 is 33 . and now we divide this number by 4 . 4 goes into 336 . goes into 33 , 8 times . 8 times 4 is 32 . 33 minus 32 is 1 , 16 . 4 . so the average is equal to 84 . so depending on what school you go to that 's either a b or a c. so , so far my average after the first four exams is an 84 . now let 's make this a little bit more difficult . we know that the average after four exams , at four exams , is equal to 84 . if i were to ask you what do i have to get on the next test to average an 88 , to average an 88 in the class . so let 's say that x is what i get on the next test . so now what we can say is , is that the first four exams , i could either list out the first four exams that i took . or i already know what the average is . so i know the sum of the first four exams is going to 4 times 84 . and now i want to add the , what i get on the 5th exam , x . and i 'm going to divide that by all five exams . so in other words , this number is the average of my first five exams . we just figured out the average of the first four exams . but now , we sum up the first four exams here . we add what i got on the fifth exam , and then we divide it by 5 , because now we 're averaging five exams . and i said that i need to get in an 88 in the class . and now we solve for x . let me make some space here . so , 5 times 88 is , let 's see . 5 times 80 is 400 , so it 's 440 . 440 equals 4 times 84 , we just saw that , is 320 plus 16 is 336 . 336 plus x is equal to 440 . well , it turns out if you subtract 336 from both sides , you get x is equal to 104 . so unless you have a exam that has some bonus problems on it , it 's probably impossible for you to get ah an 88 average in the class after just the next exam . you 'd have to get 104 on that next exam . and let 's just look at what we just did . we said , after 4 exams we had an 84 . what do i have to get on that next exam to average an 88 in the class after 5 exams ? and that 's what we solved for when we got x . now , let 's ask another question . i said after four exams , after four exams , i had an 84 average . if i said that there are 6 exams in the class , and the highest score i could get on an exam is 100 , what is the highest average i can finish in the class if i were to really study hard and get 100 on the next 2 exams ? well , once again , what we 'll want to do is assume we get 100 on the next 2 exams and then take the average . so we 'll have to solve all 6 exams . so we 're going to have the average of 6 , so in the denominator we 're going to have 6 . the first four exams , the sum , as we already learned , is 4 exams times the 84 average . and this dot is just times . plus , and there 's going to be 2 more exams , right ? because there 's 6 exams in the class . and i 'm going to get 100 in each . so that 's 200 . and what 's this average ? well , 4 times 84 , we already said , is 336 . plus 200 over 6 . so that 's 536 over 6 . 6 goes into 5 36 . i do n't know if if i gave myself enough space . but 6 goes into 53 , 8 times . 48 . 56 . 9 times . 9 times 6 is 54 . 6 minus is 20 6 goes into -- so we 'll see it 's actually 89.333333 , goes on forever . so 89.3 repeating . so no matter how hard i try in this class , the best i can do . because i only have two exams left , even if i were to get 100 on the next two exams . i can finish the class with an 89.333 average . hopefully , i think some of this might have been a little bit of a review for you . you already had kind of a sense of what an average is . and hopefully these last two problems not only taught you how to do some algebra problems involving average , but they 'll also help you figure out how well you have to do on your exams to get an a in your math class . i think you 're now ready for the average module . have fun .
|
so that 's 200 . and what 's this average ? well , 4 times 84 , we already said , is 336 .
|
how do you do average of averages ?
|
welcome to the presentation on averages . averages is probably a concept that you 've already used before , maybe not in a mathematical way . but people will talk in terms of , the average voter wants a politician to do this , or the average student in a class wants to get out early . so you 're probably already familiar with the concept of an average . and you probably already intuitively knew that an average is just a number that represents the different values that a group could have . but it can represent that as one number as opposed to giving all the different values . and let 's give a couple of examples of how to compute an average , and you might already know how to do this . so let 's say i had the numbers 1 , 3 , 5 , and 20 . and i asked you , what is the average of these four numbers ? well , what we do is , we literally just add up the numbers . and then divide by the number of numbers we have . so we say 1 plus 3 is 4 . so let me write that . 1 plus 3 plus 5 plus 20 equals , let 's see , 1 plus 3 is 4 . 4 plus 5 is 9 . 9 plus 20 is 29 . and we had 4 numbers ; one , two , three , four . so 4 goes into 29 . and it goes , 7 , 7 , 28 . and then we have 10 , i did n't have to do that decimal there , oh well . 2 , 8 , 25 . so 4 goes into 29 7.25 times . so the average of these four numbers is equal to 7.25 . and that might make sense to you because 7.25 is someplace in between these numbers . and we can kind of view this , 7.25 , as one way to represent these four numbers without having to list these four numbers . there are other representations you 'll learn later on . like the mode . you 'll also the mean , which we 'll talk about later , is actually the same thing as the average . but the average is just one number that you can use to represent a set of numbers . so let 's do some problems which i think are going to be close to your heart . let 's say on the first four tests of an exam , i got a -- let 's see , i got an 80 , an 81 . an 87 , and an 88 . what 's my average in the class so far ? well , all i have to do is add up these four numbers . so i say , 80 plus 81 plus 87 plus 88 . well , zero plus 1 is 1 . 1 plus 7 is 8 . 8 plus 8 is 16 . i just ran eight miles , so i 'm a bit tired . and , 4/8 , so that 's 32 . plus 1 is 33 . and now we divide this number by 4 . 4 goes into 336 . goes into 33 , 8 times . 8 times 4 is 32 . 33 minus 32 is 1 , 16 . 4 . so the average is equal to 84 . so depending on what school you go to that 's either a b or a c. so , so far my average after the first four exams is an 84 . now let 's make this a little bit more difficult . we know that the average after four exams , at four exams , is equal to 84 . if i were to ask you what do i have to get on the next test to average an 88 , to average an 88 in the class . so let 's say that x is what i get on the next test . so now what we can say is , is that the first four exams , i could either list out the first four exams that i took . or i already know what the average is . so i know the sum of the first four exams is going to 4 times 84 . and now i want to add the , what i get on the 5th exam , x . and i 'm going to divide that by all five exams . so in other words , this number is the average of my first five exams . we just figured out the average of the first four exams . but now , we sum up the first four exams here . we add what i got on the fifth exam , and then we divide it by 5 , because now we 're averaging five exams . and i said that i need to get in an 88 in the class . and now we solve for x . let me make some space here . so , 5 times 88 is , let 's see . 5 times 80 is 400 , so it 's 440 . 440 equals 4 times 84 , we just saw that , is 320 plus 16 is 336 . 336 plus x is equal to 440 . well , it turns out if you subtract 336 from both sides , you get x is equal to 104 . so unless you have a exam that has some bonus problems on it , it 's probably impossible for you to get ah an 88 average in the class after just the next exam . you 'd have to get 104 on that next exam . and let 's just look at what we just did . we said , after 4 exams we had an 84 . what do i have to get on that next exam to average an 88 in the class after 5 exams ? and that 's what we solved for when we got x . now , let 's ask another question . i said after four exams , after four exams , i had an 84 average . if i said that there are 6 exams in the class , and the highest score i could get on an exam is 100 , what is the highest average i can finish in the class if i were to really study hard and get 100 on the next 2 exams ? well , once again , what we 'll want to do is assume we get 100 on the next 2 exams and then take the average . so we 'll have to solve all 6 exams . so we 're going to have the average of 6 , so in the denominator we 're going to have 6 . the first four exams , the sum , as we already learned , is 4 exams times the 84 average . and this dot is just times . plus , and there 's going to be 2 more exams , right ? because there 's 6 exams in the class . and i 'm going to get 100 in each . so that 's 200 . and what 's this average ? well , 4 times 84 , we already said , is 336 . plus 200 over 6 . so that 's 536 over 6 . 6 goes into 5 36 . i do n't know if if i gave myself enough space . but 6 goes into 53 , 8 times . 48 . 56 . 9 times . 9 times 6 is 54 . 6 minus is 20 6 goes into -- so we 'll see it 's actually 89.333333 , goes on forever . so 89.3 repeating . so no matter how hard i try in this class , the best i can do . because i only have two exams left , even if i were to get 100 on the next two exams . i can finish the class with an 89.333 average . hopefully , i think some of this might have been a little bit of a review for you . you already had kind of a sense of what an average is . and hopefully these last two problems not only taught you how to do some algebra problems involving average , but they 'll also help you figure out how well you have to do on your exams to get an a in your math class . i think you 're now ready for the average module . have fun .
|
i said after four exams , after four exams , i had an 84 average . if i said that there are 6 exams in the class , and the highest score i could get on an exam is 100 , what is the highest average i can finish in the class if i were to really study hard and get 100 on the next 2 exams ? well , once again , what we 'll want to do is assume we get 100 on the next 2 exams and then take the average . so we 'll have to solve all 6 exams .
|
what is the minimum number of tests that zack has to score a 100 on to get the average score to at least 93 ?
|
welcome to the presentation on averages . averages is probably a concept that you 've already used before , maybe not in a mathematical way . but people will talk in terms of , the average voter wants a politician to do this , or the average student in a class wants to get out early . so you 're probably already familiar with the concept of an average . and you probably already intuitively knew that an average is just a number that represents the different values that a group could have . but it can represent that as one number as opposed to giving all the different values . and let 's give a couple of examples of how to compute an average , and you might already know how to do this . so let 's say i had the numbers 1 , 3 , 5 , and 20 . and i asked you , what is the average of these four numbers ? well , what we do is , we literally just add up the numbers . and then divide by the number of numbers we have . so we say 1 plus 3 is 4 . so let me write that . 1 plus 3 plus 5 plus 20 equals , let 's see , 1 plus 3 is 4 . 4 plus 5 is 9 . 9 plus 20 is 29 . and we had 4 numbers ; one , two , three , four . so 4 goes into 29 . and it goes , 7 , 7 , 28 . and then we have 10 , i did n't have to do that decimal there , oh well . 2 , 8 , 25 . so 4 goes into 29 7.25 times . so the average of these four numbers is equal to 7.25 . and that might make sense to you because 7.25 is someplace in between these numbers . and we can kind of view this , 7.25 , as one way to represent these four numbers without having to list these four numbers . there are other representations you 'll learn later on . like the mode . you 'll also the mean , which we 'll talk about later , is actually the same thing as the average . but the average is just one number that you can use to represent a set of numbers . so let 's do some problems which i think are going to be close to your heart . let 's say on the first four tests of an exam , i got a -- let 's see , i got an 80 , an 81 . an 87 , and an 88 . what 's my average in the class so far ? well , all i have to do is add up these four numbers . so i say , 80 plus 81 plus 87 plus 88 . well , zero plus 1 is 1 . 1 plus 7 is 8 . 8 plus 8 is 16 . i just ran eight miles , so i 'm a bit tired . and , 4/8 , so that 's 32 . plus 1 is 33 . and now we divide this number by 4 . 4 goes into 336 . goes into 33 , 8 times . 8 times 4 is 32 . 33 minus 32 is 1 , 16 . 4 . so the average is equal to 84 . so depending on what school you go to that 's either a b or a c. so , so far my average after the first four exams is an 84 . now let 's make this a little bit more difficult . we know that the average after four exams , at four exams , is equal to 84 . if i were to ask you what do i have to get on the next test to average an 88 , to average an 88 in the class . so let 's say that x is what i get on the next test . so now what we can say is , is that the first four exams , i could either list out the first four exams that i took . or i already know what the average is . so i know the sum of the first four exams is going to 4 times 84 . and now i want to add the , what i get on the 5th exam , x . and i 'm going to divide that by all five exams . so in other words , this number is the average of my first five exams . we just figured out the average of the first four exams . but now , we sum up the first four exams here . we add what i got on the fifth exam , and then we divide it by 5 , because now we 're averaging five exams . and i said that i need to get in an 88 in the class . and now we solve for x . let me make some space here . so , 5 times 88 is , let 's see . 5 times 80 is 400 , so it 's 440 . 440 equals 4 times 84 , we just saw that , is 320 plus 16 is 336 . 336 plus x is equal to 440 . well , it turns out if you subtract 336 from both sides , you get x is equal to 104 . so unless you have a exam that has some bonus problems on it , it 's probably impossible for you to get ah an 88 average in the class after just the next exam . you 'd have to get 104 on that next exam . and let 's just look at what we just did . we said , after 4 exams we had an 84 . what do i have to get on that next exam to average an 88 in the class after 5 exams ? and that 's what we solved for when we got x . now , let 's ask another question . i said after four exams , after four exams , i had an 84 average . if i said that there are 6 exams in the class , and the highest score i could get on an exam is 100 , what is the highest average i can finish in the class if i were to really study hard and get 100 on the next 2 exams ? well , once again , what we 'll want to do is assume we get 100 on the next 2 exams and then take the average . so we 'll have to solve all 6 exams . so we 're going to have the average of 6 , so in the denominator we 're going to have 6 . the first four exams , the sum , as we already learned , is 4 exams times the 84 average . and this dot is just times . plus , and there 's going to be 2 more exams , right ? because there 's 6 exams in the class . and i 'm going to get 100 in each . so that 's 200 . and what 's this average ? well , 4 times 84 , we already said , is 336 . plus 200 over 6 . so that 's 536 over 6 . 6 goes into 5 36 . i do n't know if if i gave myself enough space . but 6 goes into 53 , 8 times . 48 . 56 . 9 times . 9 times 6 is 54 . 6 minus is 20 6 goes into -- so we 'll see it 's actually 89.333333 , goes on forever . so 89.3 repeating . so no matter how hard i try in this class , the best i can do . because i only have two exams left , even if i were to get 100 on the next two exams . i can finish the class with an 89.333 average . hopefully , i think some of this might have been a little bit of a review for you . you already had kind of a sense of what an average is . and hopefully these last two problems not only taught you how to do some algebra problems involving average , but they 'll also help you figure out how well you have to do on your exams to get an a in your math class . i think you 're now ready for the average module . have fun .
|
440 equals 4 times 84 , we just saw that , is 320 plus 16 is 336 . 336 plus x is equal to 440 . well , it turns out if you subtract 336 from both sides , you get x is equal to 104 .
|
how did you get the 440 in the 3rd equation ?
|
welcome to the presentation on averages . averages is probably a concept that you 've already used before , maybe not in a mathematical way . but people will talk in terms of , the average voter wants a politician to do this , or the average student in a class wants to get out early . so you 're probably already familiar with the concept of an average . and you probably already intuitively knew that an average is just a number that represents the different values that a group could have . but it can represent that as one number as opposed to giving all the different values . and let 's give a couple of examples of how to compute an average , and you might already know how to do this . so let 's say i had the numbers 1 , 3 , 5 , and 20 . and i asked you , what is the average of these four numbers ? well , what we do is , we literally just add up the numbers . and then divide by the number of numbers we have . so we say 1 plus 3 is 4 . so let me write that . 1 plus 3 plus 5 plus 20 equals , let 's see , 1 plus 3 is 4 . 4 plus 5 is 9 . 9 plus 20 is 29 . and we had 4 numbers ; one , two , three , four . so 4 goes into 29 . and it goes , 7 , 7 , 28 . and then we have 10 , i did n't have to do that decimal there , oh well . 2 , 8 , 25 . so 4 goes into 29 7.25 times . so the average of these four numbers is equal to 7.25 . and that might make sense to you because 7.25 is someplace in between these numbers . and we can kind of view this , 7.25 , as one way to represent these four numbers without having to list these four numbers . there are other representations you 'll learn later on . like the mode . you 'll also the mean , which we 'll talk about later , is actually the same thing as the average . but the average is just one number that you can use to represent a set of numbers . so let 's do some problems which i think are going to be close to your heart . let 's say on the first four tests of an exam , i got a -- let 's see , i got an 80 , an 81 . an 87 , and an 88 . what 's my average in the class so far ? well , all i have to do is add up these four numbers . so i say , 80 plus 81 plus 87 plus 88 . well , zero plus 1 is 1 . 1 plus 7 is 8 . 8 plus 8 is 16 . i just ran eight miles , so i 'm a bit tired . and , 4/8 , so that 's 32 . plus 1 is 33 . and now we divide this number by 4 . 4 goes into 336 . goes into 33 , 8 times . 8 times 4 is 32 . 33 minus 32 is 1 , 16 . 4 . so the average is equal to 84 . so depending on what school you go to that 's either a b or a c. so , so far my average after the first four exams is an 84 . now let 's make this a little bit more difficult . we know that the average after four exams , at four exams , is equal to 84 . if i were to ask you what do i have to get on the next test to average an 88 , to average an 88 in the class . so let 's say that x is what i get on the next test . so now what we can say is , is that the first four exams , i could either list out the first four exams that i took . or i already know what the average is . so i know the sum of the first four exams is going to 4 times 84 . and now i want to add the , what i get on the 5th exam , x . and i 'm going to divide that by all five exams . so in other words , this number is the average of my first five exams . we just figured out the average of the first four exams . but now , we sum up the first four exams here . we add what i got on the fifth exam , and then we divide it by 5 , because now we 're averaging five exams . and i said that i need to get in an 88 in the class . and now we solve for x . let me make some space here . so , 5 times 88 is , let 's see . 5 times 80 is 400 , so it 's 440 . 440 equals 4 times 84 , we just saw that , is 320 plus 16 is 336 . 336 plus x is equal to 440 . well , it turns out if you subtract 336 from both sides , you get x is equal to 104 . so unless you have a exam that has some bonus problems on it , it 's probably impossible for you to get ah an 88 average in the class after just the next exam . you 'd have to get 104 on that next exam . and let 's just look at what we just did . we said , after 4 exams we had an 84 . what do i have to get on that next exam to average an 88 in the class after 5 exams ? and that 's what we solved for when we got x . now , let 's ask another question . i said after four exams , after four exams , i had an 84 average . if i said that there are 6 exams in the class , and the highest score i could get on an exam is 100 , what is the highest average i can finish in the class if i were to really study hard and get 100 on the next 2 exams ? well , once again , what we 'll want to do is assume we get 100 on the next 2 exams and then take the average . so we 'll have to solve all 6 exams . so we 're going to have the average of 6 , so in the denominator we 're going to have 6 . the first four exams , the sum , as we already learned , is 4 exams times the 84 average . and this dot is just times . plus , and there 's going to be 2 more exams , right ? because there 's 6 exams in the class . and i 'm going to get 100 in each . so that 's 200 . and what 's this average ? well , 4 times 84 , we already said , is 336 . plus 200 over 6 . so that 's 536 over 6 . 6 goes into 5 36 . i do n't know if if i gave myself enough space . but 6 goes into 53 , 8 times . 48 . 56 . 9 times . 9 times 6 is 54 . 6 minus is 20 6 goes into -- so we 'll see it 's actually 89.333333 , goes on forever . so 89.3 repeating . so no matter how hard i try in this class , the best i can do . because i only have two exams left , even if i were to get 100 on the next two exams . i can finish the class with an 89.333 average . hopefully , i think some of this might have been a little bit of a review for you . you already had kind of a sense of what an average is . and hopefully these last two problems not only taught you how to do some algebra problems involving average , but they 'll also help you figure out how well you have to do on your exams to get an a in your math class . i think you 're now ready for the average module . have fun .
|
so the average is equal to 84 . so depending on what school you go to that 's either a b or a c. so , so far my average after the first four exams is an 84 . now let 's make this a little bit more difficult .
|
is math mr. khan 's favorite subject is school ?
|
welcome to the presentation on averages . averages is probably a concept that you 've already used before , maybe not in a mathematical way . but people will talk in terms of , the average voter wants a politician to do this , or the average student in a class wants to get out early . so you 're probably already familiar with the concept of an average . and you probably already intuitively knew that an average is just a number that represents the different values that a group could have . but it can represent that as one number as opposed to giving all the different values . and let 's give a couple of examples of how to compute an average , and you might already know how to do this . so let 's say i had the numbers 1 , 3 , 5 , and 20 . and i asked you , what is the average of these four numbers ? well , what we do is , we literally just add up the numbers . and then divide by the number of numbers we have . so we say 1 plus 3 is 4 . so let me write that . 1 plus 3 plus 5 plus 20 equals , let 's see , 1 plus 3 is 4 . 4 plus 5 is 9 . 9 plus 20 is 29 . and we had 4 numbers ; one , two , three , four . so 4 goes into 29 . and it goes , 7 , 7 , 28 . and then we have 10 , i did n't have to do that decimal there , oh well . 2 , 8 , 25 . so 4 goes into 29 7.25 times . so the average of these four numbers is equal to 7.25 . and that might make sense to you because 7.25 is someplace in between these numbers . and we can kind of view this , 7.25 , as one way to represent these four numbers without having to list these four numbers . there are other representations you 'll learn later on . like the mode . you 'll also the mean , which we 'll talk about later , is actually the same thing as the average . but the average is just one number that you can use to represent a set of numbers . so let 's do some problems which i think are going to be close to your heart . let 's say on the first four tests of an exam , i got a -- let 's see , i got an 80 , an 81 . an 87 , and an 88 . what 's my average in the class so far ? well , all i have to do is add up these four numbers . so i say , 80 plus 81 plus 87 plus 88 . well , zero plus 1 is 1 . 1 plus 7 is 8 . 8 plus 8 is 16 . i just ran eight miles , so i 'm a bit tired . and , 4/8 , so that 's 32 . plus 1 is 33 . and now we divide this number by 4 . 4 goes into 336 . goes into 33 , 8 times . 8 times 4 is 32 . 33 minus 32 is 1 , 16 . 4 . so the average is equal to 84 . so depending on what school you go to that 's either a b or a c. so , so far my average after the first four exams is an 84 . now let 's make this a little bit more difficult . we know that the average after four exams , at four exams , is equal to 84 . if i were to ask you what do i have to get on the next test to average an 88 , to average an 88 in the class . so let 's say that x is what i get on the next test . so now what we can say is , is that the first four exams , i could either list out the first four exams that i took . or i already know what the average is . so i know the sum of the first four exams is going to 4 times 84 . and now i want to add the , what i get on the 5th exam , x . and i 'm going to divide that by all five exams . so in other words , this number is the average of my first five exams . we just figured out the average of the first four exams . but now , we sum up the first four exams here . we add what i got on the fifth exam , and then we divide it by 5 , because now we 're averaging five exams . and i said that i need to get in an 88 in the class . and now we solve for x . let me make some space here . so , 5 times 88 is , let 's see . 5 times 80 is 400 , so it 's 440 . 440 equals 4 times 84 , we just saw that , is 320 plus 16 is 336 . 336 plus x is equal to 440 . well , it turns out if you subtract 336 from both sides , you get x is equal to 104 . so unless you have a exam that has some bonus problems on it , it 's probably impossible for you to get ah an 88 average in the class after just the next exam . you 'd have to get 104 on that next exam . and let 's just look at what we just did . we said , after 4 exams we had an 84 . what do i have to get on that next exam to average an 88 in the class after 5 exams ? and that 's what we solved for when we got x . now , let 's ask another question . i said after four exams , after four exams , i had an 84 average . if i said that there are 6 exams in the class , and the highest score i could get on an exam is 100 , what is the highest average i can finish in the class if i were to really study hard and get 100 on the next 2 exams ? well , once again , what we 'll want to do is assume we get 100 on the next 2 exams and then take the average . so we 'll have to solve all 6 exams . so we 're going to have the average of 6 , so in the denominator we 're going to have 6 . the first four exams , the sum , as we already learned , is 4 exams times the 84 average . and this dot is just times . plus , and there 's going to be 2 more exams , right ? because there 's 6 exams in the class . and i 'm going to get 100 in each . so that 's 200 . and what 's this average ? well , 4 times 84 , we already said , is 336 . plus 200 over 6 . so that 's 536 over 6 . 6 goes into 5 36 . i do n't know if if i gave myself enough space . but 6 goes into 53 , 8 times . 48 . 56 . 9 times . 9 times 6 is 54 . 6 minus is 20 6 goes into -- so we 'll see it 's actually 89.333333 , goes on forever . so 89.3 repeating . so no matter how hard i try in this class , the best i can do . because i only have two exams left , even if i were to get 100 on the next two exams . i can finish the class with an 89.333 average . hopefully , i think some of this might have been a little bit of a review for you . you already had kind of a sense of what an average is . and hopefully these last two problems not only taught you how to do some algebra problems involving average , but they 'll also help you figure out how well you have to do on your exams to get an a in your math class . i think you 're now ready for the average module . have fun .
|
so now what we can say is , is that the first four exams , i could either list out the first four exams that i took . or i already know what the average is . so i know the sum of the first four exams is going to 4 times 84 .
|
why would you need to know percentages ?
|
welcome to the presentation on averages . averages is probably a concept that you 've already used before , maybe not in a mathematical way . but people will talk in terms of , the average voter wants a politician to do this , or the average student in a class wants to get out early . so you 're probably already familiar with the concept of an average . and you probably already intuitively knew that an average is just a number that represents the different values that a group could have . but it can represent that as one number as opposed to giving all the different values . and let 's give a couple of examples of how to compute an average , and you might already know how to do this . so let 's say i had the numbers 1 , 3 , 5 , and 20 . and i asked you , what is the average of these four numbers ? well , what we do is , we literally just add up the numbers . and then divide by the number of numbers we have . so we say 1 plus 3 is 4 . so let me write that . 1 plus 3 plus 5 plus 20 equals , let 's see , 1 plus 3 is 4 . 4 plus 5 is 9 . 9 plus 20 is 29 . and we had 4 numbers ; one , two , three , four . so 4 goes into 29 . and it goes , 7 , 7 , 28 . and then we have 10 , i did n't have to do that decimal there , oh well . 2 , 8 , 25 . so 4 goes into 29 7.25 times . so the average of these four numbers is equal to 7.25 . and that might make sense to you because 7.25 is someplace in between these numbers . and we can kind of view this , 7.25 , as one way to represent these four numbers without having to list these four numbers . there are other representations you 'll learn later on . like the mode . you 'll also the mean , which we 'll talk about later , is actually the same thing as the average . but the average is just one number that you can use to represent a set of numbers . so let 's do some problems which i think are going to be close to your heart . let 's say on the first four tests of an exam , i got a -- let 's see , i got an 80 , an 81 . an 87 , and an 88 . what 's my average in the class so far ? well , all i have to do is add up these four numbers . so i say , 80 plus 81 plus 87 plus 88 . well , zero plus 1 is 1 . 1 plus 7 is 8 . 8 plus 8 is 16 . i just ran eight miles , so i 'm a bit tired . and , 4/8 , so that 's 32 . plus 1 is 33 . and now we divide this number by 4 . 4 goes into 336 . goes into 33 , 8 times . 8 times 4 is 32 . 33 minus 32 is 1 , 16 . 4 . so the average is equal to 84 . so depending on what school you go to that 's either a b or a c. so , so far my average after the first four exams is an 84 . now let 's make this a little bit more difficult . we know that the average after four exams , at four exams , is equal to 84 . if i were to ask you what do i have to get on the next test to average an 88 , to average an 88 in the class . so let 's say that x is what i get on the next test . so now what we can say is , is that the first four exams , i could either list out the first four exams that i took . or i already know what the average is . so i know the sum of the first four exams is going to 4 times 84 . and now i want to add the , what i get on the 5th exam , x . and i 'm going to divide that by all five exams . so in other words , this number is the average of my first five exams . we just figured out the average of the first four exams . but now , we sum up the first four exams here . we add what i got on the fifth exam , and then we divide it by 5 , because now we 're averaging five exams . and i said that i need to get in an 88 in the class . and now we solve for x . let me make some space here . so , 5 times 88 is , let 's see . 5 times 80 is 400 , so it 's 440 . 440 equals 4 times 84 , we just saw that , is 320 plus 16 is 336 . 336 plus x is equal to 440 . well , it turns out if you subtract 336 from both sides , you get x is equal to 104 . so unless you have a exam that has some bonus problems on it , it 's probably impossible for you to get ah an 88 average in the class after just the next exam . you 'd have to get 104 on that next exam . and let 's just look at what we just did . we said , after 4 exams we had an 84 . what do i have to get on that next exam to average an 88 in the class after 5 exams ? and that 's what we solved for when we got x . now , let 's ask another question . i said after four exams , after four exams , i had an 84 average . if i said that there are 6 exams in the class , and the highest score i could get on an exam is 100 , what is the highest average i can finish in the class if i were to really study hard and get 100 on the next 2 exams ? well , once again , what we 'll want to do is assume we get 100 on the next 2 exams and then take the average . so we 'll have to solve all 6 exams . so we 're going to have the average of 6 , so in the denominator we 're going to have 6 . the first four exams , the sum , as we already learned , is 4 exams times the 84 average . and this dot is just times . plus , and there 's going to be 2 more exams , right ? because there 's 6 exams in the class . and i 'm going to get 100 in each . so that 's 200 . and what 's this average ? well , 4 times 84 , we already said , is 336 . plus 200 over 6 . so that 's 536 over 6 . 6 goes into 5 36 . i do n't know if if i gave myself enough space . but 6 goes into 53 , 8 times . 48 . 56 . 9 times . 9 times 6 is 54 . 6 minus is 20 6 goes into -- so we 'll see it 's actually 89.333333 , goes on forever . so 89.3 repeating . so no matter how hard i try in this class , the best i can do . because i only have two exams left , even if i were to get 100 on the next two exams . i can finish the class with an 89.333 average . hopefully , i think some of this might have been a little bit of a review for you . you already had kind of a sense of what an average is . and hopefully these last two problems not only taught you how to do some algebra problems involving average , but they 'll also help you figure out how well you have to do on your exams to get an a in your math class . i think you 're now ready for the average module . have fun .
|
so that 's 200 . and what 's this average ? well , 4 times 84 , we already said , is 336 .
|
can an average be negative ?
|
welcome to the presentation on averages . averages is probably a concept that you 've already used before , maybe not in a mathematical way . but people will talk in terms of , the average voter wants a politician to do this , or the average student in a class wants to get out early . so you 're probably already familiar with the concept of an average . and you probably already intuitively knew that an average is just a number that represents the different values that a group could have . but it can represent that as one number as opposed to giving all the different values . and let 's give a couple of examples of how to compute an average , and you might already know how to do this . so let 's say i had the numbers 1 , 3 , 5 , and 20 . and i asked you , what is the average of these four numbers ? well , what we do is , we literally just add up the numbers . and then divide by the number of numbers we have . so we say 1 plus 3 is 4 . so let me write that . 1 plus 3 plus 5 plus 20 equals , let 's see , 1 plus 3 is 4 . 4 plus 5 is 9 . 9 plus 20 is 29 . and we had 4 numbers ; one , two , three , four . so 4 goes into 29 . and it goes , 7 , 7 , 28 . and then we have 10 , i did n't have to do that decimal there , oh well . 2 , 8 , 25 . so 4 goes into 29 7.25 times . so the average of these four numbers is equal to 7.25 . and that might make sense to you because 7.25 is someplace in between these numbers . and we can kind of view this , 7.25 , as one way to represent these four numbers without having to list these four numbers . there are other representations you 'll learn later on . like the mode . you 'll also the mean , which we 'll talk about later , is actually the same thing as the average . but the average is just one number that you can use to represent a set of numbers . so let 's do some problems which i think are going to be close to your heart . let 's say on the first four tests of an exam , i got a -- let 's see , i got an 80 , an 81 . an 87 , and an 88 . what 's my average in the class so far ? well , all i have to do is add up these four numbers . so i say , 80 plus 81 plus 87 plus 88 . well , zero plus 1 is 1 . 1 plus 7 is 8 . 8 plus 8 is 16 . i just ran eight miles , so i 'm a bit tired . and , 4/8 , so that 's 32 . plus 1 is 33 . and now we divide this number by 4 . 4 goes into 336 . goes into 33 , 8 times . 8 times 4 is 32 . 33 minus 32 is 1 , 16 . 4 . so the average is equal to 84 . so depending on what school you go to that 's either a b or a c. so , so far my average after the first four exams is an 84 . now let 's make this a little bit more difficult . we know that the average after four exams , at four exams , is equal to 84 . if i were to ask you what do i have to get on the next test to average an 88 , to average an 88 in the class . so let 's say that x is what i get on the next test . so now what we can say is , is that the first four exams , i could either list out the first four exams that i took . or i already know what the average is . so i know the sum of the first four exams is going to 4 times 84 . and now i want to add the , what i get on the 5th exam , x . and i 'm going to divide that by all five exams . so in other words , this number is the average of my first five exams . we just figured out the average of the first four exams . but now , we sum up the first four exams here . we add what i got on the fifth exam , and then we divide it by 5 , because now we 're averaging five exams . and i said that i need to get in an 88 in the class . and now we solve for x . let me make some space here . so , 5 times 88 is , let 's see . 5 times 80 is 400 , so it 's 440 . 440 equals 4 times 84 , we just saw that , is 320 plus 16 is 336 . 336 plus x is equal to 440 . well , it turns out if you subtract 336 from both sides , you get x is equal to 104 . so unless you have a exam that has some bonus problems on it , it 's probably impossible for you to get ah an 88 average in the class after just the next exam . you 'd have to get 104 on that next exam . and let 's just look at what we just did . we said , after 4 exams we had an 84 . what do i have to get on that next exam to average an 88 in the class after 5 exams ? and that 's what we solved for when we got x . now , let 's ask another question . i said after four exams , after four exams , i had an 84 average . if i said that there are 6 exams in the class , and the highest score i could get on an exam is 100 , what is the highest average i can finish in the class if i were to really study hard and get 100 on the next 2 exams ? well , once again , what we 'll want to do is assume we get 100 on the next 2 exams and then take the average . so we 'll have to solve all 6 exams . so we 're going to have the average of 6 , so in the denominator we 're going to have 6 . the first four exams , the sum , as we already learned , is 4 exams times the 84 average . and this dot is just times . plus , and there 's going to be 2 more exams , right ? because there 's 6 exams in the class . and i 'm going to get 100 in each . so that 's 200 . and what 's this average ? well , 4 times 84 , we already said , is 336 . plus 200 over 6 . so that 's 536 over 6 . 6 goes into 5 36 . i do n't know if if i gave myself enough space . but 6 goes into 53 , 8 times . 48 . 56 . 9 times . 9 times 6 is 54 . 6 minus is 20 6 goes into -- so we 'll see it 's actually 89.333333 , goes on forever . so 89.3 repeating . so no matter how hard i try in this class , the best i can do . because i only have two exams left , even if i were to get 100 on the next two exams . i can finish the class with an 89.333 average . hopefully , i think some of this might have been a little bit of a review for you . you already had kind of a sense of what an average is . and hopefully these last two problems not only taught you how to do some algebra problems involving average , but they 'll also help you figure out how well you have to do on your exams to get an a in your math class . i think you 're now ready for the average module . have fun .
|
so that 's 200 . and what 's this average ? well , 4 times 84 , we already said , is 336 .
|
if the average weight of the cartons is 10 , the heavier carton weighs how much more than the lighter carton ?
|
welcome to the presentation on averages . averages is probably a concept that you 've already used before , maybe not in a mathematical way . but people will talk in terms of , the average voter wants a politician to do this , or the average student in a class wants to get out early . so you 're probably already familiar with the concept of an average . and you probably already intuitively knew that an average is just a number that represents the different values that a group could have . but it can represent that as one number as opposed to giving all the different values . and let 's give a couple of examples of how to compute an average , and you might already know how to do this . so let 's say i had the numbers 1 , 3 , 5 , and 20 . and i asked you , what is the average of these four numbers ? well , what we do is , we literally just add up the numbers . and then divide by the number of numbers we have . so we say 1 plus 3 is 4 . so let me write that . 1 plus 3 plus 5 plus 20 equals , let 's see , 1 plus 3 is 4 . 4 plus 5 is 9 . 9 plus 20 is 29 . and we had 4 numbers ; one , two , three , four . so 4 goes into 29 . and it goes , 7 , 7 , 28 . and then we have 10 , i did n't have to do that decimal there , oh well . 2 , 8 , 25 . so 4 goes into 29 7.25 times . so the average of these four numbers is equal to 7.25 . and that might make sense to you because 7.25 is someplace in between these numbers . and we can kind of view this , 7.25 , as one way to represent these four numbers without having to list these four numbers . there are other representations you 'll learn later on . like the mode . you 'll also the mean , which we 'll talk about later , is actually the same thing as the average . but the average is just one number that you can use to represent a set of numbers . so let 's do some problems which i think are going to be close to your heart . let 's say on the first four tests of an exam , i got a -- let 's see , i got an 80 , an 81 . an 87 , and an 88 . what 's my average in the class so far ? well , all i have to do is add up these four numbers . so i say , 80 plus 81 plus 87 plus 88 . well , zero plus 1 is 1 . 1 plus 7 is 8 . 8 plus 8 is 16 . i just ran eight miles , so i 'm a bit tired . and , 4/8 , so that 's 32 . plus 1 is 33 . and now we divide this number by 4 . 4 goes into 336 . goes into 33 , 8 times . 8 times 4 is 32 . 33 minus 32 is 1 , 16 . 4 . so the average is equal to 84 . so depending on what school you go to that 's either a b or a c. so , so far my average after the first four exams is an 84 . now let 's make this a little bit more difficult . we know that the average after four exams , at four exams , is equal to 84 . if i were to ask you what do i have to get on the next test to average an 88 , to average an 88 in the class . so let 's say that x is what i get on the next test . so now what we can say is , is that the first four exams , i could either list out the first four exams that i took . or i already know what the average is . so i know the sum of the first four exams is going to 4 times 84 . and now i want to add the , what i get on the 5th exam , x . and i 'm going to divide that by all five exams . so in other words , this number is the average of my first five exams . we just figured out the average of the first four exams . but now , we sum up the first four exams here . we add what i got on the fifth exam , and then we divide it by 5 , because now we 're averaging five exams . and i said that i need to get in an 88 in the class . and now we solve for x . let me make some space here . so , 5 times 88 is , let 's see . 5 times 80 is 400 , so it 's 440 . 440 equals 4 times 84 , we just saw that , is 320 plus 16 is 336 . 336 plus x is equal to 440 . well , it turns out if you subtract 336 from both sides , you get x is equal to 104 . so unless you have a exam that has some bonus problems on it , it 's probably impossible for you to get ah an 88 average in the class after just the next exam . you 'd have to get 104 on that next exam . and let 's just look at what we just did . we said , after 4 exams we had an 84 . what do i have to get on that next exam to average an 88 in the class after 5 exams ? and that 's what we solved for when we got x . now , let 's ask another question . i said after four exams , after four exams , i had an 84 average . if i said that there are 6 exams in the class , and the highest score i could get on an exam is 100 , what is the highest average i can finish in the class if i were to really study hard and get 100 on the next 2 exams ? well , once again , what we 'll want to do is assume we get 100 on the next 2 exams and then take the average . so we 'll have to solve all 6 exams . so we 're going to have the average of 6 , so in the denominator we 're going to have 6 . the first four exams , the sum , as we already learned , is 4 exams times the 84 average . and this dot is just times . plus , and there 's going to be 2 more exams , right ? because there 's 6 exams in the class . and i 'm going to get 100 in each . so that 's 200 . and what 's this average ? well , 4 times 84 , we already said , is 336 . plus 200 over 6 . so that 's 536 over 6 . 6 goes into 5 36 . i do n't know if if i gave myself enough space . but 6 goes into 53 , 8 times . 48 . 56 . 9 times . 9 times 6 is 54 . 6 minus is 20 6 goes into -- so we 'll see it 's actually 89.333333 , goes on forever . so 89.3 repeating . so no matter how hard i try in this class , the best i can do . because i only have two exams left , even if i were to get 100 on the next two exams . i can finish the class with an 89.333 average . hopefully , i think some of this might have been a little bit of a review for you . you already had kind of a sense of what an average is . and hopefully these last two problems not only taught you how to do some algebra problems involving average , but they 'll also help you figure out how well you have to do on your exams to get an a in your math class . i think you 're now ready for the average module . have fun .
|
so that 's 200 . and what 's this average ? well , 4 times 84 , we already said , is 336 .
|
is an average the same as the arithmatic mean ?
|
welcome to the presentation on averages . averages is probably a concept that you 've already used before , maybe not in a mathematical way . but people will talk in terms of , the average voter wants a politician to do this , or the average student in a class wants to get out early . so you 're probably already familiar with the concept of an average . and you probably already intuitively knew that an average is just a number that represents the different values that a group could have . but it can represent that as one number as opposed to giving all the different values . and let 's give a couple of examples of how to compute an average , and you might already know how to do this . so let 's say i had the numbers 1 , 3 , 5 , and 20 . and i asked you , what is the average of these four numbers ? well , what we do is , we literally just add up the numbers . and then divide by the number of numbers we have . so we say 1 plus 3 is 4 . so let me write that . 1 plus 3 plus 5 plus 20 equals , let 's see , 1 plus 3 is 4 . 4 plus 5 is 9 . 9 plus 20 is 29 . and we had 4 numbers ; one , two , three , four . so 4 goes into 29 . and it goes , 7 , 7 , 28 . and then we have 10 , i did n't have to do that decimal there , oh well . 2 , 8 , 25 . so 4 goes into 29 7.25 times . so the average of these four numbers is equal to 7.25 . and that might make sense to you because 7.25 is someplace in between these numbers . and we can kind of view this , 7.25 , as one way to represent these four numbers without having to list these four numbers . there are other representations you 'll learn later on . like the mode . you 'll also the mean , which we 'll talk about later , is actually the same thing as the average . but the average is just one number that you can use to represent a set of numbers . so let 's do some problems which i think are going to be close to your heart . let 's say on the first four tests of an exam , i got a -- let 's see , i got an 80 , an 81 . an 87 , and an 88 . what 's my average in the class so far ? well , all i have to do is add up these four numbers . so i say , 80 plus 81 plus 87 plus 88 . well , zero plus 1 is 1 . 1 plus 7 is 8 . 8 plus 8 is 16 . i just ran eight miles , so i 'm a bit tired . and , 4/8 , so that 's 32 . plus 1 is 33 . and now we divide this number by 4 . 4 goes into 336 . goes into 33 , 8 times . 8 times 4 is 32 . 33 minus 32 is 1 , 16 . 4 . so the average is equal to 84 . so depending on what school you go to that 's either a b or a c. so , so far my average after the first four exams is an 84 . now let 's make this a little bit more difficult . we know that the average after four exams , at four exams , is equal to 84 . if i were to ask you what do i have to get on the next test to average an 88 , to average an 88 in the class . so let 's say that x is what i get on the next test . so now what we can say is , is that the first four exams , i could either list out the first four exams that i took . or i already know what the average is . so i know the sum of the first four exams is going to 4 times 84 . and now i want to add the , what i get on the 5th exam , x . and i 'm going to divide that by all five exams . so in other words , this number is the average of my first five exams . we just figured out the average of the first four exams . but now , we sum up the first four exams here . we add what i got on the fifth exam , and then we divide it by 5 , because now we 're averaging five exams . and i said that i need to get in an 88 in the class . and now we solve for x . let me make some space here . so , 5 times 88 is , let 's see . 5 times 80 is 400 , so it 's 440 . 440 equals 4 times 84 , we just saw that , is 320 plus 16 is 336 . 336 plus x is equal to 440 . well , it turns out if you subtract 336 from both sides , you get x is equal to 104 . so unless you have a exam that has some bonus problems on it , it 's probably impossible for you to get ah an 88 average in the class after just the next exam . you 'd have to get 104 on that next exam . and let 's just look at what we just did . we said , after 4 exams we had an 84 . what do i have to get on that next exam to average an 88 in the class after 5 exams ? and that 's what we solved for when we got x . now , let 's ask another question . i said after four exams , after four exams , i had an 84 average . if i said that there are 6 exams in the class , and the highest score i could get on an exam is 100 , what is the highest average i can finish in the class if i were to really study hard and get 100 on the next 2 exams ? well , once again , what we 'll want to do is assume we get 100 on the next 2 exams and then take the average . so we 'll have to solve all 6 exams . so we 're going to have the average of 6 , so in the denominator we 're going to have 6 . the first four exams , the sum , as we already learned , is 4 exams times the 84 average . and this dot is just times . plus , and there 's going to be 2 more exams , right ? because there 's 6 exams in the class . and i 'm going to get 100 in each . so that 's 200 . and what 's this average ? well , 4 times 84 , we already said , is 336 . plus 200 over 6 . so that 's 536 over 6 . 6 goes into 5 36 . i do n't know if if i gave myself enough space . but 6 goes into 53 , 8 times . 48 . 56 . 9 times . 9 times 6 is 54 . 6 minus is 20 6 goes into -- so we 'll see it 's actually 89.333333 , goes on forever . so 89.3 repeating . so no matter how hard i try in this class , the best i can do . because i only have two exams left , even if i were to get 100 on the next two exams . i can finish the class with an 89.333 average . hopefully , i think some of this might have been a little bit of a review for you . you already had kind of a sense of what an average is . and hopefully these last two problems not only taught you how to do some algebra problems involving average , but they 'll also help you figure out how well you have to do on your exams to get an a in your math class . i think you 're now ready for the average module . have fun .
|
33 minus 32 is 1 , 16 . 4 . so the average is equal to 84 .
|
why does sal use `` 4 x 84 '' ?
|
welcome to the presentation on averages . averages is probably a concept that you 've already used before , maybe not in a mathematical way . but people will talk in terms of , the average voter wants a politician to do this , or the average student in a class wants to get out early . so you 're probably already familiar with the concept of an average . and you probably already intuitively knew that an average is just a number that represents the different values that a group could have . but it can represent that as one number as opposed to giving all the different values . and let 's give a couple of examples of how to compute an average , and you might already know how to do this . so let 's say i had the numbers 1 , 3 , 5 , and 20 . and i asked you , what is the average of these four numbers ? well , what we do is , we literally just add up the numbers . and then divide by the number of numbers we have . so we say 1 plus 3 is 4 . so let me write that . 1 plus 3 plus 5 plus 20 equals , let 's see , 1 plus 3 is 4 . 4 plus 5 is 9 . 9 plus 20 is 29 . and we had 4 numbers ; one , two , three , four . so 4 goes into 29 . and it goes , 7 , 7 , 28 . and then we have 10 , i did n't have to do that decimal there , oh well . 2 , 8 , 25 . so 4 goes into 29 7.25 times . so the average of these four numbers is equal to 7.25 . and that might make sense to you because 7.25 is someplace in between these numbers . and we can kind of view this , 7.25 , as one way to represent these four numbers without having to list these four numbers . there are other representations you 'll learn later on . like the mode . you 'll also the mean , which we 'll talk about later , is actually the same thing as the average . but the average is just one number that you can use to represent a set of numbers . so let 's do some problems which i think are going to be close to your heart . let 's say on the first four tests of an exam , i got a -- let 's see , i got an 80 , an 81 . an 87 , and an 88 . what 's my average in the class so far ? well , all i have to do is add up these four numbers . so i say , 80 plus 81 plus 87 plus 88 . well , zero plus 1 is 1 . 1 plus 7 is 8 . 8 plus 8 is 16 . i just ran eight miles , so i 'm a bit tired . and , 4/8 , so that 's 32 . plus 1 is 33 . and now we divide this number by 4 . 4 goes into 336 . goes into 33 , 8 times . 8 times 4 is 32 . 33 minus 32 is 1 , 16 . 4 . so the average is equal to 84 . so depending on what school you go to that 's either a b or a c. so , so far my average after the first four exams is an 84 . now let 's make this a little bit more difficult . we know that the average after four exams , at four exams , is equal to 84 . if i were to ask you what do i have to get on the next test to average an 88 , to average an 88 in the class . so let 's say that x is what i get on the next test . so now what we can say is , is that the first four exams , i could either list out the first four exams that i took . or i already know what the average is . so i know the sum of the first four exams is going to 4 times 84 . and now i want to add the , what i get on the 5th exam , x . and i 'm going to divide that by all five exams . so in other words , this number is the average of my first five exams . we just figured out the average of the first four exams . but now , we sum up the first four exams here . we add what i got on the fifth exam , and then we divide it by 5 , because now we 're averaging five exams . and i said that i need to get in an 88 in the class . and now we solve for x . let me make some space here . so , 5 times 88 is , let 's see . 5 times 80 is 400 , so it 's 440 . 440 equals 4 times 84 , we just saw that , is 320 plus 16 is 336 . 336 plus x is equal to 440 . well , it turns out if you subtract 336 from both sides , you get x is equal to 104 . so unless you have a exam that has some bonus problems on it , it 's probably impossible for you to get ah an 88 average in the class after just the next exam . you 'd have to get 104 on that next exam . and let 's just look at what we just did . we said , after 4 exams we had an 84 . what do i have to get on that next exam to average an 88 in the class after 5 exams ? and that 's what we solved for when we got x . now , let 's ask another question . i said after four exams , after four exams , i had an 84 average . if i said that there are 6 exams in the class , and the highest score i could get on an exam is 100 , what is the highest average i can finish in the class if i were to really study hard and get 100 on the next 2 exams ? well , once again , what we 'll want to do is assume we get 100 on the next 2 exams and then take the average . so we 'll have to solve all 6 exams . so we 're going to have the average of 6 , so in the denominator we 're going to have 6 . the first four exams , the sum , as we already learned , is 4 exams times the 84 average . and this dot is just times . plus , and there 's going to be 2 more exams , right ? because there 's 6 exams in the class . and i 'm going to get 100 in each . so that 's 200 . and what 's this average ? well , 4 times 84 , we already said , is 336 . plus 200 over 6 . so that 's 536 over 6 . 6 goes into 5 36 . i do n't know if if i gave myself enough space . but 6 goes into 53 , 8 times . 48 . 56 . 9 times . 9 times 6 is 54 . 6 minus is 20 6 goes into -- so we 'll see it 's actually 89.333333 , goes on forever . so 89.3 repeating . so no matter how hard i try in this class , the best i can do . because i only have two exams left , even if i were to get 100 on the next two exams . i can finish the class with an 89.333 average . hopefully , i think some of this might have been a little bit of a review for you . you already had kind of a sense of what an average is . and hopefully these last two problems not only taught you how to do some algebra problems involving average , but they 'll also help you figure out how well you have to do on your exams to get an a in your math class . i think you 're now ready for the average module . have fun .
|
1 plus 3 plus 5 plus 20 equals , let 's see , 1 plus 3 is 4 . 4 plus 5 is 9 . 9 plus 20 is 29 .
|
ex : 5 consecutive integers equals 523 , what is the smallest of these integers ?
|
welcome to the presentation on averages . averages is probably a concept that you 've already used before , maybe not in a mathematical way . but people will talk in terms of , the average voter wants a politician to do this , or the average student in a class wants to get out early . so you 're probably already familiar with the concept of an average . and you probably already intuitively knew that an average is just a number that represents the different values that a group could have . but it can represent that as one number as opposed to giving all the different values . and let 's give a couple of examples of how to compute an average , and you might already know how to do this . so let 's say i had the numbers 1 , 3 , 5 , and 20 . and i asked you , what is the average of these four numbers ? well , what we do is , we literally just add up the numbers . and then divide by the number of numbers we have . so we say 1 plus 3 is 4 . so let me write that . 1 plus 3 plus 5 plus 20 equals , let 's see , 1 plus 3 is 4 . 4 plus 5 is 9 . 9 plus 20 is 29 . and we had 4 numbers ; one , two , three , four . so 4 goes into 29 . and it goes , 7 , 7 , 28 . and then we have 10 , i did n't have to do that decimal there , oh well . 2 , 8 , 25 . so 4 goes into 29 7.25 times . so the average of these four numbers is equal to 7.25 . and that might make sense to you because 7.25 is someplace in between these numbers . and we can kind of view this , 7.25 , as one way to represent these four numbers without having to list these four numbers . there are other representations you 'll learn later on . like the mode . you 'll also the mean , which we 'll talk about later , is actually the same thing as the average . but the average is just one number that you can use to represent a set of numbers . so let 's do some problems which i think are going to be close to your heart . let 's say on the first four tests of an exam , i got a -- let 's see , i got an 80 , an 81 . an 87 , and an 88 . what 's my average in the class so far ? well , all i have to do is add up these four numbers . so i say , 80 plus 81 plus 87 plus 88 . well , zero plus 1 is 1 . 1 plus 7 is 8 . 8 plus 8 is 16 . i just ran eight miles , so i 'm a bit tired . and , 4/8 , so that 's 32 . plus 1 is 33 . and now we divide this number by 4 . 4 goes into 336 . goes into 33 , 8 times . 8 times 4 is 32 . 33 minus 32 is 1 , 16 . 4 . so the average is equal to 84 . so depending on what school you go to that 's either a b or a c. so , so far my average after the first four exams is an 84 . now let 's make this a little bit more difficult . we know that the average after four exams , at four exams , is equal to 84 . if i were to ask you what do i have to get on the next test to average an 88 , to average an 88 in the class . so let 's say that x is what i get on the next test . so now what we can say is , is that the first four exams , i could either list out the first four exams that i took . or i already know what the average is . so i know the sum of the first four exams is going to 4 times 84 . and now i want to add the , what i get on the 5th exam , x . and i 'm going to divide that by all five exams . so in other words , this number is the average of my first five exams . we just figured out the average of the first four exams . but now , we sum up the first four exams here . we add what i got on the fifth exam , and then we divide it by 5 , because now we 're averaging five exams . and i said that i need to get in an 88 in the class . and now we solve for x . let me make some space here . so , 5 times 88 is , let 's see . 5 times 80 is 400 , so it 's 440 . 440 equals 4 times 84 , we just saw that , is 320 plus 16 is 336 . 336 plus x is equal to 440 . well , it turns out if you subtract 336 from both sides , you get x is equal to 104 . so unless you have a exam that has some bonus problems on it , it 's probably impossible for you to get ah an 88 average in the class after just the next exam . you 'd have to get 104 on that next exam . and let 's just look at what we just did . we said , after 4 exams we had an 84 . what do i have to get on that next exam to average an 88 in the class after 5 exams ? and that 's what we solved for when we got x . now , let 's ask another question . i said after four exams , after four exams , i had an 84 average . if i said that there are 6 exams in the class , and the highest score i could get on an exam is 100 , what is the highest average i can finish in the class if i were to really study hard and get 100 on the next 2 exams ? well , once again , what we 'll want to do is assume we get 100 on the next 2 exams and then take the average . so we 'll have to solve all 6 exams . so we 're going to have the average of 6 , so in the denominator we 're going to have 6 . the first four exams , the sum , as we already learned , is 4 exams times the 84 average . and this dot is just times . plus , and there 's going to be 2 more exams , right ? because there 's 6 exams in the class . and i 'm going to get 100 in each . so that 's 200 . and what 's this average ? well , 4 times 84 , we already said , is 336 . plus 200 over 6 . so that 's 536 over 6 . 6 goes into 5 36 . i do n't know if if i gave myself enough space . but 6 goes into 53 , 8 times . 48 . 56 . 9 times . 9 times 6 is 54 . 6 minus is 20 6 goes into -- so we 'll see it 's actually 89.333333 , goes on forever . so 89.3 repeating . so no matter how hard i try in this class , the best i can do . because i only have two exams left , even if i were to get 100 on the next two exams . i can finish the class with an 89.333 average . hopefully , i think some of this might have been a little bit of a review for you . you already had kind of a sense of what an average is . and hopefully these last two problems not only taught you how to do some algebra problems involving average , but they 'll also help you figure out how well you have to do on your exams to get an a in your math class . i think you 're now ready for the average module . have fun .
|
1 plus 3 plus 5 plus 20 equals , let 's see , 1 plus 3 is 4 . 4 plus 5 is 9 . 9 plus 20 is 29 .
|
what does x stand for 3 ?
|
welcome to the presentation on averages . averages is probably a concept that you 've already used before , maybe not in a mathematical way . but people will talk in terms of , the average voter wants a politician to do this , or the average student in a class wants to get out early . so you 're probably already familiar with the concept of an average . and you probably already intuitively knew that an average is just a number that represents the different values that a group could have . but it can represent that as one number as opposed to giving all the different values . and let 's give a couple of examples of how to compute an average , and you might already know how to do this . so let 's say i had the numbers 1 , 3 , 5 , and 20 . and i asked you , what is the average of these four numbers ? well , what we do is , we literally just add up the numbers . and then divide by the number of numbers we have . so we say 1 plus 3 is 4 . so let me write that . 1 plus 3 plus 5 plus 20 equals , let 's see , 1 plus 3 is 4 . 4 plus 5 is 9 . 9 plus 20 is 29 . and we had 4 numbers ; one , two , three , four . so 4 goes into 29 . and it goes , 7 , 7 , 28 . and then we have 10 , i did n't have to do that decimal there , oh well . 2 , 8 , 25 . so 4 goes into 29 7.25 times . so the average of these four numbers is equal to 7.25 . and that might make sense to you because 7.25 is someplace in between these numbers . and we can kind of view this , 7.25 , as one way to represent these four numbers without having to list these four numbers . there are other representations you 'll learn later on . like the mode . you 'll also the mean , which we 'll talk about later , is actually the same thing as the average . but the average is just one number that you can use to represent a set of numbers . so let 's do some problems which i think are going to be close to your heart . let 's say on the first four tests of an exam , i got a -- let 's see , i got an 80 , an 81 . an 87 , and an 88 . what 's my average in the class so far ? well , all i have to do is add up these four numbers . so i say , 80 plus 81 plus 87 plus 88 . well , zero plus 1 is 1 . 1 plus 7 is 8 . 8 plus 8 is 16 . i just ran eight miles , so i 'm a bit tired . and , 4/8 , so that 's 32 . plus 1 is 33 . and now we divide this number by 4 . 4 goes into 336 . goes into 33 , 8 times . 8 times 4 is 32 . 33 minus 32 is 1 , 16 . 4 . so the average is equal to 84 . so depending on what school you go to that 's either a b or a c. so , so far my average after the first four exams is an 84 . now let 's make this a little bit more difficult . we know that the average after four exams , at four exams , is equal to 84 . if i were to ask you what do i have to get on the next test to average an 88 , to average an 88 in the class . so let 's say that x is what i get on the next test . so now what we can say is , is that the first four exams , i could either list out the first four exams that i took . or i already know what the average is . so i know the sum of the first four exams is going to 4 times 84 . and now i want to add the , what i get on the 5th exam , x . and i 'm going to divide that by all five exams . so in other words , this number is the average of my first five exams . we just figured out the average of the first four exams . but now , we sum up the first four exams here . we add what i got on the fifth exam , and then we divide it by 5 , because now we 're averaging five exams . and i said that i need to get in an 88 in the class . and now we solve for x . let me make some space here . so , 5 times 88 is , let 's see . 5 times 80 is 400 , so it 's 440 . 440 equals 4 times 84 , we just saw that , is 320 plus 16 is 336 . 336 plus x is equal to 440 . well , it turns out if you subtract 336 from both sides , you get x is equal to 104 . so unless you have a exam that has some bonus problems on it , it 's probably impossible for you to get ah an 88 average in the class after just the next exam . you 'd have to get 104 on that next exam . and let 's just look at what we just did . we said , after 4 exams we had an 84 . what do i have to get on that next exam to average an 88 in the class after 5 exams ? and that 's what we solved for when we got x . now , let 's ask another question . i said after four exams , after four exams , i had an 84 average . if i said that there are 6 exams in the class , and the highest score i could get on an exam is 100 , what is the highest average i can finish in the class if i were to really study hard and get 100 on the next 2 exams ? well , once again , what we 'll want to do is assume we get 100 on the next 2 exams and then take the average . so we 'll have to solve all 6 exams . so we 're going to have the average of 6 , so in the denominator we 're going to have 6 . the first four exams , the sum , as we already learned , is 4 exams times the 84 average . and this dot is just times . plus , and there 's going to be 2 more exams , right ? because there 's 6 exams in the class . and i 'm going to get 100 in each . so that 's 200 . and what 's this average ? well , 4 times 84 , we already said , is 336 . plus 200 over 6 . so that 's 536 over 6 . 6 goes into 5 36 . i do n't know if if i gave myself enough space . but 6 goes into 53 , 8 times . 48 . 56 . 9 times . 9 times 6 is 54 . 6 minus is 20 6 goes into -- so we 'll see it 's actually 89.333333 , goes on forever . so 89.3 repeating . so no matter how hard i try in this class , the best i can do . because i only have two exams left , even if i were to get 100 on the next two exams . i can finish the class with an 89.333 average . hopefully , i think some of this might have been a little bit of a review for you . you already had kind of a sense of what an average is . and hopefully these last two problems not only taught you how to do some algebra problems involving average , but they 'll also help you figure out how well you have to do on your exams to get an a in your math class . i think you 're now ready for the average module . have fun .
|
33 minus 32 is 1 , 16 . 4 . so the average is equal to 84 .
|
why is 4 ( 84 ) +x to get the average ?
|
welcome to the presentation on averages . averages is probably a concept that you 've already used before , maybe not in a mathematical way . but people will talk in terms of , the average voter wants a politician to do this , or the average student in a class wants to get out early . so you 're probably already familiar with the concept of an average . and you probably already intuitively knew that an average is just a number that represents the different values that a group could have . but it can represent that as one number as opposed to giving all the different values . and let 's give a couple of examples of how to compute an average , and you might already know how to do this . so let 's say i had the numbers 1 , 3 , 5 , and 20 . and i asked you , what is the average of these four numbers ? well , what we do is , we literally just add up the numbers . and then divide by the number of numbers we have . so we say 1 plus 3 is 4 . so let me write that . 1 plus 3 plus 5 plus 20 equals , let 's see , 1 plus 3 is 4 . 4 plus 5 is 9 . 9 plus 20 is 29 . and we had 4 numbers ; one , two , three , four . so 4 goes into 29 . and it goes , 7 , 7 , 28 . and then we have 10 , i did n't have to do that decimal there , oh well . 2 , 8 , 25 . so 4 goes into 29 7.25 times . so the average of these four numbers is equal to 7.25 . and that might make sense to you because 7.25 is someplace in between these numbers . and we can kind of view this , 7.25 , as one way to represent these four numbers without having to list these four numbers . there are other representations you 'll learn later on . like the mode . you 'll also the mean , which we 'll talk about later , is actually the same thing as the average . but the average is just one number that you can use to represent a set of numbers . so let 's do some problems which i think are going to be close to your heart . let 's say on the first four tests of an exam , i got a -- let 's see , i got an 80 , an 81 . an 87 , and an 88 . what 's my average in the class so far ? well , all i have to do is add up these four numbers . so i say , 80 plus 81 plus 87 plus 88 . well , zero plus 1 is 1 . 1 plus 7 is 8 . 8 plus 8 is 16 . i just ran eight miles , so i 'm a bit tired . and , 4/8 , so that 's 32 . plus 1 is 33 . and now we divide this number by 4 . 4 goes into 336 . goes into 33 , 8 times . 8 times 4 is 32 . 33 minus 32 is 1 , 16 . 4 . so the average is equal to 84 . so depending on what school you go to that 's either a b or a c. so , so far my average after the first four exams is an 84 . now let 's make this a little bit more difficult . we know that the average after four exams , at four exams , is equal to 84 . if i were to ask you what do i have to get on the next test to average an 88 , to average an 88 in the class . so let 's say that x is what i get on the next test . so now what we can say is , is that the first four exams , i could either list out the first four exams that i took . or i already know what the average is . so i know the sum of the first four exams is going to 4 times 84 . and now i want to add the , what i get on the 5th exam , x . and i 'm going to divide that by all five exams . so in other words , this number is the average of my first five exams . we just figured out the average of the first four exams . but now , we sum up the first four exams here . we add what i got on the fifth exam , and then we divide it by 5 , because now we 're averaging five exams . and i said that i need to get in an 88 in the class . and now we solve for x . let me make some space here . so , 5 times 88 is , let 's see . 5 times 80 is 400 , so it 's 440 . 440 equals 4 times 84 , we just saw that , is 320 plus 16 is 336 . 336 plus x is equal to 440 . well , it turns out if you subtract 336 from both sides , you get x is equal to 104 . so unless you have a exam that has some bonus problems on it , it 's probably impossible for you to get ah an 88 average in the class after just the next exam . you 'd have to get 104 on that next exam . and let 's just look at what we just did . we said , after 4 exams we had an 84 . what do i have to get on that next exam to average an 88 in the class after 5 exams ? and that 's what we solved for when we got x . now , let 's ask another question . i said after four exams , after four exams , i had an 84 average . if i said that there are 6 exams in the class , and the highest score i could get on an exam is 100 , what is the highest average i can finish in the class if i were to really study hard and get 100 on the next 2 exams ? well , once again , what we 'll want to do is assume we get 100 on the next 2 exams and then take the average . so we 'll have to solve all 6 exams . so we 're going to have the average of 6 , so in the denominator we 're going to have 6 . the first four exams , the sum , as we already learned , is 4 exams times the 84 average . and this dot is just times . plus , and there 's going to be 2 more exams , right ? because there 's 6 exams in the class . and i 'm going to get 100 in each . so that 's 200 . and what 's this average ? well , 4 times 84 , we already said , is 336 . plus 200 over 6 . so that 's 536 over 6 . 6 goes into 5 36 . i do n't know if if i gave myself enough space . but 6 goes into 53 , 8 times . 48 . 56 . 9 times . 9 times 6 is 54 . 6 minus is 20 6 goes into -- so we 'll see it 's actually 89.333333 , goes on forever . so 89.3 repeating . so no matter how hard i try in this class , the best i can do . because i only have two exams left , even if i were to get 100 on the next two exams . i can finish the class with an 89.333 average . hopefully , i think some of this might have been a little bit of a review for you . you already had kind of a sense of what an average is . and hopefully these last two problems not only taught you how to do some algebra problems involving average , but they 'll also help you figure out how well you have to do on your exams to get an a in your math class . i think you 're now ready for the average module . have fun .
|
so that 's 200 . and what 's this average ? well , 4 times 84 , we already said , is 336 .
|
what is the astigmatism of 5x -y of the average when the hypotenuse is less than the decimal of the denominator ?
|
welcome to the presentation on averages . averages is probably a concept that you 've already used before , maybe not in a mathematical way . but people will talk in terms of , the average voter wants a politician to do this , or the average student in a class wants to get out early . so you 're probably already familiar with the concept of an average . and you probably already intuitively knew that an average is just a number that represents the different values that a group could have . but it can represent that as one number as opposed to giving all the different values . and let 's give a couple of examples of how to compute an average , and you might already know how to do this . so let 's say i had the numbers 1 , 3 , 5 , and 20 . and i asked you , what is the average of these four numbers ? well , what we do is , we literally just add up the numbers . and then divide by the number of numbers we have . so we say 1 plus 3 is 4 . so let me write that . 1 plus 3 plus 5 plus 20 equals , let 's see , 1 plus 3 is 4 . 4 plus 5 is 9 . 9 plus 20 is 29 . and we had 4 numbers ; one , two , three , four . so 4 goes into 29 . and it goes , 7 , 7 , 28 . and then we have 10 , i did n't have to do that decimal there , oh well . 2 , 8 , 25 . so 4 goes into 29 7.25 times . so the average of these four numbers is equal to 7.25 . and that might make sense to you because 7.25 is someplace in between these numbers . and we can kind of view this , 7.25 , as one way to represent these four numbers without having to list these four numbers . there are other representations you 'll learn later on . like the mode . you 'll also the mean , which we 'll talk about later , is actually the same thing as the average . but the average is just one number that you can use to represent a set of numbers . so let 's do some problems which i think are going to be close to your heart . let 's say on the first four tests of an exam , i got a -- let 's see , i got an 80 , an 81 . an 87 , and an 88 . what 's my average in the class so far ? well , all i have to do is add up these four numbers . so i say , 80 plus 81 plus 87 plus 88 . well , zero plus 1 is 1 . 1 plus 7 is 8 . 8 plus 8 is 16 . i just ran eight miles , so i 'm a bit tired . and , 4/8 , so that 's 32 . plus 1 is 33 . and now we divide this number by 4 . 4 goes into 336 . goes into 33 , 8 times . 8 times 4 is 32 . 33 minus 32 is 1 , 16 . 4 . so the average is equal to 84 . so depending on what school you go to that 's either a b or a c. so , so far my average after the first four exams is an 84 . now let 's make this a little bit more difficult . we know that the average after four exams , at four exams , is equal to 84 . if i were to ask you what do i have to get on the next test to average an 88 , to average an 88 in the class . so let 's say that x is what i get on the next test . so now what we can say is , is that the first four exams , i could either list out the first four exams that i took . or i already know what the average is . so i know the sum of the first four exams is going to 4 times 84 . and now i want to add the , what i get on the 5th exam , x . and i 'm going to divide that by all five exams . so in other words , this number is the average of my first five exams . we just figured out the average of the first four exams . but now , we sum up the first four exams here . we add what i got on the fifth exam , and then we divide it by 5 , because now we 're averaging five exams . and i said that i need to get in an 88 in the class . and now we solve for x . let me make some space here . so , 5 times 88 is , let 's see . 5 times 80 is 400 , so it 's 440 . 440 equals 4 times 84 , we just saw that , is 320 plus 16 is 336 . 336 plus x is equal to 440 . well , it turns out if you subtract 336 from both sides , you get x is equal to 104 . so unless you have a exam that has some bonus problems on it , it 's probably impossible for you to get ah an 88 average in the class after just the next exam . you 'd have to get 104 on that next exam . and let 's just look at what we just did . we said , after 4 exams we had an 84 . what do i have to get on that next exam to average an 88 in the class after 5 exams ? and that 's what we solved for when we got x . now , let 's ask another question . i said after four exams , after four exams , i had an 84 average . if i said that there are 6 exams in the class , and the highest score i could get on an exam is 100 , what is the highest average i can finish in the class if i were to really study hard and get 100 on the next 2 exams ? well , once again , what we 'll want to do is assume we get 100 on the next 2 exams and then take the average . so we 'll have to solve all 6 exams . so we 're going to have the average of 6 , so in the denominator we 're going to have 6 . the first four exams , the sum , as we already learned , is 4 exams times the 84 average . and this dot is just times . plus , and there 's going to be 2 more exams , right ? because there 's 6 exams in the class . and i 'm going to get 100 in each . so that 's 200 . and what 's this average ? well , 4 times 84 , we already said , is 336 . plus 200 over 6 . so that 's 536 over 6 . 6 goes into 5 36 . i do n't know if if i gave myself enough space . but 6 goes into 53 , 8 times . 48 . 56 . 9 times . 9 times 6 is 54 . 6 minus is 20 6 goes into -- so we 'll see it 's actually 89.333333 , goes on forever . so 89.3 repeating . so no matter how hard i try in this class , the best i can do . because i only have two exams left , even if i were to get 100 on the next two exams . i can finish the class with an 89.333 average . hopefully , i think some of this might have been a little bit of a review for you . you already had kind of a sense of what an average is . and hopefully these last two problems not only taught you how to do some algebra problems involving average , but they 'll also help you figure out how well you have to do on your exams to get an a in your math class . i think you 're now ready for the average module . have fun .
|
and then we have 10 , i did n't have to do that decimal there , oh well . 2 , 8 , 25 . so 4 goes into 29 7.25 times .
|
a car travelled 25 km at an average speed of xkm/h.if ti speed was increased by 5 km/h , it would complete same journey 15 minutes earlier.what is the original speed of car ?
|
welcome to the presentation on averages . averages is probably a concept that you 've already used before , maybe not in a mathematical way . but people will talk in terms of , the average voter wants a politician to do this , or the average student in a class wants to get out early . so you 're probably already familiar with the concept of an average . and you probably already intuitively knew that an average is just a number that represents the different values that a group could have . but it can represent that as one number as opposed to giving all the different values . and let 's give a couple of examples of how to compute an average , and you might already know how to do this . so let 's say i had the numbers 1 , 3 , 5 , and 20 . and i asked you , what is the average of these four numbers ? well , what we do is , we literally just add up the numbers . and then divide by the number of numbers we have . so we say 1 plus 3 is 4 . so let me write that . 1 plus 3 plus 5 plus 20 equals , let 's see , 1 plus 3 is 4 . 4 plus 5 is 9 . 9 plus 20 is 29 . and we had 4 numbers ; one , two , three , four . so 4 goes into 29 . and it goes , 7 , 7 , 28 . and then we have 10 , i did n't have to do that decimal there , oh well . 2 , 8 , 25 . so 4 goes into 29 7.25 times . so the average of these four numbers is equal to 7.25 . and that might make sense to you because 7.25 is someplace in between these numbers . and we can kind of view this , 7.25 , as one way to represent these four numbers without having to list these four numbers . there are other representations you 'll learn later on . like the mode . you 'll also the mean , which we 'll talk about later , is actually the same thing as the average . but the average is just one number that you can use to represent a set of numbers . so let 's do some problems which i think are going to be close to your heart . let 's say on the first four tests of an exam , i got a -- let 's see , i got an 80 , an 81 . an 87 , and an 88 . what 's my average in the class so far ? well , all i have to do is add up these four numbers . so i say , 80 plus 81 plus 87 plus 88 . well , zero plus 1 is 1 . 1 plus 7 is 8 . 8 plus 8 is 16 . i just ran eight miles , so i 'm a bit tired . and , 4/8 , so that 's 32 . plus 1 is 33 . and now we divide this number by 4 . 4 goes into 336 . goes into 33 , 8 times . 8 times 4 is 32 . 33 minus 32 is 1 , 16 . 4 . so the average is equal to 84 . so depending on what school you go to that 's either a b or a c. so , so far my average after the first four exams is an 84 . now let 's make this a little bit more difficult . we know that the average after four exams , at four exams , is equal to 84 . if i were to ask you what do i have to get on the next test to average an 88 , to average an 88 in the class . so let 's say that x is what i get on the next test . so now what we can say is , is that the first four exams , i could either list out the first four exams that i took . or i already know what the average is . so i know the sum of the first four exams is going to 4 times 84 . and now i want to add the , what i get on the 5th exam , x . and i 'm going to divide that by all five exams . so in other words , this number is the average of my first five exams . we just figured out the average of the first four exams . but now , we sum up the first four exams here . we add what i got on the fifth exam , and then we divide it by 5 , because now we 're averaging five exams . and i said that i need to get in an 88 in the class . and now we solve for x . let me make some space here . so , 5 times 88 is , let 's see . 5 times 80 is 400 , so it 's 440 . 440 equals 4 times 84 , we just saw that , is 320 plus 16 is 336 . 336 plus x is equal to 440 . well , it turns out if you subtract 336 from both sides , you get x is equal to 104 . so unless you have a exam that has some bonus problems on it , it 's probably impossible for you to get ah an 88 average in the class after just the next exam . you 'd have to get 104 on that next exam . and let 's just look at what we just did . we said , after 4 exams we had an 84 . what do i have to get on that next exam to average an 88 in the class after 5 exams ? and that 's what we solved for when we got x . now , let 's ask another question . i said after four exams , after four exams , i had an 84 average . if i said that there are 6 exams in the class , and the highest score i could get on an exam is 100 , what is the highest average i can finish in the class if i were to really study hard and get 100 on the next 2 exams ? well , once again , what we 'll want to do is assume we get 100 on the next 2 exams and then take the average . so we 'll have to solve all 6 exams . so we 're going to have the average of 6 , so in the denominator we 're going to have 6 . the first four exams , the sum , as we already learned , is 4 exams times the 84 average . and this dot is just times . plus , and there 's going to be 2 more exams , right ? because there 's 6 exams in the class . and i 'm going to get 100 in each . so that 's 200 . and what 's this average ? well , 4 times 84 , we already said , is 336 . plus 200 over 6 . so that 's 536 over 6 . 6 goes into 5 36 . i do n't know if if i gave myself enough space . but 6 goes into 53 , 8 times . 48 . 56 . 9 times . 9 times 6 is 54 . 6 minus is 20 6 goes into -- so we 'll see it 's actually 89.333333 , goes on forever . so 89.3 repeating . so no matter how hard i try in this class , the best i can do . because i only have two exams left , even if i were to get 100 on the next two exams . i can finish the class with an 89.333 average . hopefully , i think some of this might have been a little bit of a review for you . you already had kind of a sense of what an average is . and hopefully these last two problems not only taught you how to do some algebra problems involving average , but they 'll also help you figure out how well you have to do on your exams to get an a in your math class . i think you 're now ready for the average module . have fun .
|
1 plus 3 plus 5 plus 20 equals , let 's see , 1 plus 3 is 4 . 4 plus 5 is 9 . 9 plus 20 is 29 .
|
: the sum of 5 consecutive odd integers is 345.what is the smallest of 5 integers ?
|
: i want to make it super clear , what the difference is between a lot of these words that sound really similar , but their subtly different from each other . so , the first one is molarity , and we know that means moles in one liter of solution . keeping the numerator the same , but tweaking the denominator you get molality with an l. now , molality is moles and here instead of a liter of solution we said it was one kilogram of solvent . that was the major difference between the two of them . right ? so , here you can see the denominator is just slightly different between these two words . if you now carry this on , let 's say we switch over to this side , and we go to osmolarity . osmolarity . we keep the same denominator . we do n't change that , so , let me write that in . we say one liter of solution . that 's still the same . what changes it the numerator is now osmoles . osmoles . you remember osmoles talks about not just what is in the solution , but how things can split apart . we talked about salt as a great example of having a different molarity from osmolarity . rounding it out , then you can imagine that there 's a fourth word osmolality . osmolality is going to be the same numerator as osmolarity , it 's going to be osmoles . except in the denominator , we 're going to pick up that one kilogram per of solvent definition . so , that 's osmolality . so , then , switching between these two , you can see the major difference here is just the numerator . right ? the numerator is different between these two . that 's four of the terms , and the last term that we picked up was tonicity . tonicity . where does that fall into the mix ? tonicity . we 've even split it up into hypotonic solutions , isontonic solutions , and hypertonic solutions . we said that 's kind of how we think of tonicity , one of those three groups usually . very broadly speaking , these terms , these four terms are really a way to define or describe one solution . if you have a solution you can describe it using these terms . these terms are used for two solutions . if you have two solutions separated by a membrane , as you probably write that with a membrane , then you can use these terms to describe how they are relative to one another . that 's the key difference between these terms . the first four are really for one solution , and the tonicity terms describe how you might talk about two solutions separated by membrane , and keep in mind that all of this stuff , all of it , let me just make a little bit of space , is really trying to refer to medical terms . we usually use these in terms of medicine , and in medicine you might have cells with a permeable membrane . actually , let me make it a little permeable membrane . this cell is usually going to be setting in some solution , and that solution could be the blood , or it could be the interstitial fluid , or some solution . when we talk about these four terms , and especially , tinicity , we 're talking about the relationship between this solution , and this solution inside of cells . that 's one way to frame all these different words that we 've talked about .
|
actually , let me make it a little permeable membrane . this cell is usually going to be setting in some solution , and that solution could be the blood , or it could be the interstitial fluid , or some solution . when we talk about these four terms , and especially , tinicity , we 're talking about the relationship between this solution , and this solution inside of cells . that 's one way to frame all these different words that we 've talked about .
|
how to calculate a 1 molar solution ?
|
: i want to make it super clear , what the difference is between a lot of these words that sound really similar , but their subtly different from each other . so , the first one is molarity , and we know that means moles in one liter of solution . keeping the numerator the same , but tweaking the denominator you get molality with an l. now , molality is moles and here instead of a liter of solution we said it was one kilogram of solvent . that was the major difference between the two of them . right ? so , here you can see the denominator is just slightly different between these two words . if you now carry this on , let 's say we switch over to this side , and we go to osmolarity . osmolarity . we keep the same denominator . we do n't change that , so , let me write that in . we say one liter of solution . that 's still the same . what changes it the numerator is now osmoles . osmoles . you remember osmoles talks about not just what is in the solution , but how things can split apart . we talked about salt as a great example of having a different molarity from osmolarity . rounding it out , then you can imagine that there 's a fourth word osmolality . osmolality is going to be the same numerator as osmolarity , it 's going to be osmoles . except in the denominator , we 're going to pick up that one kilogram per of solvent definition . so , that 's osmolality . so , then , switching between these two , you can see the major difference here is just the numerator . right ? the numerator is different between these two . that 's four of the terms , and the last term that we picked up was tonicity . tonicity . where does that fall into the mix ? tonicity . we 've even split it up into hypotonic solutions , isontonic solutions , and hypertonic solutions . we said that 's kind of how we think of tonicity , one of those three groups usually . very broadly speaking , these terms , these four terms are really a way to define or describe one solution . if you have a solution you can describe it using these terms . these terms are used for two solutions . if you have two solutions separated by a membrane , as you probably write that with a membrane , then you can use these terms to describe how they are relative to one another . that 's the key difference between these terms . the first four are really for one solution , and the tonicity terms describe how you might talk about two solutions separated by membrane , and keep in mind that all of this stuff , all of it , let me just make a little bit of space , is really trying to refer to medical terms . we usually use these in terms of medicine , and in medicine you might have cells with a permeable membrane . actually , let me make it a little permeable membrane . this cell is usually going to be setting in some solution , and that solution could be the blood , or it could be the interstitial fluid , or some solution . when we talk about these four terms , and especially , tinicity , we 're talking about the relationship between this solution , and this solution inside of cells . that 's one way to frame all these different words that we 've talked about .
|
actually , let me make it a little permeable membrane . this cell is usually going to be setting in some solution , and that solution could be the blood , or it could be the interstitial fluid , or some solution . when we talk about these four terms , and especially , tinicity , we 're talking about the relationship between this solution , and this solution inside of cells . that 's one way to frame all these different words that we 've talked about .
|
how do you find molality of solution `` x '' when the new solute is mixed in with a solution that has 2 different substances such as water and something else ?
|
: let 's see how to form imines and enamines and start with the formation of an imine we start with an aldehyde or a ketone . and you add an amine to it and you need an acid catalyst and over here on the right if your y is equal to hydrogen or an alkyl group which is a r double prime you will form an imine . so let 's go ahead an write that this would be an imine . and the y can be other things , we 'll talk about those other things in the next video . you also form water and so if you , since it reacts as [ that ] equilibrium if you wanted to shift equilibrium to the right , you could remove water as it 's formed and equilibrium will shift this way and give you more of your imine product . so let 's look at the mechanism to form imines . and i 've seen two different ways to start off , the mechanism and so i 'll present both ways and you can choose which one you want to use . and so one way would be to think about an acid being present right , so we 'll say it 's h-a plus the generic acid and you think about this as protonating your imine . so you have an imine and protons , you 're protonating your imine to form this generic acid here . and then your carbonyl oxygen is going to be protonated . a lone pair of electrons on the oxygen takes this proton , leaves these electrons behind , so let 's go ahead and show the results of that . so now we have our oxygen , with a plus one formal charge and we still have our prime group over here , so let 's show those electrons . so this lone pair of electrons here on the oxygen picks up this proton and that forms this bond right here . we 've talked about in previous videos how protonation of the carbonyl oxygen makes the carbonyl carbon more electrophilic . and so we have a good electrophile , this carbon here is a good electrophile and the imine can be a good nucleophile . so a lone pair of electrons and the nitrogen over here are going to attack our carbonyl carbon and that 's going to kick these electrons in here off onto the oxygen , so let 's draw the results of that . so now we 're going to have our nitrogen is forming a bond with our carbon , so let 's show those electrons so let 's make those blue . so this lone pair of electrons of electrons on the nitrogen forms a bond between this nitrogen and this carbon here . this nitrogen is also bonded to two hydrogens . so let 's draw those in . and still our y group like that . that gives our nitrogen a plus one formal charge . and for our carbon , it 's also bonded to this oxygen , this oxygen now has two lone pairs of electrons . so let 's show the movement of those electrons so i 'll use green here . so these electrons in green moved out onto our oxygen . the carbon is still bonded to our alkyl group so r and r prime , so i 'm saying we start with a ketone here . and so that 's one way to start off your mechanism and of course there is another way to do it . so let 's talk about just using our nitrogen as a nucleophile straightaway so the imine is a pretty good nucleophile and so we can just show our nucleophile attacking our electrophile directly . so partial negative oxygen , partial positive carbon this is our electrophile and then a lone pair of electrons on our nitrogen right makes our imine a good nucleophile so they can just attack directly . right , attack this carbon push these electrons off onto your oxygen . so let 's go ahead and draw the result of that and you would have our carbon bonded to our nitrogen . our nitrogen is bonded to two hydrogens and our y group and that would make our nitrogen a plus one formal charge and then our oxygen this time would be a negative one formal charge . so negative one formal charge here . we have our alkyl groups down here , so let 's once again show our electrons . so if nitrogen attacks directly , this lone pair in blue forms this bond right here . and then we can say that these pi electrons in here kick off onto our oxygen so it does n't really matter which lone pair you make them as long as it states that one right there . and then we can show this intermediate going to the one we already talked about , if we just protonate our negative one formal charge on our oxygen here . so let 's show that , so if we have our generic acid so h-a plus , we can show a lone pair picking up this proton , leaving these electrons behind . and so now that would give us this intermediate the same one that we had before . so whichever way you would like to start off your mechanism , once we 've reached this intermediate we can think about the base . the imine here , a lone pair of electrons picking up this proton and leaving these electrons behind on our nitrogens . so let 's show what we would make from that . let 's get a little more room down here alright . so we deprotonate and then we form this structure , a carbon bonded to a nitrogen and then we still have our y group , we still have a hydrogen on this nitrogen . we have a lone pair of electrons on this nitrogen , so that lone pair came from right in here . so you take that proton and leave those electrons in magenta behind on your nitrogen . so we also have an o-h group bonded to our carbon here , and then we also have our alkyl group , so r and r prime . so this intermediate is called a carbinolamine . so let me go ahead and write that . so this is called a carbinolamine . and for the next step we can protonate the o-h group . so we take our generic acid once again so h-a plus and lone pair of electrons on our oxygen could pick up this proton leaving these electrons behind so let 's go ahead and show that . so let 's get once again some more space in here . so if we protonate our carbinolamine , we would have our carbon here , our nitrogen , lone pair of electrons on our nitrogen , our y group , our hydrogen here , our alkyl groups . and then if we protonate the o-h , we would form water as a good leaving group . so right in here , this would be a plus one formal charge on our oxygen and let 's show those electrons . so in here , lone pair of electrons on the oxygen , pick up this proton and so forming this bond right here and now you can see we have a good leaving group . so we have water as a good leaving group . so if these electrons came off onto the oxygen these electrons in the nitrogen could move in here to form a double bond between the carbon and the nitrogen . so let 's go ahead and show that . so our next step here , this is where we lose water . so we 're going to minus h2o and let 's go ahead and show our next structure here . we have our carbon double bonded to our nitrogen this time , our nitrogen is still bonded to our y group . and our hydrogen over here . that gives the nitrogen a plus one formal charge and the carbon is still bonded to our alkyl groups and so let 's show those electrons here on our nitrogen . so i 'm going to go ahead and make those magenta . so these magenta electrons move in here to form our double bond between our carbon and our nitrogen . and this is an important structure , this is called an iminium ion , so let 's go ahead and draw that . so an iminium ion and then we lose water of course . so minus water at this stage . so we 're almost to our final product , we would just have to deprotonate our iminium ion . and so we can do that with our imine . could take that proton and leave these electrons behind on the nitrogen . so that 's going to give us our final product where we have carbon double bonded to our nitrogen and then we have nitrogen bonded to our y group here . a lone pair of electrons on that nitrogen and then we have our alkyl groups and so let 's show those electrons here in blue moving in here , off onto our nitrogen . and then once again , if our y is a hydrogen an alkyl group we have formed an imine . alright , so if that 's formation of an imine , let 's look at an example . so here is our reaction , we 're going to start with a ketone here , so cyclohexanone and react it with a primary amine . so the nitrogen is bonded to one carbon . so this is a reaction with a primary amine and then we 're going to use sulphuric acid as our catalyst here . so to figure out the product of this reaction since it 's kind of a long mechanism so it does n't really makes sense to run through the entire mechanism but we could think about part of the mechanism here . we know that the nucleophile is going to be our amine and it 's going to attack our carbonyl carbon here . so we know that in our mechanism we lose one of these protons on our amines so let 's say it 's that proton right there . and we know we 're going to lose water so minus h2o and this going to think about taking us to our iminium ion steps , so we 're going to have our ring . and so we 've already lost water so now we 're going to have a double bond to our nitrogen . and our nitrogen is going to have an ethyl group on it here and then we still have one proton on our nitrogen . so we still have the hydrogen attached here at this point . which gives us a plus one formal charge on our nitrogen . so this is our iminium ion right here . and then in our last step we know that our amine can come along and act as a base . so a lone pair of electrons on the nitrogen could take this proton right here and leave these electrons behind on our nitrogen to give us our imine products . so we go over here and we draw our product , double bond to our nitrogen and then we have our ethyl group and lone pair of electrons on our nitrogen . so if i follow those electrons , so these electrons in here in blue ended up on our nitrogen to form our imine . so a few things that i want to point out here for the iminium ion , we still have a proton attached to it . so deprotonation for our last step will form an imine . and this is once again done with a primary amine right here . and so that 's a little bit different from what we 're going to talk about next , we 're going to talk about formation of an enamine , and formation of an enamine is , it starts off the same way in terms of the mechanism but it 's this iminium ion step that changes a little bit . so let 's look at reacting a ketone with a secondary amine this time . and so let 's look at this one right here , so here we have a secondary amine and our nitrogen bonded to two other carbons so a secondary amine this time . and once again , we can kind of run through the mechanism . just thinking a little bit about what happens without going through each individual step , we know the nucleophiles going to attack here and we know we 're going to lose a proton . so this is the only proton we have left to lose and so we lose that one , we 're going to lose water in the mechanism to form our iminium ion , so let 's go ahead and draw that out we 're going to have our ring , we 're going to have a double bond to this nitrogen , we 've already lost the proton in red so now we have two alkyl groups attached to our nitrogen and a plus one formal charge on our nitrogen here . and so in the previous reaction , let me go back up to here to the previous reaction again . once again we still had a proton on our iminium ion at this step and so we 're able to deprotonate here . but that 's not the case with what we have for this ion . we do n't have a proton on our nitrogen here . and so we ca n't deprotonate in the same place , we have to pick a carbon adjacent over here . so here 's a proton over here on this carbon and our base could come along . so let 's go ahead and draw out our amine base with a lone pair of electrons right here . and a lone pair could take this proton that would push these electrons in here and then push these electrons off on to the nitrogen to give us our products , so let me go ahead and draw this out here . so now we 're going to have a double bond here and then we have our nitrogen with these ethyl groups and a lone pair of electrons on our nitrogen . so let 's follow some electrons . so these electrons in here once you deprotonate , move in here to form our double bond and then these electrons in here move out onto our nitrogen like that to form a different product . so this called an enamine , so let me go ahead and write this out here , an `` enamine . '' and the name come from `` -en '' come from the double bond . so we form double bonds , that 's the `` -en '' portion and then the `` amine '' portion comes from this part up here . we have an amine so this is an enamine . and so formation of an enamine happens because once again we do n't have the same iminium ion that we had before because of the fact that we started with a secondary amine . so that 's why it 's important to recognize the kind of amine that you were reacting with your aldehyde or a ketone . and if it 's secondary , you 're going to form an enamine here . and enamines are useful synthetic intermediates , so we 'll talk much more about them in later videos . but once again the mechanism is the same until you get to this last step here .
|
and i 've seen two different ways to start off , the mechanism and so i 'll present both ways and you can choose which one you want to use . and so one way would be to think about an acid being present right , so we 'll say it 's h-a plus the generic acid and you think about this as protonating your imine . so you have an imine and protons , you 're protonating your imine to form this generic acid here .
|
the amine is basic , wo n't it take a h+ from the acid and thereby become unreactive ?
|
: let 's see how to form imines and enamines and start with the formation of an imine we start with an aldehyde or a ketone . and you add an amine to it and you need an acid catalyst and over here on the right if your y is equal to hydrogen or an alkyl group which is a r double prime you will form an imine . so let 's go ahead an write that this would be an imine . and the y can be other things , we 'll talk about those other things in the next video . you also form water and so if you , since it reacts as [ that ] equilibrium if you wanted to shift equilibrium to the right , you could remove water as it 's formed and equilibrium will shift this way and give you more of your imine product . so let 's look at the mechanism to form imines . and i 've seen two different ways to start off , the mechanism and so i 'll present both ways and you can choose which one you want to use . and so one way would be to think about an acid being present right , so we 'll say it 's h-a plus the generic acid and you think about this as protonating your imine . so you have an imine and protons , you 're protonating your imine to form this generic acid here . and then your carbonyl oxygen is going to be protonated . a lone pair of electrons on the oxygen takes this proton , leaves these electrons behind , so let 's go ahead and show the results of that . so now we have our oxygen , with a plus one formal charge and we still have our prime group over here , so let 's show those electrons . so this lone pair of electrons here on the oxygen picks up this proton and that forms this bond right here . we 've talked about in previous videos how protonation of the carbonyl oxygen makes the carbonyl carbon more electrophilic . and so we have a good electrophile , this carbon here is a good electrophile and the imine can be a good nucleophile . so a lone pair of electrons and the nitrogen over here are going to attack our carbonyl carbon and that 's going to kick these electrons in here off onto the oxygen , so let 's draw the results of that . so now we 're going to have our nitrogen is forming a bond with our carbon , so let 's show those electrons so let 's make those blue . so this lone pair of electrons of electrons on the nitrogen forms a bond between this nitrogen and this carbon here . this nitrogen is also bonded to two hydrogens . so let 's draw those in . and still our y group like that . that gives our nitrogen a plus one formal charge . and for our carbon , it 's also bonded to this oxygen , this oxygen now has two lone pairs of electrons . so let 's show the movement of those electrons so i 'll use green here . so these electrons in green moved out onto our oxygen . the carbon is still bonded to our alkyl group so r and r prime , so i 'm saying we start with a ketone here . and so that 's one way to start off your mechanism and of course there is another way to do it . so let 's talk about just using our nitrogen as a nucleophile straightaway so the imine is a pretty good nucleophile and so we can just show our nucleophile attacking our electrophile directly . so partial negative oxygen , partial positive carbon this is our electrophile and then a lone pair of electrons on our nitrogen right makes our imine a good nucleophile so they can just attack directly . right , attack this carbon push these electrons off onto your oxygen . so let 's go ahead and draw the result of that and you would have our carbon bonded to our nitrogen . our nitrogen is bonded to two hydrogens and our y group and that would make our nitrogen a plus one formal charge and then our oxygen this time would be a negative one formal charge . so negative one formal charge here . we have our alkyl groups down here , so let 's once again show our electrons . so if nitrogen attacks directly , this lone pair in blue forms this bond right here . and then we can say that these pi electrons in here kick off onto our oxygen so it does n't really matter which lone pair you make them as long as it states that one right there . and then we can show this intermediate going to the one we already talked about , if we just protonate our negative one formal charge on our oxygen here . so let 's show that , so if we have our generic acid so h-a plus , we can show a lone pair picking up this proton , leaving these electrons behind . and so now that would give us this intermediate the same one that we had before . so whichever way you would like to start off your mechanism , once we 've reached this intermediate we can think about the base . the imine here , a lone pair of electrons picking up this proton and leaving these electrons behind on our nitrogens . so let 's show what we would make from that . let 's get a little more room down here alright . so we deprotonate and then we form this structure , a carbon bonded to a nitrogen and then we still have our y group , we still have a hydrogen on this nitrogen . we have a lone pair of electrons on this nitrogen , so that lone pair came from right in here . so you take that proton and leave those electrons in magenta behind on your nitrogen . so we also have an o-h group bonded to our carbon here , and then we also have our alkyl group , so r and r prime . so this intermediate is called a carbinolamine . so let me go ahead and write that . so this is called a carbinolamine . and for the next step we can protonate the o-h group . so we take our generic acid once again so h-a plus and lone pair of electrons on our oxygen could pick up this proton leaving these electrons behind so let 's go ahead and show that . so let 's get once again some more space in here . so if we protonate our carbinolamine , we would have our carbon here , our nitrogen , lone pair of electrons on our nitrogen , our y group , our hydrogen here , our alkyl groups . and then if we protonate the o-h , we would form water as a good leaving group . so right in here , this would be a plus one formal charge on our oxygen and let 's show those electrons . so in here , lone pair of electrons on the oxygen , pick up this proton and so forming this bond right here and now you can see we have a good leaving group . so we have water as a good leaving group . so if these electrons came off onto the oxygen these electrons in the nitrogen could move in here to form a double bond between the carbon and the nitrogen . so let 's go ahead and show that . so our next step here , this is where we lose water . so we 're going to minus h2o and let 's go ahead and show our next structure here . we have our carbon double bonded to our nitrogen this time , our nitrogen is still bonded to our y group . and our hydrogen over here . that gives the nitrogen a plus one formal charge and the carbon is still bonded to our alkyl groups and so let 's show those electrons here on our nitrogen . so i 'm going to go ahead and make those magenta . so these magenta electrons move in here to form our double bond between our carbon and our nitrogen . and this is an important structure , this is called an iminium ion , so let 's go ahead and draw that . so an iminium ion and then we lose water of course . so minus water at this stage . so we 're almost to our final product , we would just have to deprotonate our iminium ion . and so we can do that with our imine . could take that proton and leave these electrons behind on the nitrogen . so that 's going to give us our final product where we have carbon double bonded to our nitrogen and then we have nitrogen bonded to our y group here . a lone pair of electrons on that nitrogen and then we have our alkyl groups and so let 's show those electrons here in blue moving in here , off onto our nitrogen . and then once again , if our y is a hydrogen an alkyl group we have formed an imine . alright , so if that 's formation of an imine , let 's look at an example . so here is our reaction , we 're going to start with a ketone here , so cyclohexanone and react it with a primary amine . so the nitrogen is bonded to one carbon . so this is a reaction with a primary amine and then we 're going to use sulphuric acid as our catalyst here . so to figure out the product of this reaction since it 's kind of a long mechanism so it does n't really makes sense to run through the entire mechanism but we could think about part of the mechanism here . we know that the nucleophile is going to be our amine and it 's going to attack our carbonyl carbon here . so we know that in our mechanism we lose one of these protons on our amines so let 's say it 's that proton right there . and we know we 're going to lose water so minus h2o and this going to think about taking us to our iminium ion steps , so we 're going to have our ring . and so we 've already lost water so now we 're going to have a double bond to our nitrogen . and our nitrogen is going to have an ethyl group on it here and then we still have one proton on our nitrogen . so we still have the hydrogen attached here at this point . which gives us a plus one formal charge on our nitrogen . so this is our iminium ion right here . and then in our last step we know that our amine can come along and act as a base . so a lone pair of electrons on the nitrogen could take this proton right here and leave these electrons behind on our nitrogen to give us our imine products . so we go over here and we draw our product , double bond to our nitrogen and then we have our ethyl group and lone pair of electrons on our nitrogen . so if i follow those electrons , so these electrons in here in blue ended up on our nitrogen to form our imine . so a few things that i want to point out here for the iminium ion , we still have a proton attached to it . so deprotonation for our last step will form an imine . and this is once again done with a primary amine right here . and so that 's a little bit different from what we 're going to talk about next , we 're going to talk about formation of an enamine , and formation of an enamine is , it starts off the same way in terms of the mechanism but it 's this iminium ion step that changes a little bit . so let 's look at reacting a ketone with a secondary amine this time . and so let 's look at this one right here , so here we have a secondary amine and our nitrogen bonded to two other carbons so a secondary amine this time . and once again , we can kind of run through the mechanism . just thinking a little bit about what happens without going through each individual step , we know the nucleophiles going to attack here and we know we 're going to lose a proton . so this is the only proton we have left to lose and so we lose that one , we 're going to lose water in the mechanism to form our iminium ion , so let 's go ahead and draw that out we 're going to have our ring , we 're going to have a double bond to this nitrogen , we 've already lost the proton in red so now we have two alkyl groups attached to our nitrogen and a plus one formal charge on our nitrogen here . and so in the previous reaction , let me go back up to here to the previous reaction again . once again we still had a proton on our iminium ion at this step and so we 're able to deprotonate here . but that 's not the case with what we have for this ion . we do n't have a proton on our nitrogen here . and so we ca n't deprotonate in the same place , we have to pick a carbon adjacent over here . so here 's a proton over here on this carbon and our base could come along . so let 's go ahead and draw out our amine base with a lone pair of electrons right here . and a lone pair could take this proton that would push these electrons in here and then push these electrons off on to the nitrogen to give us our products , so let me go ahead and draw this out here . so now we 're going to have a double bond here and then we have our nitrogen with these ethyl groups and a lone pair of electrons on our nitrogen . so let 's follow some electrons . so these electrons in here once you deprotonate , move in here to form our double bond and then these electrons in here move out onto our nitrogen like that to form a different product . so this called an enamine , so let me go ahead and write this out here , an `` enamine . '' and the name come from `` -en '' come from the double bond . so we form double bonds , that 's the `` -en '' portion and then the `` amine '' portion comes from this part up here . we have an amine so this is an enamine . and so formation of an enamine happens because once again we do n't have the same iminium ion that we had before because of the fact that we started with a secondary amine . so that 's why it 's important to recognize the kind of amine that you were reacting with your aldehyde or a ketone . and if it 's secondary , you 're going to form an enamine here . and enamines are useful synthetic intermediates , so we 'll talk much more about them in later videos . but once again the mechanism is the same until you get to this last step here .
|
so you take that proton and leave those electrons in magenta behind on your nitrogen . so we also have an o-h group bonded to our carbon here , and then we also have our alkyl group , so r and r prime . so this intermediate is called a carbinolamine .
|
in general , what is the difference between the `` y '' group of the imine and the `` r '' groups ?
|
: let 's see how to form imines and enamines and start with the formation of an imine we start with an aldehyde or a ketone . and you add an amine to it and you need an acid catalyst and over here on the right if your y is equal to hydrogen or an alkyl group which is a r double prime you will form an imine . so let 's go ahead an write that this would be an imine . and the y can be other things , we 'll talk about those other things in the next video . you also form water and so if you , since it reacts as [ that ] equilibrium if you wanted to shift equilibrium to the right , you could remove water as it 's formed and equilibrium will shift this way and give you more of your imine product . so let 's look at the mechanism to form imines . and i 've seen two different ways to start off , the mechanism and so i 'll present both ways and you can choose which one you want to use . and so one way would be to think about an acid being present right , so we 'll say it 's h-a plus the generic acid and you think about this as protonating your imine . so you have an imine and protons , you 're protonating your imine to form this generic acid here . and then your carbonyl oxygen is going to be protonated . a lone pair of electrons on the oxygen takes this proton , leaves these electrons behind , so let 's go ahead and show the results of that . so now we have our oxygen , with a plus one formal charge and we still have our prime group over here , so let 's show those electrons . so this lone pair of electrons here on the oxygen picks up this proton and that forms this bond right here . we 've talked about in previous videos how protonation of the carbonyl oxygen makes the carbonyl carbon more electrophilic . and so we have a good electrophile , this carbon here is a good electrophile and the imine can be a good nucleophile . so a lone pair of electrons and the nitrogen over here are going to attack our carbonyl carbon and that 's going to kick these electrons in here off onto the oxygen , so let 's draw the results of that . so now we 're going to have our nitrogen is forming a bond with our carbon , so let 's show those electrons so let 's make those blue . so this lone pair of electrons of electrons on the nitrogen forms a bond between this nitrogen and this carbon here . this nitrogen is also bonded to two hydrogens . so let 's draw those in . and still our y group like that . that gives our nitrogen a plus one formal charge . and for our carbon , it 's also bonded to this oxygen , this oxygen now has two lone pairs of electrons . so let 's show the movement of those electrons so i 'll use green here . so these electrons in green moved out onto our oxygen . the carbon is still bonded to our alkyl group so r and r prime , so i 'm saying we start with a ketone here . and so that 's one way to start off your mechanism and of course there is another way to do it . so let 's talk about just using our nitrogen as a nucleophile straightaway so the imine is a pretty good nucleophile and so we can just show our nucleophile attacking our electrophile directly . so partial negative oxygen , partial positive carbon this is our electrophile and then a lone pair of electrons on our nitrogen right makes our imine a good nucleophile so they can just attack directly . right , attack this carbon push these electrons off onto your oxygen . so let 's go ahead and draw the result of that and you would have our carbon bonded to our nitrogen . our nitrogen is bonded to two hydrogens and our y group and that would make our nitrogen a plus one formal charge and then our oxygen this time would be a negative one formal charge . so negative one formal charge here . we have our alkyl groups down here , so let 's once again show our electrons . so if nitrogen attacks directly , this lone pair in blue forms this bond right here . and then we can say that these pi electrons in here kick off onto our oxygen so it does n't really matter which lone pair you make them as long as it states that one right there . and then we can show this intermediate going to the one we already talked about , if we just protonate our negative one formal charge on our oxygen here . so let 's show that , so if we have our generic acid so h-a plus , we can show a lone pair picking up this proton , leaving these electrons behind . and so now that would give us this intermediate the same one that we had before . so whichever way you would like to start off your mechanism , once we 've reached this intermediate we can think about the base . the imine here , a lone pair of electrons picking up this proton and leaving these electrons behind on our nitrogens . so let 's show what we would make from that . let 's get a little more room down here alright . so we deprotonate and then we form this structure , a carbon bonded to a nitrogen and then we still have our y group , we still have a hydrogen on this nitrogen . we have a lone pair of electrons on this nitrogen , so that lone pair came from right in here . so you take that proton and leave those electrons in magenta behind on your nitrogen . so we also have an o-h group bonded to our carbon here , and then we also have our alkyl group , so r and r prime . so this intermediate is called a carbinolamine . so let me go ahead and write that . so this is called a carbinolamine . and for the next step we can protonate the o-h group . so we take our generic acid once again so h-a plus and lone pair of electrons on our oxygen could pick up this proton leaving these electrons behind so let 's go ahead and show that . so let 's get once again some more space in here . so if we protonate our carbinolamine , we would have our carbon here , our nitrogen , lone pair of electrons on our nitrogen , our y group , our hydrogen here , our alkyl groups . and then if we protonate the o-h , we would form water as a good leaving group . so right in here , this would be a plus one formal charge on our oxygen and let 's show those electrons . so in here , lone pair of electrons on the oxygen , pick up this proton and so forming this bond right here and now you can see we have a good leaving group . so we have water as a good leaving group . so if these electrons came off onto the oxygen these electrons in the nitrogen could move in here to form a double bond between the carbon and the nitrogen . so let 's go ahead and show that . so our next step here , this is where we lose water . so we 're going to minus h2o and let 's go ahead and show our next structure here . we have our carbon double bonded to our nitrogen this time , our nitrogen is still bonded to our y group . and our hydrogen over here . that gives the nitrogen a plus one formal charge and the carbon is still bonded to our alkyl groups and so let 's show those electrons here on our nitrogen . so i 'm going to go ahead and make those magenta . so these magenta electrons move in here to form our double bond between our carbon and our nitrogen . and this is an important structure , this is called an iminium ion , so let 's go ahead and draw that . so an iminium ion and then we lose water of course . so minus water at this stage . so we 're almost to our final product , we would just have to deprotonate our iminium ion . and so we can do that with our imine . could take that proton and leave these electrons behind on the nitrogen . so that 's going to give us our final product where we have carbon double bonded to our nitrogen and then we have nitrogen bonded to our y group here . a lone pair of electrons on that nitrogen and then we have our alkyl groups and so let 's show those electrons here in blue moving in here , off onto our nitrogen . and then once again , if our y is a hydrogen an alkyl group we have formed an imine . alright , so if that 's formation of an imine , let 's look at an example . so here is our reaction , we 're going to start with a ketone here , so cyclohexanone and react it with a primary amine . so the nitrogen is bonded to one carbon . so this is a reaction with a primary amine and then we 're going to use sulphuric acid as our catalyst here . so to figure out the product of this reaction since it 's kind of a long mechanism so it does n't really makes sense to run through the entire mechanism but we could think about part of the mechanism here . we know that the nucleophile is going to be our amine and it 's going to attack our carbonyl carbon here . so we know that in our mechanism we lose one of these protons on our amines so let 's say it 's that proton right there . and we know we 're going to lose water so minus h2o and this going to think about taking us to our iminium ion steps , so we 're going to have our ring . and so we 've already lost water so now we 're going to have a double bond to our nitrogen . and our nitrogen is going to have an ethyl group on it here and then we still have one proton on our nitrogen . so we still have the hydrogen attached here at this point . which gives us a plus one formal charge on our nitrogen . so this is our iminium ion right here . and then in our last step we know that our amine can come along and act as a base . so a lone pair of electrons on the nitrogen could take this proton right here and leave these electrons behind on our nitrogen to give us our imine products . so we go over here and we draw our product , double bond to our nitrogen and then we have our ethyl group and lone pair of electrons on our nitrogen . so if i follow those electrons , so these electrons in here in blue ended up on our nitrogen to form our imine . so a few things that i want to point out here for the iminium ion , we still have a proton attached to it . so deprotonation for our last step will form an imine . and this is once again done with a primary amine right here . and so that 's a little bit different from what we 're going to talk about next , we 're going to talk about formation of an enamine , and formation of an enamine is , it starts off the same way in terms of the mechanism but it 's this iminium ion step that changes a little bit . so let 's look at reacting a ketone with a secondary amine this time . and so let 's look at this one right here , so here we have a secondary amine and our nitrogen bonded to two other carbons so a secondary amine this time . and once again , we can kind of run through the mechanism . just thinking a little bit about what happens without going through each individual step , we know the nucleophiles going to attack here and we know we 're going to lose a proton . so this is the only proton we have left to lose and so we lose that one , we 're going to lose water in the mechanism to form our iminium ion , so let 's go ahead and draw that out we 're going to have our ring , we 're going to have a double bond to this nitrogen , we 've already lost the proton in red so now we have two alkyl groups attached to our nitrogen and a plus one formal charge on our nitrogen here . and so in the previous reaction , let me go back up to here to the previous reaction again . once again we still had a proton on our iminium ion at this step and so we 're able to deprotonate here . but that 's not the case with what we have for this ion . we do n't have a proton on our nitrogen here . and so we ca n't deprotonate in the same place , we have to pick a carbon adjacent over here . so here 's a proton over here on this carbon and our base could come along . so let 's go ahead and draw out our amine base with a lone pair of electrons right here . and a lone pair could take this proton that would push these electrons in here and then push these electrons off on to the nitrogen to give us our products , so let me go ahead and draw this out here . so now we 're going to have a double bond here and then we have our nitrogen with these ethyl groups and a lone pair of electrons on our nitrogen . so let 's follow some electrons . so these electrons in here once you deprotonate , move in here to form our double bond and then these electrons in here move out onto our nitrogen like that to form a different product . so this called an enamine , so let me go ahead and write this out here , an `` enamine . '' and the name come from `` -en '' come from the double bond . so we form double bonds , that 's the `` -en '' portion and then the `` amine '' portion comes from this part up here . we have an amine so this is an enamine . and so formation of an enamine happens because once again we do n't have the same iminium ion that we had before because of the fact that we started with a secondary amine . so that 's why it 's important to recognize the kind of amine that you were reacting with your aldehyde or a ketone . and if it 's secondary , you 're going to form an enamine here . and enamines are useful synthetic intermediates , so we 'll talk much more about them in later videos . but once again the mechanism is the same until you get to this last step here .
|
so deprotonation for our last step will form an imine . and this is once again done with a primary amine right here . and so that 's a little bit different from what we 're going to talk about next , we 're going to talk about formation of an enamine , and formation of an enamine is , it starts off the same way in terms of the mechanism but it 's this iminium ion step that changes a little bit .
|
is it possible to deprotonate cyclic ring in primary amine ?
|
: let 's see how to form imines and enamines and start with the formation of an imine we start with an aldehyde or a ketone . and you add an amine to it and you need an acid catalyst and over here on the right if your y is equal to hydrogen or an alkyl group which is a r double prime you will form an imine . so let 's go ahead an write that this would be an imine . and the y can be other things , we 'll talk about those other things in the next video . you also form water and so if you , since it reacts as [ that ] equilibrium if you wanted to shift equilibrium to the right , you could remove water as it 's formed and equilibrium will shift this way and give you more of your imine product . so let 's look at the mechanism to form imines . and i 've seen two different ways to start off , the mechanism and so i 'll present both ways and you can choose which one you want to use . and so one way would be to think about an acid being present right , so we 'll say it 's h-a plus the generic acid and you think about this as protonating your imine . so you have an imine and protons , you 're protonating your imine to form this generic acid here . and then your carbonyl oxygen is going to be protonated . a lone pair of electrons on the oxygen takes this proton , leaves these electrons behind , so let 's go ahead and show the results of that . so now we have our oxygen , with a plus one formal charge and we still have our prime group over here , so let 's show those electrons . so this lone pair of electrons here on the oxygen picks up this proton and that forms this bond right here . we 've talked about in previous videos how protonation of the carbonyl oxygen makes the carbonyl carbon more electrophilic . and so we have a good electrophile , this carbon here is a good electrophile and the imine can be a good nucleophile . so a lone pair of electrons and the nitrogen over here are going to attack our carbonyl carbon and that 's going to kick these electrons in here off onto the oxygen , so let 's draw the results of that . so now we 're going to have our nitrogen is forming a bond with our carbon , so let 's show those electrons so let 's make those blue . so this lone pair of electrons of electrons on the nitrogen forms a bond between this nitrogen and this carbon here . this nitrogen is also bonded to two hydrogens . so let 's draw those in . and still our y group like that . that gives our nitrogen a plus one formal charge . and for our carbon , it 's also bonded to this oxygen , this oxygen now has two lone pairs of electrons . so let 's show the movement of those electrons so i 'll use green here . so these electrons in green moved out onto our oxygen . the carbon is still bonded to our alkyl group so r and r prime , so i 'm saying we start with a ketone here . and so that 's one way to start off your mechanism and of course there is another way to do it . so let 's talk about just using our nitrogen as a nucleophile straightaway so the imine is a pretty good nucleophile and so we can just show our nucleophile attacking our electrophile directly . so partial negative oxygen , partial positive carbon this is our electrophile and then a lone pair of electrons on our nitrogen right makes our imine a good nucleophile so they can just attack directly . right , attack this carbon push these electrons off onto your oxygen . so let 's go ahead and draw the result of that and you would have our carbon bonded to our nitrogen . our nitrogen is bonded to two hydrogens and our y group and that would make our nitrogen a plus one formal charge and then our oxygen this time would be a negative one formal charge . so negative one formal charge here . we have our alkyl groups down here , so let 's once again show our electrons . so if nitrogen attacks directly , this lone pair in blue forms this bond right here . and then we can say that these pi electrons in here kick off onto our oxygen so it does n't really matter which lone pair you make them as long as it states that one right there . and then we can show this intermediate going to the one we already talked about , if we just protonate our negative one formal charge on our oxygen here . so let 's show that , so if we have our generic acid so h-a plus , we can show a lone pair picking up this proton , leaving these electrons behind . and so now that would give us this intermediate the same one that we had before . so whichever way you would like to start off your mechanism , once we 've reached this intermediate we can think about the base . the imine here , a lone pair of electrons picking up this proton and leaving these electrons behind on our nitrogens . so let 's show what we would make from that . let 's get a little more room down here alright . so we deprotonate and then we form this structure , a carbon bonded to a nitrogen and then we still have our y group , we still have a hydrogen on this nitrogen . we have a lone pair of electrons on this nitrogen , so that lone pair came from right in here . so you take that proton and leave those electrons in magenta behind on your nitrogen . so we also have an o-h group bonded to our carbon here , and then we also have our alkyl group , so r and r prime . so this intermediate is called a carbinolamine . so let me go ahead and write that . so this is called a carbinolamine . and for the next step we can protonate the o-h group . so we take our generic acid once again so h-a plus and lone pair of electrons on our oxygen could pick up this proton leaving these electrons behind so let 's go ahead and show that . so let 's get once again some more space in here . so if we protonate our carbinolamine , we would have our carbon here , our nitrogen , lone pair of electrons on our nitrogen , our y group , our hydrogen here , our alkyl groups . and then if we protonate the o-h , we would form water as a good leaving group . so right in here , this would be a plus one formal charge on our oxygen and let 's show those electrons . so in here , lone pair of electrons on the oxygen , pick up this proton and so forming this bond right here and now you can see we have a good leaving group . so we have water as a good leaving group . so if these electrons came off onto the oxygen these electrons in the nitrogen could move in here to form a double bond between the carbon and the nitrogen . so let 's go ahead and show that . so our next step here , this is where we lose water . so we 're going to minus h2o and let 's go ahead and show our next structure here . we have our carbon double bonded to our nitrogen this time , our nitrogen is still bonded to our y group . and our hydrogen over here . that gives the nitrogen a plus one formal charge and the carbon is still bonded to our alkyl groups and so let 's show those electrons here on our nitrogen . so i 'm going to go ahead and make those magenta . so these magenta electrons move in here to form our double bond between our carbon and our nitrogen . and this is an important structure , this is called an iminium ion , so let 's go ahead and draw that . so an iminium ion and then we lose water of course . so minus water at this stage . so we 're almost to our final product , we would just have to deprotonate our iminium ion . and so we can do that with our imine . could take that proton and leave these electrons behind on the nitrogen . so that 's going to give us our final product where we have carbon double bonded to our nitrogen and then we have nitrogen bonded to our y group here . a lone pair of electrons on that nitrogen and then we have our alkyl groups and so let 's show those electrons here in blue moving in here , off onto our nitrogen . and then once again , if our y is a hydrogen an alkyl group we have formed an imine . alright , so if that 's formation of an imine , let 's look at an example . so here is our reaction , we 're going to start with a ketone here , so cyclohexanone and react it with a primary amine . so the nitrogen is bonded to one carbon . so this is a reaction with a primary amine and then we 're going to use sulphuric acid as our catalyst here . so to figure out the product of this reaction since it 's kind of a long mechanism so it does n't really makes sense to run through the entire mechanism but we could think about part of the mechanism here . we know that the nucleophile is going to be our amine and it 's going to attack our carbonyl carbon here . so we know that in our mechanism we lose one of these protons on our amines so let 's say it 's that proton right there . and we know we 're going to lose water so minus h2o and this going to think about taking us to our iminium ion steps , so we 're going to have our ring . and so we 've already lost water so now we 're going to have a double bond to our nitrogen . and our nitrogen is going to have an ethyl group on it here and then we still have one proton on our nitrogen . so we still have the hydrogen attached here at this point . which gives us a plus one formal charge on our nitrogen . so this is our iminium ion right here . and then in our last step we know that our amine can come along and act as a base . so a lone pair of electrons on the nitrogen could take this proton right here and leave these electrons behind on our nitrogen to give us our imine products . so we go over here and we draw our product , double bond to our nitrogen and then we have our ethyl group and lone pair of electrons on our nitrogen . so if i follow those electrons , so these electrons in here in blue ended up on our nitrogen to form our imine . so a few things that i want to point out here for the iminium ion , we still have a proton attached to it . so deprotonation for our last step will form an imine . and this is once again done with a primary amine right here . and so that 's a little bit different from what we 're going to talk about next , we 're going to talk about formation of an enamine , and formation of an enamine is , it starts off the same way in terms of the mechanism but it 's this iminium ion step that changes a little bit . so let 's look at reacting a ketone with a secondary amine this time . and so let 's look at this one right here , so here we have a secondary amine and our nitrogen bonded to two other carbons so a secondary amine this time . and once again , we can kind of run through the mechanism . just thinking a little bit about what happens without going through each individual step , we know the nucleophiles going to attack here and we know we 're going to lose a proton . so this is the only proton we have left to lose and so we lose that one , we 're going to lose water in the mechanism to form our iminium ion , so let 's go ahead and draw that out we 're going to have our ring , we 're going to have a double bond to this nitrogen , we 've already lost the proton in red so now we have two alkyl groups attached to our nitrogen and a plus one formal charge on our nitrogen here . and so in the previous reaction , let me go back up to here to the previous reaction again . once again we still had a proton on our iminium ion at this step and so we 're able to deprotonate here . but that 's not the case with what we have for this ion . we do n't have a proton on our nitrogen here . and so we ca n't deprotonate in the same place , we have to pick a carbon adjacent over here . so here 's a proton over here on this carbon and our base could come along . so let 's go ahead and draw out our amine base with a lone pair of electrons right here . and a lone pair could take this proton that would push these electrons in here and then push these electrons off on to the nitrogen to give us our products , so let me go ahead and draw this out here . so now we 're going to have a double bond here and then we have our nitrogen with these ethyl groups and a lone pair of electrons on our nitrogen . so let 's follow some electrons . so these electrons in here once you deprotonate , move in here to form our double bond and then these electrons in here move out onto our nitrogen like that to form a different product . so this called an enamine , so let me go ahead and write this out here , an `` enamine . '' and the name come from `` -en '' come from the double bond . so we form double bonds , that 's the `` -en '' portion and then the `` amine '' portion comes from this part up here . we have an amine so this is an enamine . and so formation of an enamine happens because once again we do n't have the same iminium ion that we had before because of the fact that we started with a secondary amine . so that 's why it 's important to recognize the kind of amine that you were reacting with your aldehyde or a ketone . and if it 's secondary , you 're going to form an enamine here . and enamines are useful synthetic intermediates , so we 'll talk much more about them in later videos . but once again the mechanism is the same until you get to this last step here .
|
so this lone pair of electrons here on the oxygen picks up this proton and that forms this bond right here . we 've talked about in previous videos how protonation of the carbonyl oxygen makes the carbonyl carbon more electrophilic . and so we have a good electrophile , this carbon here is a good electrophile and the imine can be a good nucleophile .
|
does it matter if the acid catalyst was added before or after the amine attacks the carbonyl carbon ?
|
: let 's see how to form imines and enamines and start with the formation of an imine we start with an aldehyde or a ketone . and you add an amine to it and you need an acid catalyst and over here on the right if your y is equal to hydrogen or an alkyl group which is a r double prime you will form an imine . so let 's go ahead an write that this would be an imine . and the y can be other things , we 'll talk about those other things in the next video . you also form water and so if you , since it reacts as [ that ] equilibrium if you wanted to shift equilibrium to the right , you could remove water as it 's formed and equilibrium will shift this way and give you more of your imine product . so let 's look at the mechanism to form imines . and i 've seen two different ways to start off , the mechanism and so i 'll present both ways and you can choose which one you want to use . and so one way would be to think about an acid being present right , so we 'll say it 's h-a plus the generic acid and you think about this as protonating your imine . so you have an imine and protons , you 're protonating your imine to form this generic acid here . and then your carbonyl oxygen is going to be protonated . a lone pair of electrons on the oxygen takes this proton , leaves these electrons behind , so let 's go ahead and show the results of that . so now we have our oxygen , with a plus one formal charge and we still have our prime group over here , so let 's show those electrons . so this lone pair of electrons here on the oxygen picks up this proton and that forms this bond right here . we 've talked about in previous videos how protonation of the carbonyl oxygen makes the carbonyl carbon more electrophilic . and so we have a good electrophile , this carbon here is a good electrophile and the imine can be a good nucleophile . so a lone pair of electrons and the nitrogen over here are going to attack our carbonyl carbon and that 's going to kick these electrons in here off onto the oxygen , so let 's draw the results of that . so now we 're going to have our nitrogen is forming a bond with our carbon , so let 's show those electrons so let 's make those blue . so this lone pair of electrons of electrons on the nitrogen forms a bond between this nitrogen and this carbon here . this nitrogen is also bonded to two hydrogens . so let 's draw those in . and still our y group like that . that gives our nitrogen a plus one formal charge . and for our carbon , it 's also bonded to this oxygen , this oxygen now has two lone pairs of electrons . so let 's show the movement of those electrons so i 'll use green here . so these electrons in green moved out onto our oxygen . the carbon is still bonded to our alkyl group so r and r prime , so i 'm saying we start with a ketone here . and so that 's one way to start off your mechanism and of course there is another way to do it . so let 's talk about just using our nitrogen as a nucleophile straightaway so the imine is a pretty good nucleophile and so we can just show our nucleophile attacking our electrophile directly . so partial negative oxygen , partial positive carbon this is our electrophile and then a lone pair of electrons on our nitrogen right makes our imine a good nucleophile so they can just attack directly . right , attack this carbon push these electrons off onto your oxygen . so let 's go ahead and draw the result of that and you would have our carbon bonded to our nitrogen . our nitrogen is bonded to two hydrogens and our y group and that would make our nitrogen a plus one formal charge and then our oxygen this time would be a negative one formal charge . so negative one formal charge here . we have our alkyl groups down here , so let 's once again show our electrons . so if nitrogen attacks directly , this lone pair in blue forms this bond right here . and then we can say that these pi electrons in here kick off onto our oxygen so it does n't really matter which lone pair you make them as long as it states that one right there . and then we can show this intermediate going to the one we already talked about , if we just protonate our negative one formal charge on our oxygen here . so let 's show that , so if we have our generic acid so h-a plus , we can show a lone pair picking up this proton , leaving these electrons behind . and so now that would give us this intermediate the same one that we had before . so whichever way you would like to start off your mechanism , once we 've reached this intermediate we can think about the base . the imine here , a lone pair of electrons picking up this proton and leaving these electrons behind on our nitrogens . so let 's show what we would make from that . let 's get a little more room down here alright . so we deprotonate and then we form this structure , a carbon bonded to a nitrogen and then we still have our y group , we still have a hydrogen on this nitrogen . we have a lone pair of electrons on this nitrogen , so that lone pair came from right in here . so you take that proton and leave those electrons in magenta behind on your nitrogen . so we also have an o-h group bonded to our carbon here , and then we also have our alkyl group , so r and r prime . so this intermediate is called a carbinolamine . so let me go ahead and write that . so this is called a carbinolamine . and for the next step we can protonate the o-h group . so we take our generic acid once again so h-a plus and lone pair of electrons on our oxygen could pick up this proton leaving these electrons behind so let 's go ahead and show that . so let 's get once again some more space in here . so if we protonate our carbinolamine , we would have our carbon here , our nitrogen , lone pair of electrons on our nitrogen , our y group , our hydrogen here , our alkyl groups . and then if we protonate the o-h , we would form water as a good leaving group . so right in here , this would be a plus one formal charge on our oxygen and let 's show those electrons . so in here , lone pair of electrons on the oxygen , pick up this proton and so forming this bond right here and now you can see we have a good leaving group . so we have water as a good leaving group . so if these electrons came off onto the oxygen these electrons in the nitrogen could move in here to form a double bond between the carbon and the nitrogen . so let 's go ahead and show that . so our next step here , this is where we lose water . so we 're going to minus h2o and let 's go ahead and show our next structure here . we have our carbon double bonded to our nitrogen this time , our nitrogen is still bonded to our y group . and our hydrogen over here . that gives the nitrogen a plus one formal charge and the carbon is still bonded to our alkyl groups and so let 's show those electrons here on our nitrogen . so i 'm going to go ahead and make those magenta . so these magenta electrons move in here to form our double bond between our carbon and our nitrogen . and this is an important structure , this is called an iminium ion , so let 's go ahead and draw that . so an iminium ion and then we lose water of course . so minus water at this stage . so we 're almost to our final product , we would just have to deprotonate our iminium ion . and so we can do that with our imine . could take that proton and leave these electrons behind on the nitrogen . so that 's going to give us our final product where we have carbon double bonded to our nitrogen and then we have nitrogen bonded to our y group here . a lone pair of electrons on that nitrogen and then we have our alkyl groups and so let 's show those electrons here in blue moving in here , off onto our nitrogen . and then once again , if our y is a hydrogen an alkyl group we have formed an imine . alright , so if that 's formation of an imine , let 's look at an example . so here is our reaction , we 're going to start with a ketone here , so cyclohexanone and react it with a primary amine . so the nitrogen is bonded to one carbon . so this is a reaction with a primary amine and then we 're going to use sulphuric acid as our catalyst here . so to figure out the product of this reaction since it 's kind of a long mechanism so it does n't really makes sense to run through the entire mechanism but we could think about part of the mechanism here . we know that the nucleophile is going to be our amine and it 's going to attack our carbonyl carbon here . so we know that in our mechanism we lose one of these protons on our amines so let 's say it 's that proton right there . and we know we 're going to lose water so minus h2o and this going to think about taking us to our iminium ion steps , so we 're going to have our ring . and so we 've already lost water so now we 're going to have a double bond to our nitrogen . and our nitrogen is going to have an ethyl group on it here and then we still have one proton on our nitrogen . so we still have the hydrogen attached here at this point . which gives us a plus one formal charge on our nitrogen . so this is our iminium ion right here . and then in our last step we know that our amine can come along and act as a base . so a lone pair of electrons on the nitrogen could take this proton right here and leave these electrons behind on our nitrogen to give us our imine products . so we go over here and we draw our product , double bond to our nitrogen and then we have our ethyl group and lone pair of electrons on our nitrogen . so if i follow those electrons , so these electrons in here in blue ended up on our nitrogen to form our imine . so a few things that i want to point out here for the iminium ion , we still have a proton attached to it . so deprotonation for our last step will form an imine . and this is once again done with a primary amine right here . and so that 's a little bit different from what we 're going to talk about next , we 're going to talk about formation of an enamine , and formation of an enamine is , it starts off the same way in terms of the mechanism but it 's this iminium ion step that changes a little bit . so let 's look at reacting a ketone with a secondary amine this time . and so let 's look at this one right here , so here we have a secondary amine and our nitrogen bonded to two other carbons so a secondary amine this time . and once again , we can kind of run through the mechanism . just thinking a little bit about what happens without going through each individual step , we know the nucleophiles going to attack here and we know we 're going to lose a proton . so this is the only proton we have left to lose and so we lose that one , we 're going to lose water in the mechanism to form our iminium ion , so let 's go ahead and draw that out we 're going to have our ring , we 're going to have a double bond to this nitrogen , we 've already lost the proton in red so now we have two alkyl groups attached to our nitrogen and a plus one formal charge on our nitrogen here . and so in the previous reaction , let me go back up to here to the previous reaction again . once again we still had a proton on our iminium ion at this step and so we 're able to deprotonate here . but that 's not the case with what we have for this ion . we do n't have a proton on our nitrogen here . and so we ca n't deprotonate in the same place , we have to pick a carbon adjacent over here . so here 's a proton over here on this carbon and our base could come along . so let 's go ahead and draw out our amine base with a lone pair of electrons right here . and a lone pair could take this proton that would push these electrons in here and then push these electrons off on to the nitrogen to give us our products , so let me go ahead and draw this out here . so now we 're going to have a double bond here and then we have our nitrogen with these ethyl groups and a lone pair of electrons on our nitrogen . so let 's follow some electrons . so these electrons in here once you deprotonate , move in here to form our double bond and then these electrons in here move out onto our nitrogen like that to form a different product . so this called an enamine , so let me go ahead and write this out here , an `` enamine . '' and the name come from `` -en '' come from the double bond . so we form double bonds , that 's the `` -en '' portion and then the `` amine '' portion comes from this part up here . we have an amine so this is an enamine . and so formation of an enamine happens because once again we do n't have the same iminium ion that we had before because of the fact that we started with a secondary amine . so that 's why it 's important to recognize the kind of amine that you were reacting with your aldehyde or a ketone . and if it 's secondary , you 're going to form an enamine here . and enamines are useful synthetic intermediates , so we 'll talk much more about them in later videos . but once again the mechanism is the same until you get to this last step here .
|
so partial negative oxygen , partial positive carbon this is our electrophile and then a lone pair of electrons on our nitrogen right makes our imine a good nucleophile so they can just attack directly . right , attack this carbon push these electrons off onto your oxygen . so let 's go ahead and draw the result of that and you would have our carbon bonded to our nitrogen .
|
and anyway , should n't the ammine attack the c from both sides ( from the top and from the bottom ) forming a racemic mixture ?
|
: let 's see how to form imines and enamines and start with the formation of an imine we start with an aldehyde or a ketone . and you add an amine to it and you need an acid catalyst and over here on the right if your y is equal to hydrogen or an alkyl group which is a r double prime you will form an imine . so let 's go ahead an write that this would be an imine . and the y can be other things , we 'll talk about those other things in the next video . you also form water and so if you , since it reacts as [ that ] equilibrium if you wanted to shift equilibrium to the right , you could remove water as it 's formed and equilibrium will shift this way and give you more of your imine product . so let 's look at the mechanism to form imines . and i 've seen two different ways to start off , the mechanism and so i 'll present both ways and you can choose which one you want to use . and so one way would be to think about an acid being present right , so we 'll say it 's h-a plus the generic acid and you think about this as protonating your imine . so you have an imine and protons , you 're protonating your imine to form this generic acid here . and then your carbonyl oxygen is going to be protonated . a lone pair of electrons on the oxygen takes this proton , leaves these electrons behind , so let 's go ahead and show the results of that . so now we have our oxygen , with a plus one formal charge and we still have our prime group over here , so let 's show those electrons . so this lone pair of electrons here on the oxygen picks up this proton and that forms this bond right here . we 've talked about in previous videos how protonation of the carbonyl oxygen makes the carbonyl carbon more electrophilic . and so we have a good electrophile , this carbon here is a good electrophile and the imine can be a good nucleophile . so a lone pair of electrons and the nitrogen over here are going to attack our carbonyl carbon and that 's going to kick these electrons in here off onto the oxygen , so let 's draw the results of that . so now we 're going to have our nitrogen is forming a bond with our carbon , so let 's show those electrons so let 's make those blue . so this lone pair of electrons of electrons on the nitrogen forms a bond between this nitrogen and this carbon here . this nitrogen is also bonded to two hydrogens . so let 's draw those in . and still our y group like that . that gives our nitrogen a plus one formal charge . and for our carbon , it 's also bonded to this oxygen , this oxygen now has two lone pairs of electrons . so let 's show the movement of those electrons so i 'll use green here . so these electrons in green moved out onto our oxygen . the carbon is still bonded to our alkyl group so r and r prime , so i 'm saying we start with a ketone here . and so that 's one way to start off your mechanism and of course there is another way to do it . so let 's talk about just using our nitrogen as a nucleophile straightaway so the imine is a pretty good nucleophile and so we can just show our nucleophile attacking our electrophile directly . so partial negative oxygen , partial positive carbon this is our electrophile and then a lone pair of electrons on our nitrogen right makes our imine a good nucleophile so they can just attack directly . right , attack this carbon push these electrons off onto your oxygen . so let 's go ahead and draw the result of that and you would have our carbon bonded to our nitrogen . our nitrogen is bonded to two hydrogens and our y group and that would make our nitrogen a plus one formal charge and then our oxygen this time would be a negative one formal charge . so negative one formal charge here . we have our alkyl groups down here , so let 's once again show our electrons . so if nitrogen attacks directly , this lone pair in blue forms this bond right here . and then we can say that these pi electrons in here kick off onto our oxygen so it does n't really matter which lone pair you make them as long as it states that one right there . and then we can show this intermediate going to the one we already talked about , if we just protonate our negative one formal charge on our oxygen here . so let 's show that , so if we have our generic acid so h-a plus , we can show a lone pair picking up this proton , leaving these electrons behind . and so now that would give us this intermediate the same one that we had before . so whichever way you would like to start off your mechanism , once we 've reached this intermediate we can think about the base . the imine here , a lone pair of electrons picking up this proton and leaving these electrons behind on our nitrogens . so let 's show what we would make from that . let 's get a little more room down here alright . so we deprotonate and then we form this structure , a carbon bonded to a nitrogen and then we still have our y group , we still have a hydrogen on this nitrogen . we have a lone pair of electrons on this nitrogen , so that lone pair came from right in here . so you take that proton and leave those electrons in magenta behind on your nitrogen . so we also have an o-h group bonded to our carbon here , and then we also have our alkyl group , so r and r prime . so this intermediate is called a carbinolamine . so let me go ahead and write that . so this is called a carbinolamine . and for the next step we can protonate the o-h group . so we take our generic acid once again so h-a plus and lone pair of electrons on our oxygen could pick up this proton leaving these electrons behind so let 's go ahead and show that . so let 's get once again some more space in here . so if we protonate our carbinolamine , we would have our carbon here , our nitrogen , lone pair of electrons on our nitrogen , our y group , our hydrogen here , our alkyl groups . and then if we protonate the o-h , we would form water as a good leaving group . so right in here , this would be a plus one formal charge on our oxygen and let 's show those electrons . so in here , lone pair of electrons on the oxygen , pick up this proton and so forming this bond right here and now you can see we have a good leaving group . so we have water as a good leaving group . so if these electrons came off onto the oxygen these electrons in the nitrogen could move in here to form a double bond between the carbon and the nitrogen . so let 's go ahead and show that . so our next step here , this is where we lose water . so we 're going to minus h2o and let 's go ahead and show our next structure here . we have our carbon double bonded to our nitrogen this time , our nitrogen is still bonded to our y group . and our hydrogen over here . that gives the nitrogen a plus one formal charge and the carbon is still bonded to our alkyl groups and so let 's show those electrons here on our nitrogen . so i 'm going to go ahead and make those magenta . so these magenta electrons move in here to form our double bond between our carbon and our nitrogen . and this is an important structure , this is called an iminium ion , so let 's go ahead and draw that . so an iminium ion and then we lose water of course . so minus water at this stage . so we 're almost to our final product , we would just have to deprotonate our iminium ion . and so we can do that with our imine . could take that proton and leave these electrons behind on the nitrogen . so that 's going to give us our final product where we have carbon double bonded to our nitrogen and then we have nitrogen bonded to our y group here . a lone pair of electrons on that nitrogen and then we have our alkyl groups and so let 's show those electrons here in blue moving in here , off onto our nitrogen . and then once again , if our y is a hydrogen an alkyl group we have formed an imine . alright , so if that 's formation of an imine , let 's look at an example . so here is our reaction , we 're going to start with a ketone here , so cyclohexanone and react it with a primary amine . so the nitrogen is bonded to one carbon . so this is a reaction with a primary amine and then we 're going to use sulphuric acid as our catalyst here . so to figure out the product of this reaction since it 's kind of a long mechanism so it does n't really makes sense to run through the entire mechanism but we could think about part of the mechanism here . we know that the nucleophile is going to be our amine and it 's going to attack our carbonyl carbon here . so we know that in our mechanism we lose one of these protons on our amines so let 's say it 's that proton right there . and we know we 're going to lose water so minus h2o and this going to think about taking us to our iminium ion steps , so we 're going to have our ring . and so we 've already lost water so now we 're going to have a double bond to our nitrogen . and our nitrogen is going to have an ethyl group on it here and then we still have one proton on our nitrogen . so we still have the hydrogen attached here at this point . which gives us a plus one formal charge on our nitrogen . so this is our iminium ion right here . and then in our last step we know that our amine can come along and act as a base . so a lone pair of electrons on the nitrogen could take this proton right here and leave these electrons behind on our nitrogen to give us our imine products . so we go over here and we draw our product , double bond to our nitrogen and then we have our ethyl group and lone pair of electrons on our nitrogen . so if i follow those electrons , so these electrons in here in blue ended up on our nitrogen to form our imine . so a few things that i want to point out here for the iminium ion , we still have a proton attached to it . so deprotonation for our last step will form an imine . and this is once again done with a primary amine right here . and so that 's a little bit different from what we 're going to talk about next , we 're going to talk about formation of an enamine , and formation of an enamine is , it starts off the same way in terms of the mechanism but it 's this iminium ion step that changes a little bit . so let 's look at reacting a ketone with a secondary amine this time . and so let 's look at this one right here , so here we have a secondary amine and our nitrogen bonded to two other carbons so a secondary amine this time . and once again , we can kind of run through the mechanism . just thinking a little bit about what happens without going through each individual step , we know the nucleophiles going to attack here and we know we 're going to lose a proton . so this is the only proton we have left to lose and so we lose that one , we 're going to lose water in the mechanism to form our iminium ion , so let 's go ahead and draw that out we 're going to have our ring , we 're going to have a double bond to this nitrogen , we 've already lost the proton in red so now we have two alkyl groups attached to our nitrogen and a plus one formal charge on our nitrogen here . and so in the previous reaction , let me go back up to here to the previous reaction again . once again we still had a proton on our iminium ion at this step and so we 're able to deprotonate here . but that 's not the case with what we have for this ion . we do n't have a proton on our nitrogen here . and so we ca n't deprotonate in the same place , we have to pick a carbon adjacent over here . so here 's a proton over here on this carbon and our base could come along . so let 's go ahead and draw out our amine base with a lone pair of electrons right here . and a lone pair could take this proton that would push these electrons in here and then push these electrons off on to the nitrogen to give us our products , so let me go ahead and draw this out here . so now we 're going to have a double bond here and then we have our nitrogen with these ethyl groups and a lone pair of electrons on our nitrogen . so let 's follow some electrons . so these electrons in here once you deprotonate , move in here to form our double bond and then these electrons in here move out onto our nitrogen like that to form a different product . so this called an enamine , so let me go ahead and write this out here , an `` enamine . '' and the name come from `` -en '' come from the double bond . so we form double bonds , that 's the `` -en '' portion and then the `` amine '' portion comes from this part up here . we have an amine so this is an enamine . and so formation of an enamine happens because once again we do n't have the same iminium ion that we had before because of the fact that we started with a secondary amine . so that 's why it 's important to recognize the kind of amine that you were reacting with your aldehyde or a ketone . and if it 's secondary , you 're going to form an enamine here . and enamines are useful synthetic intermediates , so we 'll talk much more about them in later videos . but once again the mechanism is the same until you get to this last step here .
|
and so one way would be to think about an acid being present right , so we 'll say it 's h-a plus the generic acid and you think about this as protonating your imine . so you have an imine and protons , you 're protonating your imine to form this generic acid here . and then your carbonyl oxygen is going to be protonated .
|
if i wanted to use an imine and form an aldehyde with an animo acid , i would simply follow these steps backwards ?
|
: let 's see how to form imines and enamines and start with the formation of an imine we start with an aldehyde or a ketone . and you add an amine to it and you need an acid catalyst and over here on the right if your y is equal to hydrogen or an alkyl group which is a r double prime you will form an imine . so let 's go ahead an write that this would be an imine . and the y can be other things , we 'll talk about those other things in the next video . you also form water and so if you , since it reacts as [ that ] equilibrium if you wanted to shift equilibrium to the right , you could remove water as it 's formed and equilibrium will shift this way and give you more of your imine product . so let 's look at the mechanism to form imines . and i 've seen two different ways to start off , the mechanism and so i 'll present both ways and you can choose which one you want to use . and so one way would be to think about an acid being present right , so we 'll say it 's h-a plus the generic acid and you think about this as protonating your imine . so you have an imine and protons , you 're protonating your imine to form this generic acid here . and then your carbonyl oxygen is going to be protonated . a lone pair of electrons on the oxygen takes this proton , leaves these electrons behind , so let 's go ahead and show the results of that . so now we have our oxygen , with a plus one formal charge and we still have our prime group over here , so let 's show those electrons . so this lone pair of electrons here on the oxygen picks up this proton and that forms this bond right here . we 've talked about in previous videos how protonation of the carbonyl oxygen makes the carbonyl carbon more electrophilic . and so we have a good electrophile , this carbon here is a good electrophile and the imine can be a good nucleophile . so a lone pair of electrons and the nitrogen over here are going to attack our carbonyl carbon and that 's going to kick these electrons in here off onto the oxygen , so let 's draw the results of that . so now we 're going to have our nitrogen is forming a bond with our carbon , so let 's show those electrons so let 's make those blue . so this lone pair of electrons of electrons on the nitrogen forms a bond between this nitrogen and this carbon here . this nitrogen is also bonded to two hydrogens . so let 's draw those in . and still our y group like that . that gives our nitrogen a plus one formal charge . and for our carbon , it 's also bonded to this oxygen , this oxygen now has two lone pairs of electrons . so let 's show the movement of those electrons so i 'll use green here . so these electrons in green moved out onto our oxygen . the carbon is still bonded to our alkyl group so r and r prime , so i 'm saying we start with a ketone here . and so that 's one way to start off your mechanism and of course there is another way to do it . so let 's talk about just using our nitrogen as a nucleophile straightaway so the imine is a pretty good nucleophile and so we can just show our nucleophile attacking our electrophile directly . so partial negative oxygen , partial positive carbon this is our electrophile and then a lone pair of electrons on our nitrogen right makes our imine a good nucleophile so they can just attack directly . right , attack this carbon push these electrons off onto your oxygen . so let 's go ahead and draw the result of that and you would have our carbon bonded to our nitrogen . our nitrogen is bonded to two hydrogens and our y group and that would make our nitrogen a plus one formal charge and then our oxygen this time would be a negative one formal charge . so negative one formal charge here . we have our alkyl groups down here , so let 's once again show our electrons . so if nitrogen attacks directly , this lone pair in blue forms this bond right here . and then we can say that these pi electrons in here kick off onto our oxygen so it does n't really matter which lone pair you make them as long as it states that one right there . and then we can show this intermediate going to the one we already talked about , if we just protonate our negative one formal charge on our oxygen here . so let 's show that , so if we have our generic acid so h-a plus , we can show a lone pair picking up this proton , leaving these electrons behind . and so now that would give us this intermediate the same one that we had before . so whichever way you would like to start off your mechanism , once we 've reached this intermediate we can think about the base . the imine here , a lone pair of electrons picking up this proton and leaving these electrons behind on our nitrogens . so let 's show what we would make from that . let 's get a little more room down here alright . so we deprotonate and then we form this structure , a carbon bonded to a nitrogen and then we still have our y group , we still have a hydrogen on this nitrogen . we have a lone pair of electrons on this nitrogen , so that lone pair came from right in here . so you take that proton and leave those electrons in magenta behind on your nitrogen . so we also have an o-h group bonded to our carbon here , and then we also have our alkyl group , so r and r prime . so this intermediate is called a carbinolamine . so let me go ahead and write that . so this is called a carbinolamine . and for the next step we can protonate the o-h group . so we take our generic acid once again so h-a plus and lone pair of electrons on our oxygen could pick up this proton leaving these electrons behind so let 's go ahead and show that . so let 's get once again some more space in here . so if we protonate our carbinolamine , we would have our carbon here , our nitrogen , lone pair of electrons on our nitrogen , our y group , our hydrogen here , our alkyl groups . and then if we protonate the o-h , we would form water as a good leaving group . so right in here , this would be a plus one formal charge on our oxygen and let 's show those electrons . so in here , lone pair of electrons on the oxygen , pick up this proton and so forming this bond right here and now you can see we have a good leaving group . so we have water as a good leaving group . so if these electrons came off onto the oxygen these electrons in the nitrogen could move in here to form a double bond between the carbon and the nitrogen . so let 's go ahead and show that . so our next step here , this is where we lose water . so we 're going to minus h2o and let 's go ahead and show our next structure here . we have our carbon double bonded to our nitrogen this time , our nitrogen is still bonded to our y group . and our hydrogen over here . that gives the nitrogen a plus one formal charge and the carbon is still bonded to our alkyl groups and so let 's show those electrons here on our nitrogen . so i 'm going to go ahead and make those magenta . so these magenta electrons move in here to form our double bond between our carbon and our nitrogen . and this is an important structure , this is called an iminium ion , so let 's go ahead and draw that . so an iminium ion and then we lose water of course . so minus water at this stage . so we 're almost to our final product , we would just have to deprotonate our iminium ion . and so we can do that with our imine . could take that proton and leave these electrons behind on the nitrogen . so that 's going to give us our final product where we have carbon double bonded to our nitrogen and then we have nitrogen bonded to our y group here . a lone pair of electrons on that nitrogen and then we have our alkyl groups and so let 's show those electrons here in blue moving in here , off onto our nitrogen . and then once again , if our y is a hydrogen an alkyl group we have formed an imine . alright , so if that 's formation of an imine , let 's look at an example . so here is our reaction , we 're going to start with a ketone here , so cyclohexanone and react it with a primary amine . so the nitrogen is bonded to one carbon . so this is a reaction with a primary amine and then we 're going to use sulphuric acid as our catalyst here . so to figure out the product of this reaction since it 's kind of a long mechanism so it does n't really makes sense to run through the entire mechanism but we could think about part of the mechanism here . we know that the nucleophile is going to be our amine and it 's going to attack our carbonyl carbon here . so we know that in our mechanism we lose one of these protons on our amines so let 's say it 's that proton right there . and we know we 're going to lose water so minus h2o and this going to think about taking us to our iminium ion steps , so we 're going to have our ring . and so we 've already lost water so now we 're going to have a double bond to our nitrogen . and our nitrogen is going to have an ethyl group on it here and then we still have one proton on our nitrogen . so we still have the hydrogen attached here at this point . which gives us a plus one formal charge on our nitrogen . so this is our iminium ion right here . and then in our last step we know that our amine can come along and act as a base . so a lone pair of electrons on the nitrogen could take this proton right here and leave these electrons behind on our nitrogen to give us our imine products . so we go over here and we draw our product , double bond to our nitrogen and then we have our ethyl group and lone pair of electrons on our nitrogen . so if i follow those electrons , so these electrons in here in blue ended up on our nitrogen to form our imine . so a few things that i want to point out here for the iminium ion , we still have a proton attached to it . so deprotonation for our last step will form an imine . and this is once again done with a primary amine right here . and so that 's a little bit different from what we 're going to talk about next , we 're going to talk about formation of an enamine , and formation of an enamine is , it starts off the same way in terms of the mechanism but it 's this iminium ion step that changes a little bit . so let 's look at reacting a ketone with a secondary amine this time . and so let 's look at this one right here , so here we have a secondary amine and our nitrogen bonded to two other carbons so a secondary amine this time . and once again , we can kind of run through the mechanism . just thinking a little bit about what happens without going through each individual step , we know the nucleophiles going to attack here and we know we 're going to lose a proton . so this is the only proton we have left to lose and so we lose that one , we 're going to lose water in the mechanism to form our iminium ion , so let 's go ahead and draw that out we 're going to have our ring , we 're going to have a double bond to this nitrogen , we 've already lost the proton in red so now we have two alkyl groups attached to our nitrogen and a plus one formal charge on our nitrogen here . and so in the previous reaction , let me go back up to here to the previous reaction again . once again we still had a proton on our iminium ion at this step and so we 're able to deprotonate here . but that 's not the case with what we have for this ion . we do n't have a proton on our nitrogen here . and so we ca n't deprotonate in the same place , we have to pick a carbon adjacent over here . so here 's a proton over here on this carbon and our base could come along . so let 's go ahead and draw out our amine base with a lone pair of electrons right here . and a lone pair could take this proton that would push these electrons in here and then push these electrons off on to the nitrogen to give us our products , so let me go ahead and draw this out here . so now we 're going to have a double bond here and then we have our nitrogen with these ethyl groups and a lone pair of electrons on our nitrogen . so let 's follow some electrons . so these electrons in here once you deprotonate , move in here to form our double bond and then these electrons in here move out onto our nitrogen like that to form a different product . so this called an enamine , so let me go ahead and write this out here , an `` enamine . '' and the name come from `` -en '' come from the double bond . so we form double bonds , that 's the `` -en '' portion and then the `` amine '' portion comes from this part up here . we have an amine so this is an enamine . and so formation of an enamine happens because once again we do n't have the same iminium ion that we had before because of the fact that we started with a secondary amine . so that 's why it 's important to recognize the kind of amine that you were reacting with your aldehyde or a ketone . and if it 's secondary , you 're going to form an enamine here . and enamines are useful synthetic intermediates , so we 'll talk much more about them in later videos . but once again the mechanism is the same until you get to this last step here .
|
and i 've seen two different ways to start off , the mechanism and so i 'll present both ways and you can choose which one you want to use . and so one way would be to think about an acid being present right , so we 'll say it 's h-a plus the generic acid and you think about this as protonating your imine . so you have an imine and protons , you 're protonating your imine to form this generic acid here .
|
why is the acid written as h-a+ ?
|
: let 's see how to form imines and enamines and start with the formation of an imine we start with an aldehyde or a ketone . and you add an amine to it and you need an acid catalyst and over here on the right if your y is equal to hydrogen or an alkyl group which is a r double prime you will form an imine . so let 's go ahead an write that this would be an imine . and the y can be other things , we 'll talk about those other things in the next video . you also form water and so if you , since it reacts as [ that ] equilibrium if you wanted to shift equilibrium to the right , you could remove water as it 's formed and equilibrium will shift this way and give you more of your imine product . so let 's look at the mechanism to form imines . and i 've seen two different ways to start off , the mechanism and so i 'll present both ways and you can choose which one you want to use . and so one way would be to think about an acid being present right , so we 'll say it 's h-a plus the generic acid and you think about this as protonating your imine . so you have an imine and protons , you 're protonating your imine to form this generic acid here . and then your carbonyl oxygen is going to be protonated . a lone pair of electrons on the oxygen takes this proton , leaves these electrons behind , so let 's go ahead and show the results of that . so now we have our oxygen , with a plus one formal charge and we still have our prime group over here , so let 's show those electrons . so this lone pair of electrons here on the oxygen picks up this proton and that forms this bond right here . we 've talked about in previous videos how protonation of the carbonyl oxygen makes the carbonyl carbon more electrophilic . and so we have a good electrophile , this carbon here is a good electrophile and the imine can be a good nucleophile . so a lone pair of electrons and the nitrogen over here are going to attack our carbonyl carbon and that 's going to kick these electrons in here off onto the oxygen , so let 's draw the results of that . so now we 're going to have our nitrogen is forming a bond with our carbon , so let 's show those electrons so let 's make those blue . so this lone pair of electrons of electrons on the nitrogen forms a bond between this nitrogen and this carbon here . this nitrogen is also bonded to two hydrogens . so let 's draw those in . and still our y group like that . that gives our nitrogen a plus one formal charge . and for our carbon , it 's also bonded to this oxygen , this oxygen now has two lone pairs of electrons . so let 's show the movement of those electrons so i 'll use green here . so these electrons in green moved out onto our oxygen . the carbon is still bonded to our alkyl group so r and r prime , so i 'm saying we start with a ketone here . and so that 's one way to start off your mechanism and of course there is another way to do it . so let 's talk about just using our nitrogen as a nucleophile straightaway so the imine is a pretty good nucleophile and so we can just show our nucleophile attacking our electrophile directly . so partial negative oxygen , partial positive carbon this is our electrophile and then a lone pair of electrons on our nitrogen right makes our imine a good nucleophile so they can just attack directly . right , attack this carbon push these electrons off onto your oxygen . so let 's go ahead and draw the result of that and you would have our carbon bonded to our nitrogen . our nitrogen is bonded to two hydrogens and our y group and that would make our nitrogen a plus one formal charge and then our oxygen this time would be a negative one formal charge . so negative one formal charge here . we have our alkyl groups down here , so let 's once again show our electrons . so if nitrogen attacks directly , this lone pair in blue forms this bond right here . and then we can say that these pi electrons in here kick off onto our oxygen so it does n't really matter which lone pair you make them as long as it states that one right there . and then we can show this intermediate going to the one we already talked about , if we just protonate our negative one formal charge on our oxygen here . so let 's show that , so if we have our generic acid so h-a plus , we can show a lone pair picking up this proton , leaving these electrons behind . and so now that would give us this intermediate the same one that we had before . so whichever way you would like to start off your mechanism , once we 've reached this intermediate we can think about the base . the imine here , a lone pair of electrons picking up this proton and leaving these electrons behind on our nitrogens . so let 's show what we would make from that . let 's get a little more room down here alright . so we deprotonate and then we form this structure , a carbon bonded to a nitrogen and then we still have our y group , we still have a hydrogen on this nitrogen . we have a lone pair of electrons on this nitrogen , so that lone pair came from right in here . so you take that proton and leave those electrons in magenta behind on your nitrogen . so we also have an o-h group bonded to our carbon here , and then we also have our alkyl group , so r and r prime . so this intermediate is called a carbinolamine . so let me go ahead and write that . so this is called a carbinolamine . and for the next step we can protonate the o-h group . so we take our generic acid once again so h-a plus and lone pair of electrons on our oxygen could pick up this proton leaving these electrons behind so let 's go ahead and show that . so let 's get once again some more space in here . so if we protonate our carbinolamine , we would have our carbon here , our nitrogen , lone pair of electrons on our nitrogen , our y group , our hydrogen here , our alkyl groups . and then if we protonate the o-h , we would form water as a good leaving group . so right in here , this would be a plus one formal charge on our oxygen and let 's show those electrons . so in here , lone pair of electrons on the oxygen , pick up this proton and so forming this bond right here and now you can see we have a good leaving group . so we have water as a good leaving group . so if these electrons came off onto the oxygen these electrons in the nitrogen could move in here to form a double bond between the carbon and the nitrogen . so let 's go ahead and show that . so our next step here , this is where we lose water . so we 're going to minus h2o and let 's go ahead and show our next structure here . we have our carbon double bonded to our nitrogen this time , our nitrogen is still bonded to our y group . and our hydrogen over here . that gives the nitrogen a plus one formal charge and the carbon is still bonded to our alkyl groups and so let 's show those electrons here on our nitrogen . so i 'm going to go ahead and make those magenta . so these magenta electrons move in here to form our double bond between our carbon and our nitrogen . and this is an important structure , this is called an iminium ion , so let 's go ahead and draw that . so an iminium ion and then we lose water of course . so minus water at this stage . so we 're almost to our final product , we would just have to deprotonate our iminium ion . and so we can do that with our imine . could take that proton and leave these electrons behind on the nitrogen . so that 's going to give us our final product where we have carbon double bonded to our nitrogen and then we have nitrogen bonded to our y group here . a lone pair of electrons on that nitrogen and then we have our alkyl groups and so let 's show those electrons here in blue moving in here , off onto our nitrogen . and then once again , if our y is a hydrogen an alkyl group we have formed an imine . alright , so if that 's formation of an imine , let 's look at an example . so here is our reaction , we 're going to start with a ketone here , so cyclohexanone and react it with a primary amine . so the nitrogen is bonded to one carbon . so this is a reaction with a primary amine and then we 're going to use sulphuric acid as our catalyst here . so to figure out the product of this reaction since it 's kind of a long mechanism so it does n't really makes sense to run through the entire mechanism but we could think about part of the mechanism here . we know that the nucleophile is going to be our amine and it 's going to attack our carbonyl carbon here . so we know that in our mechanism we lose one of these protons on our amines so let 's say it 's that proton right there . and we know we 're going to lose water so minus h2o and this going to think about taking us to our iminium ion steps , so we 're going to have our ring . and so we 've already lost water so now we 're going to have a double bond to our nitrogen . and our nitrogen is going to have an ethyl group on it here and then we still have one proton on our nitrogen . so we still have the hydrogen attached here at this point . which gives us a plus one formal charge on our nitrogen . so this is our iminium ion right here . and then in our last step we know that our amine can come along and act as a base . so a lone pair of electrons on the nitrogen could take this proton right here and leave these electrons behind on our nitrogen to give us our imine products . so we go over here and we draw our product , double bond to our nitrogen and then we have our ethyl group and lone pair of electrons on our nitrogen . so if i follow those electrons , so these electrons in here in blue ended up on our nitrogen to form our imine . so a few things that i want to point out here for the iminium ion , we still have a proton attached to it . so deprotonation for our last step will form an imine . and this is once again done with a primary amine right here . and so that 's a little bit different from what we 're going to talk about next , we 're going to talk about formation of an enamine , and formation of an enamine is , it starts off the same way in terms of the mechanism but it 's this iminium ion step that changes a little bit . so let 's look at reacting a ketone with a secondary amine this time . and so let 's look at this one right here , so here we have a secondary amine and our nitrogen bonded to two other carbons so a secondary amine this time . and once again , we can kind of run through the mechanism . just thinking a little bit about what happens without going through each individual step , we know the nucleophiles going to attack here and we know we 're going to lose a proton . so this is the only proton we have left to lose and so we lose that one , we 're going to lose water in the mechanism to form our iminium ion , so let 's go ahead and draw that out we 're going to have our ring , we 're going to have a double bond to this nitrogen , we 've already lost the proton in red so now we have two alkyl groups attached to our nitrogen and a plus one formal charge on our nitrogen here . and so in the previous reaction , let me go back up to here to the previous reaction again . once again we still had a proton on our iminium ion at this step and so we 're able to deprotonate here . but that 's not the case with what we have for this ion . we do n't have a proton on our nitrogen here . and so we ca n't deprotonate in the same place , we have to pick a carbon adjacent over here . so here 's a proton over here on this carbon and our base could come along . so let 's go ahead and draw out our amine base with a lone pair of electrons right here . and a lone pair could take this proton that would push these electrons in here and then push these electrons off on to the nitrogen to give us our products , so let me go ahead and draw this out here . so now we 're going to have a double bond here and then we have our nitrogen with these ethyl groups and a lone pair of electrons on our nitrogen . so let 's follow some electrons . so these electrons in here once you deprotonate , move in here to form our double bond and then these electrons in here move out onto our nitrogen like that to form a different product . so this called an enamine , so let me go ahead and write this out here , an `` enamine . '' and the name come from `` -en '' come from the double bond . so we form double bonds , that 's the `` -en '' portion and then the `` amine '' portion comes from this part up here . we have an amine so this is an enamine . and so formation of an enamine happens because once again we do n't have the same iminium ion that we had before because of the fact that we started with a secondary amine . so that 's why it 's important to recognize the kind of amine that you were reacting with your aldehyde or a ketone . and if it 's secondary , you 're going to form an enamine here . and enamines are useful synthetic intermediates , so we 'll talk much more about them in later videos . but once again the mechanism is the same until you get to this last step here .
|
and so that 's one way to start off your mechanism and of course there is another way to do it . so let 's talk about just using our nitrogen as a nucleophile straightaway so the imine is a pretty good nucleophile and so we can just show our nucleophile attacking our electrophile directly . so partial negative oxygen , partial positive carbon this is our electrophile and then a lone pair of electrons on our nitrogen right makes our imine a good nucleophile so they can just attack directly .
|
can an ammonium salt e.g nh4cl , ammonium succinate , ammonium oxalate etc act as a nucleophile ( in place of ammonia , amines ) & kick-on this reaction ?
|
: let 's see how to form imines and enamines and start with the formation of an imine we start with an aldehyde or a ketone . and you add an amine to it and you need an acid catalyst and over here on the right if your y is equal to hydrogen or an alkyl group which is a r double prime you will form an imine . so let 's go ahead an write that this would be an imine . and the y can be other things , we 'll talk about those other things in the next video . you also form water and so if you , since it reacts as [ that ] equilibrium if you wanted to shift equilibrium to the right , you could remove water as it 's formed and equilibrium will shift this way and give you more of your imine product . so let 's look at the mechanism to form imines . and i 've seen two different ways to start off , the mechanism and so i 'll present both ways and you can choose which one you want to use . and so one way would be to think about an acid being present right , so we 'll say it 's h-a plus the generic acid and you think about this as protonating your imine . so you have an imine and protons , you 're protonating your imine to form this generic acid here . and then your carbonyl oxygen is going to be protonated . a lone pair of electrons on the oxygen takes this proton , leaves these electrons behind , so let 's go ahead and show the results of that . so now we have our oxygen , with a plus one formal charge and we still have our prime group over here , so let 's show those electrons . so this lone pair of electrons here on the oxygen picks up this proton and that forms this bond right here . we 've talked about in previous videos how protonation of the carbonyl oxygen makes the carbonyl carbon more electrophilic . and so we have a good electrophile , this carbon here is a good electrophile and the imine can be a good nucleophile . so a lone pair of electrons and the nitrogen over here are going to attack our carbonyl carbon and that 's going to kick these electrons in here off onto the oxygen , so let 's draw the results of that . so now we 're going to have our nitrogen is forming a bond with our carbon , so let 's show those electrons so let 's make those blue . so this lone pair of electrons of electrons on the nitrogen forms a bond between this nitrogen and this carbon here . this nitrogen is also bonded to two hydrogens . so let 's draw those in . and still our y group like that . that gives our nitrogen a plus one formal charge . and for our carbon , it 's also bonded to this oxygen , this oxygen now has two lone pairs of electrons . so let 's show the movement of those electrons so i 'll use green here . so these electrons in green moved out onto our oxygen . the carbon is still bonded to our alkyl group so r and r prime , so i 'm saying we start with a ketone here . and so that 's one way to start off your mechanism and of course there is another way to do it . so let 's talk about just using our nitrogen as a nucleophile straightaway so the imine is a pretty good nucleophile and so we can just show our nucleophile attacking our electrophile directly . so partial negative oxygen , partial positive carbon this is our electrophile and then a lone pair of electrons on our nitrogen right makes our imine a good nucleophile so they can just attack directly . right , attack this carbon push these electrons off onto your oxygen . so let 's go ahead and draw the result of that and you would have our carbon bonded to our nitrogen . our nitrogen is bonded to two hydrogens and our y group and that would make our nitrogen a plus one formal charge and then our oxygen this time would be a negative one formal charge . so negative one formal charge here . we have our alkyl groups down here , so let 's once again show our electrons . so if nitrogen attacks directly , this lone pair in blue forms this bond right here . and then we can say that these pi electrons in here kick off onto our oxygen so it does n't really matter which lone pair you make them as long as it states that one right there . and then we can show this intermediate going to the one we already talked about , if we just protonate our negative one formal charge on our oxygen here . so let 's show that , so if we have our generic acid so h-a plus , we can show a lone pair picking up this proton , leaving these electrons behind . and so now that would give us this intermediate the same one that we had before . so whichever way you would like to start off your mechanism , once we 've reached this intermediate we can think about the base . the imine here , a lone pair of electrons picking up this proton and leaving these electrons behind on our nitrogens . so let 's show what we would make from that . let 's get a little more room down here alright . so we deprotonate and then we form this structure , a carbon bonded to a nitrogen and then we still have our y group , we still have a hydrogen on this nitrogen . we have a lone pair of electrons on this nitrogen , so that lone pair came from right in here . so you take that proton and leave those electrons in magenta behind on your nitrogen . so we also have an o-h group bonded to our carbon here , and then we also have our alkyl group , so r and r prime . so this intermediate is called a carbinolamine . so let me go ahead and write that . so this is called a carbinolamine . and for the next step we can protonate the o-h group . so we take our generic acid once again so h-a plus and lone pair of electrons on our oxygen could pick up this proton leaving these electrons behind so let 's go ahead and show that . so let 's get once again some more space in here . so if we protonate our carbinolamine , we would have our carbon here , our nitrogen , lone pair of electrons on our nitrogen , our y group , our hydrogen here , our alkyl groups . and then if we protonate the o-h , we would form water as a good leaving group . so right in here , this would be a plus one formal charge on our oxygen and let 's show those electrons . so in here , lone pair of electrons on the oxygen , pick up this proton and so forming this bond right here and now you can see we have a good leaving group . so we have water as a good leaving group . so if these electrons came off onto the oxygen these electrons in the nitrogen could move in here to form a double bond between the carbon and the nitrogen . so let 's go ahead and show that . so our next step here , this is where we lose water . so we 're going to minus h2o and let 's go ahead and show our next structure here . we have our carbon double bonded to our nitrogen this time , our nitrogen is still bonded to our y group . and our hydrogen over here . that gives the nitrogen a plus one formal charge and the carbon is still bonded to our alkyl groups and so let 's show those electrons here on our nitrogen . so i 'm going to go ahead and make those magenta . so these magenta electrons move in here to form our double bond between our carbon and our nitrogen . and this is an important structure , this is called an iminium ion , so let 's go ahead and draw that . so an iminium ion and then we lose water of course . so minus water at this stage . so we 're almost to our final product , we would just have to deprotonate our iminium ion . and so we can do that with our imine . could take that proton and leave these electrons behind on the nitrogen . so that 's going to give us our final product where we have carbon double bonded to our nitrogen and then we have nitrogen bonded to our y group here . a lone pair of electrons on that nitrogen and then we have our alkyl groups and so let 's show those electrons here in blue moving in here , off onto our nitrogen . and then once again , if our y is a hydrogen an alkyl group we have formed an imine . alright , so if that 's formation of an imine , let 's look at an example . so here is our reaction , we 're going to start with a ketone here , so cyclohexanone and react it with a primary amine . so the nitrogen is bonded to one carbon . so this is a reaction with a primary amine and then we 're going to use sulphuric acid as our catalyst here . so to figure out the product of this reaction since it 's kind of a long mechanism so it does n't really makes sense to run through the entire mechanism but we could think about part of the mechanism here . we know that the nucleophile is going to be our amine and it 's going to attack our carbonyl carbon here . so we know that in our mechanism we lose one of these protons on our amines so let 's say it 's that proton right there . and we know we 're going to lose water so minus h2o and this going to think about taking us to our iminium ion steps , so we 're going to have our ring . and so we 've already lost water so now we 're going to have a double bond to our nitrogen . and our nitrogen is going to have an ethyl group on it here and then we still have one proton on our nitrogen . so we still have the hydrogen attached here at this point . which gives us a plus one formal charge on our nitrogen . so this is our iminium ion right here . and then in our last step we know that our amine can come along and act as a base . so a lone pair of electrons on the nitrogen could take this proton right here and leave these electrons behind on our nitrogen to give us our imine products . so we go over here and we draw our product , double bond to our nitrogen and then we have our ethyl group and lone pair of electrons on our nitrogen . so if i follow those electrons , so these electrons in here in blue ended up on our nitrogen to form our imine . so a few things that i want to point out here for the iminium ion , we still have a proton attached to it . so deprotonation for our last step will form an imine . and this is once again done with a primary amine right here . and so that 's a little bit different from what we 're going to talk about next , we 're going to talk about formation of an enamine , and formation of an enamine is , it starts off the same way in terms of the mechanism but it 's this iminium ion step that changes a little bit . so let 's look at reacting a ketone with a secondary amine this time . and so let 's look at this one right here , so here we have a secondary amine and our nitrogen bonded to two other carbons so a secondary amine this time . and once again , we can kind of run through the mechanism . just thinking a little bit about what happens without going through each individual step , we know the nucleophiles going to attack here and we know we 're going to lose a proton . so this is the only proton we have left to lose and so we lose that one , we 're going to lose water in the mechanism to form our iminium ion , so let 's go ahead and draw that out we 're going to have our ring , we 're going to have a double bond to this nitrogen , we 've already lost the proton in red so now we have two alkyl groups attached to our nitrogen and a plus one formal charge on our nitrogen here . and so in the previous reaction , let me go back up to here to the previous reaction again . once again we still had a proton on our iminium ion at this step and so we 're able to deprotonate here . but that 's not the case with what we have for this ion . we do n't have a proton on our nitrogen here . and so we ca n't deprotonate in the same place , we have to pick a carbon adjacent over here . so here 's a proton over here on this carbon and our base could come along . so let 's go ahead and draw out our amine base with a lone pair of electrons right here . and a lone pair could take this proton that would push these electrons in here and then push these electrons off on to the nitrogen to give us our products , so let me go ahead and draw this out here . so now we 're going to have a double bond here and then we have our nitrogen with these ethyl groups and a lone pair of electrons on our nitrogen . so let 's follow some electrons . so these electrons in here once you deprotonate , move in here to form our double bond and then these electrons in here move out onto our nitrogen like that to form a different product . so this called an enamine , so let me go ahead and write this out here , an `` enamine . '' and the name come from `` -en '' come from the double bond . so we form double bonds , that 's the `` -en '' portion and then the `` amine '' portion comes from this part up here . we have an amine so this is an enamine . and so formation of an enamine happens because once again we do n't have the same iminium ion that we had before because of the fact that we started with a secondary amine . so that 's why it 's important to recognize the kind of amine that you were reacting with your aldehyde or a ketone . and if it 's secondary , you 're going to form an enamine here . and enamines are useful synthetic intermediates , so we 'll talk much more about them in later videos . but once again the mechanism is the same until you get to this last step here .
|
and so that 's a little bit different from what we 're going to talk about next , we 're going to talk about formation of an enamine , and formation of an enamine is , it starts off the same way in terms of the mechanism but it 's this iminium ion step that changes a little bit . so let 's look at reacting a ketone with a secondary amine this time . and so let 's look at this one right here , so here we have a secondary amine and our nitrogen bonded to two other carbons so a secondary amine this time . and once again , we can kind of run through the mechanism . just thinking a little bit about what happens without going through each individual step , we know the nucleophiles going to attack here and we know we 're going to lose a proton .
|
do enamines only form from ketone + 2o amine ?
|
: let 's see how to form imines and enamines and start with the formation of an imine we start with an aldehyde or a ketone . and you add an amine to it and you need an acid catalyst and over here on the right if your y is equal to hydrogen or an alkyl group which is a r double prime you will form an imine . so let 's go ahead an write that this would be an imine . and the y can be other things , we 'll talk about those other things in the next video . you also form water and so if you , since it reacts as [ that ] equilibrium if you wanted to shift equilibrium to the right , you could remove water as it 's formed and equilibrium will shift this way and give you more of your imine product . so let 's look at the mechanism to form imines . and i 've seen two different ways to start off , the mechanism and so i 'll present both ways and you can choose which one you want to use . and so one way would be to think about an acid being present right , so we 'll say it 's h-a plus the generic acid and you think about this as protonating your imine . so you have an imine and protons , you 're protonating your imine to form this generic acid here . and then your carbonyl oxygen is going to be protonated . a lone pair of electrons on the oxygen takes this proton , leaves these electrons behind , so let 's go ahead and show the results of that . so now we have our oxygen , with a plus one formal charge and we still have our prime group over here , so let 's show those electrons . so this lone pair of electrons here on the oxygen picks up this proton and that forms this bond right here . we 've talked about in previous videos how protonation of the carbonyl oxygen makes the carbonyl carbon more electrophilic . and so we have a good electrophile , this carbon here is a good electrophile and the imine can be a good nucleophile . so a lone pair of electrons and the nitrogen over here are going to attack our carbonyl carbon and that 's going to kick these electrons in here off onto the oxygen , so let 's draw the results of that . so now we 're going to have our nitrogen is forming a bond with our carbon , so let 's show those electrons so let 's make those blue . so this lone pair of electrons of electrons on the nitrogen forms a bond between this nitrogen and this carbon here . this nitrogen is also bonded to two hydrogens . so let 's draw those in . and still our y group like that . that gives our nitrogen a plus one formal charge . and for our carbon , it 's also bonded to this oxygen , this oxygen now has two lone pairs of electrons . so let 's show the movement of those electrons so i 'll use green here . so these electrons in green moved out onto our oxygen . the carbon is still bonded to our alkyl group so r and r prime , so i 'm saying we start with a ketone here . and so that 's one way to start off your mechanism and of course there is another way to do it . so let 's talk about just using our nitrogen as a nucleophile straightaway so the imine is a pretty good nucleophile and so we can just show our nucleophile attacking our electrophile directly . so partial negative oxygen , partial positive carbon this is our electrophile and then a lone pair of electrons on our nitrogen right makes our imine a good nucleophile so they can just attack directly . right , attack this carbon push these electrons off onto your oxygen . so let 's go ahead and draw the result of that and you would have our carbon bonded to our nitrogen . our nitrogen is bonded to two hydrogens and our y group and that would make our nitrogen a plus one formal charge and then our oxygen this time would be a negative one formal charge . so negative one formal charge here . we have our alkyl groups down here , so let 's once again show our electrons . so if nitrogen attacks directly , this lone pair in blue forms this bond right here . and then we can say that these pi electrons in here kick off onto our oxygen so it does n't really matter which lone pair you make them as long as it states that one right there . and then we can show this intermediate going to the one we already talked about , if we just protonate our negative one formal charge on our oxygen here . so let 's show that , so if we have our generic acid so h-a plus , we can show a lone pair picking up this proton , leaving these electrons behind . and so now that would give us this intermediate the same one that we had before . so whichever way you would like to start off your mechanism , once we 've reached this intermediate we can think about the base . the imine here , a lone pair of electrons picking up this proton and leaving these electrons behind on our nitrogens . so let 's show what we would make from that . let 's get a little more room down here alright . so we deprotonate and then we form this structure , a carbon bonded to a nitrogen and then we still have our y group , we still have a hydrogen on this nitrogen . we have a lone pair of electrons on this nitrogen , so that lone pair came from right in here . so you take that proton and leave those electrons in magenta behind on your nitrogen . so we also have an o-h group bonded to our carbon here , and then we also have our alkyl group , so r and r prime . so this intermediate is called a carbinolamine . so let me go ahead and write that . so this is called a carbinolamine . and for the next step we can protonate the o-h group . so we take our generic acid once again so h-a plus and lone pair of electrons on our oxygen could pick up this proton leaving these electrons behind so let 's go ahead and show that . so let 's get once again some more space in here . so if we protonate our carbinolamine , we would have our carbon here , our nitrogen , lone pair of electrons on our nitrogen , our y group , our hydrogen here , our alkyl groups . and then if we protonate the o-h , we would form water as a good leaving group . so right in here , this would be a plus one formal charge on our oxygen and let 's show those electrons . so in here , lone pair of electrons on the oxygen , pick up this proton and so forming this bond right here and now you can see we have a good leaving group . so we have water as a good leaving group . so if these electrons came off onto the oxygen these electrons in the nitrogen could move in here to form a double bond between the carbon and the nitrogen . so let 's go ahead and show that . so our next step here , this is where we lose water . so we 're going to minus h2o and let 's go ahead and show our next structure here . we have our carbon double bonded to our nitrogen this time , our nitrogen is still bonded to our y group . and our hydrogen over here . that gives the nitrogen a plus one formal charge and the carbon is still bonded to our alkyl groups and so let 's show those electrons here on our nitrogen . so i 'm going to go ahead and make those magenta . so these magenta electrons move in here to form our double bond between our carbon and our nitrogen . and this is an important structure , this is called an iminium ion , so let 's go ahead and draw that . so an iminium ion and then we lose water of course . so minus water at this stage . so we 're almost to our final product , we would just have to deprotonate our iminium ion . and so we can do that with our imine . could take that proton and leave these electrons behind on the nitrogen . so that 's going to give us our final product where we have carbon double bonded to our nitrogen and then we have nitrogen bonded to our y group here . a lone pair of electrons on that nitrogen and then we have our alkyl groups and so let 's show those electrons here in blue moving in here , off onto our nitrogen . and then once again , if our y is a hydrogen an alkyl group we have formed an imine . alright , so if that 's formation of an imine , let 's look at an example . so here is our reaction , we 're going to start with a ketone here , so cyclohexanone and react it with a primary amine . so the nitrogen is bonded to one carbon . so this is a reaction with a primary amine and then we 're going to use sulphuric acid as our catalyst here . so to figure out the product of this reaction since it 's kind of a long mechanism so it does n't really makes sense to run through the entire mechanism but we could think about part of the mechanism here . we know that the nucleophile is going to be our amine and it 's going to attack our carbonyl carbon here . so we know that in our mechanism we lose one of these protons on our amines so let 's say it 's that proton right there . and we know we 're going to lose water so minus h2o and this going to think about taking us to our iminium ion steps , so we 're going to have our ring . and so we 've already lost water so now we 're going to have a double bond to our nitrogen . and our nitrogen is going to have an ethyl group on it here and then we still have one proton on our nitrogen . so we still have the hydrogen attached here at this point . which gives us a plus one formal charge on our nitrogen . so this is our iminium ion right here . and then in our last step we know that our amine can come along and act as a base . so a lone pair of electrons on the nitrogen could take this proton right here and leave these electrons behind on our nitrogen to give us our imine products . so we go over here and we draw our product , double bond to our nitrogen and then we have our ethyl group and lone pair of electrons on our nitrogen . so if i follow those electrons , so these electrons in here in blue ended up on our nitrogen to form our imine . so a few things that i want to point out here for the iminium ion , we still have a proton attached to it . so deprotonation for our last step will form an imine . and this is once again done with a primary amine right here . and so that 's a little bit different from what we 're going to talk about next , we 're going to talk about formation of an enamine , and formation of an enamine is , it starts off the same way in terms of the mechanism but it 's this iminium ion step that changes a little bit . so let 's look at reacting a ketone with a secondary amine this time . and so let 's look at this one right here , so here we have a secondary amine and our nitrogen bonded to two other carbons so a secondary amine this time . and once again , we can kind of run through the mechanism . just thinking a little bit about what happens without going through each individual step , we know the nucleophiles going to attack here and we know we 're going to lose a proton . so this is the only proton we have left to lose and so we lose that one , we 're going to lose water in the mechanism to form our iminium ion , so let 's go ahead and draw that out we 're going to have our ring , we 're going to have a double bond to this nitrogen , we 've already lost the proton in red so now we have two alkyl groups attached to our nitrogen and a plus one formal charge on our nitrogen here . and so in the previous reaction , let me go back up to here to the previous reaction again . once again we still had a proton on our iminium ion at this step and so we 're able to deprotonate here . but that 's not the case with what we have for this ion . we do n't have a proton on our nitrogen here . and so we ca n't deprotonate in the same place , we have to pick a carbon adjacent over here . so here 's a proton over here on this carbon and our base could come along . so let 's go ahead and draw out our amine base with a lone pair of electrons right here . and a lone pair could take this proton that would push these electrons in here and then push these electrons off on to the nitrogen to give us our products , so let me go ahead and draw this out here . so now we 're going to have a double bond here and then we have our nitrogen with these ethyl groups and a lone pair of electrons on our nitrogen . so let 's follow some electrons . so these electrons in here once you deprotonate , move in here to form our double bond and then these electrons in here move out onto our nitrogen like that to form a different product . so this called an enamine , so let me go ahead and write this out here , an `` enamine . '' and the name come from `` -en '' come from the double bond . so we form double bonds , that 's the `` -en '' portion and then the `` amine '' portion comes from this part up here . we have an amine so this is an enamine . and so formation of an enamine happens because once again we do n't have the same iminium ion that we had before because of the fact that we started with a secondary amine . so that 's why it 's important to recognize the kind of amine that you were reacting with your aldehyde or a ketone . and if it 's secondary , you 're going to form an enamine here . and enamines are useful synthetic intermediates , so we 'll talk much more about them in later videos . but once again the mechanism is the same until you get to this last step here .
|
: let 's see how to form imines and enamines and start with the formation of an imine we start with an aldehyde or a ketone . and you add an amine to it and you need an acid catalyst and over here on the right if your y is equal to hydrogen or an alkyl group which is a r double prime you will form an imine .
|
why is the carbinoleamine not stable ?
|
: let 's see how to form imines and enamines and start with the formation of an imine we start with an aldehyde or a ketone . and you add an amine to it and you need an acid catalyst and over here on the right if your y is equal to hydrogen or an alkyl group which is a r double prime you will form an imine . so let 's go ahead an write that this would be an imine . and the y can be other things , we 'll talk about those other things in the next video . you also form water and so if you , since it reacts as [ that ] equilibrium if you wanted to shift equilibrium to the right , you could remove water as it 's formed and equilibrium will shift this way and give you more of your imine product . so let 's look at the mechanism to form imines . and i 've seen two different ways to start off , the mechanism and so i 'll present both ways and you can choose which one you want to use . and so one way would be to think about an acid being present right , so we 'll say it 's h-a plus the generic acid and you think about this as protonating your imine . so you have an imine and protons , you 're protonating your imine to form this generic acid here . and then your carbonyl oxygen is going to be protonated . a lone pair of electrons on the oxygen takes this proton , leaves these electrons behind , so let 's go ahead and show the results of that . so now we have our oxygen , with a plus one formal charge and we still have our prime group over here , so let 's show those electrons . so this lone pair of electrons here on the oxygen picks up this proton and that forms this bond right here . we 've talked about in previous videos how protonation of the carbonyl oxygen makes the carbonyl carbon more electrophilic . and so we have a good electrophile , this carbon here is a good electrophile and the imine can be a good nucleophile . so a lone pair of electrons and the nitrogen over here are going to attack our carbonyl carbon and that 's going to kick these electrons in here off onto the oxygen , so let 's draw the results of that . so now we 're going to have our nitrogen is forming a bond with our carbon , so let 's show those electrons so let 's make those blue . so this lone pair of electrons of electrons on the nitrogen forms a bond between this nitrogen and this carbon here . this nitrogen is also bonded to two hydrogens . so let 's draw those in . and still our y group like that . that gives our nitrogen a plus one formal charge . and for our carbon , it 's also bonded to this oxygen , this oxygen now has two lone pairs of electrons . so let 's show the movement of those electrons so i 'll use green here . so these electrons in green moved out onto our oxygen . the carbon is still bonded to our alkyl group so r and r prime , so i 'm saying we start with a ketone here . and so that 's one way to start off your mechanism and of course there is another way to do it . so let 's talk about just using our nitrogen as a nucleophile straightaway so the imine is a pretty good nucleophile and so we can just show our nucleophile attacking our electrophile directly . so partial negative oxygen , partial positive carbon this is our electrophile and then a lone pair of electrons on our nitrogen right makes our imine a good nucleophile so they can just attack directly . right , attack this carbon push these electrons off onto your oxygen . so let 's go ahead and draw the result of that and you would have our carbon bonded to our nitrogen . our nitrogen is bonded to two hydrogens and our y group and that would make our nitrogen a plus one formal charge and then our oxygen this time would be a negative one formal charge . so negative one formal charge here . we have our alkyl groups down here , so let 's once again show our electrons . so if nitrogen attacks directly , this lone pair in blue forms this bond right here . and then we can say that these pi electrons in here kick off onto our oxygen so it does n't really matter which lone pair you make them as long as it states that one right there . and then we can show this intermediate going to the one we already talked about , if we just protonate our negative one formal charge on our oxygen here . so let 's show that , so if we have our generic acid so h-a plus , we can show a lone pair picking up this proton , leaving these electrons behind . and so now that would give us this intermediate the same one that we had before . so whichever way you would like to start off your mechanism , once we 've reached this intermediate we can think about the base . the imine here , a lone pair of electrons picking up this proton and leaving these electrons behind on our nitrogens . so let 's show what we would make from that . let 's get a little more room down here alright . so we deprotonate and then we form this structure , a carbon bonded to a nitrogen and then we still have our y group , we still have a hydrogen on this nitrogen . we have a lone pair of electrons on this nitrogen , so that lone pair came from right in here . so you take that proton and leave those electrons in magenta behind on your nitrogen . so we also have an o-h group bonded to our carbon here , and then we also have our alkyl group , so r and r prime . so this intermediate is called a carbinolamine . so let me go ahead and write that . so this is called a carbinolamine . and for the next step we can protonate the o-h group . so we take our generic acid once again so h-a plus and lone pair of electrons on our oxygen could pick up this proton leaving these electrons behind so let 's go ahead and show that . so let 's get once again some more space in here . so if we protonate our carbinolamine , we would have our carbon here , our nitrogen , lone pair of electrons on our nitrogen , our y group , our hydrogen here , our alkyl groups . and then if we protonate the o-h , we would form water as a good leaving group . so right in here , this would be a plus one formal charge on our oxygen and let 's show those electrons . so in here , lone pair of electrons on the oxygen , pick up this proton and so forming this bond right here and now you can see we have a good leaving group . so we have water as a good leaving group . so if these electrons came off onto the oxygen these electrons in the nitrogen could move in here to form a double bond between the carbon and the nitrogen . so let 's go ahead and show that . so our next step here , this is where we lose water . so we 're going to minus h2o and let 's go ahead and show our next structure here . we have our carbon double bonded to our nitrogen this time , our nitrogen is still bonded to our y group . and our hydrogen over here . that gives the nitrogen a plus one formal charge and the carbon is still bonded to our alkyl groups and so let 's show those electrons here on our nitrogen . so i 'm going to go ahead and make those magenta . so these magenta electrons move in here to form our double bond between our carbon and our nitrogen . and this is an important structure , this is called an iminium ion , so let 's go ahead and draw that . so an iminium ion and then we lose water of course . so minus water at this stage . so we 're almost to our final product , we would just have to deprotonate our iminium ion . and so we can do that with our imine . could take that proton and leave these electrons behind on the nitrogen . so that 's going to give us our final product where we have carbon double bonded to our nitrogen and then we have nitrogen bonded to our y group here . a lone pair of electrons on that nitrogen and then we have our alkyl groups and so let 's show those electrons here in blue moving in here , off onto our nitrogen . and then once again , if our y is a hydrogen an alkyl group we have formed an imine . alright , so if that 's formation of an imine , let 's look at an example . so here is our reaction , we 're going to start with a ketone here , so cyclohexanone and react it with a primary amine . so the nitrogen is bonded to one carbon . so this is a reaction with a primary amine and then we 're going to use sulphuric acid as our catalyst here . so to figure out the product of this reaction since it 's kind of a long mechanism so it does n't really makes sense to run through the entire mechanism but we could think about part of the mechanism here . we know that the nucleophile is going to be our amine and it 's going to attack our carbonyl carbon here . so we know that in our mechanism we lose one of these protons on our amines so let 's say it 's that proton right there . and we know we 're going to lose water so minus h2o and this going to think about taking us to our iminium ion steps , so we 're going to have our ring . and so we 've already lost water so now we 're going to have a double bond to our nitrogen . and our nitrogen is going to have an ethyl group on it here and then we still have one proton on our nitrogen . so we still have the hydrogen attached here at this point . which gives us a plus one formal charge on our nitrogen . so this is our iminium ion right here . and then in our last step we know that our amine can come along and act as a base . so a lone pair of electrons on the nitrogen could take this proton right here and leave these electrons behind on our nitrogen to give us our imine products . so we go over here and we draw our product , double bond to our nitrogen and then we have our ethyl group and lone pair of electrons on our nitrogen . so if i follow those electrons , so these electrons in here in blue ended up on our nitrogen to form our imine . so a few things that i want to point out here for the iminium ion , we still have a proton attached to it . so deprotonation for our last step will form an imine . and this is once again done with a primary amine right here . and so that 's a little bit different from what we 're going to talk about next , we 're going to talk about formation of an enamine , and formation of an enamine is , it starts off the same way in terms of the mechanism but it 's this iminium ion step that changes a little bit . so let 's look at reacting a ketone with a secondary amine this time . and so let 's look at this one right here , so here we have a secondary amine and our nitrogen bonded to two other carbons so a secondary amine this time . and once again , we can kind of run through the mechanism . just thinking a little bit about what happens without going through each individual step , we know the nucleophiles going to attack here and we know we 're going to lose a proton . so this is the only proton we have left to lose and so we lose that one , we 're going to lose water in the mechanism to form our iminium ion , so let 's go ahead and draw that out we 're going to have our ring , we 're going to have a double bond to this nitrogen , we 've already lost the proton in red so now we have two alkyl groups attached to our nitrogen and a plus one formal charge on our nitrogen here . and so in the previous reaction , let me go back up to here to the previous reaction again . once again we still had a proton on our iminium ion at this step and so we 're able to deprotonate here . but that 's not the case with what we have for this ion . we do n't have a proton on our nitrogen here . and so we ca n't deprotonate in the same place , we have to pick a carbon adjacent over here . so here 's a proton over here on this carbon and our base could come along . so let 's go ahead and draw out our amine base with a lone pair of electrons right here . and a lone pair could take this proton that would push these electrons in here and then push these electrons off on to the nitrogen to give us our products , so let me go ahead and draw this out here . so now we 're going to have a double bond here and then we have our nitrogen with these ethyl groups and a lone pair of electrons on our nitrogen . so let 's follow some electrons . so these electrons in here once you deprotonate , move in here to form our double bond and then these electrons in here move out onto our nitrogen like that to form a different product . so this called an enamine , so let me go ahead and write this out here , an `` enamine . '' and the name come from `` -en '' come from the double bond . so we form double bonds , that 's the `` -en '' portion and then the `` amine '' portion comes from this part up here . we have an amine so this is an enamine . and so formation of an enamine happens because once again we do n't have the same iminium ion that we had before because of the fact that we started with a secondary amine . so that 's why it 's important to recognize the kind of amine that you were reacting with your aldehyde or a ketone . and if it 's secondary , you 're going to form an enamine here . and enamines are useful synthetic intermediates , so we 'll talk much more about them in later videos . but once again the mechanism is the same until you get to this last step here .
|
and so formation of an enamine happens because once again we do n't have the same iminium ion that we had before because of the fact that we started with a secondary amine . so that 's why it 's important to recognize the kind of amine that you were reacting with your aldehyde or a ketone . and if it 's secondary , you 're going to form an enamine here .
|
what would be the product of ammonia reacting with aldehyde or ketone ( basic conditions ) ?
|
: let 's see how to form imines and enamines and start with the formation of an imine we start with an aldehyde or a ketone . and you add an amine to it and you need an acid catalyst and over here on the right if your y is equal to hydrogen or an alkyl group which is a r double prime you will form an imine . so let 's go ahead an write that this would be an imine . and the y can be other things , we 'll talk about those other things in the next video . you also form water and so if you , since it reacts as [ that ] equilibrium if you wanted to shift equilibrium to the right , you could remove water as it 's formed and equilibrium will shift this way and give you more of your imine product . so let 's look at the mechanism to form imines . and i 've seen two different ways to start off , the mechanism and so i 'll present both ways and you can choose which one you want to use . and so one way would be to think about an acid being present right , so we 'll say it 's h-a plus the generic acid and you think about this as protonating your imine . so you have an imine and protons , you 're protonating your imine to form this generic acid here . and then your carbonyl oxygen is going to be protonated . a lone pair of electrons on the oxygen takes this proton , leaves these electrons behind , so let 's go ahead and show the results of that . so now we have our oxygen , with a plus one formal charge and we still have our prime group over here , so let 's show those electrons . so this lone pair of electrons here on the oxygen picks up this proton and that forms this bond right here . we 've talked about in previous videos how protonation of the carbonyl oxygen makes the carbonyl carbon more electrophilic . and so we have a good electrophile , this carbon here is a good electrophile and the imine can be a good nucleophile . so a lone pair of electrons and the nitrogen over here are going to attack our carbonyl carbon and that 's going to kick these electrons in here off onto the oxygen , so let 's draw the results of that . so now we 're going to have our nitrogen is forming a bond with our carbon , so let 's show those electrons so let 's make those blue . so this lone pair of electrons of electrons on the nitrogen forms a bond between this nitrogen and this carbon here . this nitrogen is also bonded to two hydrogens . so let 's draw those in . and still our y group like that . that gives our nitrogen a plus one formal charge . and for our carbon , it 's also bonded to this oxygen , this oxygen now has two lone pairs of electrons . so let 's show the movement of those electrons so i 'll use green here . so these electrons in green moved out onto our oxygen . the carbon is still bonded to our alkyl group so r and r prime , so i 'm saying we start with a ketone here . and so that 's one way to start off your mechanism and of course there is another way to do it . so let 's talk about just using our nitrogen as a nucleophile straightaway so the imine is a pretty good nucleophile and so we can just show our nucleophile attacking our electrophile directly . so partial negative oxygen , partial positive carbon this is our electrophile and then a lone pair of electrons on our nitrogen right makes our imine a good nucleophile so they can just attack directly . right , attack this carbon push these electrons off onto your oxygen . so let 's go ahead and draw the result of that and you would have our carbon bonded to our nitrogen . our nitrogen is bonded to two hydrogens and our y group and that would make our nitrogen a plus one formal charge and then our oxygen this time would be a negative one formal charge . so negative one formal charge here . we have our alkyl groups down here , so let 's once again show our electrons . so if nitrogen attacks directly , this lone pair in blue forms this bond right here . and then we can say that these pi electrons in here kick off onto our oxygen so it does n't really matter which lone pair you make them as long as it states that one right there . and then we can show this intermediate going to the one we already talked about , if we just protonate our negative one formal charge on our oxygen here . so let 's show that , so if we have our generic acid so h-a plus , we can show a lone pair picking up this proton , leaving these electrons behind . and so now that would give us this intermediate the same one that we had before . so whichever way you would like to start off your mechanism , once we 've reached this intermediate we can think about the base . the imine here , a lone pair of electrons picking up this proton and leaving these electrons behind on our nitrogens . so let 's show what we would make from that . let 's get a little more room down here alright . so we deprotonate and then we form this structure , a carbon bonded to a nitrogen and then we still have our y group , we still have a hydrogen on this nitrogen . we have a lone pair of electrons on this nitrogen , so that lone pair came from right in here . so you take that proton and leave those electrons in magenta behind on your nitrogen . so we also have an o-h group bonded to our carbon here , and then we also have our alkyl group , so r and r prime . so this intermediate is called a carbinolamine . so let me go ahead and write that . so this is called a carbinolamine . and for the next step we can protonate the o-h group . so we take our generic acid once again so h-a plus and lone pair of electrons on our oxygen could pick up this proton leaving these electrons behind so let 's go ahead and show that . so let 's get once again some more space in here . so if we protonate our carbinolamine , we would have our carbon here , our nitrogen , lone pair of electrons on our nitrogen , our y group , our hydrogen here , our alkyl groups . and then if we protonate the o-h , we would form water as a good leaving group . so right in here , this would be a plus one formal charge on our oxygen and let 's show those electrons . so in here , lone pair of electrons on the oxygen , pick up this proton and so forming this bond right here and now you can see we have a good leaving group . so we have water as a good leaving group . so if these electrons came off onto the oxygen these electrons in the nitrogen could move in here to form a double bond between the carbon and the nitrogen . so let 's go ahead and show that . so our next step here , this is where we lose water . so we 're going to minus h2o and let 's go ahead and show our next structure here . we have our carbon double bonded to our nitrogen this time , our nitrogen is still bonded to our y group . and our hydrogen over here . that gives the nitrogen a plus one formal charge and the carbon is still bonded to our alkyl groups and so let 's show those electrons here on our nitrogen . so i 'm going to go ahead and make those magenta . so these magenta electrons move in here to form our double bond between our carbon and our nitrogen . and this is an important structure , this is called an iminium ion , so let 's go ahead and draw that . so an iminium ion and then we lose water of course . so minus water at this stage . so we 're almost to our final product , we would just have to deprotonate our iminium ion . and so we can do that with our imine . could take that proton and leave these electrons behind on the nitrogen . so that 's going to give us our final product where we have carbon double bonded to our nitrogen and then we have nitrogen bonded to our y group here . a lone pair of electrons on that nitrogen and then we have our alkyl groups and so let 's show those electrons here in blue moving in here , off onto our nitrogen . and then once again , if our y is a hydrogen an alkyl group we have formed an imine . alright , so if that 's formation of an imine , let 's look at an example . so here is our reaction , we 're going to start with a ketone here , so cyclohexanone and react it with a primary amine . so the nitrogen is bonded to one carbon . so this is a reaction with a primary amine and then we 're going to use sulphuric acid as our catalyst here . so to figure out the product of this reaction since it 's kind of a long mechanism so it does n't really makes sense to run through the entire mechanism but we could think about part of the mechanism here . we know that the nucleophile is going to be our amine and it 's going to attack our carbonyl carbon here . so we know that in our mechanism we lose one of these protons on our amines so let 's say it 's that proton right there . and we know we 're going to lose water so minus h2o and this going to think about taking us to our iminium ion steps , so we 're going to have our ring . and so we 've already lost water so now we 're going to have a double bond to our nitrogen . and our nitrogen is going to have an ethyl group on it here and then we still have one proton on our nitrogen . so we still have the hydrogen attached here at this point . which gives us a plus one formal charge on our nitrogen . so this is our iminium ion right here . and then in our last step we know that our amine can come along and act as a base . so a lone pair of electrons on the nitrogen could take this proton right here and leave these electrons behind on our nitrogen to give us our imine products . so we go over here and we draw our product , double bond to our nitrogen and then we have our ethyl group and lone pair of electrons on our nitrogen . so if i follow those electrons , so these electrons in here in blue ended up on our nitrogen to form our imine . so a few things that i want to point out here for the iminium ion , we still have a proton attached to it . so deprotonation for our last step will form an imine . and this is once again done with a primary amine right here . and so that 's a little bit different from what we 're going to talk about next , we 're going to talk about formation of an enamine , and formation of an enamine is , it starts off the same way in terms of the mechanism but it 's this iminium ion step that changes a little bit . so let 's look at reacting a ketone with a secondary amine this time . and so let 's look at this one right here , so here we have a secondary amine and our nitrogen bonded to two other carbons so a secondary amine this time . and once again , we can kind of run through the mechanism . just thinking a little bit about what happens without going through each individual step , we know the nucleophiles going to attack here and we know we 're going to lose a proton . so this is the only proton we have left to lose and so we lose that one , we 're going to lose water in the mechanism to form our iminium ion , so let 's go ahead and draw that out we 're going to have our ring , we 're going to have a double bond to this nitrogen , we 've already lost the proton in red so now we have two alkyl groups attached to our nitrogen and a plus one formal charge on our nitrogen here . and so in the previous reaction , let me go back up to here to the previous reaction again . once again we still had a proton on our iminium ion at this step and so we 're able to deprotonate here . but that 's not the case with what we have for this ion . we do n't have a proton on our nitrogen here . and so we ca n't deprotonate in the same place , we have to pick a carbon adjacent over here . so here 's a proton over here on this carbon and our base could come along . so let 's go ahead and draw out our amine base with a lone pair of electrons right here . and a lone pair could take this proton that would push these electrons in here and then push these electrons off on to the nitrogen to give us our products , so let me go ahead and draw this out here . so now we 're going to have a double bond here and then we have our nitrogen with these ethyl groups and a lone pair of electrons on our nitrogen . so let 's follow some electrons . so these electrons in here once you deprotonate , move in here to form our double bond and then these electrons in here move out onto our nitrogen like that to form a different product . so this called an enamine , so let me go ahead and write this out here , an `` enamine . '' and the name come from `` -en '' come from the double bond . so we form double bonds , that 's the `` -en '' portion and then the `` amine '' portion comes from this part up here . we have an amine so this is an enamine . and so formation of an enamine happens because once again we do n't have the same iminium ion that we had before because of the fact that we started with a secondary amine . so that 's why it 's important to recognize the kind of amine that you were reacting with your aldehyde or a ketone . and if it 's secondary , you 're going to form an enamine here . and enamines are useful synthetic intermediates , so we 'll talk much more about them in later videos . but once again the mechanism is the same until you get to this last step here .
|
and so we ca n't deprotonate in the same place , we have to pick a carbon adjacent over here . so here 's a proton over here on this carbon and our base could come along . so let 's go ahead and draw out our amine base with a lone pair of electrons right here .
|
when the base picks up that specific proton , could have taken even the other one ?
|
: let 's see how to form imines and enamines and start with the formation of an imine we start with an aldehyde or a ketone . and you add an amine to it and you need an acid catalyst and over here on the right if your y is equal to hydrogen or an alkyl group which is a r double prime you will form an imine . so let 's go ahead an write that this would be an imine . and the y can be other things , we 'll talk about those other things in the next video . you also form water and so if you , since it reacts as [ that ] equilibrium if you wanted to shift equilibrium to the right , you could remove water as it 's formed and equilibrium will shift this way and give you more of your imine product . so let 's look at the mechanism to form imines . and i 've seen two different ways to start off , the mechanism and so i 'll present both ways and you can choose which one you want to use . and so one way would be to think about an acid being present right , so we 'll say it 's h-a plus the generic acid and you think about this as protonating your imine . so you have an imine and protons , you 're protonating your imine to form this generic acid here . and then your carbonyl oxygen is going to be protonated . a lone pair of electrons on the oxygen takes this proton , leaves these electrons behind , so let 's go ahead and show the results of that . so now we have our oxygen , with a plus one formal charge and we still have our prime group over here , so let 's show those electrons . so this lone pair of electrons here on the oxygen picks up this proton and that forms this bond right here . we 've talked about in previous videos how protonation of the carbonyl oxygen makes the carbonyl carbon more electrophilic . and so we have a good electrophile , this carbon here is a good electrophile and the imine can be a good nucleophile . so a lone pair of electrons and the nitrogen over here are going to attack our carbonyl carbon and that 's going to kick these electrons in here off onto the oxygen , so let 's draw the results of that . so now we 're going to have our nitrogen is forming a bond with our carbon , so let 's show those electrons so let 's make those blue . so this lone pair of electrons of electrons on the nitrogen forms a bond between this nitrogen and this carbon here . this nitrogen is also bonded to two hydrogens . so let 's draw those in . and still our y group like that . that gives our nitrogen a plus one formal charge . and for our carbon , it 's also bonded to this oxygen , this oxygen now has two lone pairs of electrons . so let 's show the movement of those electrons so i 'll use green here . so these electrons in green moved out onto our oxygen . the carbon is still bonded to our alkyl group so r and r prime , so i 'm saying we start with a ketone here . and so that 's one way to start off your mechanism and of course there is another way to do it . so let 's talk about just using our nitrogen as a nucleophile straightaway so the imine is a pretty good nucleophile and so we can just show our nucleophile attacking our electrophile directly . so partial negative oxygen , partial positive carbon this is our electrophile and then a lone pair of electrons on our nitrogen right makes our imine a good nucleophile so they can just attack directly . right , attack this carbon push these electrons off onto your oxygen . so let 's go ahead and draw the result of that and you would have our carbon bonded to our nitrogen . our nitrogen is bonded to two hydrogens and our y group and that would make our nitrogen a plus one formal charge and then our oxygen this time would be a negative one formal charge . so negative one formal charge here . we have our alkyl groups down here , so let 's once again show our electrons . so if nitrogen attacks directly , this lone pair in blue forms this bond right here . and then we can say that these pi electrons in here kick off onto our oxygen so it does n't really matter which lone pair you make them as long as it states that one right there . and then we can show this intermediate going to the one we already talked about , if we just protonate our negative one formal charge on our oxygen here . so let 's show that , so if we have our generic acid so h-a plus , we can show a lone pair picking up this proton , leaving these electrons behind . and so now that would give us this intermediate the same one that we had before . so whichever way you would like to start off your mechanism , once we 've reached this intermediate we can think about the base . the imine here , a lone pair of electrons picking up this proton and leaving these electrons behind on our nitrogens . so let 's show what we would make from that . let 's get a little more room down here alright . so we deprotonate and then we form this structure , a carbon bonded to a nitrogen and then we still have our y group , we still have a hydrogen on this nitrogen . we have a lone pair of electrons on this nitrogen , so that lone pair came from right in here . so you take that proton and leave those electrons in magenta behind on your nitrogen . so we also have an o-h group bonded to our carbon here , and then we also have our alkyl group , so r and r prime . so this intermediate is called a carbinolamine . so let me go ahead and write that . so this is called a carbinolamine . and for the next step we can protonate the o-h group . so we take our generic acid once again so h-a plus and lone pair of electrons on our oxygen could pick up this proton leaving these electrons behind so let 's go ahead and show that . so let 's get once again some more space in here . so if we protonate our carbinolamine , we would have our carbon here , our nitrogen , lone pair of electrons on our nitrogen , our y group , our hydrogen here , our alkyl groups . and then if we protonate the o-h , we would form water as a good leaving group . so right in here , this would be a plus one formal charge on our oxygen and let 's show those electrons . so in here , lone pair of electrons on the oxygen , pick up this proton and so forming this bond right here and now you can see we have a good leaving group . so we have water as a good leaving group . so if these electrons came off onto the oxygen these electrons in the nitrogen could move in here to form a double bond between the carbon and the nitrogen . so let 's go ahead and show that . so our next step here , this is where we lose water . so we 're going to minus h2o and let 's go ahead and show our next structure here . we have our carbon double bonded to our nitrogen this time , our nitrogen is still bonded to our y group . and our hydrogen over here . that gives the nitrogen a plus one formal charge and the carbon is still bonded to our alkyl groups and so let 's show those electrons here on our nitrogen . so i 'm going to go ahead and make those magenta . so these magenta electrons move in here to form our double bond between our carbon and our nitrogen . and this is an important structure , this is called an iminium ion , so let 's go ahead and draw that . so an iminium ion and then we lose water of course . so minus water at this stage . so we 're almost to our final product , we would just have to deprotonate our iminium ion . and so we can do that with our imine . could take that proton and leave these electrons behind on the nitrogen . so that 's going to give us our final product where we have carbon double bonded to our nitrogen and then we have nitrogen bonded to our y group here . a lone pair of electrons on that nitrogen and then we have our alkyl groups and so let 's show those electrons here in blue moving in here , off onto our nitrogen . and then once again , if our y is a hydrogen an alkyl group we have formed an imine . alright , so if that 's formation of an imine , let 's look at an example . so here is our reaction , we 're going to start with a ketone here , so cyclohexanone and react it with a primary amine . so the nitrogen is bonded to one carbon . so this is a reaction with a primary amine and then we 're going to use sulphuric acid as our catalyst here . so to figure out the product of this reaction since it 's kind of a long mechanism so it does n't really makes sense to run through the entire mechanism but we could think about part of the mechanism here . we know that the nucleophile is going to be our amine and it 's going to attack our carbonyl carbon here . so we know that in our mechanism we lose one of these protons on our amines so let 's say it 's that proton right there . and we know we 're going to lose water so minus h2o and this going to think about taking us to our iminium ion steps , so we 're going to have our ring . and so we 've already lost water so now we 're going to have a double bond to our nitrogen . and our nitrogen is going to have an ethyl group on it here and then we still have one proton on our nitrogen . so we still have the hydrogen attached here at this point . which gives us a plus one formal charge on our nitrogen . so this is our iminium ion right here . and then in our last step we know that our amine can come along and act as a base . so a lone pair of electrons on the nitrogen could take this proton right here and leave these electrons behind on our nitrogen to give us our imine products . so we go over here and we draw our product , double bond to our nitrogen and then we have our ethyl group and lone pair of electrons on our nitrogen . so if i follow those electrons , so these electrons in here in blue ended up on our nitrogen to form our imine . so a few things that i want to point out here for the iminium ion , we still have a proton attached to it . so deprotonation for our last step will form an imine . and this is once again done with a primary amine right here . and so that 's a little bit different from what we 're going to talk about next , we 're going to talk about formation of an enamine , and formation of an enamine is , it starts off the same way in terms of the mechanism but it 's this iminium ion step that changes a little bit . so let 's look at reacting a ketone with a secondary amine this time . and so let 's look at this one right here , so here we have a secondary amine and our nitrogen bonded to two other carbons so a secondary amine this time . and once again , we can kind of run through the mechanism . just thinking a little bit about what happens without going through each individual step , we know the nucleophiles going to attack here and we know we 're going to lose a proton . so this is the only proton we have left to lose and so we lose that one , we 're going to lose water in the mechanism to form our iminium ion , so let 's go ahead and draw that out we 're going to have our ring , we 're going to have a double bond to this nitrogen , we 've already lost the proton in red so now we have two alkyl groups attached to our nitrogen and a plus one formal charge on our nitrogen here . and so in the previous reaction , let me go back up to here to the previous reaction again . once again we still had a proton on our iminium ion at this step and so we 're able to deprotonate here . but that 's not the case with what we have for this ion . we do n't have a proton on our nitrogen here . and so we ca n't deprotonate in the same place , we have to pick a carbon adjacent over here . so here 's a proton over here on this carbon and our base could come along . so let 's go ahead and draw out our amine base with a lone pair of electrons right here . and a lone pair could take this proton that would push these electrons in here and then push these electrons off on to the nitrogen to give us our products , so let me go ahead and draw this out here . so now we 're going to have a double bond here and then we have our nitrogen with these ethyl groups and a lone pair of electrons on our nitrogen . so let 's follow some electrons . so these electrons in here once you deprotonate , move in here to form our double bond and then these electrons in here move out onto our nitrogen like that to form a different product . so this called an enamine , so let me go ahead and write this out here , an `` enamine . '' and the name come from `` -en '' come from the double bond . so we form double bonds , that 's the `` -en '' portion and then the `` amine '' portion comes from this part up here . we have an amine so this is an enamine . and so formation of an enamine happens because once again we do n't have the same iminium ion that we had before because of the fact that we started with a secondary amine . so that 's why it 's important to recognize the kind of amine that you were reacting with your aldehyde or a ketone . and if it 's secondary , you 're going to form an enamine here . and enamines are useful synthetic intermediates , so we 'll talk much more about them in later videos . but once again the mechanism is the same until you get to this last step here .
|
so the nitrogen is bonded to one carbon . so this is a reaction with a primary amine and then we 're going to use sulphuric acid as our catalyst here . so to figure out the product of this reaction since it 's kind of a long mechanism so it does n't really makes sense to run through the entire mechanism but we could think about part of the mechanism here .
|
can you use the protonated acid or the amine as a source of h ?
|
: let 's see how to form imines and enamines and start with the formation of an imine we start with an aldehyde or a ketone . and you add an amine to it and you need an acid catalyst and over here on the right if your y is equal to hydrogen or an alkyl group which is a r double prime you will form an imine . so let 's go ahead an write that this would be an imine . and the y can be other things , we 'll talk about those other things in the next video . you also form water and so if you , since it reacts as [ that ] equilibrium if you wanted to shift equilibrium to the right , you could remove water as it 's formed and equilibrium will shift this way and give you more of your imine product . so let 's look at the mechanism to form imines . and i 've seen two different ways to start off , the mechanism and so i 'll present both ways and you can choose which one you want to use . and so one way would be to think about an acid being present right , so we 'll say it 's h-a plus the generic acid and you think about this as protonating your imine . so you have an imine and protons , you 're protonating your imine to form this generic acid here . and then your carbonyl oxygen is going to be protonated . a lone pair of electrons on the oxygen takes this proton , leaves these electrons behind , so let 's go ahead and show the results of that . so now we have our oxygen , with a plus one formal charge and we still have our prime group over here , so let 's show those electrons . so this lone pair of electrons here on the oxygen picks up this proton and that forms this bond right here . we 've talked about in previous videos how protonation of the carbonyl oxygen makes the carbonyl carbon more electrophilic . and so we have a good electrophile , this carbon here is a good electrophile and the imine can be a good nucleophile . so a lone pair of electrons and the nitrogen over here are going to attack our carbonyl carbon and that 's going to kick these electrons in here off onto the oxygen , so let 's draw the results of that . so now we 're going to have our nitrogen is forming a bond with our carbon , so let 's show those electrons so let 's make those blue . so this lone pair of electrons of electrons on the nitrogen forms a bond between this nitrogen and this carbon here . this nitrogen is also bonded to two hydrogens . so let 's draw those in . and still our y group like that . that gives our nitrogen a plus one formal charge . and for our carbon , it 's also bonded to this oxygen , this oxygen now has two lone pairs of electrons . so let 's show the movement of those electrons so i 'll use green here . so these electrons in green moved out onto our oxygen . the carbon is still bonded to our alkyl group so r and r prime , so i 'm saying we start with a ketone here . and so that 's one way to start off your mechanism and of course there is another way to do it . so let 's talk about just using our nitrogen as a nucleophile straightaway so the imine is a pretty good nucleophile and so we can just show our nucleophile attacking our electrophile directly . so partial negative oxygen , partial positive carbon this is our electrophile and then a lone pair of electrons on our nitrogen right makes our imine a good nucleophile so they can just attack directly . right , attack this carbon push these electrons off onto your oxygen . so let 's go ahead and draw the result of that and you would have our carbon bonded to our nitrogen . our nitrogen is bonded to two hydrogens and our y group and that would make our nitrogen a plus one formal charge and then our oxygen this time would be a negative one formal charge . so negative one formal charge here . we have our alkyl groups down here , so let 's once again show our electrons . so if nitrogen attacks directly , this lone pair in blue forms this bond right here . and then we can say that these pi electrons in here kick off onto our oxygen so it does n't really matter which lone pair you make them as long as it states that one right there . and then we can show this intermediate going to the one we already talked about , if we just protonate our negative one formal charge on our oxygen here . so let 's show that , so if we have our generic acid so h-a plus , we can show a lone pair picking up this proton , leaving these electrons behind . and so now that would give us this intermediate the same one that we had before . so whichever way you would like to start off your mechanism , once we 've reached this intermediate we can think about the base . the imine here , a lone pair of electrons picking up this proton and leaving these electrons behind on our nitrogens . so let 's show what we would make from that . let 's get a little more room down here alright . so we deprotonate and then we form this structure , a carbon bonded to a nitrogen and then we still have our y group , we still have a hydrogen on this nitrogen . we have a lone pair of electrons on this nitrogen , so that lone pair came from right in here . so you take that proton and leave those electrons in magenta behind on your nitrogen . so we also have an o-h group bonded to our carbon here , and then we also have our alkyl group , so r and r prime . so this intermediate is called a carbinolamine . so let me go ahead and write that . so this is called a carbinolamine . and for the next step we can protonate the o-h group . so we take our generic acid once again so h-a plus and lone pair of electrons on our oxygen could pick up this proton leaving these electrons behind so let 's go ahead and show that . so let 's get once again some more space in here . so if we protonate our carbinolamine , we would have our carbon here , our nitrogen , lone pair of electrons on our nitrogen , our y group , our hydrogen here , our alkyl groups . and then if we protonate the o-h , we would form water as a good leaving group . so right in here , this would be a plus one formal charge on our oxygen and let 's show those electrons . so in here , lone pair of electrons on the oxygen , pick up this proton and so forming this bond right here and now you can see we have a good leaving group . so we have water as a good leaving group . so if these electrons came off onto the oxygen these electrons in the nitrogen could move in here to form a double bond between the carbon and the nitrogen . so let 's go ahead and show that . so our next step here , this is where we lose water . so we 're going to minus h2o and let 's go ahead and show our next structure here . we have our carbon double bonded to our nitrogen this time , our nitrogen is still bonded to our y group . and our hydrogen over here . that gives the nitrogen a plus one formal charge and the carbon is still bonded to our alkyl groups and so let 's show those electrons here on our nitrogen . so i 'm going to go ahead and make those magenta . so these magenta electrons move in here to form our double bond between our carbon and our nitrogen . and this is an important structure , this is called an iminium ion , so let 's go ahead and draw that . so an iminium ion and then we lose water of course . so minus water at this stage . so we 're almost to our final product , we would just have to deprotonate our iminium ion . and so we can do that with our imine . could take that proton and leave these electrons behind on the nitrogen . so that 's going to give us our final product where we have carbon double bonded to our nitrogen and then we have nitrogen bonded to our y group here . a lone pair of electrons on that nitrogen and then we have our alkyl groups and so let 's show those electrons here in blue moving in here , off onto our nitrogen . and then once again , if our y is a hydrogen an alkyl group we have formed an imine . alright , so if that 's formation of an imine , let 's look at an example . so here is our reaction , we 're going to start with a ketone here , so cyclohexanone and react it with a primary amine . so the nitrogen is bonded to one carbon . so this is a reaction with a primary amine and then we 're going to use sulphuric acid as our catalyst here . so to figure out the product of this reaction since it 's kind of a long mechanism so it does n't really makes sense to run through the entire mechanism but we could think about part of the mechanism here . we know that the nucleophile is going to be our amine and it 's going to attack our carbonyl carbon here . so we know that in our mechanism we lose one of these protons on our amines so let 's say it 's that proton right there . and we know we 're going to lose water so minus h2o and this going to think about taking us to our iminium ion steps , so we 're going to have our ring . and so we 've already lost water so now we 're going to have a double bond to our nitrogen . and our nitrogen is going to have an ethyl group on it here and then we still have one proton on our nitrogen . so we still have the hydrogen attached here at this point . which gives us a plus one formal charge on our nitrogen . so this is our iminium ion right here . and then in our last step we know that our amine can come along and act as a base . so a lone pair of electrons on the nitrogen could take this proton right here and leave these electrons behind on our nitrogen to give us our imine products . so we go over here and we draw our product , double bond to our nitrogen and then we have our ethyl group and lone pair of electrons on our nitrogen . so if i follow those electrons , so these electrons in here in blue ended up on our nitrogen to form our imine . so a few things that i want to point out here for the iminium ion , we still have a proton attached to it . so deprotonation for our last step will form an imine . and this is once again done with a primary amine right here . and so that 's a little bit different from what we 're going to talk about next , we 're going to talk about formation of an enamine , and formation of an enamine is , it starts off the same way in terms of the mechanism but it 's this iminium ion step that changes a little bit . so let 's look at reacting a ketone with a secondary amine this time . and so let 's look at this one right here , so here we have a secondary amine and our nitrogen bonded to two other carbons so a secondary amine this time . and once again , we can kind of run through the mechanism . just thinking a little bit about what happens without going through each individual step , we know the nucleophiles going to attack here and we know we 're going to lose a proton . so this is the only proton we have left to lose and so we lose that one , we 're going to lose water in the mechanism to form our iminium ion , so let 's go ahead and draw that out we 're going to have our ring , we 're going to have a double bond to this nitrogen , we 've already lost the proton in red so now we have two alkyl groups attached to our nitrogen and a plus one formal charge on our nitrogen here . and so in the previous reaction , let me go back up to here to the previous reaction again . once again we still had a proton on our iminium ion at this step and so we 're able to deprotonate here . but that 's not the case with what we have for this ion . we do n't have a proton on our nitrogen here . and so we ca n't deprotonate in the same place , we have to pick a carbon adjacent over here . so here 's a proton over here on this carbon and our base could come along . so let 's go ahead and draw out our amine base with a lone pair of electrons right here . and a lone pair could take this proton that would push these electrons in here and then push these electrons off on to the nitrogen to give us our products , so let me go ahead and draw this out here . so now we 're going to have a double bond here and then we have our nitrogen with these ethyl groups and a lone pair of electrons on our nitrogen . so let 's follow some electrons . so these electrons in here once you deprotonate , move in here to form our double bond and then these electrons in here move out onto our nitrogen like that to form a different product . so this called an enamine , so let me go ahead and write this out here , an `` enamine . '' and the name come from `` -en '' come from the double bond . so we form double bonds , that 's the `` -en '' portion and then the `` amine '' portion comes from this part up here . we have an amine so this is an enamine . and so formation of an enamine happens because once again we do n't have the same iminium ion that we had before because of the fact that we started with a secondary amine . so that 's why it 's important to recognize the kind of amine that you were reacting with your aldehyde or a ketone . and if it 's secondary , you 're going to form an enamine here . and enamines are useful synthetic intermediates , so we 'll talk much more about them in later videos . but once again the mechanism is the same until you get to this last step here .
|
our nitrogen is bonded to two hydrogens and our y group and that would make our nitrogen a plus one formal charge and then our oxygen this time would be a negative one formal charge . so negative one formal charge here . we have our alkyl groups down here , so let 's once again show our electrons .
|
does a formal charge of -1 mean an actual charge of -2 ?
|
: let 's see how to form imines and enamines and start with the formation of an imine we start with an aldehyde or a ketone . and you add an amine to it and you need an acid catalyst and over here on the right if your y is equal to hydrogen or an alkyl group which is a r double prime you will form an imine . so let 's go ahead an write that this would be an imine . and the y can be other things , we 'll talk about those other things in the next video . you also form water and so if you , since it reacts as [ that ] equilibrium if you wanted to shift equilibrium to the right , you could remove water as it 's formed and equilibrium will shift this way and give you more of your imine product . so let 's look at the mechanism to form imines . and i 've seen two different ways to start off , the mechanism and so i 'll present both ways and you can choose which one you want to use . and so one way would be to think about an acid being present right , so we 'll say it 's h-a plus the generic acid and you think about this as protonating your imine . so you have an imine and protons , you 're protonating your imine to form this generic acid here . and then your carbonyl oxygen is going to be protonated . a lone pair of electrons on the oxygen takes this proton , leaves these electrons behind , so let 's go ahead and show the results of that . so now we have our oxygen , with a plus one formal charge and we still have our prime group over here , so let 's show those electrons . so this lone pair of electrons here on the oxygen picks up this proton and that forms this bond right here . we 've talked about in previous videos how protonation of the carbonyl oxygen makes the carbonyl carbon more electrophilic . and so we have a good electrophile , this carbon here is a good electrophile and the imine can be a good nucleophile . so a lone pair of electrons and the nitrogen over here are going to attack our carbonyl carbon and that 's going to kick these electrons in here off onto the oxygen , so let 's draw the results of that . so now we 're going to have our nitrogen is forming a bond with our carbon , so let 's show those electrons so let 's make those blue . so this lone pair of electrons of electrons on the nitrogen forms a bond between this nitrogen and this carbon here . this nitrogen is also bonded to two hydrogens . so let 's draw those in . and still our y group like that . that gives our nitrogen a plus one formal charge . and for our carbon , it 's also bonded to this oxygen , this oxygen now has two lone pairs of electrons . so let 's show the movement of those electrons so i 'll use green here . so these electrons in green moved out onto our oxygen . the carbon is still bonded to our alkyl group so r and r prime , so i 'm saying we start with a ketone here . and so that 's one way to start off your mechanism and of course there is another way to do it . so let 's talk about just using our nitrogen as a nucleophile straightaway so the imine is a pretty good nucleophile and so we can just show our nucleophile attacking our electrophile directly . so partial negative oxygen , partial positive carbon this is our electrophile and then a lone pair of electrons on our nitrogen right makes our imine a good nucleophile so they can just attack directly . right , attack this carbon push these electrons off onto your oxygen . so let 's go ahead and draw the result of that and you would have our carbon bonded to our nitrogen . our nitrogen is bonded to two hydrogens and our y group and that would make our nitrogen a plus one formal charge and then our oxygen this time would be a negative one formal charge . so negative one formal charge here . we have our alkyl groups down here , so let 's once again show our electrons . so if nitrogen attacks directly , this lone pair in blue forms this bond right here . and then we can say that these pi electrons in here kick off onto our oxygen so it does n't really matter which lone pair you make them as long as it states that one right there . and then we can show this intermediate going to the one we already talked about , if we just protonate our negative one formal charge on our oxygen here . so let 's show that , so if we have our generic acid so h-a plus , we can show a lone pair picking up this proton , leaving these electrons behind . and so now that would give us this intermediate the same one that we had before . so whichever way you would like to start off your mechanism , once we 've reached this intermediate we can think about the base . the imine here , a lone pair of electrons picking up this proton and leaving these electrons behind on our nitrogens . so let 's show what we would make from that . let 's get a little more room down here alright . so we deprotonate and then we form this structure , a carbon bonded to a nitrogen and then we still have our y group , we still have a hydrogen on this nitrogen . we have a lone pair of electrons on this nitrogen , so that lone pair came from right in here . so you take that proton and leave those electrons in magenta behind on your nitrogen . so we also have an o-h group bonded to our carbon here , and then we also have our alkyl group , so r and r prime . so this intermediate is called a carbinolamine . so let me go ahead and write that . so this is called a carbinolamine . and for the next step we can protonate the o-h group . so we take our generic acid once again so h-a plus and lone pair of electrons on our oxygen could pick up this proton leaving these electrons behind so let 's go ahead and show that . so let 's get once again some more space in here . so if we protonate our carbinolamine , we would have our carbon here , our nitrogen , lone pair of electrons on our nitrogen , our y group , our hydrogen here , our alkyl groups . and then if we protonate the o-h , we would form water as a good leaving group . so right in here , this would be a plus one formal charge on our oxygen and let 's show those electrons . so in here , lone pair of electrons on the oxygen , pick up this proton and so forming this bond right here and now you can see we have a good leaving group . so we have water as a good leaving group . so if these electrons came off onto the oxygen these electrons in the nitrogen could move in here to form a double bond between the carbon and the nitrogen . so let 's go ahead and show that . so our next step here , this is where we lose water . so we 're going to minus h2o and let 's go ahead and show our next structure here . we have our carbon double bonded to our nitrogen this time , our nitrogen is still bonded to our y group . and our hydrogen over here . that gives the nitrogen a plus one formal charge and the carbon is still bonded to our alkyl groups and so let 's show those electrons here on our nitrogen . so i 'm going to go ahead and make those magenta . so these magenta electrons move in here to form our double bond between our carbon and our nitrogen . and this is an important structure , this is called an iminium ion , so let 's go ahead and draw that . so an iminium ion and then we lose water of course . so minus water at this stage . so we 're almost to our final product , we would just have to deprotonate our iminium ion . and so we can do that with our imine . could take that proton and leave these electrons behind on the nitrogen . so that 's going to give us our final product where we have carbon double bonded to our nitrogen and then we have nitrogen bonded to our y group here . a lone pair of electrons on that nitrogen and then we have our alkyl groups and so let 's show those electrons here in blue moving in here , off onto our nitrogen . and then once again , if our y is a hydrogen an alkyl group we have formed an imine . alright , so if that 's formation of an imine , let 's look at an example . so here is our reaction , we 're going to start with a ketone here , so cyclohexanone and react it with a primary amine . so the nitrogen is bonded to one carbon . so this is a reaction with a primary amine and then we 're going to use sulphuric acid as our catalyst here . so to figure out the product of this reaction since it 's kind of a long mechanism so it does n't really makes sense to run through the entire mechanism but we could think about part of the mechanism here . we know that the nucleophile is going to be our amine and it 's going to attack our carbonyl carbon here . so we know that in our mechanism we lose one of these protons on our amines so let 's say it 's that proton right there . and we know we 're going to lose water so minus h2o and this going to think about taking us to our iminium ion steps , so we 're going to have our ring . and so we 've already lost water so now we 're going to have a double bond to our nitrogen . and our nitrogen is going to have an ethyl group on it here and then we still have one proton on our nitrogen . so we still have the hydrogen attached here at this point . which gives us a plus one formal charge on our nitrogen . so this is our iminium ion right here . and then in our last step we know that our amine can come along and act as a base . so a lone pair of electrons on the nitrogen could take this proton right here and leave these electrons behind on our nitrogen to give us our imine products . so we go over here and we draw our product , double bond to our nitrogen and then we have our ethyl group and lone pair of electrons on our nitrogen . so if i follow those electrons , so these electrons in here in blue ended up on our nitrogen to form our imine . so a few things that i want to point out here for the iminium ion , we still have a proton attached to it . so deprotonation for our last step will form an imine . and this is once again done with a primary amine right here . and so that 's a little bit different from what we 're going to talk about next , we 're going to talk about formation of an enamine , and formation of an enamine is , it starts off the same way in terms of the mechanism but it 's this iminium ion step that changes a little bit . so let 's look at reacting a ketone with a secondary amine this time . and so let 's look at this one right here , so here we have a secondary amine and our nitrogen bonded to two other carbons so a secondary amine this time . and once again , we can kind of run through the mechanism . just thinking a little bit about what happens without going through each individual step , we know the nucleophiles going to attack here and we know we 're going to lose a proton . so this is the only proton we have left to lose and so we lose that one , we 're going to lose water in the mechanism to form our iminium ion , so let 's go ahead and draw that out we 're going to have our ring , we 're going to have a double bond to this nitrogen , we 've already lost the proton in red so now we have two alkyl groups attached to our nitrogen and a plus one formal charge on our nitrogen here . and so in the previous reaction , let me go back up to here to the previous reaction again . once again we still had a proton on our iminium ion at this step and so we 're able to deprotonate here . but that 's not the case with what we have for this ion . we do n't have a proton on our nitrogen here . and so we ca n't deprotonate in the same place , we have to pick a carbon adjacent over here . so here 's a proton over here on this carbon and our base could come along . so let 's go ahead and draw out our amine base with a lone pair of electrons right here . and a lone pair could take this proton that would push these electrons in here and then push these electrons off on to the nitrogen to give us our products , so let me go ahead and draw this out here . so now we 're going to have a double bond here and then we have our nitrogen with these ethyl groups and a lone pair of electrons on our nitrogen . so let 's follow some electrons . so these electrons in here once you deprotonate , move in here to form our double bond and then these electrons in here move out onto our nitrogen like that to form a different product . so this called an enamine , so let me go ahead and write this out here , an `` enamine . '' and the name come from `` -en '' come from the double bond . so we form double bonds , that 's the `` -en '' portion and then the `` amine '' portion comes from this part up here . we have an amine so this is an enamine . and so formation of an enamine happens because once again we do n't have the same iminium ion that we had before because of the fact that we started with a secondary amine . so that 's why it 's important to recognize the kind of amine that you were reacting with your aldehyde or a ketone . and if it 's secondary , you 're going to form an enamine here . and enamines are useful synthetic intermediates , so we 'll talk much more about them in later videos . but once again the mechanism is the same until you get to this last step here .
|
and i 've seen two different ways to start off , the mechanism and so i 'll present both ways and you can choose which one you want to use . and so one way would be to think about an acid being present right , so we 'll say it 's h-a plus the generic acid and you think about this as protonating your imine . so you have an imine and protons , you 're protonating your imine to form this generic acid here .
|
why is the acid written as h-a+ ?
|
in this video we 're going to introduce ourselves to the idea of adding decimals . and i encourage you as we work through these problems to keep pausing the video and seeing if you can think about it on your own before we work through it together . now we 're going to build up slowly , and in future videos we 're gon na find out faster ways of doing it . but the way we 're learning it in this video in the next view is to really make sure we understand what is happening . so let 's say we wanted to add 0.1 to 0.8 . or you could say we 're adding 1/10 to 8/10 . pause this video and see if you can figure that out . well there 's a couple of ways to think about it . you could say , hey look 0.1 , that is 1/10 , and 0.8 , that is 8/10 . and so if i have one of something and i add eight more of that something , so i have 1/10 , and i 'm gon na add eight more tenths , well i 'm gon na end up with nine of that something , in this case we 're talking about tenths . so that is going to be equal to 9/10 . that 's one way to think about it . another way , we could think about it visually . so let 's say we take a whole , and we were to divide it into tenths , which we have right over here . so if we say this whole square is a whole , we divide it into 10 equal sections . so each of these white bars you can view as a tenth . so we have 1/10 , so let me fill that in . so 1/10 , woops that 's not what i wanted to do . we have 1/10 right over there . and to that we want to add 8/10 . so one , two , three , four , five , six , seven , eight . and so how many total tenths do i now have ? well let 's just count 'em up . we have this one here . one , two , three , four , five , six , seven , eight , nine . these are really saying the same thing . all of this together , is going to be , let me do that a little neater . all of this together is going to be 9/10 . now in either case , how do we write 9/10 in decimal form ? well we go to the tenth 's place , which is one space on the right side of the decimal . we say hey we have 9/10 . this is the tenth 's space right over here . so that 's just saying we have 9/10 . we have nine of these tenths right over here . so let 's keep building . let 's do another example . so let 's say that we , let me clear all of this out . so let 's say that we want to add , do these with different colors . so let 's say we want to add , i have trouble because my pen is n't working . let 's see . let 's say we want to add ... my pen is , oh here we go . let 's say we want to add 3/10 , and to that , we want to add 9/10 . what is that going to be ? well you could use the same idea . if you say this is 3/10 , and this is 9/10 , plus 9/10 , well if i have three of something and i add nine of them , well that 's going to be 12 . three plus nine is 12 . so we could say this is going to be equal to 12/10 . now this one might be a little bit counterintuitive . 12/10 , what does that mean ? well one way to think about it , this is 10/10 , plus 2/10 . and what are 10/10 ? well if i have 10/10 , this right over here is one whole . so that is going to be one . so we have one and 2/10 . so how do we write one and 2/10 ? well we could write it as in the one 's place , we just write a one . and then in the tenth 's place , we write our 2/10 . so you could say it 's equal to 1.2 , or you could say it 's equal to one and 2/10 , which is the same thing as 12/10 . now if we want to see that visually , let 's get our diagram out again . so actually i 'm gon na put two of these here . so one , and then a second one . and we want to add , so let 's start with the 3/10 . so let me color these in really fast . use that light blue color . that is 1/10 . this is 2/10 . just coloring 'em in really fast . and this is 3/10 . and then to that we 're gon na add 9/10 . so to that we 're gon na add one , two , three . i 'm not coloring them in fully . four , you get the idea . five , almost there . six . i need to color faster . seven . seven . eight . nine . so there you have it . i have added 9/10 . you notice i 've colored in nine , i 've colored in yellow , nine of the tenths , and before i had three of the tenths colored in . and when you add 'em all together , what happens ? well the 3/10 plus the 7/10 right over here , they made a hole . so this right over here is our one . and then we also have another 2/10 left over . and so this is where , this is our 0.2 , or 2/10 . so it 's gon na be one plus 2/10 , which is 1.2 . so hopefully this gives you a good sense of how we think about adding decimals . and even though in the future we 're gon na figure out faster ways of doing it , or more systematic ways of doing it , this is still the way that i still do it in my head if someone walks up to me on the street and says hey , add 0.3 to 0.9 . that 's how i think about it .
|
but the way we 're learning it in this video in the next view is to really make sure we understand what is happening . so let 's say we wanted to add 0.1 to 0.8 . or you could say we 're adding 1/10 to 8/10 .
|
do we say like 0 point 1 or 0 and 1 because my teacher would say and ?
|
in this video we 're going to introduce ourselves to the idea of adding decimals . and i encourage you as we work through these problems to keep pausing the video and seeing if you can think about it on your own before we work through it together . now we 're going to build up slowly , and in future videos we 're gon na find out faster ways of doing it . but the way we 're learning it in this video in the next view is to really make sure we understand what is happening . so let 's say we wanted to add 0.1 to 0.8 . or you could say we 're adding 1/10 to 8/10 . pause this video and see if you can figure that out . well there 's a couple of ways to think about it . you could say , hey look 0.1 , that is 1/10 , and 0.8 , that is 8/10 . and so if i have one of something and i add eight more of that something , so i have 1/10 , and i 'm gon na add eight more tenths , well i 'm gon na end up with nine of that something , in this case we 're talking about tenths . so that is going to be equal to 9/10 . that 's one way to think about it . another way , we could think about it visually . so let 's say we take a whole , and we were to divide it into tenths , which we have right over here . so if we say this whole square is a whole , we divide it into 10 equal sections . so each of these white bars you can view as a tenth . so we have 1/10 , so let me fill that in . so 1/10 , woops that 's not what i wanted to do . we have 1/10 right over there . and to that we want to add 8/10 . so one , two , three , four , five , six , seven , eight . and so how many total tenths do i now have ? well let 's just count 'em up . we have this one here . one , two , three , four , five , six , seven , eight , nine . these are really saying the same thing . all of this together , is going to be , let me do that a little neater . all of this together is going to be 9/10 . now in either case , how do we write 9/10 in decimal form ? well we go to the tenth 's place , which is one space on the right side of the decimal . we say hey we have 9/10 . this is the tenth 's space right over here . so that 's just saying we have 9/10 . we have nine of these tenths right over here . so let 's keep building . let 's do another example . so let 's say that we , let me clear all of this out . so let 's say that we want to add , do these with different colors . so let 's say we want to add , i have trouble because my pen is n't working . let 's see . let 's say we want to add ... my pen is , oh here we go . let 's say we want to add 3/10 , and to that , we want to add 9/10 . what is that going to be ? well you could use the same idea . if you say this is 3/10 , and this is 9/10 , plus 9/10 , well if i have three of something and i add nine of them , well that 's going to be 12 . three plus nine is 12 . so we could say this is going to be equal to 12/10 . now this one might be a little bit counterintuitive . 12/10 , what does that mean ? well one way to think about it , this is 10/10 , plus 2/10 . and what are 10/10 ? well if i have 10/10 , this right over here is one whole . so that is going to be one . so we have one and 2/10 . so how do we write one and 2/10 ? well we could write it as in the one 's place , we just write a one . and then in the tenth 's place , we write our 2/10 . so you could say it 's equal to 1.2 , or you could say it 's equal to one and 2/10 , which is the same thing as 12/10 . now if we want to see that visually , let 's get our diagram out again . so actually i 'm gon na put two of these here . so one , and then a second one . and we want to add , so let 's start with the 3/10 . so let me color these in really fast . use that light blue color . that is 1/10 . this is 2/10 . just coloring 'em in really fast . and this is 3/10 . and then to that we 're gon na add 9/10 . so to that we 're gon na add one , two , three . i 'm not coloring them in fully . four , you get the idea . five , almost there . six . i need to color faster . seven . seven . eight . nine . so there you have it . i have added 9/10 . you notice i 've colored in nine , i 've colored in yellow , nine of the tenths , and before i had three of the tenths colored in . and when you add 'em all together , what happens ? well the 3/10 plus the 7/10 right over here , they made a hole . so this right over here is our one . and then we also have another 2/10 left over . and so this is where , this is our 0.2 , or 2/10 . so it 's gon na be one plus 2/10 , which is 1.2 . so hopefully this gives you a good sense of how we think about adding decimals . and even though in the future we 're gon na figure out faster ways of doing it , or more systematic ways of doing it , this is still the way that i still do it in my head if someone walks up to me on the street and says hey , add 0.3 to 0.9 . that 's how i think about it .
|
so let me color these in really fast . use that light blue color . that is 1/10 .
|
why did sal use his pen to color the parts instead of the color tool he used last time ?
|
imagine if it was pitch black in your room . would you be able to walk ? as long as nothing got in your way , you 'd probably be able to walk perfectly fine . but why is this ? when walking in a pitch black room , you rely on your sense of balance . you know exactly where you are in space . you know whether or not you 're standing straight up or if you 're sitting down . and as you 're walking , you know if your right foot is in front of your left or if your left foot is in front of your right . so how is it that we know exactly where our body is in space without actually having a look at it ? well , this is known as our sense of propioception . and proprioception is basically defined as our ability to sense exactly where our body is in space . in other words , this is our sense of position . and this sense actually originates from a bunch of tiny little sensors that are located throughout our entire body in almost all of our muscles . so let 's imagine that this right here is a muscle in your leg or in your arm . let 's just say it 's a muscle in your arm . so there 's a tiny little receptor in it , inside of the muscle . and this receptor will actually go up to your spinal cord and then eventually to your brand . and this receptor , if we kind of zoom in on it , we zoom in on this receptor , is sensitive to stretching . so as this muscle contracts , so if i was going to lift something really heavy , the muscle would contract , and it would get thinner , so it would look like this . it would get a bit thinner . and so this sensor , which is known as a spindle -- i 'll write that down here . this is a spindle . this sensor can sense that the muscle has been stretched out , and it too will also stretch . so it will go from this conformation to this conformation . it will actually get stretched out . and so we can draw a little spring-like structure inside , which is actually quite similar to what 's actually inside the spindle . there 's actually a protein inside that gets stretched . when that protein gets stretched , it fires a signal to the brain . this is the basic principle behind proprioception . so we 're able to tell exactly how contracted or how relaxed every single muscle is in our entire body , and this allows us to know exactly where our body is in space . there 's another word that 's commonly used to talk about your body 's movements . and this is known as kinesthesia . so let me write that down -- kinesthesia . and so if we talk about proprioception to include your body 's position in space as well as your ability to balance , so your sense of balance would be included under proprioception , kinesthesia is talking more about the movement of your body . so one final way to differentiate between the two is that proprioception can be thought of as a cognitive awareness of your body in space . so it 's more cognitive . so one way to think about this is that it 's a little bit more subconscious . so you 're not always thinking about exactly where your body is in space , exactly how you 're orientated , if you 're walking , or if you 're running . your main concern is n't , oh , i hope -- it might be , but your main concern is n't i hope i 'm not going to fall down . pretty much your sense of balance , your sense of position , are pretty well taken care of by your proprioceptive sense . so it 's more cognitive . it 's more something that 's under the surface . you 're not overtly thinking about it . on the other hand , kinesthesia is a little bit more behavioral . and what i mean by that is let 's imagine that you 're playing golf or you 're trying to hit a baseball , constantly you 're swinging . sometimes you miss , sometimes you hit the ball . but every single time you actually swing the bat or swing the golf club , your body is able to detect exactly how it 's moving . and so over time , if you learn that , ok , if i move in this certain direction , i 'm able to hit the golf ball , or if i move in this direction , i 'm able to hit the baseball , then over time , your body is able to detect exactly what that movement is and start to undergo that movement more and more often . so you 're able to actually teach yourself exactly how you should move in order to successfully complete whatever task is at hand . so that 's just another way to distinguish between the two terms . but just keep in mind that proprioception and kinesthesia are not the same . so they are not the same . so i 'll draw a big x here . but they do share a lot in common . and what they do share in common is inferring movement , and position , and where your body is in space . and the big difference is , just to kind of summarize , is that proprioception is concerned with position while kinesthesia is concerned with movement . and proprioception includes your sense of balance , while kinesthesia does not .
|
and this is known as kinesthesia . so let me write that down -- kinesthesia . and so if we talk about proprioception to include your body 's position in space as well as your ability to balance , so your sense of balance would be included under proprioception , kinesthesia is talking more about the movement of your body . so one final way to differentiate between the two is that proprioception can be thought of as a cognitive awareness of your body in space .
|
so , both proprioception and kinesthesia rely on spindles in the muscles to determine where the body is in space , balance , and movement ?
|
imagine if it was pitch black in your room . would you be able to walk ? as long as nothing got in your way , you 'd probably be able to walk perfectly fine . but why is this ? when walking in a pitch black room , you rely on your sense of balance . you know exactly where you are in space . you know whether or not you 're standing straight up or if you 're sitting down . and as you 're walking , you know if your right foot is in front of your left or if your left foot is in front of your right . so how is it that we know exactly where our body is in space without actually having a look at it ? well , this is known as our sense of propioception . and proprioception is basically defined as our ability to sense exactly where our body is in space . in other words , this is our sense of position . and this sense actually originates from a bunch of tiny little sensors that are located throughout our entire body in almost all of our muscles . so let 's imagine that this right here is a muscle in your leg or in your arm . let 's just say it 's a muscle in your arm . so there 's a tiny little receptor in it , inside of the muscle . and this receptor will actually go up to your spinal cord and then eventually to your brand . and this receptor , if we kind of zoom in on it , we zoom in on this receptor , is sensitive to stretching . so as this muscle contracts , so if i was going to lift something really heavy , the muscle would contract , and it would get thinner , so it would look like this . it would get a bit thinner . and so this sensor , which is known as a spindle -- i 'll write that down here . this is a spindle . this sensor can sense that the muscle has been stretched out , and it too will also stretch . so it will go from this conformation to this conformation . it will actually get stretched out . and so we can draw a little spring-like structure inside , which is actually quite similar to what 's actually inside the spindle . there 's actually a protein inside that gets stretched . when that protein gets stretched , it fires a signal to the brain . this is the basic principle behind proprioception . so we 're able to tell exactly how contracted or how relaxed every single muscle is in our entire body , and this allows us to know exactly where our body is in space . there 's another word that 's commonly used to talk about your body 's movements . and this is known as kinesthesia . so let me write that down -- kinesthesia . and so if we talk about proprioception to include your body 's position in space as well as your ability to balance , so your sense of balance would be included under proprioception , kinesthesia is talking more about the movement of your body . so one final way to differentiate between the two is that proprioception can be thought of as a cognitive awareness of your body in space . so it 's more cognitive . so one way to think about this is that it 's a little bit more subconscious . so you 're not always thinking about exactly where your body is in space , exactly how you 're orientated , if you 're walking , or if you 're running . your main concern is n't , oh , i hope -- it might be , but your main concern is n't i hope i 'm not going to fall down . pretty much your sense of balance , your sense of position , are pretty well taken care of by your proprioceptive sense . so it 's more cognitive . it 's more something that 's under the surface . you 're not overtly thinking about it . on the other hand , kinesthesia is a little bit more behavioral . and what i mean by that is let 's imagine that you 're playing golf or you 're trying to hit a baseball , constantly you 're swinging . sometimes you miss , sometimes you hit the ball . but every single time you actually swing the bat or swing the golf club , your body is able to detect exactly how it 's moving . and so over time , if you learn that , ok , if i move in this certain direction , i 'm able to hit the golf ball , or if i move in this direction , i 'm able to hit the baseball , then over time , your body is able to detect exactly what that movement is and start to undergo that movement more and more often . so you 're able to actually teach yourself exactly how you should move in order to successfully complete whatever task is at hand . so that 's just another way to distinguish between the two terms . but just keep in mind that proprioception and kinesthesia are not the same . so they are not the same . so i 'll draw a big x here . but they do share a lot in common . and what they do share in common is inferring movement , and position , and where your body is in space . and the big difference is , just to kind of summarize , is that proprioception is concerned with position while kinesthesia is concerned with movement . and proprioception includes your sense of balance , while kinesthesia does not .
|
there 's another word that 's commonly used to talk about your body 's movements . and this is known as kinesthesia . so let me write that down -- kinesthesia .
|
can kinesthesia be thought of muscle memory ?
|
imagine if it was pitch black in your room . would you be able to walk ? as long as nothing got in your way , you 'd probably be able to walk perfectly fine . but why is this ? when walking in a pitch black room , you rely on your sense of balance . you know exactly where you are in space . you know whether or not you 're standing straight up or if you 're sitting down . and as you 're walking , you know if your right foot is in front of your left or if your left foot is in front of your right . so how is it that we know exactly where our body is in space without actually having a look at it ? well , this is known as our sense of propioception . and proprioception is basically defined as our ability to sense exactly where our body is in space . in other words , this is our sense of position . and this sense actually originates from a bunch of tiny little sensors that are located throughout our entire body in almost all of our muscles . so let 's imagine that this right here is a muscle in your leg or in your arm . let 's just say it 's a muscle in your arm . so there 's a tiny little receptor in it , inside of the muscle . and this receptor will actually go up to your spinal cord and then eventually to your brand . and this receptor , if we kind of zoom in on it , we zoom in on this receptor , is sensitive to stretching . so as this muscle contracts , so if i was going to lift something really heavy , the muscle would contract , and it would get thinner , so it would look like this . it would get a bit thinner . and so this sensor , which is known as a spindle -- i 'll write that down here . this is a spindle . this sensor can sense that the muscle has been stretched out , and it too will also stretch . so it will go from this conformation to this conformation . it will actually get stretched out . and so we can draw a little spring-like structure inside , which is actually quite similar to what 's actually inside the spindle . there 's actually a protein inside that gets stretched . when that protein gets stretched , it fires a signal to the brain . this is the basic principle behind proprioception . so we 're able to tell exactly how contracted or how relaxed every single muscle is in our entire body , and this allows us to know exactly where our body is in space . there 's another word that 's commonly used to talk about your body 's movements . and this is known as kinesthesia . so let me write that down -- kinesthesia . and so if we talk about proprioception to include your body 's position in space as well as your ability to balance , so your sense of balance would be included under proprioception , kinesthesia is talking more about the movement of your body . so one final way to differentiate between the two is that proprioception can be thought of as a cognitive awareness of your body in space . so it 's more cognitive . so one way to think about this is that it 's a little bit more subconscious . so you 're not always thinking about exactly where your body is in space , exactly how you 're orientated , if you 're walking , or if you 're running . your main concern is n't , oh , i hope -- it might be , but your main concern is n't i hope i 'm not going to fall down . pretty much your sense of balance , your sense of position , are pretty well taken care of by your proprioceptive sense . so it 's more cognitive . it 's more something that 's under the surface . you 're not overtly thinking about it . on the other hand , kinesthesia is a little bit more behavioral . and what i mean by that is let 's imagine that you 're playing golf or you 're trying to hit a baseball , constantly you 're swinging . sometimes you miss , sometimes you hit the ball . but every single time you actually swing the bat or swing the golf club , your body is able to detect exactly how it 's moving . and so over time , if you learn that , ok , if i move in this certain direction , i 'm able to hit the golf ball , or if i move in this direction , i 'm able to hit the baseball , then over time , your body is able to detect exactly what that movement is and start to undergo that movement more and more often . so you 're able to actually teach yourself exactly how you should move in order to successfully complete whatever task is at hand . so that 's just another way to distinguish between the two terms . but just keep in mind that proprioception and kinesthesia are not the same . so they are not the same . so i 'll draw a big x here . but they do share a lot in common . and what they do share in common is inferring movement , and position , and where your body is in space . and the big difference is , just to kind of summarize , is that proprioception is concerned with position while kinesthesia is concerned with movement . and proprioception includes your sense of balance , while kinesthesia does not .
|
and the big difference is , just to kind of summarize , is that proprioception is concerned with position while kinesthesia is concerned with movement . and proprioception includes your sense of balance , while kinesthesia does not .
|
is it possible for proprioception sense to adapt to kinesthesia sense ?
|
imagine if it was pitch black in your room . would you be able to walk ? as long as nothing got in your way , you 'd probably be able to walk perfectly fine . but why is this ? when walking in a pitch black room , you rely on your sense of balance . you know exactly where you are in space . you know whether or not you 're standing straight up or if you 're sitting down . and as you 're walking , you know if your right foot is in front of your left or if your left foot is in front of your right . so how is it that we know exactly where our body is in space without actually having a look at it ? well , this is known as our sense of propioception . and proprioception is basically defined as our ability to sense exactly where our body is in space . in other words , this is our sense of position . and this sense actually originates from a bunch of tiny little sensors that are located throughout our entire body in almost all of our muscles . so let 's imagine that this right here is a muscle in your leg or in your arm . let 's just say it 's a muscle in your arm . so there 's a tiny little receptor in it , inside of the muscle . and this receptor will actually go up to your spinal cord and then eventually to your brand . and this receptor , if we kind of zoom in on it , we zoom in on this receptor , is sensitive to stretching . so as this muscle contracts , so if i was going to lift something really heavy , the muscle would contract , and it would get thinner , so it would look like this . it would get a bit thinner . and so this sensor , which is known as a spindle -- i 'll write that down here . this is a spindle . this sensor can sense that the muscle has been stretched out , and it too will also stretch . so it will go from this conformation to this conformation . it will actually get stretched out . and so we can draw a little spring-like structure inside , which is actually quite similar to what 's actually inside the spindle . there 's actually a protein inside that gets stretched . when that protein gets stretched , it fires a signal to the brain . this is the basic principle behind proprioception . so we 're able to tell exactly how contracted or how relaxed every single muscle is in our entire body , and this allows us to know exactly where our body is in space . there 's another word that 's commonly used to talk about your body 's movements . and this is known as kinesthesia . so let me write that down -- kinesthesia . and so if we talk about proprioception to include your body 's position in space as well as your ability to balance , so your sense of balance would be included under proprioception , kinesthesia is talking more about the movement of your body . so one final way to differentiate between the two is that proprioception can be thought of as a cognitive awareness of your body in space . so it 's more cognitive . so one way to think about this is that it 's a little bit more subconscious . so you 're not always thinking about exactly where your body is in space , exactly how you 're orientated , if you 're walking , or if you 're running . your main concern is n't , oh , i hope -- it might be , but your main concern is n't i hope i 'm not going to fall down . pretty much your sense of balance , your sense of position , are pretty well taken care of by your proprioceptive sense . so it 's more cognitive . it 's more something that 's under the surface . you 're not overtly thinking about it . on the other hand , kinesthesia is a little bit more behavioral . and what i mean by that is let 's imagine that you 're playing golf or you 're trying to hit a baseball , constantly you 're swinging . sometimes you miss , sometimes you hit the ball . but every single time you actually swing the bat or swing the golf club , your body is able to detect exactly how it 's moving . and so over time , if you learn that , ok , if i move in this certain direction , i 'm able to hit the golf ball , or if i move in this direction , i 'm able to hit the baseball , then over time , your body is able to detect exactly what that movement is and start to undergo that movement more and more often . so you 're able to actually teach yourself exactly how you should move in order to successfully complete whatever task is at hand . so that 's just another way to distinguish between the two terms . but just keep in mind that proprioception and kinesthesia are not the same . so they are not the same . so i 'll draw a big x here . but they do share a lot in common . and what they do share in common is inferring movement , and position , and where your body is in space . and the big difference is , just to kind of summarize , is that proprioception is concerned with position while kinesthesia is concerned with movement . and proprioception includes your sense of balance , while kinesthesia does not .
|
and the big difference is , just to kind of summarize , is that proprioception is concerned with position while kinesthesia is concerned with movement . and proprioception includes your sense of balance , while kinesthesia does not .
|
are the receptors for proprioception and kinesthesia the same ?
|
let 's see if we might be able to make some use of the divergence theorem . so i have this region , this simple solid right over here . x can go between negative 1 and 1. z , this kind of arch part right over here , is going to be a function of x . that 's the upper bound on z . the lower bound on z is just 0 . and then y could go anywhere between 0 , and then it 's bounded here by this plane , where we can express y as a function of z. y is 2 minus z along this plane right over here . and we 're given this crazy vector field . it has natural logs and tangents in it . and we 're asked to evaluate the surface integral , or actually , i should say the flux of our vector field across the boundary of this region , across the surface of this region right over here . and surface integrals are messy as is , especially when you have a crazy vector field like this . but you could imagine that there might be a way to simplify this , perhaps using the divergence theorem . the divergence theorem tells us that the flux across the boundary of this simple solid region is going to be the same thing as the triple integral over the volume of it , or i 'll just call it over the region , of the divergence of f dv , where dv is some combination of dx , dy , dz . the divergence times each little cubic volume , infinitesimal cubic volume , so times dv . so let 's see if this simplifies things a bit . so let 's calculate the divergence of f first . so the divergence of f is going to be the partial of the x component , or the partial of the -- you could say the i component or the x component with respect to x . well , the derivative of this with respect to x is just x . the derivative of this with respect to x , luckily , is just 0 . this is a constant in terms of x . now let 's go over here , the partial of this with respect to y . the partial of this with respect to y is just x . and then this is just a constant in terms of y , so it 's just going to be 0 when you take the derivative with respect to y . and then , finally , the partial of this with respect to z , well , this is just a constant in terms of z . does n't change when z changes . so the partial with respect to z is just going to be 0 here . and so taking the divergence really , really , really simplified things . the divergence of f simplified down to 2x . and so now we can restate the flux across the surface as a triple integral of 2x . so let me just write 2x here . and let 's think about the ordering . so y can go between 0 and this plane that is a function of z . so let 's write that down . so y is bounded below by 0 and above by this plane 2 minus z. z is bounded below by 0 and above by -- you could call them these parabolas of 1 minus x squared . and then x is bounded below by negative 1 and bounded above by 1 . so negative 1 is less than or equal to x is less than or equal to 1 . and so this is probably a good order of integration . we can integrate with respect to y first , and then we 'll get a function of z . then we can integrate with respect to z , and we 'll get a function of x . and then we can integrate with respect to x . so let 's do it in that order . so first we 'll integrate with respect to y , so we have dy . y is bounded below at 0 and above by the plane 2 minus z . so this right over here is a plane y is equal to 0 . and this up over here is the plane y is equal to 2 minus z . then we can integrate with respect to z . and z , once again , is bounded below by 0 and bounded above by these parabolas , 1 minus x squared . and then , finally , we can integrate with respect to x . and x is bounded below by negative 1 and bounded above by 1 . so let 's do some integration here . so the first thing , when we 're integrating with respect to x -- sorry , when we 're integrating with respect to y , 2x is just a constant . so this expression right over here is just going to be 2x times y , and then we 're going to evaluate it from 0 to 2 minus z . so it 's going to be 2x times 2 minus z minus 2x times 0 . well , that second part 's just going to be 0 . so this is going to simplify as -- i 'll write it this way -- 2x times 2 minus z . and actually , i 'll just leave it like that . and then we 're going to integrate this with respect to z . and that 's going to go from 0 to 1 minus x squared , and then we have our dz there . and then after that , we 're going to integrate with respect to x , negative 1 to 1 dx . so let 's take the antiderivative here with respect to z . this you really can just view as a constant . we can actually even bring it out front , but i 'll leave it there . so this piece right over here -- i 'll do it in z 's color -- this piece right over here , see , we can leave the 2x out front . actually , i 'll leave the 2x out front of the whole thing . it 's going to be 2x times -- so the antiderivative of this with respect to z is going to be 2z . antiderivative of this is negative z squared over 2 , and we are going to evaluate this from 0 to 1 minus x squared . when we evaluate them at 0 , we 're just going to get 0 right over here . and so we really just have to worry about when z is equal to 1 minus x squared . did i do that right ? yep . 2z , and then minus z squared over 2 . you take the derivative , you get negative z . take the derivative here , you just get 2 . so that 's right . so this is going to be equal to 2x -- let me do that same color -- it 's going to be equal to 2x times -- let me get this right , let me go into that pink color -- 2x times 2z . well , z is going to be 1 minus x squared , so it 's going to be 2 minus 2x squared . that was just 2 times that . and then minus -- i 'll just write 1/2 times this quantity squared . so this quantity squared is going to be 1 minus 2x squared plus x to the fourth . that 's just some basic algebra right over there . and then from that , you 're going to subtract this thing evaluated at 0 , which is just going to be 0 . so [ ? y , you ? ] just wo n't even think about that . and now we need to simplify this a little bit . and we are going to get , if we simplify this , we get 2 minus 2x squared minus 1/2 , and then plus -- so this is negative 1/2 times negative 2x squared . so it 's going to be positive x squared minus 1/2 x to the fourth . now , let 's see , can we simplify this part ? let me just make sure we know what we 're doing here . so we have this 2x right over there . i want to make sure i got the signs right . yep , looks like i did . and now let 's look at this . so let 's see , can i simplify a little bit ? i have 2 minus 1/2 , which is 3/2 . so i have 3/2 . that 's that term and that term take into account . and then i have negative 2x squared plus x squared . so that 's just going to result in negative x squared , if i take that term and that term . and then i have negative 1/2 x to the fourth , and i 'm multiplying this whole thing by 2x . and so that 's going to give us -- we have , let 's see , 2x times 3/2 . and i want to make sure . i 'm doing this slowly , so i do n't make any careless mistakes . the 2 's cancel out . you get 3x , and then 2x times negative x squared is negative 2x to the third . and then 2x times this right over here . the 2 cancels out with the negative 1/2 , you have negative x to the fifth . so all of this simplifies to this right over here . so our whole thing simplifies to an integral with respect to x. x will go from negative 1 to 1 of this business of 3x minus 2x to the third minus x to the fifth , and then we have dx . and now we just take the antiderivative with respect to x , which is going to be 3/2 x squared minus -- let 's see , x to the fourth power -- minus 1/2 , because it 's going to be 2/4 , x to the fourth . is that right ? because if you multiply it , you 're going to 2 . yep , x to the third , and then minus x to the sixth over 6 . and it 's going to go from 1 to negative 1 or negative 1 to 1 . so when you evaluate it at 1 -- i 'll just write it out real fast . so first , when you evaluate it at 1 , you get 3/2 minus 1/2 minus 1/6 . and then from that , we are going to subtract 3/2 minus 1/2 plus 1/6 . or actually , no , they 're actually all going to cancel out . is that right ? are they all going to cancel out ? yep , i think that 's right . they all cancel out . so it 's actually going to be plus , or i should say minus 1/6 right over here . and then all of these cancel out . that cancels with that . that cancels with that because we 're subtracting the negative 1/2 . and that cancels with that . and so we are actually left with 0 . so after doing all of that work , this whole thing evaluates to 0 , which was actually kind of a neat simplification . so this whole thing right over here evaluated , very conveniently , evaluated to be equal to 0 .
|
so let 's see , can i simplify a little bit ? i have 2 minus 1/2 , which is 3/2 . so i have 3/2 .
|
is n't the height ( z ) of the region not always z=1-x^2 ?
|
let 's see if we might be able to make some use of the divergence theorem . so i have this region , this simple solid right over here . x can go between negative 1 and 1. z , this kind of arch part right over here , is going to be a function of x . that 's the upper bound on z . the lower bound on z is just 0 . and then y could go anywhere between 0 , and then it 's bounded here by this plane , where we can express y as a function of z. y is 2 minus z along this plane right over here . and we 're given this crazy vector field . it has natural logs and tangents in it . and we 're asked to evaluate the surface integral , or actually , i should say the flux of our vector field across the boundary of this region , across the surface of this region right over here . and surface integrals are messy as is , especially when you have a crazy vector field like this . but you could imagine that there might be a way to simplify this , perhaps using the divergence theorem . the divergence theorem tells us that the flux across the boundary of this simple solid region is going to be the same thing as the triple integral over the volume of it , or i 'll just call it over the region , of the divergence of f dv , where dv is some combination of dx , dy , dz . the divergence times each little cubic volume , infinitesimal cubic volume , so times dv . so let 's see if this simplifies things a bit . so let 's calculate the divergence of f first . so the divergence of f is going to be the partial of the x component , or the partial of the -- you could say the i component or the x component with respect to x . well , the derivative of this with respect to x is just x . the derivative of this with respect to x , luckily , is just 0 . this is a constant in terms of x . now let 's go over here , the partial of this with respect to y . the partial of this with respect to y is just x . and then this is just a constant in terms of y , so it 's just going to be 0 when you take the derivative with respect to y . and then , finally , the partial of this with respect to z , well , this is just a constant in terms of z . does n't change when z changes . so the partial with respect to z is just going to be 0 here . and so taking the divergence really , really , really simplified things . the divergence of f simplified down to 2x . and so now we can restate the flux across the surface as a triple integral of 2x . so let me just write 2x here . and let 's think about the ordering . so y can go between 0 and this plane that is a function of z . so let 's write that down . so y is bounded below by 0 and above by this plane 2 minus z. z is bounded below by 0 and above by -- you could call them these parabolas of 1 minus x squared . and then x is bounded below by negative 1 and bounded above by 1 . so negative 1 is less than or equal to x is less than or equal to 1 . and so this is probably a good order of integration . we can integrate with respect to y first , and then we 'll get a function of z . then we can integrate with respect to z , and we 'll get a function of x . and then we can integrate with respect to x . so let 's do it in that order . so first we 'll integrate with respect to y , so we have dy . y is bounded below at 0 and above by the plane 2 minus z . so this right over here is a plane y is equal to 0 . and this up over here is the plane y is equal to 2 minus z . then we can integrate with respect to z . and z , once again , is bounded below by 0 and bounded above by these parabolas , 1 minus x squared . and then , finally , we can integrate with respect to x . and x is bounded below by negative 1 and bounded above by 1 . so let 's do some integration here . so the first thing , when we 're integrating with respect to x -- sorry , when we 're integrating with respect to y , 2x is just a constant . so this expression right over here is just going to be 2x times y , and then we 're going to evaluate it from 0 to 2 minus z . so it 's going to be 2x times 2 minus z minus 2x times 0 . well , that second part 's just going to be 0 . so this is going to simplify as -- i 'll write it this way -- 2x times 2 minus z . and actually , i 'll just leave it like that . and then we 're going to integrate this with respect to z . and that 's going to go from 0 to 1 minus x squared , and then we have our dz there . and then after that , we 're going to integrate with respect to x , negative 1 to 1 dx . so let 's take the antiderivative here with respect to z . this you really can just view as a constant . we can actually even bring it out front , but i 'll leave it there . so this piece right over here -- i 'll do it in z 's color -- this piece right over here , see , we can leave the 2x out front . actually , i 'll leave the 2x out front of the whole thing . it 's going to be 2x times -- so the antiderivative of this with respect to z is going to be 2z . antiderivative of this is negative z squared over 2 , and we are going to evaluate this from 0 to 1 minus x squared . when we evaluate them at 0 , we 're just going to get 0 right over here . and so we really just have to worry about when z is equal to 1 minus x squared . did i do that right ? yep . 2z , and then minus z squared over 2 . you take the derivative , you get negative z . take the derivative here , you just get 2 . so that 's right . so this is going to be equal to 2x -- let me do that same color -- it 's going to be equal to 2x times -- let me get this right , let me go into that pink color -- 2x times 2z . well , z is going to be 1 minus x squared , so it 's going to be 2 minus 2x squared . that was just 2 times that . and then minus -- i 'll just write 1/2 times this quantity squared . so this quantity squared is going to be 1 minus 2x squared plus x to the fourth . that 's just some basic algebra right over there . and then from that , you 're going to subtract this thing evaluated at 0 , which is just going to be 0 . so [ ? y , you ? ] just wo n't even think about that . and now we need to simplify this a little bit . and we are going to get , if we simplify this , we get 2 minus 2x squared minus 1/2 , and then plus -- so this is negative 1/2 times negative 2x squared . so it 's going to be positive x squared minus 1/2 x to the fourth . now , let 's see , can we simplify this part ? let me just make sure we know what we 're doing here . so we have this 2x right over there . i want to make sure i got the signs right . yep , looks like i did . and now let 's look at this . so let 's see , can i simplify a little bit ? i have 2 minus 1/2 , which is 3/2 . so i have 3/2 . that 's that term and that term take into account . and then i have negative 2x squared plus x squared . so that 's just going to result in negative x squared , if i take that term and that term . and then i have negative 1/2 x to the fourth , and i 'm multiplying this whole thing by 2x . and so that 's going to give us -- we have , let 's see , 2x times 3/2 . and i want to make sure . i 'm doing this slowly , so i do n't make any careless mistakes . the 2 's cancel out . you get 3x , and then 2x times negative x squared is negative 2x to the third . and then 2x times this right over here . the 2 cancels out with the negative 1/2 , you have negative x to the fifth . so all of this simplifies to this right over here . so our whole thing simplifies to an integral with respect to x. x will go from negative 1 to 1 of this business of 3x minus 2x to the third minus x to the fifth , and then we have dx . and now we just take the antiderivative with respect to x , which is going to be 3/2 x squared minus -- let 's see , x to the fourth power -- minus 1/2 , because it 's going to be 2/4 , x to the fourth . is that right ? because if you multiply it , you 're going to 2 . yep , x to the third , and then minus x to the sixth over 6 . and it 's going to go from 1 to negative 1 or negative 1 to 1 . so when you evaluate it at 1 -- i 'll just write it out real fast . so first , when you evaluate it at 1 , you get 3/2 minus 1/2 minus 1/6 . and then from that , we are going to subtract 3/2 minus 1/2 plus 1/6 . or actually , no , they 're actually all going to cancel out . is that right ? are they all going to cancel out ? yep , i think that 's right . they all cancel out . so it 's actually going to be plus , or i should say minus 1/6 right over here . and then all of these cancel out . that cancels with that . that cancels with that because we 're subtracting the negative 1/2 . and that cancels with that . and so we are actually left with 0 . so after doing all of that work , this whole thing evaluates to 0 , which was actually kind of a neat simplification . so this whole thing right over here evaluated , very conveniently , evaluated to be equal to 0 .
|
so let 's see , can i simplify a little bit ? i have 2 minus 1/2 , which is 3/2 . so i have 3/2 .
|
sometimes it is z=1-x^2 and sometimes it is the plane y=2-z ?
|
let 's see if we might be able to make some use of the divergence theorem . so i have this region , this simple solid right over here . x can go between negative 1 and 1. z , this kind of arch part right over here , is going to be a function of x . that 's the upper bound on z . the lower bound on z is just 0 . and then y could go anywhere between 0 , and then it 's bounded here by this plane , where we can express y as a function of z. y is 2 minus z along this plane right over here . and we 're given this crazy vector field . it has natural logs and tangents in it . and we 're asked to evaluate the surface integral , or actually , i should say the flux of our vector field across the boundary of this region , across the surface of this region right over here . and surface integrals are messy as is , especially when you have a crazy vector field like this . but you could imagine that there might be a way to simplify this , perhaps using the divergence theorem . the divergence theorem tells us that the flux across the boundary of this simple solid region is going to be the same thing as the triple integral over the volume of it , or i 'll just call it over the region , of the divergence of f dv , where dv is some combination of dx , dy , dz . the divergence times each little cubic volume , infinitesimal cubic volume , so times dv . so let 's see if this simplifies things a bit . so let 's calculate the divergence of f first . so the divergence of f is going to be the partial of the x component , or the partial of the -- you could say the i component or the x component with respect to x . well , the derivative of this with respect to x is just x . the derivative of this with respect to x , luckily , is just 0 . this is a constant in terms of x . now let 's go over here , the partial of this with respect to y . the partial of this with respect to y is just x . and then this is just a constant in terms of y , so it 's just going to be 0 when you take the derivative with respect to y . and then , finally , the partial of this with respect to z , well , this is just a constant in terms of z . does n't change when z changes . so the partial with respect to z is just going to be 0 here . and so taking the divergence really , really , really simplified things . the divergence of f simplified down to 2x . and so now we can restate the flux across the surface as a triple integral of 2x . so let me just write 2x here . and let 's think about the ordering . so y can go between 0 and this plane that is a function of z . so let 's write that down . so y is bounded below by 0 and above by this plane 2 minus z. z is bounded below by 0 and above by -- you could call them these parabolas of 1 minus x squared . and then x is bounded below by negative 1 and bounded above by 1 . so negative 1 is less than or equal to x is less than or equal to 1 . and so this is probably a good order of integration . we can integrate with respect to y first , and then we 'll get a function of z . then we can integrate with respect to z , and we 'll get a function of x . and then we can integrate with respect to x . so let 's do it in that order . so first we 'll integrate with respect to y , so we have dy . y is bounded below at 0 and above by the plane 2 minus z . so this right over here is a plane y is equal to 0 . and this up over here is the plane y is equal to 2 minus z . then we can integrate with respect to z . and z , once again , is bounded below by 0 and bounded above by these parabolas , 1 minus x squared . and then , finally , we can integrate with respect to x . and x is bounded below by negative 1 and bounded above by 1 . so let 's do some integration here . so the first thing , when we 're integrating with respect to x -- sorry , when we 're integrating with respect to y , 2x is just a constant . so this expression right over here is just going to be 2x times y , and then we 're going to evaluate it from 0 to 2 minus z . so it 's going to be 2x times 2 minus z minus 2x times 0 . well , that second part 's just going to be 0 . so this is going to simplify as -- i 'll write it this way -- 2x times 2 minus z . and actually , i 'll just leave it like that . and then we 're going to integrate this with respect to z . and that 's going to go from 0 to 1 minus x squared , and then we have our dz there . and then after that , we 're going to integrate with respect to x , negative 1 to 1 dx . so let 's take the antiderivative here with respect to z . this you really can just view as a constant . we can actually even bring it out front , but i 'll leave it there . so this piece right over here -- i 'll do it in z 's color -- this piece right over here , see , we can leave the 2x out front . actually , i 'll leave the 2x out front of the whole thing . it 's going to be 2x times -- so the antiderivative of this with respect to z is going to be 2z . antiderivative of this is negative z squared over 2 , and we are going to evaluate this from 0 to 1 minus x squared . when we evaluate them at 0 , we 're just going to get 0 right over here . and so we really just have to worry about when z is equal to 1 minus x squared . did i do that right ? yep . 2z , and then minus z squared over 2 . you take the derivative , you get negative z . take the derivative here , you just get 2 . so that 's right . so this is going to be equal to 2x -- let me do that same color -- it 's going to be equal to 2x times -- let me get this right , let me go into that pink color -- 2x times 2z . well , z is going to be 1 minus x squared , so it 's going to be 2 minus 2x squared . that was just 2 times that . and then minus -- i 'll just write 1/2 times this quantity squared . so this quantity squared is going to be 1 minus 2x squared plus x to the fourth . that 's just some basic algebra right over there . and then from that , you 're going to subtract this thing evaluated at 0 , which is just going to be 0 . so [ ? y , you ? ] just wo n't even think about that . and now we need to simplify this a little bit . and we are going to get , if we simplify this , we get 2 minus 2x squared minus 1/2 , and then plus -- so this is negative 1/2 times negative 2x squared . so it 's going to be positive x squared minus 1/2 x to the fourth . now , let 's see , can we simplify this part ? let me just make sure we know what we 're doing here . so we have this 2x right over there . i want to make sure i got the signs right . yep , looks like i did . and now let 's look at this . so let 's see , can i simplify a little bit ? i have 2 minus 1/2 , which is 3/2 . so i have 3/2 . that 's that term and that term take into account . and then i have negative 2x squared plus x squared . so that 's just going to result in negative x squared , if i take that term and that term . and then i have negative 1/2 x to the fourth , and i 'm multiplying this whole thing by 2x . and so that 's going to give us -- we have , let 's see , 2x times 3/2 . and i want to make sure . i 'm doing this slowly , so i do n't make any careless mistakes . the 2 's cancel out . you get 3x , and then 2x times negative x squared is negative 2x to the third . and then 2x times this right over here . the 2 cancels out with the negative 1/2 , you have negative x to the fifth . so all of this simplifies to this right over here . so our whole thing simplifies to an integral with respect to x. x will go from negative 1 to 1 of this business of 3x minus 2x to the third minus x to the fifth , and then we have dx . and now we just take the antiderivative with respect to x , which is going to be 3/2 x squared minus -- let 's see , x to the fourth power -- minus 1/2 , because it 's going to be 2/4 , x to the fourth . is that right ? because if you multiply it , you 're going to 2 . yep , x to the third , and then minus x to the sixth over 6 . and it 's going to go from 1 to negative 1 or negative 1 to 1 . so when you evaluate it at 1 -- i 'll just write it out real fast . so first , when you evaluate it at 1 , you get 3/2 minus 1/2 minus 1/6 . and then from that , we are going to subtract 3/2 minus 1/2 plus 1/6 . or actually , no , they 're actually all going to cancel out . is that right ? are they all going to cancel out ? yep , i think that 's right . they all cancel out . so it 's actually going to be plus , or i should say minus 1/6 right over here . and then all of these cancel out . that cancels with that . that cancels with that because we 're subtracting the negative 1/2 . and that cancels with that . and so we are actually left with 0 . so after doing all of that work , this whole thing evaluates to 0 , which was actually kind of a neat simplification . so this whole thing right over here evaluated , very conveniently , evaluated to be equal to 0 .
|
and we are going to get , if we simplify this , we get 2 minus 2x squared minus 1/2 , and then plus -- so this is negative 1/2 times negative 2x squared . so it 's going to be positive x squared minus 1/2 x to the fourth . now , let 's see , can we simplify this part ?
|
because the function ( a*x^2 + b*x^4 + c*x^6 ) is even , should n't we multiply it by 2 and change the limit of integration of x to be from 0 to 1 ?
|
let 's see if we might be able to make some use of the divergence theorem . so i have this region , this simple solid right over here . x can go between negative 1 and 1. z , this kind of arch part right over here , is going to be a function of x . that 's the upper bound on z . the lower bound on z is just 0 . and then y could go anywhere between 0 , and then it 's bounded here by this plane , where we can express y as a function of z. y is 2 minus z along this plane right over here . and we 're given this crazy vector field . it has natural logs and tangents in it . and we 're asked to evaluate the surface integral , or actually , i should say the flux of our vector field across the boundary of this region , across the surface of this region right over here . and surface integrals are messy as is , especially when you have a crazy vector field like this . but you could imagine that there might be a way to simplify this , perhaps using the divergence theorem . the divergence theorem tells us that the flux across the boundary of this simple solid region is going to be the same thing as the triple integral over the volume of it , or i 'll just call it over the region , of the divergence of f dv , where dv is some combination of dx , dy , dz . the divergence times each little cubic volume , infinitesimal cubic volume , so times dv . so let 's see if this simplifies things a bit . so let 's calculate the divergence of f first . so the divergence of f is going to be the partial of the x component , or the partial of the -- you could say the i component or the x component with respect to x . well , the derivative of this with respect to x is just x . the derivative of this with respect to x , luckily , is just 0 . this is a constant in terms of x . now let 's go over here , the partial of this with respect to y . the partial of this with respect to y is just x . and then this is just a constant in terms of y , so it 's just going to be 0 when you take the derivative with respect to y . and then , finally , the partial of this with respect to z , well , this is just a constant in terms of z . does n't change when z changes . so the partial with respect to z is just going to be 0 here . and so taking the divergence really , really , really simplified things . the divergence of f simplified down to 2x . and so now we can restate the flux across the surface as a triple integral of 2x . so let me just write 2x here . and let 's think about the ordering . so y can go between 0 and this plane that is a function of z . so let 's write that down . so y is bounded below by 0 and above by this plane 2 minus z. z is bounded below by 0 and above by -- you could call them these parabolas of 1 minus x squared . and then x is bounded below by negative 1 and bounded above by 1 . so negative 1 is less than or equal to x is less than or equal to 1 . and so this is probably a good order of integration . we can integrate with respect to y first , and then we 'll get a function of z . then we can integrate with respect to z , and we 'll get a function of x . and then we can integrate with respect to x . so let 's do it in that order . so first we 'll integrate with respect to y , so we have dy . y is bounded below at 0 and above by the plane 2 minus z . so this right over here is a plane y is equal to 0 . and this up over here is the plane y is equal to 2 minus z . then we can integrate with respect to z . and z , once again , is bounded below by 0 and bounded above by these parabolas , 1 minus x squared . and then , finally , we can integrate with respect to x . and x is bounded below by negative 1 and bounded above by 1 . so let 's do some integration here . so the first thing , when we 're integrating with respect to x -- sorry , when we 're integrating with respect to y , 2x is just a constant . so this expression right over here is just going to be 2x times y , and then we 're going to evaluate it from 0 to 2 minus z . so it 's going to be 2x times 2 minus z minus 2x times 0 . well , that second part 's just going to be 0 . so this is going to simplify as -- i 'll write it this way -- 2x times 2 minus z . and actually , i 'll just leave it like that . and then we 're going to integrate this with respect to z . and that 's going to go from 0 to 1 minus x squared , and then we have our dz there . and then after that , we 're going to integrate with respect to x , negative 1 to 1 dx . so let 's take the antiderivative here with respect to z . this you really can just view as a constant . we can actually even bring it out front , but i 'll leave it there . so this piece right over here -- i 'll do it in z 's color -- this piece right over here , see , we can leave the 2x out front . actually , i 'll leave the 2x out front of the whole thing . it 's going to be 2x times -- so the antiderivative of this with respect to z is going to be 2z . antiderivative of this is negative z squared over 2 , and we are going to evaluate this from 0 to 1 minus x squared . when we evaluate them at 0 , we 're just going to get 0 right over here . and so we really just have to worry about when z is equal to 1 minus x squared . did i do that right ? yep . 2z , and then minus z squared over 2 . you take the derivative , you get negative z . take the derivative here , you just get 2 . so that 's right . so this is going to be equal to 2x -- let me do that same color -- it 's going to be equal to 2x times -- let me get this right , let me go into that pink color -- 2x times 2z . well , z is going to be 1 minus x squared , so it 's going to be 2 minus 2x squared . that was just 2 times that . and then minus -- i 'll just write 1/2 times this quantity squared . so this quantity squared is going to be 1 minus 2x squared plus x to the fourth . that 's just some basic algebra right over there . and then from that , you 're going to subtract this thing evaluated at 0 , which is just going to be 0 . so [ ? y , you ? ] just wo n't even think about that . and now we need to simplify this a little bit . and we are going to get , if we simplify this , we get 2 minus 2x squared minus 1/2 , and then plus -- so this is negative 1/2 times negative 2x squared . so it 's going to be positive x squared minus 1/2 x to the fourth . now , let 's see , can we simplify this part ? let me just make sure we know what we 're doing here . so we have this 2x right over there . i want to make sure i got the signs right . yep , looks like i did . and now let 's look at this . so let 's see , can i simplify a little bit ? i have 2 minus 1/2 , which is 3/2 . so i have 3/2 . that 's that term and that term take into account . and then i have negative 2x squared plus x squared . so that 's just going to result in negative x squared , if i take that term and that term . and then i have negative 1/2 x to the fourth , and i 'm multiplying this whole thing by 2x . and so that 's going to give us -- we have , let 's see , 2x times 3/2 . and i want to make sure . i 'm doing this slowly , so i do n't make any careless mistakes . the 2 's cancel out . you get 3x , and then 2x times negative x squared is negative 2x to the third . and then 2x times this right over here . the 2 cancels out with the negative 1/2 , you have negative x to the fifth . so all of this simplifies to this right over here . so our whole thing simplifies to an integral with respect to x. x will go from negative 1 to 1 of this business of 3x minus 2x to the third minus x to the fifth , and then we have dx . and now we just take the antiderivative with respect to x , which is going to be 3/2 x squared minus -- let 's see , x to the fourth power -- minus 1/2 , because it 's going to be 2/4 , x to the fourth . is that right ? because if you multiply it , you 're going to 2 . yep , x to the third , and then minus x to the sixth over 6 . and it 's going to go from 1 to negative 1 or negative 1 to 1 . so when you evaluate it at 1 -- i 'll just write it out real fast . so first , when you evaluate it at 1 , you get 3/2 minus 1/2 minus 1/6 . and then from that , we are going to subtract 3/2 minus 1/2 plus 1/6 . or actually , no , they 're actually all going to cancel out . is that right ? are they all going to cancel out ? yep , i think that 's right . they all cancel out . so it 's actually going to be plus , or i should say minus 1/6 right over here . and then all of these cancel out . that cancels with that . that cancels with that because we 're subtracting the negative 1/2 . and that cancels with that . and so we are actually left with 0 . so after doing all of that work , this whole thing evaluates to 0 , which was actually kind of a neat simplification . so this whole thing right over here evaluated , very conveniently , evaluated to be equal to 0 .
|
the divergence of f simplified down to 2x . and so now we can restate the flux across the surface as a triple integral of 2x . so let me just write 2x here .
|
at the beginning of the video , should n't the surface integral also be dotted with the normal vector of the surface in order to get the flux ?
|
let 's see if we might be able to make some use of the divergence theorem . so i have this region , this simple solid right over here . x can go between negative 1 and 1. z , this kind of arch part right over here , is going to be a function of x . that 's the upper bound on z . the lower bound on z is just 0 . and then y could go anywhere between 0 , and then it 's bounded here by this plane , where we can express y as a function of z. y is 2 minus z along this plane right over here . and we 're given this crazy vector field . it has natural logs and tangents in it . and we 're asked to evaluate the surface integral , or actually , i should say the flux of our vector field across the boundary of this region , across the surface of this region right over here . and surface integrals are messy as is , especially when you have a crazy vector field like this . but you could imagine that there might be a way to simplify this , perhaps using the divergence theorem . the divergence theorem tells us that the flux across the boundary of this simple solid region is going to be the same thing as the triple integral over the volume of it , or i 'll just call it over the region , of the divergence of f dv , where dv is some combination of dx , dy , dz . the divergence times each little cubic volume , infinitesimal cubic volume , so times dv . so let 's see if this simplifies things a bit . so let 's calculate the divergence of f first . so the divergence of f is going to be the partial of the x component , or the partial of the -- you could say the i component or the x component with respect to x . well , the derivative of this with respect to x is just x . the derivative of this with respect to x , luckily , is just 0 . this is a constant in terms of x . now let 's go over here , the partial of this with respect to y . the partial of this with respect to y is just x . and then this is just a constant in terms of y , so it 's just going to be 0 when you take the derivative with respect to y . and then , finally , the partial of this with respect to z , well , this is just a constant in terms of z . does n't change when z changes . so the partial with respect to z is just going to be 0 here . and so taking the divergence really , really , really simplified things . the divergence of f simplified down to 2x . and so now we can restate the flux across the surface as a triple integral of 2x . so let me just write 2x here . and let 's think about the ordering . so y can go between 0 and this plane that is a function of z . so let 's write that down . so y is bounded below by 0 and above by this plane 2 minus z. z is bounded below by 0 and above by -- you could call them these parabolas of 1 minus x squared . and then x is bounded below by negative 1 and bounded above by 1 . so negative 1 is less than or equal to x is less than or equal to 1 . and so this is probably a good order of integration . we can integrate with respect to y first , and then we 'll get a function of z . then we can integrate with respect to z , and we 'll get a function of x . and then we can integrate with respect to x . so let 's do it in that order . so first we 'll integrate with respect to y , so we have dy . y is bounded below at 0 and above by the plane 2 minus z . so this right over here is a plane y is equal to 0 . and this up over here is the plane y is equal to 2 minus z . then we can integrate with respect to z . and z , once again , is bounded below by 0 and bounded above by these parabolas , 1 minus x squared . and then , finally , we can integrate with respect to x . and x is bounded below by negative 1 and bounded above by 1 . so let 's do some integration here . so the first thing , when we 're integrating with respect to x -- sorry , when we 're integrating with respect to y , 2x is just a constant . so this expression right over here is just going to be 2x times y , and then we 're going to evaluate it from 0 to 2 minus z . so it 's going to be 2x times 2 minus z minus 2x times 0 . well , that second part 's just going to be 0 . so this is going to simplify as -- i 'll write it this way -- 2x times 2 minus z . and actually , i 'll just leave it like that . and then we 're going to integrate this with respect to z . and that 's going to go from 0 to 1 minus x squared , and then we have our dz there . and then after that , we 're going to integrate with respect to x , negative 1 to 1 dx . so let 's take the antiderivative here with respect to z . this you really can just view as a constant . we can actually even bring it out front , but i 'll leave it there . so this piece right over here -- i 'll do it in z 's color -- this piece right over here , see , we can leave the 2x out front . actually , i 'll leave the 2x out front of the whole thing . it 's going to be 2x times -- so the antiderivative of this with respect to z is going to be 2z . antiderivative of this is negative z squared over 2 , and we are going to evaluate this from 0 to 1 minus x squared . when we evaluate them at 0 , we 're just going to get 0 right over here . and so we really just have to worry about when z is equal to 1 minus x squared . did i do that right ? yep . 2z , and then minus z squared over 2 . you take the derivative , you get negative z . take the derivative here , you just get 2 . so that 's right . so this is going to be equal to 2x -- let me do that same color -- it 's going to be equal to 2x times -- let me get this right , let me go into that pink color -- 2x times 2z . well , z is going to be 1 minus x squared , so it 's going to be 2 minus 2x squared . that was just 2 times that . and then minus -- i 'll just write 1/2 times this quantity squared . so this quantity squared is going to be 1 minus 2x squared plus x to the fourth . that 's just some basic algebra right over there . and then from that , you 're going to subtract this thing evaluated at 0 , which is just going to be 0 . so [ ? y , you ? ] just wo n't even think about that . and now we need to simplify this a little bit . and we are going to get , if we simplify this , we get 2 minus 2x squared minus 1/2 , and then plus -- so this is negative 1/2 times negative 2x squared . so it 's going to be positive x squared minus 1/2 x to the fourth . now , let 's see , can we simplify this part ? let me just make sure we know what we 're doing here . so we have this 2x right over there . i want to make sure i got the signs right . yep , looks like i did . and now let 's look at this . so let 's see , can i simplify a little bit ? i have 2 minus 1/2 , which is 3/2 . so i have 3/2 . that 's that term and that term take into account . and then i have negative 2x squared plus x squared . so that 's just going to result in negative x squared , if i take that term and that term . and then i have negative 1/2 x to the fourth , and i 'm multiplying this whole thing by 2x . and so that 's going to give us -- we have , let 's see , 2x times 3/2 . and i want to make sure . i 'm doing this slowly , so i do n't make any careless mistakes . the 2 's cancel out . you get 3x , and then 2x times negative x squared is negative 2x to the third . and then 2x times this right over here . the 2 cancels out with the negative 1/2 , you have negative x to the fifth . so all of this simplifies to this right over here . so our whole thing simplifies to an integral with respect to x. x will go from negative 1 to 1 of this business of 3x minus 2x to the third minus x to the fifth , and then we have dx . and now we just take the antiderivative with respect to x , which is going to be 3/2 x squared minus -- let 's see , x to the fourth power -- minus 1/2 , because it 's going to be 2/4 , x to the fourth . is that right ? because if you multiply it , you 're going to 2 . yep , x to the third , and then minus x to the sixth over 6 . and it 's going to go from 1 to negative 1 or negative 1 to 1 . so when you evaluate it at 1 -- i 'll just write it out real fast . so first , when you evaluate it at 1 , you get 3/2 minus 1/2 minus 1/6 . and then from that , we are going to subtract 3/2 minus 1/2 plus 1/6 . or actually , no , they 're actually all going to cancel out . is that right ? are they all going to cancel out ? yep , i think that 's right . they all cancel out . so it 's actually going to be plus , or i should say minus 1/6 right over here . and then all of these cancel out . that cancels with that . that cancels with that because we 're subtracting the negative 1/2 . and that cancels with that . and so we are actually left with 0 . so after doing all of that work , this whole thing evaluates to 0 , which was actually kind of a neat simplification . so this whole thing right over here evaluated , very conveniently , evaluated to be equal to 0 .
|
so the divergence of f is going to be the partial of the x component , or the partial of the -- you could say the i component or the x component with respect to x . well , the derivative of this with respect to x is just x . the derivative of this with respect to x , luckily , is just 0 .
|
so should n't div f be x+y ?
|
let 's see if we might be able to make some use of the divergence theorem . so i have this region , this simple solid right over here . x can go between negative 1 and 1. z , this kind of arch part right over here , is going to be a function of x . that 's the upper bound on z . the lower bound on z is just 0 . and then y could go anywhere between 0 , and then it 's bounded here by this plane , where we can express y as a function of z. y is 2 minus z along this plane right over here . and we 're given this crazy vector field . it has natural logs and tangents in it . and we 're asked to evaluate the surface integral , or actually , i should say the flux of our vector field across the boundary of this region , across the surface of this region right over here . and surface integrals are messy as is , especially when you have a crazy vector field like this . but you could imagine that there might be a way to simplify this , perhaps using the divergence theorem . the divergence theorem tells us that the flux across the boundary of this simple solid region is going to be the same thing as the triple integral over the volume of it , or i 'll just call it over the region , of the divergence of f dv , where dv is some combination of dx , dy , dz . the divergence times each little cubic volume , infinitesimal cubic volume , so times dv . so let 's see if this simplifies things a bit . so let 's calculate the divergence of f first . so the divergence of f is going to be the partial of the x component , or the partial of the -- you could say the i component or the x component with respect to x . well , the derivative of this with respect to x is just x . the derivative of this with respect to x , luckily , is just 0 . this is a constant in terms of x . now let 's go over here , the partial of this with respect to y . the partial of this with respect to y is just x . and then this is just a constant in terms of y , so it 's just going to be 0 when you take the derivative with respect to y . and then , finally , the partial of this with respect to z , well , this is just a constant in terms of z . does n't change when z changes . so the partial with respect to z is just going to be 0 here . and so taking the divergence really , really , really simplified things . the divergence of f simplified down to 2x . and so now we can restate the flux across the surface as a triple integral of 2x . so let me just write 2x here . and let 's think about the ordering . so y can go between 0 and this plane that is a function of z . so let 's write that down . so y is bounded below by 0 and above by this plane 2 minus z. z is bounded below by 0 and above by -- you could call them these parabolas of 1 minus x squared . and then x is bounded below by negative 1 and bounded above by 1 . so negative 1 is less than or equal to x is less than or equal to 1 . and so this is probably a good order of integration . we can integrate with respect to y first , and then we 'll get a function of z . then we can integrate with respect to z , and we 'll get a function of x . and then we can integrate with respect to x . so let 's do it in that order . so first we 'll integrate with respect to y , so we have dy . y is bounded below at 0 and above by the plane 2 minus z . so this right over here is a plane y is equal to 0 . and this up over here is the plane y is equal to 2 minus z . then we can integrate with respect to z . and z , once again , is bounded below by 0 and bounded above by these parabolas , 1 minus x squared . and then , finally , we can integrate with respect to x . and x is bounded below by negative 1 and bounded above by 1 . so let 's do some integration here . so the first thing , when we 're integrating with respect to x -- sorry , when we 're integrating with respect to y , 2x is just a constant . so this expression right over here is just going to be 2x times y , and then we 're going to evaluate it from 0 to 2 minus z . so it 's going to be 2x times 2 minus z minus 2x times 0 . well , that second part 's just going to be 0 . so this is going to simplify as -- i 'll write it this way -- 2x times 2 minus z . and actually , i 'll just leave it like that . and then we 're going to integrate this with respect to z . and that 's going to go from 0 to 1 minus x squared , and then we have our dz there . and then after that , we 're going to integrate with respect to x , negative 1 to 1 dx . so let 's take the antiderivative here with respect to z . this you really can just view as a constant . we can actually even bring it out front , but i 'll leave it there . so this piece right over here -- i 'll do it in z 's color -- this piece right over here , see , we can leave the 2x out front . actually , i 'll leave the 2x out front of the whole thing . it 's going to be 2x times -- so the antiderivative of this with respect to z is going to be 2z . antiderivative of this is negative z squared over 2 , and we are going to evaluate this from 0 to 1 minus x squared . when we evaluate them at 0 , we 're just going to get 0 right over here . and so we really just have to worry about when z is equal to 1 minus x squared . did i do that right ? yep . 2z , and then minus z squared over 2 . you take the derivative , you get negative z . take the derivative here , you just get 2 . so that 's right . so this is going to be equal to 2x -- let me do that same color -- it 's going to be equal to 2x times -- let me get this right , let me go into that pink color -- 2x times 2z . well , z is going to be 1 minus x squared , so it 's going to be 2 minus 2x squared . that was just 2 times that . and then minus -- i 'll just write 1/2 times this quantity squared . so this quantity squared is going to be 1 minus 2x squared plus x to the fourth . that 's just some basic algebra right over there . and then from that , you 're going to subtract this thing evaluated at 0 , which is just going to be 0 . so [ ? y , you ? ] just wo n't even think about that . and now we need to simplify this a little bit . and we are going to get , if we simplify this , we get 2 minus 2x squared minus 1/2 , and then plus -- so this is negative 1/2 times negative 2x squared . so it 's going to be positive x squared minus 1/2 x to the fourth . now , let 's see , can we simplify this part ? let me just make sure we know what we 're doing here . so we have this 2x right over there . i want to make sure i got the signs right . yep , looks like i did . and now let 's look at this . so let 's see , can i simplify a little bit ? i have 2 minus 1/2 , which is 3/2 . so i have 3/2 . that 's that term and that term take into account . and then i have negative 2x squared plus x squared . so that 's just going to result in negative x squared , if i take that term and that term . and then i have negative 1/2 x to the fourth , and i 'm multiplying this whole thing by 2x . and so that 's going to give us -- we have , let 's see , 2x times 3/2 . and i want to make sure . i 'm doing this slowly , so i do n't make any careless mistakes . the 2 's cancel out . you get 3x , and then 2x times negative x squared is negative 2x to the third . and then 2x times this right over here . the 2 cancels out with the negative 1/2 , you have negative x to the fifth . so all of this simplifies to this right over here . so our whole thing simplifies to an integral with respect to x. x will go from negative 1 to 1 of this business of 3x minus 2x to the third minus x to the fifth , and then we have dx . and now we just take the antiderivative with respect to x , which is going to be 3/2 x squared minus -- let 's see , x to the fourth power -- minus 1/2 , because it 's going to be 2/4 , x to the fourth . is that right ? because if you multiply it , you 're going to 2 . yep , x to the third , and then minus x to the sixth over 6 . and it 's going to go from 1 to negative 1 or negative 1 to 1 . so when you evaluate it at 1 -- i 'll just write it out real fast . so first , when you evaluate it at 1 , you get 3/2 minus 1/2 minus 1/6 . and then from that , we are going to subtract 3/2 minus 1/2 plus 1/6 . or actually , no , they 're actually all going to cancel out . is that right ? are they all going to cancel out ? yep , i think that 's right . they all cancel out . so it 's actually going to be plus , or i should say minus 1/6 right over here . and then all of these cancel out . that cancels with that . that cancels with that because we 're subtracting the negative 1/2 . and that cancels with that . and so we are actually left with 0 . so after doing all of that work , this whole thing evaluates to 0 , which was actually kind of a neat simplification . so this whole thing right over here evaluated , very conveniently , evaluated to be equal to 0 .
|
and surface integrals are messy as is , especially when you have a crazy vector field like this . but you could imagine that there might be a way to simplify this , perhaps using the divergence theorem . the divergence theorem tells us that the flux across the boundary of this simple solid region is going to be the same thing as the triple integral over the volume of it , or i 'll just call it over the region , of the divergence of f dv , where dv is some combination of dx , dy , dz . the divergence times each little cubic volume , infinitesimal cubic volume , so times dv . so let 's see if this simplifies things a bit .
|
does that triple integral in the divergence theorem count as a volume integral ?
|
in the last video , we figured out the formulas for the observed period and frequency for an observer sitting in the path of the source . so the source is moving towards the observer . so this is the example where the train is moving towards you , and you perceive the train 's horn as having a higher pitch or a higher frequency . and we were able to do that by doing a thought experiment . saying , ok , my object starts here . after one period -- a period is just a measure of time , but it 's the measure of time over which the source emits a cycle , so it emits a cycle every period . but after one period , we said , ok , where is that first wave front , or that first pulse , or that first crest ? and where is the source ? because exactly one period has passed by , and the source will be ready to emit another crest or another cycle . so the distance between where the source is and that front of the crest , or that first crest , that is going to be the wavelength . because this next thing that emits is going to be traveling at the exact same velocity , and it 's going to be separated by that distance , which we saw is this expression . we said how long will it take it to travel that distance ? well , it 's traveling at a speed of v sub w. that 'll tell you what the observed period would be for this dude over here . we calculated it right here , and then the observed frequency is just the inverse of that . now let 's think about the situation where the observer is over here . so these equations , or these formulas that we came up with right here , this is observer -- or let me say source traveling in direction of observer . now let 's think about the opposite case , where the source is traveling away from the observer , and in this case , the observer is that guy over there . maybe i 'll do it in a different color . he 'll be blue . so this is the observer . so when we started off , our source was right here . after exactly one period from the source 's point of view , that first crest emitted has traveled radially outward that far . this is the distance . the velocity of the wave times the amount of time that passed -- velocity times time is going to give you distance -- and where the source of the wave will have traveled to the right exactly this distance . it 's velocity times the amount of time that 's gone by . now , in the last video , we said , ok , that wave is just passing this guy . how long will it take for that pulse that 's being emitted right then to also reach him ? and then that tells us the period between two pulses or between two crests . now let 's think about that exact same situation here . that first crest is just passing this guy . and a period , or the t sub s , which is a period of the emitted wave has just passed by . so this guy is just about to emit another wave , so ? that other wave is going to be right here . so what is the distance between the crest , or the cycle , or however you want to think about it , the pulse that is passing him by right now , what is the distance between that and the pulse , the front edge of that pulse that is being emitted right at that moment ? right at that moment . well , it 's going to be this radius , which is this value . it 'll be v sub w times the period . that is that distance plus the amount of distance that our source has traveled away from this guy . so plus v sub s times the period . so that 's what this distance is , and this is how far apart this wave pulse is going to be from that one or this crest is going to be from that one . so if he 's seeing this first crest right now , right at this moment , how long will it take him to see the crest that 's being emitted right now , that 's this far away from him ? well , it 's that far away from him . so let me write this down . so the amount of time it takes for him to see the next crest or the same point in the next cycle , that 's the period . that 's the observed period . that 's going to be equal to this distance . the velocity of the wave times the period from the perspective of the source plus the velocity of the source , because the source has gotten that much further away from him , velocity of the source times the period of the source , so that 's how far the next crest is . and then you divide it by the speed of the wave , by the speed of each of the crests , which is just the velocity of the wave . and we can just factor out the t sub s 's here and say this is t sub s times v sub w , the velocity of the wave plus the velocity of the source divided by the velocity the wave . so this will be a larger observed period than if this guy was stationary and especially if the observer was in the path of the guy . that make sense , because every time this guy issues a cycle , he is moving a little bit further away . so every crest the same point in the cycle is going to be further and further apart , so you 're going to have longer wavelengths , longer periods . and if you want the observed frequency for that guy over there -- i 'll do it in the same color . the observed frequency for a guy where the source is traveling away from him is just the inverse of that . so one over the period , one over the period , same argument we did there , one over the period from the point of view of the source is the frequency of the source . let me color code it . so 1 over t sub s is equal to the frequency of the source . this is the inverse of that . so i 'm just taking one over everything here . so 1 over t sub s is the frequency of the source , and then we take the inverse of this over here : the velocity of the wave divided by the velocity of the wave plus the velocity of the source . so we 're done , at least for the simple cases . obviously , it becomes a little more interesting when someone is n't exactly in the direction of the source or exactly being moved away from , but these are kind of the two extreme cases . so this is the situation when it is moving away from you . now , just to check our math and maybe make it a little bit concrete in relation to the video we did where we introduced to the idea of the doppler effect , let 's actually apply those numbers . so on that video , two videos ago , we had a situation where the velocity of our source was 5 meters per second to the right , and the velocity of the wave was 10 meters per second radially outward , and the period of our wave -- i 'll call it the period of our -- let me do it in another color -- from the point of view of the source was 1 second per cycle , and the frequency was just the inverse of that . so 1 cycle per second , or 1 hertz , which is a cycle per second . so using those numbers , let 's see if we get to the exact same answer we got in that first video where we first learned about the doppler effect . so let 's look at the frequency from the point of view of this gentleman right here . so the frequency of the source is going to be 1 cycle per second , 1 hertz . the velocity of the wave is 10 meters per second . let me write this . 1 cycle per second . the velocity of the wave is 10 meters per second . the velocity of the wave is 10 meters per second minus the velocity of the source is 5 meters per second . so what 's this going to be equal to ? the observed frequency for this guy right there , is going to be 1 cycle per second times 10 over 10 minus 5 , and the meters per second cancel out . meters per second in the numerator , meters per second in the denominator . so 10 divided by 10 minus 5 , or 10 divided by 5 , is going to be 2 . so it 's going to be 2 cycles per second . and if you want the observed period for this guy , its going to be the inverse of that , or it 's going to be 1/2 second per cycle . and this is exactly what we got in the previous video , or actually , it was two videos ago . now what about the guy who this guy 's running away from ? well , we 'll do the exact same thing . you have 1 cycle per second , or 1 hertz . that 's the emitted frequency from the point of view of the source times the velocity of the wave divided by the velocity of the wave plus the velocity of the source , because it 's moving away from him . so it 's 10 over 10 plus 5 . that 's 10 over 15 . that is 2/3 , so this is equal to 2/3 . the units over here all cancel out . this was cycles per second . so 2/3 cycle per second , which confirms the numbers we got in that first video , so that should make us feel good and it also makes a lot of sense . this guy is going to see the wave crest more frequently . he 's going to observe a higher frequency . if this is a sound , a higher pitch . this guy , since each crest or the cycles are getting spread out , he 's going to see them less frequently , and if this is sound , he 's going to observe a lower pitch .
|
in the last video , we figured out the formulas for the observed period and frequency for an observer sitting in the path of the source . so the source is moving towards the observer . so this is the example where the train is moving towards you , and you perceive the train 's horn as having a higher pitch or a higher frequency .
|
so , what will the formula be if the observer is in an angle from the source 's direction of motion ( not exactly in front of or behind the source ) ?
|
in the last video , we figured out the formulas for the observed period and frequency for an observer sitting in the path of the source . so the source is moving towards the observer . so this is the example where the train is moving towards you , and you perceive the train 's horn as having a higher pitch or a higher frequency . and we were able to do that by doing a thought experiment . saying , ok , my object starts here . after one period -- a period is just a measure of time , but it 's the measure of time over which the source emits a cycle , so it emits a cycle every period . but after one period , we said , ok , where is that first wave front , or that first pulse , or that first crest ? and where is the source ? because exactly one period has passed by , and the source will be ready to emit another crest or another cycle . so the distance between where the source is and that front of the crest , or that first crest , that is going to be the wavelength . because this next thing that emits is going to be traveling at the exact same velocity , and it 's going to be separated by that distance , which we saw is this expression . we said how long will it take it to travel that distance ? well , it 's traveling at a speed of v sub w. that 'll tell you what the observed period would be for this dude over here . we calculated it right here , and then the observed frequency is just the inverse of that . now let 's think about the situation where the observer is over here . so these equations , or these formulas that we came up with right here , this is observer -- or let me say source traveling in direction of observer . now let 's think about the opposite case , where the source is traveling away from the observer , and in this case , the observer is that guy over there . maybe i 'll do it in a different color . he 'll be blue . so this is the observer . so when we started off , our source was right here . after exactly one period from the source 's point of view , that first crest emitted has traveled radially outward that far . this is the distance . the velocity of the wave times the amount of time that passed -- velocity times time is going to give you distance -- and where the source of the wave will have traveled to the right exactly this distance . it 's velocity times the amount of time that 's gone by . now , in the last video , we said , ok , that wave is just passing this guy . how long will it take for that pulse that 's being emitted right then to also reach him ? and then that tells us the period between two pulses or between two crests . now let 's think about that exact same situation here . that first crest is just passing this guy . and a period , or the t sub s , which is a period of the emitted wave has just passed by . so this guy is just about to emit another wave , so ? that other wave is going to be right here . so what is the distance between the crest , or the cycle , or however you want to think about it , the pulse that is passing him by right now , what is the distance between that and the pulse , the front edge of that pulse that is being emitted right at that moment ? right at that moment . well , it 's going to be this radius , which is this value . it 'll be v sub w times the period . that is that distance plus the amount of distance that our source has traveled away from this guy . so plus v sub s times the period . so that 's what this distance is , and this is how far apart this wave pulse is going to be from that one or this crest is going to be from that one . so if he 's seeing this first crest right now , right at this moment , how long will it take him to see the crest that 's being emitted right now , that 's this far away from him ? well , it 's that far away from him . so let me write this down . so the amount of time it takes for him to see the next crest or the same point in the next cycle , that 's the period . that 's the observed period . that 's going to be equal to this distance . the velocity of the wave times the period from the perspective of the source plus the velocity of the source , because the source has gotten that much further away from him , velocity of the source times the period of the source , so that 's how far the next crest is . and then you divide it by the speed of the wave , by the speed of each of the crests , which is just the velocity of the wave . and we can just factor out the t sub s 's here and say this is t sub s times v sub w , the velocity of the wave plus the velocity of the source divided by the velocity the wave . so this will be a larger observed period than if this guy was stationary and especially if the observer was in the path of the guy . that make sense , because every time this guy issues a cycle , he is moving a little bit further away . so every crest the same point in the cycle is going to be further and further apart , so you 're going to have longer wavelengths , longer periods . and if you want the observed frequency for that guy over there -- i 'll do it in the same color . the observed frequency for a guy where the source is traveling away from him is just the inverse of that . so one over the period , one over the period , same argument we did there , one over the period from the point of view of the source is the frequency of the source . let me color code it . so 1 over t sub s is equal to the frequency of the source . this is the inverse of that . so i 'm just taking one over everything here . so 1 over t sub s is the frequency of the source , and then we take the inverse of this over here : the velocity of the wave divided by the velocity of the wave plus the velocity of the source . so we 're done , at least for the simple cases . obviously , it becomes a little more interesting when someone is n't exactly in the direction of the source or exactly being moved away from , but these are kind of the two extreme cases . so this is the situation when it is moving away from you . now , just to check our math and maybe make it a little bit concrete in relation to the video we did where we introduced to the idea of the doppler effect , let 's actually apply those numbers . so on that video , two videos ago , we had a situation where the velocity of our source was 5 meters per second to the right , and the velocity of the wave was 10 meters per second radially outward , and the period of our wave -- i 'll call it the period of our -- let me do it in another color -- from the point of view of the source was 1 second per cycle , and the frequency was just the inverse of that . so 1 cycle per second , or 1 hertz , which is a cycle per second . so using those numbers , let 's see if we get to the exact same answer we got in that first video where we first learned about the doppler effect . so let 's look at the frequency from the point of view of this gentleman right here . so the frequency of the source is going to be 1 cycle per second , 1 hertz . the velocity of the wave is 10 meters per second . let me write this . 1 cycle per second . the velocity of the wave is 10 meters per second . the velocity of the wave is 10 meters per second minus the velocity of the source is 5 meters per second . so what 's this going to be equal to ? the observed frequency for this guy right there , is going to be 1 cycle per second times 10 over 10 minus 5 , and the meters per second cancel out . meters per second in the numerator , meters per second in the denominator . so 10 divided by 10 minus 5 , or 10 divided by 5 , is going to be 2 . so it 's going to be 2 cycles per second . and if you want the observed period for this guy , its going to be the inverse of that , or it 's going to be 1/2 second per cycle . and this is exactly what we got in the previous video , or actually , it was two videos ago . now what about the guy who this guy 's running away from ? well , we 'll do the exact same thing . you have 1 cycle per second , or 1 hertz . that 's the emitted frequency from the point of view of the source times the velocity of the wave divided by the velocity of the wave plus the velocity of the source , because it 's moving away from him . so it 's 10 over 10 plus 5 . that 's 10 over 15 . that is 2/3 , so this is equal to 2/3 . the units over here all cancel out . this was cycles per second . so 2/3 cycle per second , which confirms the numbers we got in that first video , so that should make us feel good and it also makes a lot of sense . this guy is going to see the wave crest more frequently . he 's going to observe a higher frequency . if this is a sound , a higher pitch . this guy , since each crest or the cycles are getting spread out , he 's going to see them less frequently , and if this is sound , he 's going to observe a lower pitch .
|
in the last video , we figured out the formulas for the observed period and frequency for an observer sitting in the path of the source . so the source is moving towards the observer . so this is the example where the train is moving towards you , and you perceive the train 's horn as having a higher pitch or a higher frequency .
|
can the observer sitting in the source emitting sound , if traveling faster than the speed of sound , hear the source sound normally ?
|
in the last video , we figured out the formulas for the observed period and frequency for an observer sitting in the path of the source . so the source is moving towards the observer . so this is the example where the train is moving towards you , and you perceive the train 's horn as having a higher pitch or a higher frequency . and we were able to do that by doing a thought experiment . saying , ok , my object starts here . after one period -- a period is just a measure of time , but it 's the measure of time over which the source emits a cycle , so it emits a cycle every period . but after one period , we said , ok , where is that first wave front , or that first pulse , or that first crest ? and where is the source ? because exactly one period has passed by , and the source will be ready to emit another crest or another cycle . so the distance between where the source is and that front of the crest , or that first crest , that is going to be the wavelength . because this next thing that emits is going to be traveling at the exact same velocity , and it 's going to be separated by that distance , which we saw is this expression . we said how long will it take it to travel that distance ? well , it 's traveling at a speed of v sub w. that 'll tell you what the observed period would be for this dude over here . we calculated it right here , and then the observed frequency is just the inverse of that . now let 's think about the situation where the observer is over here . so these equations , or these formulas that we came up with right here , this is observer -- or let me say source traveling in direction of observer . now let 's think about the opposite case , where the source is traveling away from the observer , and in this case , the observer is that guy over there . maybe i 'll do it in a different color . he 'll be blue . so this is the observer . so when we started off , our source was right here . after exactly one period from the source 's point of view , that first crest emitted has traveled radially outward that far . this is the distance . the velocity of the wave times the amount of time that passed -- velocity times time is going to give you distance -- and where the source of the wave will have traveled to the right exactly this distance . it 's velocity times the amount of time that 's gone by . now , in the last video , we said , ok , that wave is just passing this guy . how long will it take for that pulse that 's being emitted right then to also reach him ? and then that tells us the period between two pulses or between two crests . now let 's think about that exact same situation here . that first crest is just passing this guy . and a period , or the t sub s , which is a period of the emitted wave has just passed by . so this guy is just about to emit another wave , so ? that other wave is going to be right here . so what is the distance between the crest , or the cycle , or however you want to think about it , the pulse that is passing him by right now , what is the distance between that and the pulse , the front edge of that pulse that is being emitted right at that moment ? right at that moment . well , it 's going to be this radius , which is this value . it 'll be v sub w times the period . that is that distance plus the amount of distance that our source has traveled away from this guy . so plus v sub s times the period . so that 's what this distance is , and this is how far apart this wave pulse is going to be from that one or this crest is going to be from that one . so if he 's seeing this first crest right now , right at this moment , how long will it take him to see the crest that 's being emitted right now , that 's this far away from him ? well , it 's that far away from him . so let me write this down . so the amount of time it takes for him to see the next crest or the same point in the next cycle , that 's the period . that 's the observed period . that 's going to be equal to this distance . the velocity of the wave times the period from the perspective of the source plus the velocity of the source , because the source has gotten that much further away from him , velocity of the source times the period of the source , so that 's how far the next crest is . and then you divide it by the speed of the wave , by the speed of each of the crests , which is just the velocity of the wave . and we can just factor out the t sub s 's here and say this is t sub s times v sub w , the velocity of the wave plus the velocity of the source divided by the velocity the wave . so this will be a larger observed period than if this guy was stationary and especially if the observer was in the path of the guy . that make sense , because every time this guy issues a cycle , he is moving a little bit further away . so every crest the same point in the cycle is going to be further and further apart , so you 're going to have longer wavelengths , longer periods . and if you want the observed frequency for that guy over there -- i 'll do it in the same color . the observed frequency for a guy where the source is traveling away from him is just the inverse of that . so one over the period , one over the period , same argument we did there , one over the period from the point of view of the source is the frequency of the source . let me color code it . so 1 over t sub s is equal to the frequency of the source . this is the inverse of that . so i 'm just taking one over everything here . so 1 over t sub s is the frequency of the source , and then we take the inverse of this over here : the velocity of the wave divided by the velocity of the wave plus the velocity of the source . so we 're done , at least for the simple cases . obviously , it becomes a little more interesting when someone is n't exactly in the direction of the source or exactly being moved away from , but these are kind of the two extreme cases . so this is the situation when it is moving away from you . now , just to check our math and maybe make it a little bit concrete in relation to the video we did where we introduced to the idea of the doppler effect , let 's actually apply those numbers . so on that video , two videos ago , we had a situation where the velocity of our source was 5 meters per second to the right , and the velocity of the wave was 10 meters per second radially outward , and the period of our wave -- i 'll call it the period of our -- let me do it in another color -- from the point of view of the source was 1 second per cycle , and the frequency was just the inverse of that . so 1 cycle per second , or 1 hertz , which is a cycle per second . so using those numbers , let 's see if we get to the exact same answer we got in that first video where we first learned about the doppler effect . so let 's look at the frequency from the point of view of this gentleman right here . so the frequency of the source is going to be 1 cycle per second , 1 hertz . the velocity of the wave is 10 meters per second . let me write this . 1 cycle per second . the velocity of the wave is 10 meters per second . the velocity of the wave is 10 meters per second minus the velocity of the source is 5 meters per second . so what 's this going to be equal to ? the observed frequency for this guy right there , is going to be 1 cycle per second times 10 over 10 minus 5 , and the meters per second cancel out . meters per second in the numerator , meters per second in the denominator . so 10 divided by 10 minus 5 , or 10 divided by 5 , is going to be 2 . so it 's going to be 2 cycles per second . and if you want the observed period for this guy , its going to be the inverse of that , or it 's going to be 1/2 second per cycle . and this is exactly what we got in the previous video , or actually , it was two videos ago . now what about the guy who this guy 's running away from ? well , we 'll do the exact same thing . you have 1 cycle per second , or 1 hertz . that 's the emitted frequency from the point of view of the source times the velocity of the wave divided by the velocity of the wave plus the velocity of the source , because it 's moving away from him . so it 's 10 over 10 plus 5 . that 's 10 over 15 . that is 2/3 , so this is equal to 2/3 . the units over here all cancel out . this was cycles per second . so 2/3 cycle per second , which confirms the numbers we got in that first video , so that should make us feel good and it also makes a lot of sense . this guy is going to see the wave crest more frequently . he 's going to observe a higher frequency . if this is a sound , a higher pitch . this guy , since each crest or the cycles are getting spread out , he 's going to see them less frequently , and if this is sound , he 's going to observe a lower pitch .
|
in the last video , we figured out the formulas for the observed period and frequency for an observer sitting in the path of the source . so the source is moving towards the observer . so this is the example where the train is moving towards you , and you perceive the train 's horn as having a higher pitch or a higher frequency .
|
what happens if the observer is traveling at the exact same speed as the wave from the source or the source itself ?
|
in the last video , we figured out the formulas for the observed period and frequency for an observer sitting in the path of the source . so the source is moving towards the observer . so this is the example where the train is moving towards you , and you perceive the train 's horn as having a higher pitch or a higher frequency . and we were able to do that by doing a thought experiment . saying , ok , my object starts here . after one period -- a period is just a measure of time , but it 's the measure of time over which the source emits a cycle , so it emits a cycle every period . but after one period , we said , ok , where is that first wave front , or that first pulse , or that first crest ? and where is the source ? because exactly one period has passed by , and the source will be ready to emit another crest or another cycle . so the distance between where the source is and that front of the crest , or that first crest , that is going to be the wavelength . because this next thing that emits is going to be traveling at the exact same velocity , and it 's going to be separated by that distance , which we saw is this expression . we said how long will it take it to travel that distance ? well , it 's traveling at a speed of v sub w. that 'll tell you what the observed period would be for this dude over here . we calculated it right here , and then the observed frequency is just the inverse of that . now let 's think about the situation where the observer is over here . so these equations , or these formulas that we came up with right here , this is observer -- or let me say source traveling in direction of observer . now let 's think about the opposite case , where the source is traveling away from the observer , and in this case , the observer is that guy over there . maybe i 'll do it in a different color . he 'll be blue . so this is the observer . so when we started off , our source was right here . after exactly one period from the source 's point of view , that first crest emitted has traveled radially outward that far . this is the distance . the velocity of the wave times the amount of time that passed -- velocity times time is going to give you distance -- and where the source of the wave will have traveled to the right exactly this distance . it 's velocity times the amount of time that 's gone by . now , in the last video , we said , ok , that wave is just passing this guy . how long will it take for that pulse that 's being emitted right then to also reach him ? and then that tells us the period between two pulses or between two crests . now let 's think about that exact same situation here . that first crest is just passing this guy . and a period , or the t sub s , which is a period of the emitted wave has just passed by . so this guy is just about to emit another wave , so ? that other wave is going to be right here . so what is the distance between the crest , or the cycle , or however you want to think about it , the pulse that is passing him by right now , what is the distance between that and the pulse , the front edge of that pulse that is being emitted right at that moment ? right at that moment . well , it 's going to be this radius , which is this value . it 'll be v sub w times the period . that is that distance plus the amount of distance that our source has traveled away from this guy . so plus v sub s times the period . so that 's what this distance is , and this is how far apart this wave pulse is going to be from that one or this crest is going to be from that one . so if he 's seeing this first crest right now , right at this moment , how long will it take him to see the crest that 's being emitted right now , that 's this far away from him ? well , it 's that far away from him . so let me write this down . so the amount of time it takes for him to see the next crest or the same point in the next cycle , that 's the period . that 's the observed period . that 's going to be equal to this distance . the velocity of the wave times the period from the perspective of the source plus the velocity of the source , because the source has gotten that much further away from him , velocity of the source times the period of the source , so that 's how far the next crest is . and then you divide it by the speed of the wave , by the speed of each of the crests , which is just the velocity of the wave . and we can just factor out the t sub s 's here and say this is t sub s times v sub w , the velocity of the wave plus the velocity of the source divided by the velocity the wave . so this will be a larger observed period than if this guy was stationary and especially if the observer was in the path of the guy . that make sense , because every time this guy issues a cycle , he is moving a little bit further away . so every crest the same point in the cycle is going to be further and further apart , so you 're going to have longer wavelengths , longer periods . and if you want the observed frequency for that guy over there -- i 'll do it in the same color . the observed frequency for a guy where the source is traveling away from him is just the inverse of that . so one over the period , one over the period , same argument we did there , one over the period from the point of view of the source is the frequency of the source . let me color code it . so 1 over t sub s is equal to the frequency of the source . this is the inverse of that . so i 'm just taking one over everything here . so 1 over t sub s is the frequency of the source , and then we take the inverse of this over here : the velocity of the wave divided by the velocity of the wave plus the velocity of the source . so we 're done , at least for the simple cases . obviously , it becomes a little more interesting when someone is n't exactly in the direction of the source or exactly being moved away from , but these are kind of the two extreme cases . so this is the situation when it is moving away from you . now , just to check our math and maybe make it a little bit concrete in relation to the video we did where we introduced to the idea of the doppler effect , let 's actually apply those numbers . so on that video , two videos ago , we had a situation where the velocity of our source was 5 meters per second to the right , and the velocity of the wave was 10 meters per second radially outward , and the period of our wave -- i 'll call it the period of our -- let me do it in another color -- from the point of view of the source was 1 second per cycle , and the frequency was just the inverse of that . so 1 cycle per second , or 1 hertz , which is a cycle per second . so using those numbers , let 's see if we get to the exact same answer we got in that first video where we first learned about the doppler effect . so let 's look at the frequency from the point of view of this gentleman right here . so the frequency of the source is going to be 1 cycle per second , 1 hertz . the velocity of the wave is 10 meters per second . let me write this . 1 cycle per second . the velocity of the wave is 10 meters per second . the velocity of the wave is 10 meters per second minus the velocity of the source is 5 meters per second . so what 's this going to be equal to ? the observed frequency for this guy right there , is going to be 1 cycle per second times 10 over 10 minus 5 , and the meters per second cancel out . meters per second in the numerator , meters per second in the denominator . so 10 divided by 10 minus 5 , or 10 divided by 5 , is going to be 2 . so it 's going to be 2 cycles per second . and if you want the observed period for this guy , its going to be the inverse of that , or it 's going to be 1/2 second per cycle . and this is exactly what we got in the previous video , or actually , it was two videos ago . now what about the guy who this guy 's running away from ? well , we 'll do the exact same thing . you have 1 cycle per second , or 1 hertz . that 's the emitted frequency from the point of view of the source times the velocity of the wave divided by the velocity of the wave plus the velocity of the source , because it 's moving away from him . so it 's 10 over 10 plus 5 . that 's 10 over 15 . that is 2/3 , so this is equal to 2/3 . the units over here all cancel out . this was cycles per second . so 2/3 cycle per second , which confirms the numbers we got in that first video , so that should make us feel good and it also makes a lot of sense . this guy is going to see the wave crest more frequently . he 's going to observe a higher frequency . if this is a sound , a higher pitch . this guy , since each crest or the cycles are getting spread out , he 's going to see them less frequently , and if this is sound , he 's going to observe a lower pitch .
|
he 'll be blue . so this is the observer . so when we started off , our source was right here .
|
what does the observer experience ?
|
in the last video , we figured out the formulas for the observed period and frequency for an observer sitting in the path of the source . so the source is moving towards the observer . so this is the example where the train is moving towards you , and you perceive the train 's horn as having a higher pitch or a higher frequency . and we were able to do that by doing a thought experiment . saying , ok , my object starts here . after one period -- a period is just a measure of time , but it 's the measure of time over which the source emits a cycle , so it emits a cycle every period . but after one period , we said , ok , where is that first wave front , or that first pulse , or that first crest ? and where is the source ? because exactly one period has passed by , and the source will be ready to emit another crest or another cycle . so the distance between where the source is and that front of the crest , or that first crest , that is going to be the wavelength . because this next thing that emits is going to be traveling at the exact same velocity , and it 's going to be separated by that distance , which we saw is this expression . we said how long will it take it to travel that distance ? well , it 's traveling at a speed of v sub w. that 'll tell you what the observed period would be for this dude over here . we calculated it right here , and then the observed frequency is just the inverse of that . now let 's think about the situation where the observer is over here . so these equations , or these formulas that we came up with right here , this is observer -- or let me say source traveling in direction of observer . now let 's think about the opposite case , where the source is traveling away from the observer , and in this case , the observer is that guy over there . maybe i 'll do it in a different color . he 'll be blue . so this is the observer . so when we started off , our source was right here . after exactly one period from the source 's point of view , that first crest emitted has traveled radially outward that far . this is the distance . the velocity of the wave times the amount of time that passed -- velocity times time is going to give you distance -- and where the source of the wave will have traveled to the right exactly this distance . it 's velocity times the amount of time that 's gone by . now , in the last video , we said , ok , that wave is just passing this guy . how long will it take for that pulse that 's being emitted right then to also reach him ? and then that tells us the period between two pulses or between two crests . now let 's think about that exact same situation here . that first crest is just passing this guy . and a period , or the t sub s , which is a period of the emitted wave has just passed by . so this guy is just about to emit another wave , so ? that other wave is going to be right here . so what is the distance between the crest , or the cycle , or however you want to think about it , the pulse that is passing him by right now , what is the distance between that and the pulse , the front edge of that pulse that is being emitted right at that moment ? right at that moment . well , it 's going to be this radius , which is this value . it 'll be v sub w times the period . that is that distance plus the amount of distance that our source has traveled away from this guy . so plus v sub s times the period . so that 's what this distance is , and this is how far apart this wave pulse is going to be from that one or this crest is going to be from that one . so if he 's seeing this first crest right now , right at this moment , how long will it take him to see the crest that 's being emitted right now , that 's this far away from him ? well , it 's that far away from him . so let me write this down . so the amount of time it takes for him to see the next crest or the same point in the next cycle , that 's the period . that 's the observed period . that 's going to be equal to this distance . the velocity of the wave times the period from the perspective of the source plus the velocity of the source , because the source has gotten that much further away from him , velocity of the source times the period of the source , so that 's how far the next crest is . and then you divide it by the speed of the wave , by the speed of each of the crests , which is just the velocity of the wave . and we can just factor out the t sub s 's here and say this is t sub s times v sub w , the velocity of the wave plus the velocity of the source divided by the velocity the wave . so this will be a larger observed period than if this guy was stationary and especially if the observer was in the path of the guy . that make sense , because every time this guy issues a cycle , he is moving a little bit further away . so every crest the same point in the cycle is going to be further and further apart , so you 're going to have longer wavelengths , longer periods . and if you want the observed frequency for that guy over there -- i 'll do it in the same color . the observed frequency for a guy where the source is traveling away from him is just the inverse of that . so one over the period , one over the period , same argument we did there , one over the period from the point of view of the source is the frequency of the source . let me color code it . so 1 over t sub s is equal to the frequency of the source . this is the inverse of that . so i 'm just taking one over everything here . so 1 over t sub s is the frequency of the source , and then we take the inverse of this over here : the velocity of the wave divided by the velocity of the wave plus the velocity of the source . so we 're done , at least for the simple cases . obviously , it becomes a little more interesting when someone is n't exactly in the direction of the source or exactly being moved away from , but these are kind of the two extreme cases . so this is the situation when it is moving away from you . now , just to check our math and maybe make it a little bit concrete in relation to the video we did where we introduced to the idea of the doppler effect , let 's actually apply those numbers . so on that video , two videos ago , we had a situation where the velocity of our source was 5 meters per second to the right , and the velocity of the wave was 10 meters per second radially outward , and the period of our wave -- i 'll call it the period of our -- let me do it in another color -- from the point of view of the source was 1 second per cycle , and the frequency was just the inverse of that . so 1 cycle per second , or 1 hertz , which is a cycle per second . so using those numbers , let 's see if we get to the exact same answer we got in that first video where we first learned about the doppler effect . so let 's look at the frequency from the point of view of this gentleman right here . so the frequency of the source is going to be 1 cycle per second , 1 hertz . the velocity of the wave is 10 meters per second . let me write this . 1 cycle per second . the velocity of the wave is 10 meters per second . the velocity of the wave is 10 meters per second minus the velocity of the source is 5 meters per second . so what 's this going to be equal to ? the observed frequency for this guy right there , is going to be 1 cycle per second times 10 over 10 minus 5 , and the meters per second cancel out . meters per second in the numerator , meters per second in the denominator . so 10 divided by 10 minus 5 , or 10 divided by 5 , is going to be 2 . so it 's going to be 2 cycles per second . and if you want the observed period for this guy , its going to be the inverse of that , or it 's going to be 1/2 second per cycle . and this is exactly what we got in the previous video , or actually , it was two videos ago . now what about the guy who this guy 's running away from ? well , we 'll do the exact same thing . you have 1 cycle per second , or 1 hertz . that 's the emitted frequency from the point of view of the source times the velocity of the wave divided by the velocity of the wave plus the velocity of the source , because it 's moving away from him . so it 's 10 over 10 plus 5 . that 's 10 over 15 . that is 2/3 , so this is equal to 2/3 . the units over here all cancel out . this was cycles per second . so 2/3 cycle per second , which confirms the numbers we got in that first video , so that should make us feel good and it also makes a lot of sense . this guy is going to see the wave crest more frequently . he 's going to observe a higher frequency . if this is a sound , a higher pitch . this guy , since each crest or the cycles are getting spread out , he 's going to see them less frequently , and if this is sound , he 's going to observe a lower pitch .
|
this guy is going to see the wave crest more frequently . he 's going to observe a higher frequency . if this is a sound , a higher pitch .
|
how can we observe the doppler effect in objects moving faster than the speed of sound ?
|
in the last video , we figured out the formulas for the observed period and frequency for an observer sitting in the path of the source . so the source is moving towards the observer . so this is the example where the train is moving towards you , and you perceive the train 's horn as having a higher pitch or a higher frequency . and we were able to do that by doing a thought experiment . saying , ok , my object starts here . after one period -- a period is just a measure of time , but it 's the measure of time over which the source emits a cycle , so it emits a cycle every period . but after one period , we said , ok , where is that first wave front , or that first pulse , or that first crest ? and where is the source ? because exactly one period has passed by , and the source will be ready to emit another crest or another cycle . so the distance between where the source is and that front of the crest , or that first crest , that is going to be the wavelength . because this next thing that emits is going to be traveling at the exact same velocity , and it 's going to be separated by that distance , which we saw is this expression . we said how long will it take it to travel that distance ? well , it 's traveling at a speed of v sub w. that 'll tell you what the observed period would be for this dude over here . we calculated it right here , and then the observed frequency is just the inverse of that . now let 's think about the situation where the observer is over here . so these equations , or these formulas that we came up with right here , this is observer -- or let me say source traveling in direction of observer . now let 's think about the opposite case , where the source is traveling away from the observer , and in this case , the observer is that guy over there . maybe i 'll do it in a different color . he 'll be blue . so this is the observer . so when we started off , our source was right here . after exactly one period from the source 's point of view , that first crest emitted has traveled radially outward that far . this is the distance . the velocity of the wave times the amount of time that passed -- velocity times time is going to give you distance -- and where the source of the wave will have traveled to the right exactly this distance . it 's velocity times the amount of time that 's gone by . now , in the last video , we said , ok , that wave is just passing this guy . how long will it take for that pulse that 's being emitted right then to also reach him ? and then that tells us the period between two pulses or between two crests . now let 's think about that exact same situation here . that first crest is just passing this guy . and a period , or the t sub s , which is a period of the emitted wave has just passed by . so this guy is just about to emit another wave , so ? that other wave is going to be right here . so what is the distance between the crest , or the cycle , or however you want to think about it , the pulse that is passing him by right now , what is the distance between that and the pulse , the front edge of that pulse that is being emitted right at that moment ? right at that moment . well , it 's going to be this radius , which is this value . it 'll be v sub w times the period . that is that distance plus the amount of distance that our source has traveled away from this guy . so plus v sub s times the period . so that 's what this distance is , and this is how far apart this wave pulse is going to be from that one or this crest is going to be from that one . so if he 's seeing this first crest right now , right at this moment , how long will it take him to see the crest that 's being emitted right now , that 's this far away from him ? well , it 's that far away from him . so let me write this down . so the amount of time it takes for him to see the next crest or the same point in the next cycle , that 's the period . that 's the observed period . that 's going to be equal to this distance . the velocity of the wave times the period from the perspective of the source plus the velocity of the source , because the source has gotten that much further away from him , velocity of the source times the period of the source , so that 's how far the next crest is . and then you divide it by the speed of the wave , by the speed of each of the crests , which is just the velocity of the wave . and we can just factor out the t sub s 's here and say this is t sub s times v sub w , the velocity of the wave plus the velocity of the source divided by the velocity the wave . so this will be a larger observed period than if this guy was stationary and especially if the observer was in the path of the guy . that make sense , because every time this guy issues a cycle , he is moving a little bit further away . so every crest the same point in the cycle is going to be further and further apart , so you 're going to have longer wavelengths , longer periods . and if you want the observed frequency for that guy over there -- i 'll do it in the same color . the observed frequency for a guy where the source is traveling away from him is just the inverse of that . so one over the period , one over the period , same argument we did there , one over the period from the point of view of the source is the frequency of the source . let me color code it . so 1 over t sub s is equal to the frequency of the source . this is the inverse of that . so i 'm just taking one over everything here . so 1 over t sub s is the frequency of the source , and then we take the inverse of this over here : the velocity of the wave divided by the velocity of the wave plus the velocity of the source . so we 're done , at least for the simple cases . obviously , it becomes a little more interesting when someone is n't exactly in the direction of the source or exactly being moved away from , but these are kind of the two extreme cases . so this is the situation when it is moving away from you . now , just to check our math and maybe make it a little bit concrete in relation to the video we did where we introduced to the idea of the doppler effect , let 's actually apply those numbers . so on that video , two videos ago , we had a situation where the velocity of our source was 5 meters per second to the right , and the velocity of the wave was 10 meters per second radially outward , and the period of our wave -- i 'll call it the period of our -- let me do it in another color -- from the point of view of the source was 1 second per cycle , and the frequency was just the inverse of that . so 1 cycle per second , or 1 hertz , which is a cycle per second . so using those numbers , let 's see if we get to the exact same answer we got in that first video where we first learned about the doppler effect . so let 's look at the frequency from the point of view of this gentleman right here . so the frequency of the source is going to be 1 cycle per second , 1 hertz . the velocity of the wave is 10 meters per second . let me write this . 1 cycle per second . the velocity of the wave is 10 meters per second . the velocity of the wave is 10 meters per second minus the velocity of the source is 5 meters per second . so what 's this going to be equal to ? the observed frequency for this guy right there , is going to be 1 cycle per second times 10 over 10 minus 5 , and the meters per second cancel out . meters per second in the numerator , meters per second in the denominator . so 10 divided by 10 minus 5 , or 10 divided by 5 , is going to be 2 . so it 's going to be 2 cycles per second . and if you want the observed period for this guy , its going to be the inverse of that , or it 's going to be 1/2 second per cycle . and this is exactly what we got in the previous video , or actually , it was two videos ago . now what about the guy who this guy 's running away from ? well , we 'll do the exact same thing . you have 1 cycle per second , or 1 hertz . that 's the emitted frequency from the point of view of the source times the velocity of the wave divided by the velocity of the wave plus the velocity of the source , because it 's moving away from him . so it 's 10 over 10 plus 5 . that 's 10 over 15 . that is 2/3 , so this is equal to 2/3 . the units over here all cancel out . this was cycles per second . so 2/3 cycle per second , which confirms the numbers we got in that first video , so that should make us feel good and it also makes a lot of sense . this guy is going to see the wave crest more frequently . he 's going to observe a higher frequency . if this is a sound , a higher pitch . this guy , since each crest or the cycles are getting spread out , he 's going to see them less frequently , and if this is sound , he 's going to observe a lower pitch .
|
the velocity of the wave times the period from the perspective of the source plus the velocity of the source , because the source has gotten that much further away from him , velocity of the source times the period of the source , so that 's how far the next crest is . and then you divide it by the speed of the wave , by the speed of each of the crests , which is just the velocity of the wave . and we can just factor out the t sub s 's here and say this is t sub s times v sub w , the velocity of the wave plus the velocity of the source divided by the velocity the wave .
|
why speed of the source does n't affect speed of the wave ?
|
in the last video , we figured out the formulas for the observed period and frequency for an observer sitting in the path of the source . so the source is moving towards the observer . so this is the example where the train is moving towards you , and you perceive the train 's horn as having a higher pitch or a higher frequency . and we were able to do that by doing a thought experiment . saying , ok , my object starts here . after one period -- a period is just a measure of time , but it 's the measure of time over which the source emits a cycle , so it emits a cycle every period . but after one period , we said , ok , where is that first wave front , or that first pulse , or that first crest ? and where is the source ? because exactly one period has passed by , and the source will be ready to emit another crest or another cycle . so the distance between where the source is and that front of the crest , or that first crest , that is going to be the wavelength . because this next thing that emits is going to be traveling at the exact same velocity , and it 's going to be separated by that distance , which we saw is this expression . we said how long will it take it to travel that distance ? well , it 's traveling at a speed of v sub w. that 'll tell you what the observed period would be for this dude over here . we calculated it right here , and then the observed frequency is just the inverse of that . now let 's think about the situation where the observer is over here . so these equations , or these formulas that we came up with right here , this is observer -- or let me say source traveling in direction of observer . now let 's think about the opposite case , where the source is traveling away from the observer , and in this case , the observer is that guy over there . maybe i 'll do it in a different color . he 'll be blue . so this is the observer . so when we started off , our source was right here . after exactly one period from the source 's point of view , that first crest emitted has traveled radially outward that far . this is the distance . the velocity of the wave times the amount of time that passed -- velocity times time is going to give you distance -- and where the source of the wave will have traveled to the right exactly this distance . it 's velocity times the amount of time that 's gone by . now , in the last video , we said , ok , that wave is just passing this guy . how long will it take for that pulse that 's being emitted right then to also reach him ? and then that tells us the period between two pulses or between two crests . now let 's think about that exact same situation here . that first crest is just passing this guy . and a period , or the t sub s , which is a period of the emitted wave has just passed by . so this guy is just about to emit another wave , so ? that other wave is going to be right here . so what is the distance between the crest , or the cycle , or however you want to think about it , the pulse that is passing him by right now , what is the distance between that and the pulse , the front edge of that pulse that is being emitted right at that moment ? right at that moment . well , it 's going to be this radius , which is this value . it 'll be v sub w times the period . that is that distance plus the amount of distance that our source has traveled away from this guy . so plus v sub s times the period . so that 's what this distance is , and this is how far apart this wave pulse is going to be from that one or this crest is going to be from that one . so if he 's seeing this first crest right now , right at this moment , how long will it take him to see the crest that 's being emitted right now , that 's this far away from him ? well , it 's that far away from him . so let me write this down . so the amount of time it takes for him to see the next crest or the same point in the next cycle , that 's the period . that 's the observed period . that 's going to be equal to this distance . the velocity of the wave times the period from the perspective of the source plus the velocity of the source , because the source has gotten that much further away from him , velocity of the source times the period of the source , so that 's how far the next crest is . and then you divide it by the speed of the wave , by the speed of each of the crests , which is just the velocity of the wave . and we can just factor out the t sub s 's here and say this is t sub s times v sub w , the velocity of the wave plus the velocity of the source divided by the velocity the wave . so this will be a larger observed period than if this guy was stationary and especially if the observer was in the path of the guy . that make sense , because every time this guy issues a cycle , he is moving a little bit further away . so every crest the same point in the cycle is going to be further and further apart , so you 're going to have longer wavelengths , longer periods . and if you want the observed frequency for that guy over there -- i 'll do it in the same color . the observed frequency for a guy where the source is traveling away from him is just the inverse of that . so one over the period , one over the period , same argument we did there , one over the period from the point of view of the source is the frequency of the source . let me color code it . so 1 over t sub s is equal to the frequency of the source . this is the inverse of that . so i 'm just taking one over everything here . so 1 over t sub s is the frequency of the source , and then we take the inverse of this over here : the velocity of the wave divided by the velocity of the wave plus the velocity of the source . so we 're done , at least for the simple cases . obviously , it becomes a little more interesting when someone is n't exactly in the direction of the source or exactly being moved away from , but these are kind of the two extreme cases . so this is the situation when it is moving away from you . now , just to check our math and maybe make it a little bit concrete in relation to the video we did where we introduced to the idea of the doppler effect , let 's actually apply those numbers . so on that video , two videos ago , we had a situation where the velocity of our source was 5 meters per second to the right , and the velocity of the wave was 10 meters per second radially outward , and the period of our wave -- i 'll call it the period of our -- let me do it in another color -- from the point of view of the source was 1 second per cycle , and the frequency was just the inverse of that . so 1 cycle per second , or 1 hertz , which is a cycle per second . so using those numbers , let 's see if we get to the exact same answer we got in that first video where we first learned about the doppler effect . so let 's look at the frequency from the point of view of this gentleman right here . so the frequency of the source is going to be 1 cycle per second , 1 hertz . the velocity of the wave is 10 meters per second . let me write this . 1 cycle per second . the velocity of the wave is 10 meters per second . the velocity of the wave is 10 meters per second minus the velocity of the source is 5 meters per second . so what 's this going to be equal to ? the observed frequency for this guy right there , is going to be 1 cycle per second times 10 over 10 minus 5 , and the meters per second cancel out . meters per second in the numerator , meters per second in the denominator . so 10 divided by 10 minus 5 , or 10 divided by 5 , is going to be 2 . so it 's going to be 2 cycles per second . and if you want the observed period for this guy , its going to be the inverse of that , or it 's going to be 1/2 second per cycle . and this is exactly what we got in the previous video , or actually , it was two videos ago . now what about the guy who this guy 's running away from ? well , we 'll do the exact same thing . you have 1 cycle per second , or 1 hertz . that 's the emitted frequency from the point of view of the source times the velocity of the wave divided by the velocity of the wave plus the velocity of the source , because it 's moving away from him . so it 's 10 over 10 plus 5 . that 's 10 over 15 . that is 2/3 , so this is equal to 2/3 . the units over here all cancel out . this was cycles per second . so 2/3 cycle per second , which confirms the numbers we got in that first video , so that should make us feel good and it also makes a lot of sense . this guy is going to see the wave crest more frequently . he 's going to observe a higher frequency . if this is a sound , a higher pitch . this guy , since each crest or the cycles are getting spread out , he 's going to see them less frequently , and if this is sound , he 's going to observe a lower pitch .
|
and we were able to do that by doing a thought experiment . saying , ok , my object starts here . after one period -- a period is just a measure of time , but it 's the measure of time over which the source emits a cycle , so it emits a cycle every period .
|
why does the frequency of an approaching object seem to be increasing gradually , and then decrease gradually ?
|
in the last video , we figured out the formulas for the observed period and frequency for an observer sitting in the path of the source . so the source is moving towards the observer . so this is the example where the train is moving towards you , and you perceive the train 's horn as having a higher pitch or a higher frequency . and we were able to do that by doing a thought experiment . saying , ok , my object starts here . after one period -- a period is just a measure of time , but it 's the measure of time over which the source emits a cycle , so it emits a cycle every period . but after one period , we said , ok , where is that first wave front , or that first pulse , or that first crest ? and where is the source ? because exactly one period has passed by , and the source will be ready to emit another crest or another cycle . so the distance between where the source is and that front of the crest , or that first crest , that is going to be the wavelength . because this next thing that emits is going to be traveling at the exact same velocity , and it 's going to be separated by that distance , which we saw is this expression . we said how long will it take it to travel that distance ? well , it 's traveling at a speed of v sub w. that 'll tell you what the observed period would be for this dude over here . we calculated it right here , and then the observed frequency is just the inverse of that . now let 's think about the situation where the observer is over here . so these equations , or these formulas that we came up with right here , this is observer -- or let me say source traveling in direction of observer . now let 's think about the opposite case , where the source is traveling away from the observer , and in this case , the observer is that guy over there . maybe i 'll do it in a different color . he 'll be blue . so this is the observer . so when we started off , our source was right here . after exactly one period from the source 's point of view , that first crest emitted has traveled radially outward that far . this is the distance . the velocity of the wave times the amount of time that passed -- velocity times time is going to give you distance -- and where the source of the wave will have traveled to the right exactly this distance . it 's velocity times the amount of time that 's gone by . now , in the last video , we said , ok , that wave is just passing this guy . how long will it take for that pulse that 's being emitted right then to also reach him ? and then that tells us the period between two pulses or between two crests . now let 's think about that exact same situation here . that first crest is just passing this guy . and a period , or the t sub s , which is a period of the emitted wave has just passed by . so this guy is just about to emit another wave , so ? that other wave is going to be right here . so what is the distance between the crest , or the cycle , or however you want to think about it , the pulse that is passing him by right now , what is the distance between that and the pulse , the front edge of that pulse that is being emitted right at that moment ? right at that moment . well , it 's going to be this radius , which is this value . it 'll be v sub w times the period . that is that distance plus the amount of distance that our source has traveled away from this guy . so plus v sub s times the period . so that 's what this distance is , and this is how far apart this wave pulse is going to be from that one or this crest is going to be from that one . so if he 's seeing this first crest right now , right at this moment , how long will it take him to see the crest that 's being emitted right now , that 's this far away from him ? well , it 's that far away from him . so let me write this down . so the amount of time it takes for him to see the next crest or the same point in the next cycle , that 's the period . that 's the observed period . that 's going to be equal to this distance . the velocity of the wave times the period from the perspective of the source plus the velocity of the source , because the source has gotten that much further away from him , velocity of the source times the period of the source , so that 's how far the next crest is . and then you divide it by the speed of the wave , by the speed of each of the crests , which is just the velocity of the wave . and we can just factor out the t sub s 's here and say this is t sub s times v sub w , the velocity of the wave plus the velocity of the source divided by the velocity the wave . so this will be a larger observed period than if this guy was stationary and especially if the observer was in the path of the guy . that make sense , because every time this guy issues a cycle , he is moving a little bit further away . so every crest the same point in the cycle is going to be further and further apart , so you 're going to have longer wavelengths , longer periods . and if you want the observed frequency for that guy over there -- i 'll do it in the same color . the observed frequency for a guy where the source is traveling away from him is just the inverse of that . so one over the period , one over the period , same argument we did there , one over the period from the point of view of the source is the frequency of the source . let me color code it . so 1 over t sub s is equal to the frequency of the source . this is the inverse of that . so i 'm just taking one over everything here . so 1 over t sub s is the frequency of the source , and then we take the inverse of this over here : the velocity of the wave divided by the velocity of the wave plus the velocity of the source . so we 're done , at least for the simple cases . obviously , it becomes a little more interesting when someone is n't exactly in the direction of the source or exactly being moved away from , but these are kind of the two extreme cases . so this is the situation when it is moving away from you . now , just to check our math and maybe make it a little bit concrete in relation to the video we did where we introduced to the idea of the doppler effect , let 's actually apply those numbers . so on that video , two videos ago , we had a situation where the velocity of our source was 5 meters per second to the right , and the velocity of the wave was 10 meters per second radially outward , and the period of our wave -- i 'll call it the period of our -- let me do it in another color -- from the point of view of the source was 1 second per cycle , and the frequency was just the inverse of that . so 1 cycle per second , or 1 hertz , which is a cycle per second . so using those numbers , let 's see if we get to the exact same answer we got in that first video where we first learned about the doppler effect . so let 's look at the frequency from the point of view of this gentleman right here . so the frequency of the source is going to be 1 cycle per second , 1 hertz . the velocity of the wave is 10 meters per second . let me write this . 1 cycle per second . the velocity of the wave is 10 meters per second . the velocity of the wave is 10 meters per second minus the velocity of the source is 5 meters per second . so what 's this going to be equal to ? the observed frequency for this guy right there , is going to be 1 cycle per second times 10 over 10 minus 5 , and the meters per second cancel out . meters per second in the numerator , meters per second in the denominator . so 10 divided by 10 minus 5 , or 10 divided by 5 , is going to be 2 . so it 's going to be 2 cycles per second . and if you want the observed period for this guy , its going to be the inverse of that , or it 's going to be 1/2 second per cycle . and this is exactly what we got in the previous video , or actually , it was two videos ago . now what about the guy who this guy 's running away from ? well , we 'll do the exact same thing . you have 1 cycle per second , or 1 hertz . that 's the emitted frequency from the point of view of the source times the velocity of the wave divided by the velocity of the wave plus the velocity of the source , because it 's moving away from him . so it 's 10 over 10 plus 5 . that 's 10 over 15 . that is 2/3 , so this is equal to 2/3 . the units over here all cancel out . this was cycles per second . so 2/3 cycle per second , which confirms the numbers we got in that first video , so that should make us feel good and it also makes a lot of sense . this guy is going to see the wave crest more frequently . he 's going to observe a higher frequency . if this is a sound , a higher pitch . this guy , since each crest or the cycles are getting spread out , he 's going to see them less frequently , and if this is sound , he 's going to observe a lower pitch .
|
in the last video , we figured out the formulas for the observed period and frequency for an observer sitting in the path of the source . so the source is moving towards the observer . so this is the example where the train is moving towards you , and you perceive the train 's horn as having a higher pitch or a higher frequency .
|
why is the doppler effect of sound different in cases in which observer is moving and in which source is moving ?
|
in the last video , we figured out the formulas for the observed period and frequency for an observer sitting in the path of the source . so the source is moving towards the observer . so this is the example where the train is moving towards you , and you perceive the train 's horn as having a higher pitch or a higher frequency . and we were able to do that by doing a thought experiment . saying , ok , my object starts here . after one period -- a period is just a measure of time , but it 's the measure of time over which the source emits a cycle , so it emits a cycle every period . but after one period , we said , ok , where is that first wave front , or that first pulse , or that first crest ? and where is the source ? because exactly one period has passed by , and the source will be ready to emit another crest or another cycle . so the distance between where the source is and that front of the crest , or that first crest , that is going to be the wavelength . because this next thing that emits is going to be traveling at the exact same velocity , and it 's going to be separated by that distance , which we saw is this expression . we said how long will it take it to travel that distance ? well , it 's traveling at a speed of v sub w. that 'll tell you what the observed period would be for this dude over here . we calculated it right here , and then the observed frequency is just the inverse of that . now let 's think about the situation where the observer is over here . so these equations , or these formulas that we came up with right here , this is observer -- or let me say source traveling in direction of observer . now let 's think about the opposite case , where the source is traveling away from the observer , and in this case , the observer is that guy over there . maybe i 'll do it in a different color . he 'll be blue . so this is the observer . so when we started off , our source was right here . after exactly one period from the source 's point of view , that first crest emitted has traveled radially outward that far . this is the distance . the velocity of the wave times the amount of time that passed -- velocity times time is going to give you distance -- and where the source of the wave will have traveled to the right exactly this distance . it 's velocity times the amount of time that 's gone by . now , in the last video , we said , ok , that wave is just passing this guy . how long will it take for that pulse that 's being emitted right then to also reach him ? and then that tells us the period between two pulses or between two crests . now let 's think about that exact same situation here . that first crest is just passing this guy . and a period , or the t sub s , which is a period of the emitted wave has just passed by . so this guy is just about to emit another wave , so ? that other wave is going to be right here . so what is the distance between the crest , or the cycle , or however you want to think about it , the pulse that is passing him by right now , what is the distance between that and the pulse , the front edge of that pulse that is being emitted right at that moment ? right at that moment . well , it 's going to be this radius , which is this value . it 'll be v sub w times the period . that is that distance plus the amount of distance that our source has traveled away from this guy . so plus v sub s times the period . so that 's what this distance is , and this is how far apart this wave pulse is going to be from that one or this crest is going to be from that one . so if he 's seeing this first crest right now , right at this moment , how long will it take him to see the crest that 's being emitted right now , that 's this far away from him ? well , it 's that far away from him . so let me write this down . so the amount of time it takes for him to see the next crest or the same point in the next cycle , that 's the period . that 's the observed period . that 's going to be equal to this distance . the velocity of the wave times the period from the perspective of the source plus the velocity of the source , because the source has gotten that much further away from him , velocity of the source times the period of the source , so that 's how far the next crest is . and then you divide it by the speed of the wave , by the speed of each of the crests , which is just the velocity of the wave . and we can just factor out the t sub s 's here and say this is t sub s times v sub w , the velocity of the wave plus the velocity of the source divided by the velocity the wave . so this will be a larger observed period than if this guy was stationary and especially if the observer was in the path of the guy . that make sense , because every time this guy issues a cycle , he is moving a little bit further away . so every crest the same point in the cycle is going to be further and further apart , so you 're going to have longer wavelengths , longer periods . and if you want the observed frequency for that guy over there -- i 'll do it in the same color . the observed frequency for a guy where the source is traveling away from him is just the inverse of that . so one over the period , one over the period , same argument we did there , one over the period from the point of view of the source is the frequency of the source . let me color code it . so 1 over t sub s is equal to the frequency of the source . this is the inverse of that . so i 'm just taking one over everything here . so 1 over t sub s is the frequency of the source , and then we take the inverse of this over here : the velocity of the wave divided by the velocity of the wave plus the velocity of the source . so we 're done , at least for the simple cases . obviously , it becomes a little more interesting when someone is n't exactly in the direction of the source or exactly being moved away from , but these are kind of the two extreme cases . so this is the situation when it is moving away from you . now , just to check our math and maybe make it a little bit concrete in relation to the video we did where we introduced to the idea of the doppler effect , let 's actually apply those numbers . so on that video , two videos ago , we had a situation where the velocity of our source was 5 meters per second to the right , and the velocity of the wave was 10 meters per second radially outward , and the period of our wave -- i 'll call it the period of our -- let me do it in another color -- from the point of view of the source was 1 second per cycle , and the frequency was just the inverse of that . so 1 cycle per second , or 1 hertz , which is a cycle per second . so using those numbers , let 's see if we get to the exact same answer we got in that first video where we first learned about the doppler effect . so let 's look at the frequency from the point of view of this gentleman right here . so the frequency of the source is going to be 1 cycle per second , 1 hertz . the velocity of the wave is 10 meters per second . let me write this . 1 cycle per second . the velocity of the wave is 10 meters per second . the velocity of the wave is 10 meters per second minus the velocity of the source is 5 meters per second . so what 's this going to be equal to ? the observed frequency for this guy right there , is going to be 1 cycle per second times 10 over 10 minus 5 , and the meters per second cancel out . meters per second in the numerator , meters per second in the denominator . so 10 divided by 10 minus 5 , or 10 divided by 5 , is going to be 2 . so it 's going to be 2 cycles per second . and if you want the observed period for this guy , its going to be the inverse of that , or it 's going to be 1/2 second per cycle . and this is exactly what we got in the previous video , or actually , it was two videos ago . now what about the guy who this guy 's running away from ? well , we 'll do the exact same thing . you have 1 cycle per second , or 1 hertz . that 's the emitted frequency from the point of view of the source times the velocity of the wave divided by the velocity of the wave plus the velocity of the source , because it 's moving away from him . so it 's 10 over 10 plus 5 . that 's 10 over 15 . that is 2/3 , so this is equal to 2/3 . the units over here all cancel out . this was cycles per second . so 2/3 cycle per second , which confirms the numbers we got in that first video , so that should make us feel good and it also makes a lot of sense . this guy is going to see the wave crest more frequently . he 's going to observe a higher frequency . if this is a sound , a higher pitch . this guy , since each crest or the cycles are getting spread out , he 's going to see them less frequently , and if this is sound , he 's going to observe a lower pitch .
|
that 's going to be equal to this distance . the velocity of the wave times the period from the perspective of the source plus the velocity of the source , because the source has gotten that much further away from him , velocity of the source times the period of the source , so that 's how far the next crest is . and then you divide it by the speed of the wave , by the speed of each of the crests , which is just the velocity of the wave .
|
we can predict the doppler effect if we are straight in the path of the wave source , but what about when we are at an angle to the direction of the velocity of the wave source ?
|
in the last video , we figured out the formulas for the observed period and frequency for an observer sitting in the path of the source . so the source is moving towards the observer . so this is the example where the train is moving towards you , and you perceive the train 's horn as having a higher pitch or a higher frequency . and we were able to do that by doing a thought experiment . saying , ok , my object starts here . after one period -- a period is just a measure of time , but it 's the measure of time over which the source emits a cycle , so it emits a cycle every period . but after one period , we said , ok , where is that first wave front , or that first pulse , or that first crest ? and where is the source ? because exactly one period has passed by , and the source will be ready to emit another crest or another cycle . so the distance between where the source is and that front of the crest , or that first crest , that is going to be the wavelength . because this next thing that emits is going to be traveling at the exact same velocity , and it 's going to be separated by that distance , which we saw is this expression . we said how long will it take it to travel that distance ? well , it 's traveling at a speed of v sub w. that 'll tell you what the observed period would be for this dude over here . we calculated it right here , and then the observed frequency is just the inverse of that . now let 's think about the situation where the observer is over here . so these equations , or these formulas that we came up with right here , this is observer -- or let me say source traveling in direction of observer . now let 's think about the opposite case , where the source is traveling away from the observer , and in this case , the observer is that guy over there . maybe i 'll do it in a different color . he 'll be blue . so this is the observer . so when we started off , our source was right here . after exactly one period from the source 's point of view , that first crest emitted has traveled radially outward that far . this is the distance . the velocity of the wave times the amount of time that passed -- velocity times time is going to give you distance -- and where the source of the wave will have traveled to the right exactly this distance . it 's velocity times the amount of time that 's gone by . now , in the last video , we said , ok , that wave is just passing this guy . how long will it take for that pulse that 's being emitted right then to also reach him ? and then that tells us the period between two pulses or between two crests . now let 's think about that exact same situation here . that first crest is just passing this guy . and a period , or the t sub s , which is a period of the emitted wave has just passed by . so this guy is just about to emit another wave , so ? that other wave is going to be right here . so what is the distance between the crest , or the cycle , or however you want to think about it , the pulse that is passing him by right now , what is the distance between that and the pulse , the front edge of that pulse that is being emitted right at that moment ? right at that moment . well , it 's going to be this radius , which is this value . it 'll be v sub w times the period . that is that distance plus the amount of distance that our source has traveled away from this guy . so plus v sub s times the period . so that 's what this distance is , and this is how far apart this wave pulse is going to be from that one or this crest is going to be from that one . so if he 's seeing this first crest right now , right at this moment , how long will it take him to see the crest that 's being emitted right now , that 's this far away from him ? well , it 's that far away from him . so let me write this down . so the amount of time it takes for him to see the next crest or the same point in the next cycle , that 's the period . that 's the observed period . that 's going to be equal to this distance . the velocity of the wave times the period from the perspective of the source plus the velocity of the source , because the source has gotten that much further away from him , velocity of the source times the period of the source , so that 's how far the next crest is . and then you divide it by the speed of the wave , by the speed of each of the crests , which is just the velocity of the wave . and we can just factor out the t sub s 's here and say this is t sub s times v sub w , the velocity of the wave plus the velocity of the source divided by the velocity the wave . so this will be a larger observed period than if this guy was stationary and especially if the observer was in the path of the guy . that make sense , because every time this guy issues a cycle , he is moving a little bit further away . so every crest the same point in the cycle is going to be further and further apart , so you 're going to have longer wavelengths , longer periods . and if you want the observed frequency for that guy over there -- i 'll do it in the same color . the observed frequency for a guy where the source is traveling away from him is just the inverse of that . so one over the period , one over the period , same argument we did there , one over the period from the point of view of the source is the frequency of the source . let me color code it . so 1 over t sub s is equal to the frequency of the source . this is the inverse of that . so i 'm just taking one over everything here . so 1 over t sub s is the frequency of the source , and then we take the inverse of this over here : the velocity of the wave divided by the velocity of the wave plus the velocity of the source . so we 're done , at least for the simple cases . obviously , it becomes a little more interesting when someone is n't exactly in the direction of the source or exactly being moved away from , but these are kind of the two extreme cases . so this is the situation when it is moving away from you . now , just to check our math and maybe make it a little bit concrete in relation to the video we did where we introduced to the idea of the doppler effect , let 's actually apply those numbers . so on that video , two videos ago , we had a situation where the velocity of our source was 5 meters per second to the right , and the velocity of the wave was 10 meters per second radially outward , and the period of our wave -- i 'll call it the period of our -- let me do it in another color -- from the point of view of the source was 1 second per cycle , and the frequency was just the inverse of that . so 1 cycle per second , or 1 hertz , which is a cycle per second . so using those numbers , let 's see if we get to the exact same answer we got in that first video where we first learned about the doppler effect . so let 's look at the frequency from the point of view of this gentleman right here . so the frequency of the source is going to be 1 cycle per second , 1 hertz . the velocity of the wave is 10 meters per second . let me write this . 1 cycle per second . the velocity of the wave is 10 meters per second . the velocity of the wave is 10 meters per second minus the velocity of the source is 5 meters per second . so what 's this going to be equal to ? the observed frequency for this guy right there , is going to be 1 cycle per second times 10 over 10 minus 5 , and the meters per second cancel out . meters per second in the numerator , meters per second in the denominator . so 10 divided by 10 minus 5 , or 10 divided by 5 , is going to be 2 . so it 's going to be 2 cycles per second . and if you want the observed period for this guy , its going to be the inverse of that , or it 's going to be 1/2 second per cycle . and this is exactly what we got in the previous video , or actually , it was two videos ago . now what about the guy who this guy 's running away from ? well , we 'll do the exact same thing . you have 1 cycle per second , or 1 hertz . that 's the emitted frequency from the point of view of the source times the velocity of the wave divided by the velocity of the wave plus the velocity of the source , because it 's moving away from him . so it 's 10 over 10 plus 5 . that 's 10 over 15 . that is 2/3 , so this is equal to 2/3 . the units over here all cancel out . this was cycles per second . so 2/3 cycle per second , which confirms the numbers we got in that first video , so that should make us feel good and it also makes a lot of sense . this guy is going to see the wave crest more frequently . he 's going to observe a higher frequency . if this is a sound , a higher pitch . this guy , since each crest or the cycles are getting spread out , he 's going to see them less frequently , and if this is sound , he 's going to observe a lower pitch .
|
so the frequency of the source is going to be 1 cycle per second , 1 hertz . the velocity of the wave is 10 meters per second . let me write this .
|
is velocity of the wave the same thing as velocity of the obsever ?
|
in the last video , we figured out the formulas for the observed period and frequency for an observer sitting in the path of the source . so the source is moving towards the observer . so this is the example where the train is moving towards you , and you perceive the train 's horn as having a higher pitch or a higher frequency . and we were able to do that by doing a thought experiment . saying , ok , my object starts here . after one period -- a period is just a measure of time , but it 's the measure of time over which the source emits a cycle , so it emits a cycle every period . but after one period , we said , ok , where is that first wave front , or that first pulse , or that first crest ? and where is the source ? because exactly one period has passed by , and the source will be ready to emit another crest or another cycle . so the distance between where the source is and that front of the crest , or that first crest , that is going to be the wavelength . because this next thing that emits is going to be traveling at the exact same velocity , and it 's going to be separated by that distance , which we saw is this expression . we said how long will it take it to travel that distance ? well , it 's traveling at a speed of v sub w. that 'll tell you what the observed period would be for this dude over here . we calculated it right here , and then the observed frequency is just the inverse of that . now let 's think about the situation where the observer is over here . so these equations , or these formulas that we came up with right here , this is observer -- or let me say source traveling in direction of observer . now let 's think about the opposite case , where the source is traveling away from the observer , and in this case , the observer is that guy over there . maybe i 'll do it in a different color . he 'll be blue . so this is the observer . so when we started off , our source was right here . after exactly one period from the source 's point of view , that first crest emitted has traveled radially outward that far . this is the distance . the velocity of the wave times the amount of time that passed -- velocity times time is going to give you distance -- and where the source of the wave will have traveled to the right exactly this distance . it 's velocity times the amount of time that 's gone by . now , in the last video , we said , ok , that wave is just passing this guy . how long will it take for that pulse that 's being emitted right then to also reach him ? and then that tells us the period between two pulses or between two crests . now let 's think about that exact same situation here . that first crest is just passing this guy . and a period , or the t sub s , which is a period of the emitted wave has just passed by . so this guy is just about to emit another wave , so ? that other wave is going to be right here . so what is the distance between the crest , or the cycle , or however you want to think about it , the pulse that is passing him by right now , what is the distance between that and the pulse , the front edge of that pulse that is being emitted right at that moment ? right at that moment . well , it 's going to be this radius , which is this value . it 'll be v sub w times the period . that is that distance plus the amount of distance that our source has traveled away from this guy . so plus v sub s times the period . so that 's what this distance is , and this is how far apart this wave pulse is going to be from that one or this crest is going to be from that one . so if he 's seeing this first crest right now , right at this moment , how long will it take him to see the crest that 's being emitted right now , that 's this far away from him ? well , it 's that far away from him . so let me write this down . so the amount of time it takes for him to see the next crest or the same point in the next cycle , that 's the period . that 's the observed period . that 's going to be equal to this distance . the velocity of the wave times the period from the perspective of the source plus the velocity of the source , because the source has gotten that much further away from him , velocity of the source times the period of the source , so that 's how far the next crest is . and then you divide it by the speed of the wave , by the speed of each of the crests , which is just the velocity of the wave . and we can just factor out the t sub s 's here and say this is t sub s times v sub w , the velocity of the wave plus the velocity of the source divided by the velocity the wave . so this will be a larger observed period than if this guy was stationary and especially if the observer was in the path of the guy . that make sense , because every time this guy issues a cycle , he is moving a little bit further away . so every crest the same point in the cycle is going to be further and further apart , so you 're going to have longer wavelengths , longer periods . and if you want the observed frequency for that guy over there -- i 'll do it in the same color . the observed frequency for a guy where the source is traveling away from him is just the inverse of that . so one over the period , one over the period , same argument we did there , one over the period from the point of view of the source is the frequency of the source . let me color code it . so 1 over t sub s is equal to the frequency of the source . this is the inverse of that . so i 'm just taking one over everything here . so 1 over t sub s is the frequency of the source , and then we take the inverse of this over here : the velocity of the wave divided by the velocity of the wave plus the velocity of the source . so we 're done , at least for the simple cases . obviously , it becomes a little more interesting when someone is n't exactly in the direction of the source or exactly being moved away from , but these are kind of the two extreme cases . so this is the situation when it is moving away from you . now , just to check our math and maybe make it a little bit concrete in relation to the video we did where we introduced to the idea of the doppler effect , let 's actually apply those numbers . so on that video , two videos ago , we had a situation where the velocity of our source was 5 meters per second to the right , and the velocity of the wave was 10 meters per second radially outward , and the period of our wave -- i 'll call it the period of our -- let me do it in another color -- from the point of view of the source was 1 second per cycle , and the frequency was just the inverse of that . so 1 cycle per second , or 1 hertz , which is a cycle per second . so using those numbers , let 's see if we get to the exact same answer we got in that first video where we first learned about the doppler effect . so let 's look at the frequency from the point of view of this gentleman right here . so the frequency of the source is going to be 1 cycle per second , 1 hertz . the velocity of the wave is 10 meters per second . let me write this . 1 cycle per second . the velocity of the wave is 10 meters per second . the velocity of the wave is 10 meters per second minus the velocity of the source is 5 meters per second . so what 's this going to be equal to ? the observed frequency for this guy right there , is going to be 1 cycle per second times 10 over 10 minus 5 , and the meters per second cancel out . meters per second in the numerator , meters per second in the denominator . so 10 divided by 10 minus 5 , or 10 divided by 5 , is going to be 2 . so it 's going to be 2 cycles per second . and if you want the observed period for this guy , its going to be the inverse of that , or it 's going to be 1/2 second per cycle . and this is exactly what we got in the previous video , or actually , it was two videos ago . now what about the guy who this guy 's running away from ? well , we 'll do the exact same thing . you have 1 cycle per second , or 1 hertz . that 's the emitted frequency from the point of view of the source times the velocity of the wave divided by the velocity of the wave plus the velocity of the source , because it 's moving away from him . so it 's 10 over 10 plus 5 . that 's 10 over 15 . that is 2/3 , so this is equal to 2/3 . the units over here all cancel out . this was cycles per second . so 2/3 cycle per second , which confirms the numbers we got in that first video , so that should make us feel good and it also makes a lot of sense . this guy is going to see the wave crest more frequently . he 's going to observe a higher frequency . if this is a sound , a higher pitch . this guy , since each crest or the cycles are getting spread out , he 's going to see them less frequently , and if this is sound , he 's going to observe a lower pitch .
|
in the last video , we figured out the formulas for the observed period and frequency for an observer sitting in the path of the source . so the source is moving towards the observer . so this is the example where the train is moving towards you , and you perceive the train 's horn as having a higher pitch or a higher frequency .
|
is n't the source constantly moving ?
|
in the last video , we figured out the formulas for the observed period and frequency for an observer sitting in the path of the source . so the source is moving towards the observer . so this is the example where the train is moving towards you , and you perceive the train 's horn as having a higher pitch or a higher frequency . and we were able to do that by doing a thought experiment . saying , ok , my object starts here . after one period -- a period is just a measure of time , but it 's the measure of time over which the source emits a cycle , so it emits a cycle every period . but after one period , we said , ok , where is that first wave front , or that first pulse , or that first crest ? and where is the source ? because exactly one period has passed by , and the source will be ready to emit another crest or another cycle . so the distance between where the source is and that front of the crest , or that first crest , that is going to be the wavelength . because this next thing that emits is going to be traveling at the exact same velocity , and it 's going to be separated by that distance , which we saw is this expression . we said how long will it take it to travel that distance ? well , it 's traveling at a speed of v sub w. that 'll tell you what the observed period would be for this dude over here . we calculated it right here , and then the observed frequency is just the inverse of that . now let 's think about the situation where the observer is over here . so these equations , or these formulas that we came up with right here , this is observer -- or let me say source traveling in direction of observer . now let 's think about the opposite case , where the source is traveling away from the observer , and in this case , the observer is that guy over there . maybe i 'll do it in a different color . he 'll be blue . so this is the observer . so when we started off , our source was right here . after exactly one period from the source 's point of view , that first crest emitted has traveled radially outward that far . this is the distance . the velocity of the wave times the amount of time that passed -- velocity times time is going to give you distance -- and where the source of the wave will have traveled to the right exactly this distance . it 's velocity times the amount of time that 's gone by . now , in the last video , we said , ok , that wave is just passing this guy . how long will it take for that pulse that 's being emitted right then to also reach him ? and then that tells us the period between two pulses or between two crests . now let 's think about that exact same situation here . that first crest is just passing this guy . and a period , or the t sub s , which is a period of the emitted wave has just passed by . so this guy is just about to emit another wave , so ? that other wave is going to be right here . so what is the distance between the crest , or the cycle , or however you want to think about it , the pulse that is passing him by right now , what is the distance between that and the pulse , the front edge of that pulse that is being emitted right at that moment ? right at that moment . well , it 's going to be this radius , which is this value . it 'll be v sub w times the period . that is that distance plus the amount of distance that our source has traveled away from this guy . so plus v sub s times the period . so that 's what this distance is , and this is how far apart this wave pulse is going to be from that one or this crest is going to be from that one . so if he 's seeing this first crest right now , right at this moment , how long will it take him to see the crest that 's being emitted right now , that 's this far away from him ? well , it 's that far away from him . so let me write this down . so the amount of time it takes for him to see the next crest or the same point in the next cycle , that 's the period . that 's the observed period . that 's going to be equal to this distance . the velocity of the wave times the period from the perspective of the source plus the velocity of the source , because the source has gotten that much further away from him , velocity of the source times the period of the source , so that 's how far the next crest is . and then you divide it by the speed of the wave , by the speed of each of the crests , which is just the velocity of the wave . and we can just factor out the t sub s 's here and say this is t sub s times v sub w , the velocity of the wave plus the velocity of the source divided by the velocity the wave . so this will be a larger observed period than if this guy was stationary and especially if the observer was in the path of the guy . that make sense , because every time this guy issues a cycle , he is moving a little bit further away . so every crest the same point in the cycle is going to be further and further apart , so you 're going to have longer wavelengths , longer periods . and if you want the observed frequency for that guy over there -- i 'll do it in the same color . the observed frequency for a guy where the source is traveling away from him is just the inverse of that . so one over the period , one over the period , same argument we did there , one over the period from the point of view of the source is the frequency of the source . let me color code it . so 1 over t sub s is equal to the frequency of the source . this is the inverse of that . so i 'm just taking one over everything here . so 1 over t sub s is the frequency of the source , and then we take the inverse of this over here : the velocity of the wave divided by the velocity of the wave plus the velocity of the source . so we 're done , at least for the simple cases . obviously , it becomes a little more interesting when someone is n't exactly in the direction of the source or exactly being moved away from , but these are kind of the two extreme cases . so this is the situation when it is moving away from you . now , just to check our math and maybe make it a little bit concrete in relation to the video we did where we introduced to the idea of the doppler effect , let 's actually apply those numbers . so on that video , two videos ago , we had a situation where the velocity of our source was 5 meters per second to the right , and the velocity of the wave was 10 meters per second radially outward , and the period of our wave -- i 'll call it the period of our -- let me do it in another color -- from the point of view of the source was 1 second per cycle , and the frequency was just the inverse of that . so 1 cycle per second , or 1 hertz , which is a cycle per second . so using those numbers , let 's see if we get to the exact same answer we got in that first video where we first learned about the doppler effect . so let 's look at the frequency from the point of view of this gentleman right here . so the frequency of the source is going to be 1 cycle per second , 1 hertz . the velocity of the wave is 10 meters per second . let me write this . 1 cycle per second . the velocity of the wave is 10 meters per second . the velocity of the wave is 10 meters per second minus the velocity of the source is 5 meters per second . so what 's this going to be equal to ? the observed frequency for this guy right there , is going to be 1 cycle per second times 10 over 10 minus 5 , and the meters per second cancel out . meters per second in the numerator , meters per second in the denominator . so 10 divided by 10 minus 5 , or 10 divided by 5 , is going to be 2 . so it 's going to be 2 cycles per second . and if you want the observed period for this guy , its going to be the inverse of that , or it 's going to be 1/2 second per cycle . and this is exactly what we got in the previous video , or actually , it was two videos ago . now what about the guy who this guy 's running away from ? well , we 'll do the exact same thing . you have 1 cycle per second , or 1 hertz . that 's the emitted frequency from the point of view of the source times the velocity of the wave divided by the velocity of the wave plus the velocity of the source , because it 's moving away from him . so it 's 10 over 10 plus 5 . that 's 10 over 15 . that is 2/3 , so this is equal to 2/3 . the units over here all cancel out . this was cycles per second . so 2/3 cycle per second , which confirms the numbers we got in that first video , so that should make us feel good and it also makes a lot of sense . this guy is going to see the wave crest more frequently . he 's going to observe a higher frequency . if this is a sound , a higher pitch . this guy , since each crest or the cycles are getting spread out , he 's going to see them less frequently , and if this is sound , he 's going to observe a lower pitch .
|
in the last video , we figured out the formulas for the observed period and frequency for an observer sitting in the path of the source . so the source is moving towards the observer . so this is the example where the train is moving towards you , and you perceive the train 's horn as having a higher pitch or a higher frequency .
|
why is the apparent frequency greater than the actual frequency in case the source is moving towards the stationary object ?
|
in the last video , we figured out the formulas for the observed period and frequency for an observer sitting in the path of the source . so the source is moving towards the observer . so this is the example where the train is moving towards you , and you perceive the train 's horn as having a higher pitch or a higher frequency . and we were able to do that by doing a thought experiment . saying , ok , my object starts here . after one period -- a period is just a measure of time , but it 's the measure of time over which the source emits a cycle , so it emits a cycle every period . but after one period , we said , ok , where is that first wave front , or that first pulse , or that first crest ? and where is the source ? because exactly one period has passed by , and the source will be ready to emit another crest or another cycle . so the distance between where the source is and that front of the crest , or that first crest , that is going to be the wavelength . because this next thing that emits is going to be traveling at the exact same velocity , and it 's going to be separated by that distance , which we saw is this expression . we said how long will it take it to travel that distance ? well , it 's traveling at a speed of v sub w. that 'll tell you what the observed period would be for this dude over here . we calculated it right here , and then the observed frequency is just the inverse of that . now let 's think about the situation where the observer is over here . so these equations , or these formulas that we came up with right here , this is observer -- or let me say source traveling in direction of observer . now let 's think about the opposite case , where the source is traveling away from the observer , and in this case , the observer is that guy over there . maybe i 'll do it in a different color . he 'll be blue . so this is the observer . so when we started off , our source was right here . after exactly one period from the source 's point of view , that first crest emitted has traveled radially outward that far . this is the distance . the velocity of the wave times the amount of time that passed -- velocity times time is going to give you distance -- and where the source of the wave will have traveled to the right exactly this distance . it 's velocity times the amount of time that 's gone by . now , in the last video , we said , ok , that wave is just passing this guy . how long will it take for that pulse that 's being emitted right then to also reach him ? and then that tells us the period between two pulses or between two crests . now let 's think about that exact same situation here . that first crest is just passing this guy . and a period , or the t sub s , which is a period of the emitted wave has just passed by . so this guy is just about to emit another wave , so ? that other wave is going to be right here . so what is the distance between the crest , or the cycle , or however you want to think about it , the pulse that is passing him by right now , what is the distance between that and the pulse , the front edge of that pulse that is being emitted right at that moment ? right at that moment . well , it 's going to be this radius , which is this value . it 'll be v sub w times the period . that is that distance plus the amount of distance that our source has traveled away from this guy . so plus v sub s times the period . so that 's what this distance is , and this is how far apart this wave pulse is going to be from that one or this crest is going to be from that one . so if he 's seeing this first crest right now , right at this moment , how long will it take him to see the crest that 's being emitted right now , that 's this far away from him ? well , it 's that far away from him . so let me write this down . so the amount of time it takes for him to see the next crest or the same point in the next cycle , that 's the period . that 's the observed period . that 's going to be equal to this distance . the velocity of the wave times the period from the perspective of the source plus the velocity of the source , because the source has gotten that much further away from him , velocity of the source times the period of the source , so that 's how far the next crest is . and then you divide it by the speed of the wave , by the speed of each of the crests , which is just the velocity of the wave . and we can just factor out the t sub s 's here and say this is t sub s times v sub w , the velocity of the wave plus the velocity of the source divided by the velocity the wave . so this will be a larger observed period than if this guy was stationary and especially if the observer was in the path of the guy . that make sense , because every time this guy issues a cycle , he is moving a little bit further away . so every crest the same point in the cycle is going to be further and further apart , so you 're going to have longer wavelengths , longer periods . and if you want the observed frequency for that guy over there -- i 'll do it in the same color . the observed frequency for a guy where the source is traveling away from him is just the inverse of that . so one over the period , one over the period , same argument we did there , one over the period from the point of view of the source is the frequency of the source . let me color code it . so 1 over t sub s is equal to the frequency of the source . this is the inverse of that . so i 'm just taking one over everything here . so 1 over t sub s is the frequency of the source , and then we take the inverse of this over here : the velocity of the wave divided by the velocity of the wave plus the velocity of the source . so we 're done , at least for the simple cases . obviously , it becomes a little more interesting when someone is n't exactly in the direction of the source or exactly being moved away from , but these are kind of the two extreme cases . so this is the situation when it is moving away from you . now , just to check our math and maybe make it a little bit concrete in relation to the video we did where we introduced to the idea of the doppler effect , let 's actually apply those numbers . so on that video , two videos ago , we had a situation where the velocity of our source was 5 meters per second to the right , and the velocity of the wave was 10 meters per second radially outward , and the period of our wave -- i 'll call it the period of our -- let me do it in another color -- from the point of view of the source was 1 second per cycle , and the frequency was just the inverse of that . so 1 cycle per second , or 1 hertz , which is a cycle per second . so using those numbers , let 's see if we get to the exact same answer we got in that first video where we first learned about the doppler effect . so let 's look at the frequency from the point of view of this gentleman right here . so the frequency of the source is going to be 1 cycle per second , 1 hertz . the velocity of the wave is 10 meters per second . let me write this . 1 cycle per second . the velocity of the wave is 10 meters per second . the velocity of the wave is 10 meters per second minus the velocity of the source is 5 meters per second . so what 's this going to be equal to ? the observed frequency for this guy right there , is going to be 1 cycle per second times 10 over 10 minus 5 , and the meters per second cancel out . meters per second in the numerator , meters per second in the denominator . so 10 divided by 10 minus 5 , or 10 divided by 5 , is going to be 2 . so it 's going to be 2 cycles per second . and if you want the observed period for this guy , its going to be the inverse of that , or it 's going to be 1/2 second per cycle . and this is exactly what we got in the previous video , or actually , it was two videos ago . now what about the guy who this guy 's running away from ? well , we 'll do the exact same thing . you have 1 cycle per second , or 1 hertz . that 's the emitted frequency from the point of view of the source times the velocity of the wave divided by the velocity of the wave plus the velocity of the source , because it 's moving away from him . so it 's 10 over 10 plus 5 . that 's 10 over 15 . that is 2/3 , so this is equal to 2/3 . the units over here all cancel out . this was cycles per second . so 2/3 cycle per second , which confirms the numbers we got in that first video , so that should make us feel good and it also makes a lot of sense . this guy is going to see the wave crest more frequently . he 's going to observe a higher frequency . if this is a sound , a higher pitch . this guy , since each crest or the cycles are getting spread out , he 's going to see them less frequently , and if this is sound , he 's going to observe a lower pitch .
|
in the last video , we figured out the formulas for the observed period and frequency for an observer sitting in the path of the source . so the source is moving towards the observer . so this is the example where the train is moving towards you , and you perceive the train 's horn as having a higher pitch or a higher frequency .
|
will the same formula hold true if the observer is not exactly in the path of source but sideways ?
|
in the last video , we figured out the formulas for the observed period and frequency for an observer sitting in the path of the source . so the source is moving towards the observer . so this is the example where the train is moving towards you , and you perceive the train 's horn as having a higher pitch or a higher frequency . and we were able to do that by doing a thought experiment . saying , ok , my object starts here . after one period -- a period is just a measure of time , but it 's the measure of time over which the source emits a cycle , so it emits a cycle every period . but after one period , we said , ok , where is that first wave front , or that first pulse , or that first crest ? and where is the source ? because exactly one period has passed by , and the source will be ready to emit another crest or another cycle . so the distance between where the source is and that front of the crest , or that first crest , that is going to be the wavelength . because this next thing that emits is going to be traveling at the exact same velocity , and it 's going to be separated by that distance , which we saw is this expression . we said how long will it take it to travel that distance ? well , it 's traveling at a speed of v sub w. that 'll tell you what the observed period would be for this dude over here . we calculated it right here , and then the observed frequency is just the inverse of that . now let 's think about the situation where the observer is over here . so these equations , or these formulas that we came up with right here , this is observer -- or let me say source traveling in direction of observer . now let 's think about the opposite case , where the source is traveling away from the observer , and in this case , the observer is that guy over there . maybe i 'll do it in a different color . he 'll be blue . so this is the observer . so when we started off , our source was right here . after exactly one period from the source 's point of view , that first crest emitted has traveled radially outward that far . this is the distance . the velocity of the wave times the amount of time that passed -- velocity times time is going to give you distance -- and where the source of the wave will have traveled to the right exactly this distance . it 's velocity times the amount of time that 's gone by . now , in the last video , we said , ok , that wave is just passing this guy . how long will it take for that pulse that 's being emitted right then to also reach him ? and then that tells us the period between two pulses or between two crests . now let 's think about that exact same situation here . that first crest is just passing this guy . and a period , or the t sub s , which is a period of the emitted wave has just passed by . so this guy is just about to emit another wave , so ? that other wave is going to be right here . so what is the distance between the crest , or the cycle , or however you want to think about it , the pulse that is passing him by right now , what is the distance between that and the pulse , the front edge of that pulse that is being emitted right at that moment ? right at that moment . well , it 's going to be this radius , which is this value . it 'll be v sub w times the period . that is that distance plus the amount of distance that our source has traveled away from this guy . so plus v sub s times the period . so that 's what this distance is , and this is how far apart this wave pulse is going to be from that one or this crest is going to be from that one . so if he 's seeing this first crest right now , right at this moment , how long will it take him to see the crest that 's being emitted right now , that 's this far away from him ? well , it 's that far away from him . so let me write this down . so the amount of time it takes for him to see the next crest or the same point in the next cycle , that 's the period . that 's the observed period . that 's going to be equal to this distance . the velocity of the wave times the period from the perspective of the source plus the velocity of the source , because the source has gotten that much further away from him , velocity of the source times the period of the source , so that 's how far the next crest is . and then you divide it by the speed of the wave , by the speed of each of the crests , which is just the velocity of the wave . and we can just factor out the t sub s 's here and say this is t sub s times v sub w , the velocity of the wave plus the velocity of the source divided by the velocity the wave . so this will be a larger observed period than if this guy was stationary and especially if the observer was in the path of the guy . that make sense , because every time this guy issues a cycle , he is moving a little bit further away . so every crest the same point in the cycle is going to be further and further apart , so you 're going to have longer wavelengths , longer periods . and if you want the observed frequency for that guy over there -- i 'll do it in the same color . the observed frequency for a guy where the source is traveling away from him is just the inverse of that . so one over the period , one over the period , same argument we did there , one over the period from the point of view of the source is the frequency of the source . let me color code it . so 1 over t sub s is equal to the frequency of the source . this is the inverse of that . so i 'm just taking one over everything here . so 1 over t sub s is the frequency of the source , and then we take the inverse of this over here : the velocity of the wave divided by the velocity of the wave plus the velocity of the source . so we 're done , at least for the simple cases . obviously , it becomes a little more interesting when someone is n't exactly in the direction of the source or exactly being moved away from , but these are kind of the two extreme cases . so this is the situation when it is moving away from you . now , just to check our math and maybe make it a little bit concrete in relation to the video we did where we introduced to the idea of the doppler effect , let 's actually apply those numbers . so on that video , two videos ago , we had a situation where the velocity of our source was 5 meters per second to the right , and the velocity of the wave was 10 meters per second radially outward , and the period of our wave -- i 'll call it the period of our -- let me do it in another color -- from the point of view of the source was 1 second per cycle , and the frequency was just the inverse of that . so 1 cycle per second , or 1 hertz , which is a cycle per second . so using those numbers , let 's see if we get to the exact same answer we got in that first video where we first learned about the doppler effect . so let 's look at the frequency from the point of view of this gentleman right here . so the frequency of the source is going to be 1 cycle per second , 1 hertz . the velocity of the wave is 10 meters per second . let me write this . 1 cycle per second . the velocity of the wave is 10 meters per second . the velocity of the wave is 10 meters per second minus the velocity of the source is 5 meters per second . so what 's this going to be equal to ? the observed frequency for this guy right there , is going to be 1 cycle per second times 10 over 10 minus 5 , and the meters per second cancel out . meters per second in the numerator , meters per second in the denominator . so 10 divided by 10 minus 5 , or 10 divided by 5 , is going to be 2 . so it 's going to be 2 cycles per second . and if you want the observed period for this guy , its going to be the inverse of that , or it 's going to be 1/2 second per cycle . and this is exactly what we got in the previous video , or actually , it was two videos ago . now what about the guy who this guy 's running away from ? well , we 'll do the exact same thing . you have 1 cycle per second , or 1 hertz . that 's the emitted frequency from the point of view of the source times the velocity of the wave divided by the velocity of the wave plus the velocity of the source , because it 's moving away from him . so it 's 10 over 10 plus 5 . that 's 10 over 15 . that is 2/3 , so this is equal to 2/3 . the units over here all cancel out . this was cycles per second . so 2/3 cycle per second , which confirms the numbers we got in that first video , so that should make us feel good and it also makes a lot of sense . this guy is going to see the wave crest more frequently . he 's going to observe a higher frequency . if this is a sound , a higher pitch . this guy , since each crest or the cycles are getting spread out , he 's going to see them less frequently , and if this is sound , he 's going to observe a lower pitch .
|
1 cycle per second . the velocity of the wave is 10 meters per second . the velocity of the wave is 10 meters per second minus the velocity of the source is 5 meters per second . so what 's this going to be equal to ?
|
but why are n't the velocities added to get a net velocity ?
|
in the last video , we figured out the formulas for the observed period and frequency for an observer sitting in the path of the source . so the source is moving towards the observer . so this is the example where the train is moving towards you , and you perceive the train 's horn as having a higher pitch or a higher frequency . and we were able to do that by doing a thought experiment . saying , ok , my object starts here . after one period -- a period is just a measure of time , but it 's the measure of time over which the source emits a cycle , so it emits a cycle every period . but after one period , we said , ok , where is that first wave front , or that first pulse , or that first crest ? and where is the source ? because exactly one period has passed by , and the source will be ready to emit another crest or another cycle . so the distance between where the source is and that front of the crest , or that first crest , that is going to be the wavelength . because this next thing that emits is going to be traveling at the exact same velocity , and it 's going to be separated by that distance , which we saw is this expression . we said how long will it take it to travel that distance ? well , it 's traveling at a speed of v sub w. that 'll tell you what the observed period would be for this dude over here . we calculated it right here , and then the observed frequency is just the inverse of that . now let 's think about the situation where the observer is over here . so these equations , or these formulas that we came up with right here , this is observer -- or let me say source traveling in direction of observer . now let 's think about the opposite case , where the source is traveling away from the observer , and in this case , the observer is that guy over there . maybe i 'll do it in a different color . he 'll be blue . so this is the observer . so when we started off , our source was right here . after exactly one period from the source 's point of view , that first crest emitted has traveled radially outward that far . this is the distance . the velocity of the wave times the amount of time that passed -- velocity times time is going to give you distance -- and where the source of the wave will have traveled to the right exactly this distance . it 's velocity times the amount of time that 's gone by . now , in the last video , we said , ok , that wave is just passing this guy . how long will it take for that pulse that 's being emitted right then to also reach him ? and then that tells us the period between two pulses or between two crests . now let 's think about that exact same situation here . that first crest is just passing this guy . and a period , or the t sub s , which is a period of the emitted wave has just passed by . so this guy is just about to emit another wave , so ? that other wave is going to be right here . so what is the distance between the crest , or the cycle , or however you want to think about it , the pulse that is passing him by right now , what is the distance between that and the pulse , the front edge of that pulse that is being emitted right at that moment ? right at that moment . well , it 's going to be this radius , which is this value . it 'll be v sub w times the period . that is that distance plus the amount of distance that our source has traveled away from this guy . so plus v sub s times the period . so that 's what this distance is , and this is how far apart this wave pulse is going to be from that one or this crest is going to be from that one . so if he 's seeing this first crest right now , right at this moment , how long will it take him to see the crest that 's being emitted right now , that 's this far away from him ? well , it 's that far away from him . so let me write this down . so the amount of time it takes for him to see the next crest or the same point in the next cycle , that 's the period . that 's the observed period . that 's going to be equal to this distance . the velocity of the wave times the period from the perspective of the source plus the velocity of the source , because the source has gotten that much further away from him , velocity of the source times the period of the source , so that 's how far the next crest is . and then you divide it by the speed of the wave , by the speed of each of the crests , which is just the velocity of the wave . and we can just factor out the t sub s 's here and say this is t sub s times v sub w , the velocity of the wave plus the velocity of the source divided by the velocity the wave . so this will be a larger observed period than if this guy was stationary and especially if the observer was in the path of the guy . that make sense , because every time this guy issues a cycle , he is moving a little bit further away . so every crest the same point in the cycle is going to be further and further apart , so you 're going to have longer wavelengths , longer periods . and if you want the observed frequency for that guy over there -- i 'll do it in the same color . the observed frequency for a guy where the source is traveling away from him is just the inverse of that . so one over the period , one over the period , same argument we did there , one over the period from the point of view of the source is the frequency of the source . let me color code it . so 1 over t sub s is equal to the frequency of the source . this is the inverse of that . so i 'm just taking one over everything here . so 1 over t sub s is the frequency of the source , and then we take the inverse of this over here : the velocity of the wave divided by the velocity of the wave plus the velocity of the source . so we 're done , at least for the simple cases . obviously , it becomes a little more interesting when someone is n't exactly in the direction of the source or exactly being moved away from , but these are kind of the two extreme cases . so this is the situation when it is moving away from you . now , just to check our math and maybe make it a little bit concrete in relation to the video we did where we introduced to the idea of the doppler effect , let 's actually apply those numbers . so on that video , two videos ago , we had a situation where the velocity of our source was 5 meters per second to the right , and the velocity of the wave was 10 meters per second radially outward , and the period of our wave -- i 'll call it the period of our -- let me do it in another color -- from the point of view of the source was 1 second per cycle , and the frequency was just the inverse of that . so 1 cycle per second , or 1 hertz , which is a cycle per second . so using those numbers , let 's see if we get to the exact same answer we got in that first video where we first learned about the doppler effect . so let 's look at the frequency from the point of view of this gentleman right here . so the frequency of the source is going to be 1 cycle per second , 1 hertz . the velocity of the wave is 10 meters per second . let me write this . 1 cycle per second . the velocity of the wave is 10 meters per second . the velocity of the wave is 10 meters per second minus the velocity of the source is 5 meters per second . so what 's this going to be equal to ? the observed frequency for this guy right there , is going to be 1 cycle per second times 10 over 10 minus 5 , and the meters per second cancel out . meters per second in the numerator , meters per second in the denominator . so 10 divided by 10 minus 5 , or 10 divided by 5 , is going to be 2 . so it 's going to be 2 cycles per second . and if you want the observed period for this guy , its going to be the inverse of that , or it 's going to be 1/2 second per cycle . and this is exactly what we got in the previous video , or actually , it was two videos ago . now what about the guy who this guy 's running away from ? well , we 'll do the exact same thing . you have 1 cycle per second , or 1 hertz . that 's the emitted frequency from the point of view of the source times the velocity of the wave divided by the velocity of the wave plus the velocity of the source , because it 's moving away from him . so it 's 10 over 10 plus 5 . that 's 10 over 15 . that is 2/3 , so this is equal to 2/3 . the units over here all cancel out . this was cycles per second . so 2/3 cycle per second , which confirms the numbers we got in that first video , so that should make us feel good and it also makes a lot of sense . this guy is going to see the wave crest more frequently . he 's going to observe a higher frequency . if this is a sound , a higher pitch . this guy , since each crest or the cycles are getting spread out , he 's going to see them less frequently , and if this is sound , he 's going to observe a lower pitch .
|
he 's going to observe a higher frequency . if this is a sound , a higher pitch . this guy , since each crest or the cycles are getting spread out , he 's going to see them less frequently , and if this is sound , he 's going to observe a lower pitch .
|
do sound waves not act like physical objects where velocities can be added ?
|
in the last video , we figured out the formulas for the observed period and frequency for an observer sitting in the path of the source . so the source is moving towards the observer . so this is the example where the train is moving towards you , and you perceive the train 's horn as having a higher pitch or a higher frequency . and we were able to do that by doing a thought experiment . saying , ok , my object starts here . after one period -- a period is just a measure of time , but it 's the measure of time over which the source emits a cycle , so it emits a cycle every period . but after one period , we said , ok , where is that first wave front , or that first pulse , or that first crest ? and where is the source ? because exactly one period has passed by , and the source will be ready to emit another crest or another cycle . so the distance between where the source is and that front of the crest , or that first crest , that is going to be the wavelength . because this next thing that emits is going to be traveling at the exact same velocity , and it 's going to be separated by that distance , which we saw is this expression . we said how long will it take it to travel that distance ? well , it 's traveling at a speed of v sub w. that 'll tell you what the observed period would be for this dude over here . we calculated it right here , and then the observed frequency is just the inverse of that . now let 's think about the situation where the observer is over here . so these equations , or these formulas that we came up with right here , this is observer -- or let me say source traveling in direction of observer . now let 's think about the opposite case , where the source is traveling away from the observer , and in this case , the observer is that guy over there . maybe i 'll do it in a different color . he 'll be blue . so this is the observer . so when we started off , our source was right here . after exactly one period from the source 's point of view , that first crest emitted has traveled radially outward that far . this is the distance . the velocity of the wave times the amount of time that passed -- velocity times time is going to give you distance -- and where the source of the wave will have traveled to the right exactly this distance . it 's velocity times the amount of time that 's gone by . now , in the last video , we said , ok , that wave is just passing this guy . how long will it take for that pulse that 's being emitted right then to also reach him ? and then that tells us the period between two pulses or between two crests . now let 's think about that exact same situation here . that first crest is just passing this guy . and a period , or the t sub s , which is a period of the emitted wave has just passed by . so this guy is just about to emit another wave , so ? that other wave is going to be right here . so what is the distance between the crest , or the cycle , or however you want to think about it , the pulse that is passing him by right now , what is the distance between that and the pulse , the front edge of that pulse that is being emitted right at that moment ? right at that moment . well , it 's going to be this radius , which is this value . it 'll be v sub w times the period . that is that distance plus the amount of distance that our source has traveled away from this guy . so plus v sub s times the period . so that 's what this distance is , and this is how far apart this wave pulse is going to be from that one or this crest is going to be from that one . so if he 's seeing this first crest right now , right at this moment , how long will it take him to see the crest that 's being emitted right now , that 's this far away from him ? well , it 's that far away from him . so let me write this down . so the amount of time it takes for him to see the next crest or the same point in the next cycle , that 's the period . that 's the observed period . that 's going to be equal to this distance . the velocity of the wave times the period from the perspective of the source plus the velocity of the source , because the source has gotten that much further away from him , velocity of the source times the period of the source , so that 's how far the next crest is . and then you divide it by the speed of the wave , by the speed of each of the crests , which is just the velocity of the wave . and we can just factor out the t sub s 's here and say this is t sub s times v sub w , the velocity of the wave plus the velocity of the source divided by the velocity the wave . so this will be a larger observed period than if this guy was stationary and especially if the observer was in the path of the guy . that make sense , because every time this guy issues a cycle , he is moving a little bit further away . so every crest the same point in the cycle is going to be further and further apart , so you 're going to have longer wavelengths , longer periods . and if you want the observed frequency for that guy over there -- i 'll do it in the same color . the observed frequency for a guy where the source is traveling away from him is just the inverse of that . so one over the period , one over the period , same argument we did there , one over the period from the point of view of the source is the frequency of the source . let me color code it . so 1 over t sub s is equal to the frequency of the source . this is the inverse of that . so i 'm just taking one over everything here . so 1 over t sub s is the frequency of the source , and then we take the inverse of this over here : the velocity of the wave divided by the velocity of the wave plus the velocity of the source . so we 're done , at least for the simple cases . obviously , it becomes a little more interesting when someone is n't exactly in the direction of the source or exactly being moved away from , but these are kind of the two extreme cases . so this is the situation when it is moving away from you . now , just to check our math and maybe make it a little bit concrete in relation to the video we did where we introduced to the idea of the doppler effect , let 's actually apply those numbers . so on that video , two videos ago , we had a situation where the velocity of our source was 5 meters per second to the right , and the velocity of the wave was 10 meters per second radially outward , and the period of our wave -- i 'll call it the period of our -- let me do it in another color -- from the point of view of the source was 1 second per cycle , and the frequency was just the inverse of that . so 1 cycle per second , or 1 hertz , which is a cycle per second . so using those numbers , let 's see if we get to the exact same answer we got in that first video where we first learned about the doppler effect . so let 's look at the frequency from the point of view of this gentleman right here . so the frequency of the source is going to be 1 cycle per second , 1 hertz . the velocity of the wave is 10 meters per second . let me write this . 1 cycle per second . the velocity of the wave is 10 meters per second . the velocity of the wave is 10 meters per second minus the velocity of the source is 5 meters per second . so what 's this going to be equal to ? the observed frequency for this guy right there , is going to be 1 cycle per second times 10 over 10 minus 5 , and the meters per second cancel out . meters per second in the numerator , meters per second in the denominator . so 10 divided by 10 minus 5 , or 10 divided by 5 , is going to be 2 . so it 's going to be 2 cycles per second . and if you want the observed period for this guy , its going to be the inverse of that , or it 's going to be 1/2 second per cycle . and this is exactly what we got in the previous video , or actually , it was two videos ago . now what about the guy who this guy 's running away from ? well , we 'll do the exact same thing . you have 1 cycle per second , or 1 hertz . that 's the emitted frequency from the point of view of the source times the velocity of the wave divided by the velocity of the wave plus the velocity of the source , because it 's moving away from him . so it 's 10 over 10 plus 5 . that 's 10 over 15 . that is 2/3 , so this is equal to 2/3 . the units over here all cancel out . this was cycles per second . so 2/3 cycle per second , which confirms the numbers we got in that first video , so that should make us feel good and it also makes a lot of sense . this guy is going to see the wave crest more frequently . he 's going to observe a higher frequency . if this is a sound , a higher pitch . this guy , since each crest or the cycles are getting spread out , he 's going to see them less frequently , and if this is sound , he 's going to observe a lower pitch .
|
in the last video , we figured out the formulas for the observed period and frequency for an observer sitting in the path of the source . so the source is moving towards the observer . so this is the example where the train is moving towards you , and you perceive the train 's horn as having a higher pitch or a higher frequency .
|
here in case of source moving , we are considering the movement of source first wave next ... but they take place at-a-time ... so wave must be in an elliptical path ... is n't it ?
|
in the last video , we figured out the formulas for the observed period and frequency for an observer sitting in the path of the source . so the source is moving towards the observer . so this is the example where the train is moving towards you , and you perceive the train 's horn as having a higher pitch or a higher frequency . and we were able to do that by doing a thought experiment . saying , ok , my object starts here . after one period -- a period is just a measure of time , but it 's the measure of time over which the source emits a cycle , so it emits a cycle every period . but after one period , we said , ok , where is that first wave front , or that first pulse , or that first crest ? and where is the source ? because exactly one period has passed by , and the source will be ready to emit another crest or another cycle . so the distance between where the source is and that front of the crest , or that first crest , that is going to be the wavelength . because this next thing that emits is going to be traveling at the exact same velocity , and it 's going to be separated by that distance , which we saw is this expression . we said how long will it take it to travel that distance ? well , it 's traveling at a speed of v sub w. that 'll tell you what the observed period would be for this dude over here . we calculated it right here , and then the observed frequency is just the inverse of that . now let 's think about the situation where the observer is over here . so these equations , or these formulas that we came up with right here , this is observer -- or let me say source traveling in direction of observer . now let 's think about the opposite case , where the source is traveling away from the observer , and in this case , the observer is that guy over there . maybe i 'll do it in a different color . he 'll be blue . so this is the observer . so when we started off , our source was right here . after exactly one period from the source 's point of view , that first crest emitted has traveled radially outward that far . this is the distance . the velocity of the wave times the amount of time that passed -- velocity times time is going to give you distance -- and where the source of the wave will have traveled to the right exactly this distance . it 's velocity times the amount of time that 's gone by . now , in the last video , we said , ok , that wave is just passing this guy . how long will it take for that pulse that 's being emitted right then to also reach him ? and then that tells us the period between two pulses or between two crests . now let 's think about that exact same situation here . that first crest is just passing this guy . and a period , or the t sub s , which is a period of the emitted wave has just passed by . so this guy is just about to emit another wave , so ? that other wave is going to be right here . so what is the distance between the crest , or the cycle , or however you want to think about it , the pulse that is passing him by right now , what is the distance between that and the pulse , the front edge of that pulse that is being emitted right at that moment ? right at that moment . well , it 's going to be this radius , which is this value . it 'll be v sub w times the period . that is that distance plus the amount of distance that our source has traveled away from this guy . so plus v sub s times the period . so that 's what this distance is , and this is how far apart this wave pulse is going to be from that one or this crest is going to be from that one . so if he 's seeing this first crest right now , right at this moment , how long will it take him to see the crest that 's being emitted right now , that 's this far away from him ? well , it 's that far away from him . so let me write this down . so the amount of time it takes for him to see the next crest or the same point in the next cycle , that 's the period . that 's the observed period . that 's going to be equal to this distance . the velocity of the wave times the period from the perspective of the source plus the velocity of the source , because the source has gotten that much further away from him , velocity of the source times the period of the source , so that 's how far the next crest is . and then you divide it by the speed of the wave , by the speed of each of the crests , which is just the velocity of the wave . and we can just factor out the t sub s 's here and say this is t sub s times v sub w , the velocity of the wave plus the velocity of the source divided by the velocity the wave . so this will be a larger observed period than if this guy was stationary and especially if the observer was in the path of the guy . that make sense , because every time this guy issues a cycle , he is moving a little bit further away . so every crest the same point in the cycle is going to be further and further apart , so you 're going to have longer wavelengths , longer periods . and if you want the observed frequency for that guy over there -- i 'll do it in the same color . the observed frequency for a guy where the source is traveling away from him is just the inverse of that . so one over the period , one over the period , same argument we did there , one over the period from the point of view of the source is the frequency of the source . let me color code it . so 1 over t sub s is equal to the frequency of the source . this is the inverse of that . so i 'm just taking one over everything here . so 1 over t sub s is the frequency of the source , and then we take the inverse of this over here : the velocity of the wave divided by the velocity of the wave plus the velocity of the source . so we 're done , at least for the simple cases . obviously , it becomes a little more interesting when someone is n't exactly in the direction of the source or exactly being moved away from , but these are kind of the two extreme cases . so this is the situation when it is moving away from you . now , just to check our math and maybe make it a little bit concrete in relation to the video we did where we introduced to the idea of the doppler effect , let 's actually apply those numbers . so on that video , two videos ago , we had a situation where the velocity of our source was 5 meters per second to the right , and the velocity of the wave was 10 meters per second radially outward , and the period of our wave -- i 'll call it the period of our -- let me do it in another color -- from the point of view of the source was 1 second per cycle , and the frequency was just the inverse of that . so 1 cycle per second , or 1 hertz , which is a cycle per second . so using those numbers , let 's see if we get to the exact same answer we got in that first video where we first learned about the doppler effect . so let 's look at the frequency from the point of view of this gentleman right here . so the frequency of the source is going to be 1 cycle per second , 1 hertz . the velocity of the wave is 10 meters per second . let me write this . 1 cycle per second . the velocity of the wave is 10 meters per second . the velocity of the wave is 10 meters per second minus the velocity of the source is 5 meters per second . so what 's this going to be equal to ? the observed frequency for this guy right there , is going to be 1 cycle per second times 10 over 10 minus 5 , and the meters per second cancel out . meters per second in the numerator , meters per second in the denominator . so 10 divided by 10 minus 5 , or 10 divided by 5 , is going to be 2 . so it 's going to be 2 cycles per second . and if you want the observed period for this guy , its going to be the inverse of that , or it 's going to be 1/2 second per cycle . and this is exactly what we got in the previous video , or actually , it was two videos ago . now what about the guy who this guy 's running away from ? well , we 'll do the exact same thing . you have 1 cycle per second , or 1 hertz . that 's the emitted frequency from the point of view of the source times the velocity of the wave divided by the velocity of the wave plus the velocity of the source , because it 's moving away from him . so it 's 10 over 10 plus 5 . that 's 10 over 15 . that is 2/3 , so this is equal to 2/3 . the units over here all cancel out . this was cycles per second . so 2/3 cycle per second , which confirms the numbers we got in that first video , so that should make us feel good and it also makes a lot of sense . this guy is going to see the wave crest more frequently . he 's going to observe a higher frequency . if this is a sound , a higher pitch . this guy , since each crest or the cycles are getting spread out , he 's going to see them less frequently , and if this is sound , he 's going to observe a lower pitch .
|
so the amount of time it takes for him to see the next crest or the same point in the next cycle , that 's the period . that 's the observed period . that 's going to be equal to this distance .
|
would every observed period be shorter if the frequency be different ?
|
in the last video , we figured out the formulas for the observed period and frequency for an observer sitting in the path of the source . so the source is moving towards the observer . so this is the example where the train is moving towards you , and you perceive the train 's horn as having a higher pitch or a higher frequency . and we were able to do that by doing a thought experiment . saying , ok , my object starts here . after one period -- a period is just a measure of time , but it 's the measure of time over which the source emits a cycle , so it emits a cycle every period . but after one period , we said , ok , where is that first wave front , or that first pulse , or that first crest ? and where is the source ? because exactly one period has passed by , and the source will be ready to emit another crest or another cycle . so the distance between where the source is and that front of the crest , or that first crest , that is going to be the wavelength . because this next thing that emits is going to be traveling at the exact same velocity , and it 's going to be separated by that distance , which we saw is this expression . we said how long will it take it to travel that distance ? well , it 's traveling at a speed of v sub w. that 'll tell you what the observed period would be for this dude over here . we calculated it right here , and then the observed frequency is just the inverse of that . now let 's think about the situation where the observer is over here . so these equations , or these formulas that we came up with right here , this is observer -- or let me say source traveling in direction of observer . now let 's think about the opposite case , where the source is traveling away from the observer , and in this case , the observer is that guy over there . maybe i 'll do it in a different color . he 'll be blue . so this is the observer . so when we started off , our source was right here . after exactly one period from the source 's point of view , that first crest emitted has traveled radially outward that far . this is the distance . the velocity of the wave times the amount of time that passed -- velocity times time is going to give you distance -- and where the source of the wave will have traveled to the right exactly this distance . it 's velocity times the amount of time that 's gone by . now , in the last video , we said , ok , that wave is just passing this guy . how long will it take for that pulse that 's being emitted right then to also reach him ? and then that tells us the period between two pulses or between two crests . now let 's think about that exact same situation here . that first crest is just passing this guy . and a period , or the t sub s , which is a period of the emitted wave has just passed by . so this guy is just about to emit another wave , so ? that other wave is going to be right here . so what is the distance between the crest , or the cycle , or however you want to think about it , the pulse that is passing him by right now , what is the distance between that and the pulse , the front edge of that pulse that is being emitted right at that moment ? right at that moment . well , it 's going to be this radius , which is this value . it 'll be v sub w times the period . that is that distance plus the amount of distance that our source has traveled away from this guy . so plus v sub s times the period . so that 's what this distance is , and this is how far apart this wave pulse is going to be from that one or this crest is going to be from that one . so if he 's seeing this first crest right now , right at this moment , how long will it take him to see the crest that 's being emitted right now , that 's this far away from him ? well , it 's that far away from him . so let me write this down . so the amount of time it takes for him to see the next crest or the same point in the next cycle , that 's the period . that 's the observed period . that 's going to be equal to this distance . the velocity of the wave times the period from the perspective of the source plus the velocity of the source , because the source has gotten that much further away from him , velocity of the source times the period of the source , so that 's how far the next crest is . and then you divide it by the speed of the wave , by the speed of each of the crests , which is just the velocity of the wave . and we can just factor out the t sub s 's here and say this is t sub s times v sub w , the velocity of the wave plus the velocity of the source divided by the velocity the wave . so this will be a larger observed period than if this guy was stationary and especially if the observer was in the path of the guy . that make sense , because every time this guy issues a cycle , he is moving a little bit further away . so every crest the same point in the cycle is going to be further and further apart , so you 're going to have longer wavelengths , longer periods . and if you want the observed frequency for that guy over there -- i 'll do it in the same color . the observed frequency for a guy where the source is traveling away from him is just the inverse of that . so one over the period , one over the period , same argument we did there , one over the period from the point of view of the source is the frequency of the source . let me color code it . so 1 over t sub s is equal to the frequency of the source . this is the inverse of that . so i 'm just taking one over everything here . so 1 over t sub s is the frequency of the source , and then we take the inverse of this over here : the velocity of the wave divided by the velocity of the wave plus the velocity of the source . so we 're done , at least for the simple cases . obviously , it becomes a little more interesting when someone is n't exactly in the direction of the source or exactly being moved away from , but these are kind of the two extreme cases . so this is the situation when it is moving away from you . now , just to check our math and maybe make it a little bit concrete in relation to the video we did where we introduced to the idea of the doppler effect , let 's actually apply those numbers . so on that video , two videos ago , we had a situation where the velocity of our source was 5 meters per second to the right , and the velocity of the wave was 10 meters per second radially outward , and the period of our wave -- i 'll call it the period of our -- let me do it in another color -- from the point of view of the source was 1 second per cycle , and the frequency was just the inverse of that . so 1 cycle per second , or 1 hertz , which is a cycle per second . so using those numbers , let 's see if we get to the exact same answer we got in that first video where we first learned about the doppler effect . so let 's look at the frequency from the point of view of this gentleman right here . so the frequency of the source is going to be 1 cycle per second , 1 hertz . the velocity of the wave is 10 meters per second . let me write this . 1 cycle per second . the velocity of the wave is 10 meters per second . the velocity of the wave is 10 meters per second minus the velocity of the source is 5 meters per second . so what 's this going to be equal to ? the observed frequency for this guy right there , is going to be 1 cycle per second times 10 over 10 minus 5 , and the meters per second cancel out . meters per second in the numerator , meters per second in the denominator . so 10 divided by 10 minus 5 , or 10 divided by 5 , is going to be 2 . so it 's going to be 2 cycles per second . and if you want the observed period for this guy , its going to be the inverse of that , or it 's going to be 1/2 second per cycle . and this is exactly what we got in the previous video , or actually , it was two videos ago . now what about the guy who this guy 's running away from ? well , we 'll do the exact same thing . you have 1 cycle per second , or 1 hertz . that 's the emitted frequency from the point of view of the source times the velocity of the wave divided by the velocity of the wave plus the velocity of the source , because it 's moving away from him . so it 's 10 over 10 plus 5 . that 's 10 over 15 . that is 2/3 , so this is equal to 2/3 . the units over here all cancel out . this was cycles per second . so 2/3 cycle per second , which confirms the numbers we got in that first video , so that should make us feel good and it also makes a lot of sense . this guy is going to see the wave crest more frequently . he 's going to observe a higher frequency . if this is a sound , a higher pitch . this guy , since each crest or the cycles are getting spread out , he 's going to see them less frequently , and if this is sound , he 's going to observe a lower pitch .
|
1 cycle per second . the velocity of the wave is 10 meters per second . the velocity of the wave is 10 meters per second minus the velocity of the source is 5 meters per second . so what 's this going to be equal to ?
|
when the source emits a wave , wo n't the velocity of the source be added to the velocity of the wave ?
|
in the last video , we figured out the formulas for the observed period and frequency for an observer sitting in the path of the source . so the source is moving towards the observer . so this is the example where the train is moving towards you , and you perceive the train 's horn as having a higher pitch or a higher frequency . and we were able to do that by doing a thought experiment . saying , ok , my object starts here . after one period -- a period is just a measure of time , but it 's the measure of time over which the source emits a cycle , so it emits a cycle every period . but after one period , we said , ok , where is that first wave front , or that first pulse , or that first crest ? and where is the source ? because exactly one period has passed by , and the source will be ready to emit another crest or another cycle . so the distance between where the source is and that front of the crest , or that first crest , that is going to be the wavelength . because this next thing that emits is going to be traveling at the exact same velocity , and it 's going to be separated by that distance , which we saw is this expression . we said how long will it take it to travel that distance ? well , it 's traveling at a speed of v sub w. that 'll tell you what the observed period would be for this dude over here . we calculated it right here , and then the observed frequency is just the inverse of that . now let 's think about the situation where the observer is over here . so these equations , or these formulas that we came up with right here , this is observer -- or let me say source traveling in direction of observer . now let 's think about the opposite case , where the source is traveling away from the observer , and in this case , the observer is that guy over there . maybe i 'll do it in a different color . he 'll be blue . so this is the observer . so when we started off , our source was right here . after exactly one period from the source 's point of view , that first crest emitted has traveled radially outward that far . this is the distance . the velocity of the wave times the amount of time that passed -- velocity times time is going to give you distance -- and where the source of the wave will have traveled to the right exactly this distance . it 's velocity times the amount of time that 's gone by . now , in the last video , we said , ok , that wave is just passing this guy . how long will it take for that pulse that 's being emitted right then to also reach him ? and then that tells us the period between two pulses or between two crests . now let 's think about that exact same situation here . that first crest is just passing this guy . and a period , or the t sub s , which is a period of the emitted wave has just passed by . so this guy is just about to emit another wave , so ? that other wave is going to be right here . so what is the distance between the crest , or the cycle , or however you want to think about it , the pulse that is passing him by right now , what is the distance between that and the pulse , the front edge of that pulse that is being emitted right at that moment ? right at that moment . well , it 's going to be this radius , which is this value . it 'll be v sub w times the period . that is that distance plus the amount of distance that our source has traveled away from this guy . so plus v sub s times the period . so that 's what this distance is , and this is how far apart this wave pulse is going to be from that one or this crest is going to be from that one . so if he 's seeing this first crest right now , right at this moment , how long will it take him to see the crest that 's being emitted right now , that 's this far away from him ? well , it 's that far away from him . so let me write this down . so the amount of time it takes for him to see the next crest or the same point in the next cycle , that 's the period . that 's the observed period . that 's going to be equal to this distance . the velocity of the wave times the period from the perspective of the source plus the velocity of the source , because the source has gotten that much further away from him , velocity of the source times the period of the source , so that 's how far the next crest is . and then you divide it by the speed of the wave , by the speed of each of the crests , which is just the velocity of the wave . and we can just factor out the t sub s 's here and say this is t sub s times v sub w , the velocity of the wave plus the velocity of the source divided by the velocity the wave . so this will be a larger observed period than if this guy was stationary and especially if the observer was in the path of the guy . that make sense , because every time this guy issues a cycle , he is moving a little bit further away . so every crest the same point in the cycle is going to be further and further apart , so you 're going to have longer wavelengths , longer periods . and if you want the observed frequency for that guy over there -- i 'll do it in the same color . the observed frequency for a guy where the source is traveling away from him is just the inverse of that . so one over the period , one over the period , same argument we did there , one over the period from the point of view of the source is the frequency of the source . let me color code it . so 1 over t sub s is equal to the frequency of the source . this is the inverse of that . so i 'm just taking one over everything here . so 1 over t sub s is the frequency of the source , and then we take the inverse of this over here : the velocity of the wave divided by the velocity of the wave plus the velocity of the source . so we 're done , at least for the simple cases . obviously , it becomes a little more interesting when someone is n't exactly in the direction of the source or exactly being moved away from , but these are kind of the two extreme cases . so this is the situation when it is moving away from you . now , just to check our math and maybe make it a little bit concrete in relation to the video we did where we introduced to the idea of the doppler effect , let 's actually apply those numbers . so on that video , two videos ago , we had a situation where the velocity of our source was 5 meters per second to the right , and the velocity of the wave was 10 meters per second radially outward , and the period of our wave -- i 'll call it the period of our -- let me do it in another color -- from the point of view of the source was 1 second per cycle , and the frequency was just the inverse of that . so 1 cycle per second , or 1 hertz , which is a cycle per second . so using those numbers , let 's see if we get to the exact same answer we got in that first video where we first learned about the doppler effect . so let 's look at the frequency from the point of view of this gentleman right here . so the frequency of the source is going to be 1 cycle per second , 1 hertz . the velocity of the wave is 10 meters per second . let me write this . 1 cycle per second . the velocity of the wave is 10 meters per second . the velocity of the wave is 10 meters per second minus the velocity of the source is 5 meters per second . so what 's this going to be equal to ? the observed frequency for this guy right there , is going to be 1 cycle per second times 10 over 10 minus 5 , and the meters per second cancel out . meters per second in the numerator , meters per second in the denominator . so 10 divided by 10 minus 5 , or 10 divided by 5 , is going to be 2 . so it 's going to be 2 cycles per second . and if you want the observed period for this guy , its going to be the inverse of that , or it 's going to be 1/2 second per cycle . and this is exactly what we got in the previous video , or actually , it was two videos ago . now what about the guy who this guy 's running away from ? well , we 'll do the exact same thing . you have 1 cycle per second , or 1 hertz . that 's the emitted frequency from the point of view of the source times the velocity of the wave divided by the velocity of the wave plus the velocity of the source , because it 's moving away from him . so it 's 10 over 10 plus 5 . that 's 10 over 15 . that is 2/3 , so this is equal to 2/3 . the units over here all cancel out . this was cycles per second . so 2/3 cycle per second , which confirms the numbers we got in that first video , so that should make us feel good and it also makes a lot of sense . this guy is going to see the wave crest more frequently . he 's going to observe a higher frequency . if this is a sound , a higher pitch . this guy , since each crest or the cycles are getting spread out , he 's going to see them less frequently , and if this is sound , he 's going to observe a lower pitch .
|
but after one period , we said , ok , where is that first wave front , or that first pulse , or that first crest ? and where is the source ? because exactly one period has passed by , and the source will be ready to emit another crest or another cycle .
|
how would you rearrange the doppler equation to find the speed of the source ?
|
in the last video , we figured out the formulas for the observed period and frequency for an observer sitting in the path of the source . so the source is moving towards the observer . so this is the example where the train is moving towards you , and you perceive the train 's horn as having a higher pitch or a higher frequency . and we were able to do that by doing a thought experiment . saying , ok , my object starts here . after one period -- a period is just a measure of time , but it 's the measure of time over which the source emits a cycle , so it emits a cycle every period . but after one period , we said , ok , where is that first wave front , or that first pulse , or that first crest ? and where is the source ? because exactly one period has passed by , and the source will be ready to emit another crest or another cycle . so the distance between where the source is and that front of the crest , or that first crest , that is going to be the wavelength . because this next thing that emits is going to be traveling at the exact same velocity , and it 's going to be separated by that distance , which we saw is this expression . we said how long will it take it to travel that distance ? well , it 's traveling at a speed of v sub w. that 'll tell you what the observed period would be for this dude over here . we calculated it right here , and then the observed frequency is just the inverse of that . now let 's think about the situation where the observer is over here . so these equations , or these formulas that we came up with right here , this is observer -- or let me say source traveling in direction of observer . now let 's think about the opposite case , where the source is traveling away from the observer , and in this case , the observer is that guy over there . maybe i 'll do it in a different color . he 'll be blue . so this is the observer . so when we started off , our source was right here . after exactly one period from the source 's point of view , that first crest emitted has traveled radially outward that far . this is the distance . the velocity of the wave times the amount of time that passed -- velocity times time is going to give you distance -- and where the source of the wave will have traveled to the right exactly this distance . it 's velocity times the amount of time that 's gone by . now , in the last video , we said , ok , that wave is just passing this guy . how long will it take for that pulse that 's being emitted right then to also reach him ? and then that tells us the period between two pulses or between two crests . now let 's think about that exact same situation here . that first crest is just passing this guy . and a period , or the t sub s , which is a period of the emitted wave has just passed by . so this guy is just about to emit another wave , so ? that other wave is going to be right here . so what is the distance between the crest , or the cycle , or however you want to think about it , the pulse that is passing him by right now , what is the distance between that and the pulse , the front edge of that pulse that is being emitted right at that moment ? right at that moment . well , it 's going to be this radius , which is this value . it 'll be v sub w times the period . that is that distance plus the amount of distance that our source has traveled away from this guy . so plus v sub s times the period . so that 's what this distance is , and this is how far apart this wave pulse is going to be from that one or this crest is going to be from that one . so if he 's seeing this first crest right now , right at this moment , how long will it take him to see the crest that 's being emitted right now , that 's this far away from him ? well , it 's that far away from him . so let me write this down . so the amount of time it takes for him to see the next crest or the same point in the next cycle , that 's the period . that 's the observed period . that 's going to be equal to this distance . the velocity of the wave times the period from the perspective of the source plus the velocity of the source , because the source has gotten that much further away from him , velocity of the source times the period of the source , so that 's how far the next crest is . and then you divide it by the speed of the wave , by the speed of each of the crests , which is just the velocity of the wave . and we can just factor out the t sub s 's here and say this is t sub s times v sub w , the velocity of the wave plus the velocity of the source divided by the velocity the wave . so this will be a larger observed period than if this guy was stationary and especially if the observer was in the path of the guy . that make sense , because every time this guy issues a cycle , he is moving a little bit further away . so every crest the same point in the cycle is going to be further and further apart , so you 're going to have longer wavelengths , longer periods . and if you want the observed frequency for that guy over there -- i 'll do it in the same color . the observed frequency for a guy where the source is traveling away from him is just the inverse of that . so one over the period , one over the period , same argument we did there , one over the period from the point of view of the source is the frequency of the source . let me color code it . so 1 over t sub s is equal to the frequency of the source . this is the inverse of that . so i 'm just taking one over everything here . so 1 over t sub s is the frequency of the source , and then we take the inverse of this over here : the velocity of the wave divided by the velocity of the wave plus the velocity of the source . so we 're done , at least for the simple cases . obviously , it becomes a little more interesting when someone is n't exactly in the direction of the source or exactly being moved away from , but these are kind of the two extreme cases . so this is the situation when it is moving away from you . now , just to check our math and maybe make it a little bit concrete in relation to the video we did where we introduced to the idea of the doppler effect , let 's actually apply those numbers . so on that video , two videos ago , we had a situation where the velocity of our source was 5 meters per second to the right , and the velocity of the wave was 10 meters per second radially outward , and the period of our wave -- i 'll call it the period of our -- let me do it in another color -- from the point of view of the source was 1 second per cycle , and the frequency was just the inverse of that . so 1 cycle per second , or 1 hertz , which is a cycle per second . so using those numbers , let 's see if we get to the exact same answer we got in that first video where we first learned about the doppler effect . so let 's look at the frequency from the point of view of this gentleman right here . so the frequency of the source is going to be 1 cycle per second , 1 hertz . the velocity of the wave is 10 meters per second . let me write this . 1 cycle per second . the velocity of the wave is 10 meters per second . the velocity of the wave is 10 meters per second minus the velocity of the source is 5 meters per second . so what 's this going to be equal to ? the observed frequency for this guy right there , is going to be 1 cycle per second times 10 over 10 minus 5 , and the meters per second cancel out . meters per second in the numerator , meters per second in the denominator . so 10 divided by 10 minus 5 , or 10 divided by 5 , is going to be 2 . so it 's going to be 2 cycles per second . and if you want the observed period for this guy , its going to be the inverse of that , or it 's going to be 1/2 second per cycle . and this is exactly what we got in the previous video , or actually , it was two videos ago . now what about the guy who this guy 's running away from ? well , we 'll do the exact same thing . you have 1 cycle per second , or 1 hertz . that 's the emitted frequency from the point of view of the source times the velocity of the wave divided by the velocity of the wave plus the velocity of the source , because it 's moving away from him . so it 's 10 over 10 plus 5 . that 's 10 over 15 . that is 2/3 , so this is equal to 2/3 . the units over here all cancel out . this was cycles per second . so 2/3 cycle per second , which confirms the numbers we got in that first video , so that should make us feel good and it also makes a lot of sense . this guy is going to see the wave crest more frequently . he 's going to observe a higher frequency . if this is a sound , a higher pitch . this guy , since each crest or the cycles are getting spread out , he 's going to see them less frequently , and if this is sound , he 's going to observe a lower pitch .
|
so 1 cycle per second , or 1 hertz , which is a cycle per second . so using those numbers , let 's see if we get to the exact same answer we got in that first video where we first learned about the doppler effect . so let 's look at the frequency from the point of view of this gentleman right here .
|
how does doppler effects related to temperature ?
|
in the last video , we figured out the formulas for the observed period and frequency for an observer sitting in the path of the source . so the source is moving towards the observer . so this is the example where the train is moving towards you , and you perceive the train 's horn as having a higher pitch or a higher frequency . and we were able to do that by doing a thought experiment . saying , ok , my object starts here . after one period -- a period is just a measure of time , but it 's the measure of time over which the source emits a cycle , so it emits a cycle every period . but after one period , we said , ok , where is that first wave front , or that first pulse , or that first crest ? and where is the source ? because exactly one period has passed by , and the source will be ready to emit another crest or another cycle . so the distance between where the source is and that front of the crest , or that first crest , that is going to be the wavelength . because this next thing that emits is going to be traveling at the exact same velocity , and it 's going to be separated by that distance , which we saw is this expression . we said how long will it take it to travel that distance ? well , it 's traveling at a speed of v sub w. that 'll tell you what the observed period would be for this dude over here . we calculated it right here , and then the observed frequency is just the inverse of that . now let 's think about the situation where the observer is over here . so these equations , or these formulas that we came up with right here , this is observer -- or let me say source traveling in direction of observer . now let 's think about the opposite case , where the source is traveling away from the observer , and in this case , the observer is that guy over there . maybe i 'll do it in a different color . he 'll be blue . so this is the observer . so when we started off , our source was right here . after exactly one period from the source 's point of view , that first crest emitted has traveled radially outward that far . this is the distance . the velocity of the wave times the amount of time that passed -- velocity times time is going to give you distance -- and where the source of the wave will have traveled to the right exactly this distance . it 's velocity times the amount of time that 's gone by . now , in the last video , we said , ok , that wave is just passing this guy . how long will it take for that pulse that 's being emitted right then to also reach him ? and then that tells us the period between two pulses or between two crests . now let 's think about that exact same situation here . that first crest is just passing this guy . and a period , or the t sub s , which is a period of the emitted wave has just passed by . so this guy is just about to emit another wave , so ? that other wave is going to be right here . so what is the distance between the crest , or the cycle , or however you want to think about it , the pulse that is passing him by right now , what is the distance between that and the pulse , the front edge of that pulse that is being emitted right at that moment ? right at that moment . well , it 's going to be this radius , which is this value . it 'll be v sub w times the period . that is that distance plus the amount of distance that our source has traveled away from this guy . so plus v sub s times the period . so that 's what this distance is , and this is how far apart this wave pulse is going to be from that one or this crest is going to be from that one . so if he 's seeing this first crest right now , right at this moment , how long will it take him to see the crest that 's being emitted right now , that 's this far away from him ? well , it 's that far away from him . so let me write this down . so the amount of time it takes for him to see the next crest or the same point in the next cycle , that 's the period . that 's the observed period . that 's going to be equal to this distance . the velocity of the wave times the period from the perspective of the source plus the velocity of the source , because the source has gotten that much further away from him , velocity of the source times the period of the source , so that 's how far the next crest is . and then you divide it by the speed of the wave , by the speed of each of the crests , which is just the velocity of the wave . and we can just factor out the t sub s 's here and say this is t sub s times v sub w , the velocity of the wave plus the velocity of the source divided by the velocity the wave . so this will be a larger observed period than if this guy was stationary and especially if the observer was in the path of the guy . that make sense , because every time this guy issues a cycle , he is moving a little bit further away . so every crest the same point in the cycle is going to be further and further apart , so you 're going to have longer wavelengths , longer periods . and if you want the observed frequency for that guy over there -- i 'll do it in the same color . the observed frequency for a guy where the source is traveling away from him is just the inverse of that . so one over the period , one over the period , same argument we did there , one over the period from the point of view of the source is the frequency of the source . let me color code it . so 1 over t sub s is equal to the frequency of the source . this is the inverse of that . so i 'm just taking one over everything here . so 1 over t sub s is the frequency of the source , and then we take the inverse of this over here : the velocity of the wave divided by the velocity of the wave plus the velocity of the source . so we 're done , at least for the simple cases . obviously , it becomes a little more interesting when someone is n't exactly in the direction of the source or exactly being moved away from , but these are kind of the two extreme cases . so this is the situation when it is moving away from you . now , just to check our math and maybe make it a little bit concrete in relation to the video we did where we introduced to the idea of the doppler effect , let 's actually apply those numbers . so on that video , two videos ago , we had a situation where the velocity of our source was 5 meters per second to the right , and the velocity of the wave was 10 meters per second radially outward , and the period of our wave -- i 'll call it the period of our -- let me do it in another color -- from the point of view of the source was 1 second per cycle , and the frequency was just the inverse of that . so 1 cycle per second , or 1 hertz , which is a cycle per second . so using those numbers , let 's see if we get to the exact same answer we got in that first video where we first learned about the doppler effect . so let 's look at the frequency from the point of view of this gentleman right here . so the frequency of the source is going to be 1 cycle per second , 1 hertz . the velocity of the wave is 10 meters per second . let me write this . 1 cycle per second . the velocity of the wave is 10 meters per second . the velocity of the wave is 10 meters per second minus the velocity of the source is 5 meters per second . so what 's this going to be equal to ? the observed frequency for this guy right there , is going to be 1 cycle per second times 10 over 10 minus 5 , and the meters per second cancel out . meters per second in the numerator , meters per second in the denominator . so 10 divided by 10 minus 5 , or 10 divided by 5 , is going to be 2 . so it 's going to be 2 cycles per second . and if you want the observed period for this guy , its going to be the inverse of that , or it 's going to be 1/2 second per cycle . and this is exactly what we got in the previous video , or actually , it was two videos ago . now what about the guy who this guy 's running away from ? well , we 'll do the exact same thing . you have 1 cycle per second , or 1 hertz . that 's the emitted frequency from the point of view of the source times the velocity of the wave divided by the velocity of the wave plus the velocity of the source , because it 's moving away from him . so it 's 10 over 10 plus 5 . that 's 10 over 15 . that is 2/3 , so this is equal to 2/3 . the units over here all cancel out . this was cycles per second . so 2/3 cycle per second , which confirms the numbers we got in that first video , so that should make us feel good and it also makes a lot of sense . this guy is going to see the wave crest more frequently . he 's going to observe a higher frequency . if this is a sound , a higher pitch . this guy , since each crest or the cycles are getting spread out , he 's going to see them less frequently , and if this is sound , he 's going to observe a lower pitch .
|
in the last video , we figured out the formulas for the observed period and frequency for an observer sitting in the path of the source . so the source is moving towards the observer . so this is the example where the train is moving towards you , and you perceive the train 's horn as having a higher pitch or a higher frequency .
|
what happens when both the source and observer are moving ?
|
in the last video , we figured out the formulas for the observed period and frequency for an observer sitting in the path of the source . so the source is moving towards the observer . so this is the example where the train is moving towards you , and you perceive the train 's horn as having a higher pitch or a higher frequency . and we were able to do that by doing a thought experiment . saying , ok , my object starts here . after one period -- a period is just a measure of time , but it 's the measure of time over which the source emits a cycle , so it emits a cycle every period . but after one period , we said , ok , where is that first wave front , or that first pulse , or that first crest ? and where is the source ? because exactly one period has passed by , and the source will be ready to emit another crest or another cycle . so the distance between where the source is and that front of the crest , or that first crest , that is going to be the wavelength . because this next thing that emits is going to be traveling at the exact same velocity , and it 's going to be separated by that distance , which we saw is this expression . we said how long will it take it to travel that distance ? well , it 's traveling at a speed of v sub w. that 'll tell you what the observed period would be for this dude over here . we calculated it right here , and then the observed frequency is just the inverse of that . now let 's think about the situation where the observer is over here . so these equations , or these formulas that we came up with right here , this is observer -- or let me say source traveling in direction of observer . now let 's think about the opposite case , where the source is traveling away from the observer , and in this case , the observer is that guy over there . maybe i 'll do it in a different color . he 'll be blue . so this is the observer . so when we started off , our source was right here . after exactly one period from the source 's point of view , that first crest emitted has traveled radially outward that far . this is the distance . the velocity of the wave times the amount of time that passed -- velocity times time is going to give you distance -- and where the source of the wave will have traveled to the right exactly this distance . it 's velocity times the amount of time that 's gone by . now , in the last video , we said , ok , that wave is just passing this guy . how long will it take for that pulse that 's being emitted right then to also reach him ? and then that tells us the period between two pulses or between two crests . now let 's think about that exact same situation here . that first crest is just passing this guy . and a period , or the t sub s , which is a period of the emitted wave has just passed by . so this guy is just about to emit another wave , so ? that other wave is going to be right here . so what is the distance between the crest , or the cycle , or however you want to think about it , the pulse that is passing him by right now , what is the distance between that and the pulse , the front edge of that pulse that is being emitted right at that moment ? right at that moment . well , it 's going to be this radius , which is this value . it 'll be v sub w times the period . that is that distance plus the amount of distance that our source has traveled away from this guy . so plus v sub s times the period . so that 's what this distance is , and this is how far apart this wave pulse is going to be from that one or this crest is going to be from that one . so if he 's seeing this first crest right now , right at this moment , how long will it take him to see the crest that 's being emitted right now , that 's this far away from him ? well , it 's that far away from him . so let me write this down . so the amount of time it takes for him to see the next crest or the same point in the next cycle , that 's the period . that 's the observed period . that 's going to be equal to this distance . the velocity of the wave times the period from the perspective of the source plus the velocity of the source , because the source has gotten that much further away from him , velocity of the source times the period of the source , so that 's how far the next crest is . and then you divide it by the speed of the wave , by the speed of each of the crests , which is just the velocity of the wave . and we can just factor out the t sub s 's here and say this is t sub s times v sub w , the velocity of the wave plus the velocity of the source divided by the velocity the wave . so this will be a larger observed period than if this guy was stationary and especially if the observer was in the path of the guy . that make sense , because every time this guy issues a cycle , he is moving a little bit further away . so every crest the same point in the cycle is going to be further and further apart , so you 're going to have longer wavelengths , longer periods . and if you want the observed frequency for that guy over there -- i 'll do it in the same color . the observed frequency for a guy where the source is traveling away from him is just the inverse of that . so one over the period , one over the period , same argument we did there , one over the period from the point of view of the source is the frequency of the source . let me color code it . so 1 over t sub s is equal to the frequency of the source . this is the inverse of that . so i 'm just taking one over everything here . so 1 over t sub s is the frequency of the source , and then we take the inverse of this over here : the velocity of the wave divided by the velocity of the wave plus the velocity of the source . so we 're done , at least for the simple cases . obviously , it becomes a little more interesting when someone is n't exactly in the direction of the source or exactly being moved away from , but these are kind of the two extreme cases . so this is the situation when it is moving away from you . now , just to check our math and maybe make it a little bit concrete in relation to the video we did where we introduced to the idea of the doppler effect , let 's actually apply those numbers . so on that video , two videos ago , we had a situation where the velocity of our source was 5 meters per second to the right , and the velocity of the wave was 10 meters per second radially outward , and the period of our wave -- i 'll call it the period of our -- let me do it in another color -- from the point of view of the source was 1 second per cycle , and the frequency was just the inverse of that . so 1 cycle per second , or 1 hertz , which is a cycle per second . so using those numbers , let 's see if we get to the exact same answer we got in that first video where we first learned about the doppler effect . so let 's look at the frequency from the point of view of this gentleman right here . so the frequency of the source is going to be 1 cycle per second , 1 hertz . the velocity of the wave is 10 meters per second . let me write this . 1 cycle per second . the velocity of the wave is 10 meters per second . the velocity of the wave is 10 meters per second minus the velocity of the source is 5 meters per second . so what 's this going to be equal to ? the observed frequency for this guy right there , is going to be 1 cycle per second times 10 over 10 minus 5 , and the meters per second cancel out . meters per second in the numerator , meters per second in the denominator . so 10 divided by 10 minus 5 , or 10 divided by 5 , is going to be 2 . so it 's going to be 2 cycles per second . and if you want the observed period for this guy , its going to be the inverse of that , or it 's going to be 1/2 second per cycle . and this is exactly what we got in the previous video , or actually , it was two videos ago . now what about the guy who this guy 's running away from ? well , we 'll do the exact same thing . you have 1 cycle per second , or 1 hertz . that 's the emitted frequency from the point of view of the source times the velocity of the wave divided by the velocity of the wave plus the velocity of the source , because it 's moving away from him . so it 's 10 over 10 plus 5 . that 's 10 over 15 . that is 2/3 , so this is equal to 2/3 . the units over here all cancel out . this was cycles per second . so 2/3 cycle per second , which confirms the numbers we got in that first video , so that should make us feel good and it also makes a lot of sense . this guy is going to see the wave crest more frequently . he 's going to observe a higher frequency . if this is a sound , a higher pitch . this guy , since each crest or the cycles are getting spread out , he 's going to see them less frequently , and if this is sound , he 's going to observe a lower pitch .
|
and if you want the observed frequency for that guy over there -- i 'll do it in the same color . the observed frequency for a guy where the source is traveling away from him is just the inverse of that . so one over the period , one over the period , same argument we did there , one over the period from the point of view of the source is the frequency of the source .
|
if the frequency observed by someone who is getting further and further away from the source of the wave is always the same , why do people hear lower and lower frequency ( volume ) when an ambulance is getting away from them ?
|
in the last video , we figured out the formulas for the observed period and frequency for an observer sitting in the path of the source . so the source is moving towards the observer . so this is the example where the train is moving towards you , and you perceive the train 's horn as having a higher pitch or a higher frequency . and we were able to do that by doing a thought experiment . saying , ok , my object starts here . after one period -- a period is just a measure of time , but it 's the measure of time over which the source emits a cycle , so it emits a cycle every period . but after one period , we said , ok , where is that first wave front , or that first pulse , or that first crest ? and where is the source ? because exactly one period has passed by , and the source will be ready to emit another crest or another cycle . so the distance between where the source is and that front of the crest , or that first crest , that is going to be the wavelength . because this next thing that emits is going to be traveling at the exact same velocity , and it 's going to be separated by that distance , which we saw is this expression . we said how long will it take it to travel that distance ? well , it 's traveling at a speed of v sub w. that 'll tell you what the observed period would be for this dude over here . we calculated it right here , and then the observed frequency is just the inverse of that . now let 's think about the situation where the observer is over here . so these equations , or these formulas that we came up with right here , this is observer -- or let me say source traveling in direction of observer . now let 's think about the opposite case , where the source is traveling away from the observer , and in this case , the observer is that guy over there . maybe i 'll do it in a different color . he 'll be blue . so this is the observer . so when we started off , our source was right here . after exactly one period from the source 's point of view , that first crest emitted has traveled radially outward that far . this is the distance . the velocity of the wave times the amount of time that passed -- velocity times time is going to give you distance -- and where the source of the wave will have traveled to the right exactly this distance . it 's velocity times the amount of time that 's gone by . now , in the last video , we said , ok , that wave is just passing this guy . how long will it take for that pulse that 's being emitted right then to also reach him ? and then that tells us the period between two pulses or between two crests . now let 's think about that exact same situation here . that first crest is just passing this guy . and a period , or the t sub s , which is a period of the emitted wave has just passed by . so this guy is just about to emit another wave , so ? that other wave is going to be right here . so what is the distance between the crest , or the cycle , or however you want to think about it , the pulse that is passing him by right now , what is the distance between that and the pulse , the front edge of that pulse that is being emitted right at that moment ? right at that moment . well , it 's going to be this radius , which is this value . it 'll be v sub w times the period . that is that distance plus the amount of distance that our source has traveled away from this guy . so plus v sub s times the period . so that 's what this distance is , and this is how far apart this wave pulse is going to be from that one or this crest is going to be from that one . so if he 's seeing this first crest right now , right at this moment , how long will it take him to see the crest that 's being emitted right now , that 's this far away from him ? well , it 's that far away from him . so let me write this down . so the amount of time it takes for him to see the next crest or the same point in the next cycle , that 's the period . that 's the observed period . that 's going to be equal to this distance . the velocity of the wave times the period from the perspective of the source plus the velocity of the source , because the source has gotten that much further away from him , velocity of the source times the period of the source , so that 's how far the next crest is . and then you divide it by the speed of the wave , by the speed of each of the crests , which is just the velocity of the wave . and we can just factor out the t sub s 's here and say this is t sub s times v sub w , the velocity of the wave plus the velocity of the source divided by the velocity the wave . so this will be a larger observed period than if this guy was stationary and especially if the observer was in the path of the guy . that make sense , because every time this guy issues a cycle , he is moving a little bit further away . so every crest the same point in the cycle is going to be further and further apart , so you 're going to have longer wavelengths , longer periods . and if you want the observed frequency for that guy over there -- i 'll do it in the same color . the observed frequency for a guy where the source is traveling away from him is just the inverse of that . so one over the period , one over the period , same argument we did there , one over the period from the point of view of the source is the frequency of the source . let me color code it . so 1 over t sub s is equal to the frequency of the source . this is the inverse of that . so i 'm just taking one over everything here . so 1 over t sub s is the frequency of the source , and then we take the inverse of this over here : the velocity of the wave divided by the velocity of the wave plus the velocity of the source . so we 're done , at least for the simple cases . obviously , it becomes a little more interesting when someone is n't exactly in the direction of the source or exactly being moved away from , but these are kind of the two extreme cases . so this is the situation when it is moving away from you . now , just to check our math and maybe make it a little bit concrete in relation to the video we did where we introduced to the idea of the doppler effect , let 's actually apply those numbers . so on that video , two videos ago , we had a situation where the velocity of our source was 5 meters per second to the right , and the velocity of the wave was 10 meters per second radially outward , and the period of our wave -- i 'll call it the period of our -- let me do it in another color -- from the point of view of the source was 1 second per cycle , and the frequency was just the inverse of that . so 1 cycle per second , or 1 hertz , which is a cycle per second . so using those numbers , let 's see if we get to the exact same answer we got in that first video where we first learned about the doppler effect . so let 's look at the frequency from the point of view of this gentleman right here . so the frequency of the source is going to be 1 cycle per second , 1 hertz . the velocity of the wave is 10 meters per second . let me write this . 1 cycle per second . the velocity of the wave is 10 meters per second . the velocity of the wave is 10 meters per second minus the velocity of the source is 5 meters per second . so what 's this going to be equal to ? the observed frequency for this guy right there , is going to be 1 cycle per second times 10 over 10 minus 5 , and the meters per second cancel out . meters per second in the numerator , meters per second in the denominator . so 10 divided by 10 minus 5 , or 10 divided by 5 , is going to be 2 . so it 's going to be 2 cycles per second . and if you want the observed period for this guy , its going to be the inverse of that , or it 's going to be 1/2 second per cycle . and this is exactly what we got in the previous video , or actually , it was two videos ago . now what about the guy who this guy 's running away from ? well , we 'll do the exact same thing . you have 1 cycle per second , or 1 hertz . that 's the emitted frequency from the point of view of the source times the velocity of the wave divided by the velocity of the wave plus the velocity of the source , because it 's moving away from him . so it 's 10 over 10 plus 5 . that 's 10 over 15 . that is 2/3 , so this is equal to 2/3 . the units over here all cancel out . this was cycles per second . so 2/3 cycle per second , which confirms the numbers we got in that first video , so that should make us feel good and it also makes a lot of sense . this guy is going to see the wave crest more frequently . he 's going to observe a higher frequency . if this is a sound , a higher pitch . this guy , since each crest or the cycles are getting spread out , he 's going to see them less frequently , and if this is sound , he 's going to observe a lower pitch .
|
that 's going to be equal to this distance . the velocity of the wave times the period from the perspective of the source plus the velocity of the source , because the source has gotten that much further away from him , velocity of the source times the period of the source , so that 's how far the next crest is . and then you divide it by the speed of the wave , by the speed of each of the crests , which is just the velocity of the wave .
|
what will be when source is faster 5 times of sound ?
|
in the last video , we figured out the formulas for the observed period and frequency for an observer sitting in the path of the source . so the source is moving towards the observer . so this is the example where the train is moving towards you , and you perceive the train 's horn as having a higher pitch or a higher frequency . and we were able to do that by doing a thought experiment . saying , ok , my object starts here . after one period -- a period is just a measure of time , but it 's the measure of time over which the source emits a cycle , so it emits a cycle every period . but after one period , we said , ok , where is that first wave front , or that first pulse , or that first crest ? and where is the source ? because exactly one period has passed by , and the source will be ready to emit another crest or another cycle . so the distance between where the source is and that front of the crest , or that first crest , that is going to be the wavelength . because this next thing that emits is going to be traveling at the exact same velocity , and it 's going to be separated by that distance , which we saw is this expression . we said how long will it take it to travel that distance ? well , it 's traveling at a speed of v sub w. that 'll tell you what the observed period would be for this dude over here . we calculated it right here , and then the observed frequency is just the inverse of that . now let 's think about the situation where the observer is over here . so these equations , or these formulas that we came up with right here , this is observer -- or let me say source traveling in direction of observer . now let 's think about the opposite case , where the source is traveling away from the observer , and in this case , the observer is that guy over there . maybe i 'll do it in a different color . he 'll be blue . so this is the observer . so when we started off , our source was right here . after exactly one period from the source 's point of view , that first crest emitted has traveled radially outward that far . this is the distance . the velocity of the wave times the amount of time that passed -- velocity times time is going to give you distance -- and where the source of the wave will have traveled to the right exactly this distance . it 's velocity times the amount of time that 's gone by . now , in the last video , we said , ok , that wave is just passing this guy . how long will it take for that pulse that 's being emitted right then to also reach him ? and then that tells us the period between two pulses or between two crests . now let 's think about that exact same situation here . that first crest is just passing this guy . and a period , or the t sub s , which is a period of the emitted wave has just passed by . so this guy is just about to emit another wave , so ? that other wave is going to be right here . so what is the distance between the crest , or the cycle , or however you want to think about it , the pulse that is passing him by right now , what is the distance between that and the pulse , the front edge of that pulse that is being emitted right at that moment ? right at that moment . well , it 's going to be this radius , which is this value . it 'll be v sub w times the period . that is that distance plus the amount of distance that our source has traveled away from this guy . so plus v sub s times the period . so that 's what this distance is , and this is how far apart this wave pulse is going to be from that one or this crest is going to be from that one . so if he 's seeing this first crest right now , right at this moment , how long will it take him to see the crest that 's being emitted right now , that 's this far away from him ? well , it 's that far away from him . so let me write this down . so the amount of time it takes for him to see the next crest or the same point in the next cycle , that 's the period . that 's the observed period . that 's going to be equal to this distance . the velocity of the wave times the period from the perspective of the source plus the velocity of the source , because the source has gotten that much further away from him , velocity of the source times the period of the source , so that 's how far the next crest is . and then you divide it by the speed of the wave , by the speed of each of the crests , which is just the velocity of the wave . and we can just factor out the t sub s 's here and say this is t sub s times v sub w , the velocity of the wave plus the velocity of the source divided by the velocity the wave . so this will be a larger observed period than if this guy was stationary and especially if the observer was in the path of the guy . that make sense , because every time this guy issues a cycle , he is moving a little bit further away . so every crest the same point in the cycle is going to be further and further apart , so you 're going to have longer wavelengths , longer periods . and if you want the observed frequency for that guy over there -- i 'll do it in the same color . the observed frequency for a guy where the source is traveling away from him is just the inverse of that . so one over the period , one over the period , same argument we did there , one over the period from the point of view of the source is the frequency of the source . let me color code it . so 1 over t sub s is equal to the frequency of the source . this is the inverse of that . so i 'm just taking one over everything here . so 1 over t sub s is the frequency of the source , and then we take the inverse of this over here : the velocity of the wave divided by the velocity of the wave plus the velocity of the source . so we 're done , at least for the simple cases . obviously , it becomes a little more interesting when someone is n't exactly in the direction of the source or exactly being moved away from , but these are kind of the two extreme cases . so this is the situation when it is moving away from you . now , just to check our math and maybe make it a little bit concrete in relation to the video we did where we introduced to the idea of the doppler effect , let 's actually apply those numbers . so on that video , two videos ago , we had a situation where the velocity of our source was 5 meters per second to the right , and the velocity of the wave was 10 meters per second radially outward , and the period of our wave -- i 'll call it the period of our -- let me do it in another color -- from the point of view of the source was 1 second per cycle , and the frequency was just the inverse of that . so 1 cycle per second , or 1 hertz , which is a cycle per second . so using those numbers , let 's see if we get to the exact same answer we got in that first video where we first learned about the doppler effect . so let 's look at the frequency from the point of view of this gentleman right here . so the frequency of the source is going to be 1 cycle per second , 1 hertz . the velocity of the wave is 10 meters per second . let me write this . 1 cycle per second . the velocity of the wave is 10 meters per second . the velocity of the wave is 10 meters per second minus the velocity of the source is 5 meters per second . so what 's this going to be equal to ? the observed frequency for this guy right there , is going to be 1 cycle per second times 10 over 10 minus 5 , and the meters per second cancel out . meters per second in the numerator , meters per second in the denominator . so 10 divided by 10 minus 5 , or 10 divided by 5 , is going to be 2 . so it 's going to be 2 cycles per second . and if you want the observed period for this guy , its going to be the inverse of that , or it 's going to be 1/2 second per cycle . and this is exactly what we got in the previous video , or actually , it was two videos ago . now what about the guy who this guy 's running away from ? well , we 'll do the exact same thing . you have 1 cycle per second , or 1 hertz . that 's the emitted frequency from the point of view of the source times the velocity of the wave divided by the velocity of the wave plus the velocity of the source , because it 's moving away from him . so it 's 10 over 10 plus 5 . that 's 10 over 15 . that is 2/3 , so this is equal to 2/3 . the units over here all cancel out . this was cycles per second . so 2/3 cycle per second , which confirms the numbers we got in that first video , so that should make us feel good and it also makes a lot of sense . this guy is going to see the wave crest more frequently . he 's going to observe a higher frequency . if this is a sound , a higher pitch . this guy , since each crest or the cycles are getting spread out , he 's going to see them less frequently , and if this is sound , he 's going to observe a lower pitch .
|
how long will it take for that pulse that 's being emitted right then to also reach him ? and then that tells us the period between two pulses or between two crests . now let 's think about that exact same situation here .
|
what happens when train is out of circumference and two guys are in motion ?
|
in the last video , we figured out the formulas for the observed period and frequency for an observer sitting in the path of the source . so the source is moving towards the observer . so this is the example where the train is moving towards you , and you perceive the train 's horn as having a higher pitch or a higher frequency . and we were able to do that by doing a thought experiment . saying , ok , my object starts here . after one period -- a period is just a measure of time , but it 's the measure of time over which the source emits a cycle , so it emits a cycle every period . but after one period , we said , ok , where is that first wave front , or that first pulse , or that first crest ? and where is the source ? because exactly one period has passed by , and the source will be ready to emit another crest or another cycle . so the distance between where the source is and that front of the crest , or that first crest , that is going to be the wavelength . because this next thing that emits is going to be traveling at the exact same velocity , and it 's going to be separated by that distance , which we saw is this expression . we said how long will it take it to travel that distance ? well , it 's traveling at a speed of v sub w. that 'll tell you what the observed period would be for this dude over here . we calculated it right here , and then the observed frequency is just the inverse of that . now let 's think about the situation where the observer is over here . so these equations , or these formulas that we came up with right here , this is observer -- or let me say source traveling in direction of observer . now let 's think about the opposite case , where the source is traveling away from the observer , and in this case , the observer is that guy over there . maybe i 'll do it in a different color . he 'll be blue . so this is the observer . so when we started off , our source was right here . after exactly one period from the source 's point of view , that first crest emitted has traveled radially outward that far . this is the distance . the velocity of the wave times the amount of time that passed -- velocity times time is going to give you distance -- and where the source of the wave will have traveled to the right exactly this distance . it 's velocity times the amount of time that 's gone by . now , in the last video , we said , ok , that wave is just passing this guy . how long will it take for that pulse that 's being emitted right then to also reach him ? and then that tells us the period between two pulses or between two crests . now let 's think about that exact same situation here . that first crest is just passing this guy . and a period , or the t sub s , which is a period of the emitted wave has just passed by . so this guy is just about to emit another wave , so ? that other wave is going to be right here . so what is the distance between the crest , or the cycle , or however you want to think about it , the pulse that is passing him by right now , what is the distance between that and the pulse , the front edge of that pulse that is being emitted right at that moment ? right at that moment . well , it 's going to be this radius , which is this value . it 'll be v sub w times the period . that is that distance plus the amount of distance that our source has traveled away from this guy . so plus v sub s times the period . so that 's what this distance is , and this is how far apart this wave pulse is going to be from that one or this crest is going to be from that one . so if he 's seeing this first crest right now , right at this moment , how long will it take him to see the crest that 's being emitted right now , that 's this far away from him ? well , it 's that far away from him . so let me write this down . so the amount of time it takes for him to see the next crest or the same point in the next cycle , that 's the period . that 's the observed period . that 's going to be equal to this distance . the velocity of the wave times the period from the perspective of the source plus the velocity of the source , because the source has gotten that much further away from him , velocity of the source times the period of the source , so that 's how far the next crest is . and then you divide it by the speed of the wave , by the speed of each of the crests , which is just the velocity of the wave . and we can just factor out the t sub s 's here and say this is t sub s times v sub w , the velocity of the wave plus the velocity of the source divided by the velocity the wave . so this will be a larger observed period than if this guy was stationary and especially if the observer was in the path of the guy . that make sense , because every time this guy issues a cycle , he is moving a little bit further away . so every crest the same point in the cycle is going to be further and further apart , so you 're going to have longer wavelengths , longer periods . and if you want the observed frequency for that guy over there -- i 'll do it in the same color . the observed frequency for a guy where the source is traveling away from him is just the inverse of that . so one over the period , one over the period , same argument we did there , one over the period from the point of view of the source is the frequency of the source . let me color code it . so 1 over t sub s is equal to the frequency of the source . this is the inverse of that . so i 'm just taking one over everything here . so 1 over t sub s is the frequency of the source , and then we take the inverse of this over here : the velocity of the wave divided by the velocity of the wave plus the velocity of the source . so we 're done , at least for the simple cases . obviously , it becomes a little more interesting when someone is n't exactly in the direction of the source or exactly being moved away from , but these are kind of the two extreme cases . so this is the situation when it is moving away from you . now , just to check our math and maybe make it a little bit concrete in relation to the video we did where we introduced to the idea of the doppler effect , let 's actually apply those numbers . so on that video , two videos ago , we had a situation where the velocity of our source was 5 meters per second to the right , and the velocity of the wave was 10 meters per second radially outward , and the period of our wave -- i 'll call it the period of our -- let me do it in another color -- from the point of view of the source was 1 second per cycle , and the frequency was just the inverse of that . so 1 cycle per second , or 1 hertz , which is a cycle per second . so using those numbers , let 's see if we get to the exact same answer we got in that first video where we first learned about the doppler effect . so let 's look at the frequency from the point of view of this gentleman right here . so the frequency of the source is going to be 1 cycle per second , 1 hertz . the velocity of the wave is 10 meters per second . let me write this . 1 cycle per second . the velocity of the wave is 10 meters per second . the velocity of the wave is 10 meters per second minus the velocity of the source is 5 meters per second . so what 's this going to be equal to ? the observed frequency for this guy right there , is going to be 1 cycle per second times 10 over 10 minus 5 , and the meters per second cancel out . meters per second in the numerator , meters per second in the denominator . so 10 divided by 10 minus 5 , or 10 divided by 5 , is going to be 2 . so it 's going to be 2 cycles per second . and if you want the observed period for this guy , its going to be the inverse of that , or it 's going to be 1/2 second per cycle . and this is exactly what we got in the previous video , or actually , it was two videos ago . now what about the guy who this guy 's running away from ? well , we 'll do the exact same thing . you have 1 cycle per second , or 1 hertz . that 's the emitted frequency from the point of view of the source times the velocity of the wave divided by the velocity of the wave plus the velocity of the source , because it 's moving away from him . so it 's 10 over 10 plus 5 . that 's 10 over 15 . that is 2/3 , so this is equal to 2/3 . the units over here all cancel out . this was cycles per second . so 2/3 cycle per second , which confirms the numbers we got in that first video , so that should make us feel good and it also makes a lot of sense . this guy is going to see the wave crest more frequently . he 's going to observe a higher frequency . if this is a sound , a higher pitch . this guy , since each crest or the cycles are getting spread out , he 's going to see them less frequently , and if this is sound , he 's going to observe a lower pitch .
|
in the last video , we figured out the formulas for the observed period and frequency for an observer sitting in the path of the source . so the source is moving towards the observer . so this is the example where the train is moving towards you , and you perceive the train 's horn as having a higher pitch or a higher frequency .
|
what happens to the observer when the source is moving away ?
|
in the last video , we figured out the formulas for the observed period and frequency for an observer sitting in the path of the source . so the source is moving towards the observer . so this is the example where the train is moving towards you , and you perceive the train 's horn as having a higher pitch or a higher frequency . and we were able to do that by doing a thought experiment . saying , ok , my object starts here . after one period -- a period is just a measure of time , but it 's the measure of time over which the source emits a cycle , so it emits a cycle every period . but after one period , we said , ok , where is that first wave front , or that first pulse , or that first crest ? and where is the source ? because exactly one period has passed by , and the source will be ready to emit another crest or another cycle . so the distance between where the source is and that front of the crest , or that first crest , that is going to be the wavelength . because this next thing that emits is going to be traveling at the exact same velocity , and it 's going to be separated by that distance , which we saw is this expression . we said how long will it take it to travel that distance ? well , it 's traveling at a speed of v sub w. that 'll tell you what the observed period would be for this dude over here . we calculated it right here , and then the observed frequency is just the inverse of that . now let 's think about the situation where the observer is over here . so these equations , or these formulas that we came up with right here , this is observer -- or let me say source traveling in direction of observer . now let 's think about the opposite case , where the source is traveling away from the observer , and in this case , the observer is that guy over there . maybe i 'll do it in a different color . he 'll be blue . so this is the observer . so when we started off , our source was right here . after exactly one period from the source 's point of view , that first crest emitted has traveled radially outward that far . this is the distance . the velocity of the wave times the amount of time that passed -- velocity times time is going to give you distance -- and where the source of the wave will have traveled to the right exactly this distance . it 's velocity times the amount of time that 's gone by . now , in the last video , we said , ok , that wave is just passing this guy . how long will it take for that pulse that 's being emitted right then to also reach him ? and then that tells us the period between two pulses or between two crests . now let 's think about that exact same situation here . that first crest is just passing this guy . and a period , or the t sub s , which is a period of the emitted wave has just passed by . so this guy is just about to emit another wave , so ? that other wave is going to be right here . so what is the distance between the crest , or the cycle , or however you want to think about it , the pulse that is passing him by right now , what is the distance between that and the pulse , the front edge of that pulse that is being emitted right at that moment ? right at that moment . well , it 's going to be this radius , which is this value . it 'll be v sub w times the period . that is that distance plus the amount of distance that our source has traveled away from this guy . so plus v sub s times the period . so that 's what this distance is , and this is how far apart this wave pulse is going to be from that one or this crest is going to be from that one . so if he 's seeing this first crest right now , right at this moment , how long will it take him to see the crest that 's being emitted right now , that 's this far away from him ? well , it 's that far away from him . so let me write this down . so the amount of time it takes for him to see the next crest or the same point in the next cycle , that 's the period . that 's the observed period . that 's going to be equal to this distance . the velocity of the wave times the period from the perspective of the source plus the velocity of the source , because the source has gotten that much further away from him , velocity of the source times the period of the source , so that 's how far the next crest is . and then you divide it by the speed of the wave , by the speed of each of the crests , which is just the velocity of the wave . and we can just factor out the t sub s 's here and say this is t sub s times v sub w , the velocity of the wave plus the velocity of the source divided by the velocity the wave . so this will be a larger observed period than if this guy was stationary and especially if the observer was in the path of the guy . that make sense , because every time this guy issues a cycle , he is moving a little bit further away . so every crest the same point in the cycle is going to be further and further apart , so you 're going to have longer wavelengths , longer periods . and if you want the observed frequency for that guy over there -- i 'll do it in the same color . the observed frequency for a guy where the source is traveling away from him is just the inverse of that . so one over the period , one over the period , same argument we did there , one over the period from the point of view of the source is the frequency of the source . let me color code it . so 1 over t sub s is equal to the frequency of the source . this is the inverse of that . so i 'm just taking one over everything here . so 1 over t sub s is the frequency of the source , and then we take the inverse of this over here : the velocity of the wave divided by the velocity of the wave plus the velocity of the source . so we 're done , at least for the simple cases . obviously , it becomes a little more interesting when someone is n't exactly in the direction of the source or exactly being moved away from , but these are kind of the two extreme cases . so this is the situation when it is moving away from you . now , just to check our math and maybe make it a little bit concrete in relation to the video we did where we introduced to the idea of the doppler effect , let 's actually apply those numbers . so on that video , two videos ago , we had a situation where the velocity of our source was 5 meters per second to the right , and the velocity of the wave was 10 meters per second radially outward , and the period of our wave -- i 'll call it the period of our -- let me do it in another color -- from the point of view of the source was 1 second per cycle , and the frequency was just the inverse of that . so 1 cycle per second , or 1 hertz , which is a cycle per second . so using those numbers , let 's see if we get to the exact same answer we got in that first video where we first learned about the doppler effect . so let 's look at the frequency from the point of view of this gentleman right here . so the frequency of the source is going to be 1 cycle per second , 1 hertz . the velocity of the wave is 10 meters per second . let me write this . 1 cycle per second . the velocity of the wave is 10 meters per second . the velocity of the wave is 10 meters per second minus the velocity of the source is 5 meters per second . so what 's this going to be equal to ? the observed frequency for this guy right there , is going to be 1 cycle per second times 10 over 10 minus 5 , and the meters per second cancel out . meters per second in the numerator , meters per second in the denominator . so 10 divided by 10 minus 5 , or 10 divided by 5 , is going to be 2 . so it 's going to be 2 cycles per second . and if you want the observed period for this guy , its going to be the inverse of that , or it 's going to be 1/2 second per cycle . and this is exactly what we got in the previous video , or actually , it was two videos ago . now what about the guy who this guy 's running away from ? well , we 'll do the exact same thing . you have 1 cycle per second , or 1 hertz . that 's the emitted frequency from the point of view of the source times the velocity of the wave divided by the velocity of the wave plus the velocity of the source , because it 's moving away from him . so it 's 10 over 10 plus 5 . that 's 10 over 15 . that is 2/3 , so this is equal to 2/3 . the units over here all cancel out . this was cycles per second . so 2/3 cycle per second , which confirms the numbers we got in that first video , so that should make us feel good and it also makes a lot of sense . this guy is going to see the wave crest more frequently . he 's going to observe a higher frequency . if this is a sound , a higher pitch . this guy , since each crest or the cycles are getting spread out , he 's going to see them less frequently , and if this is sound , he 's going to observe a lower pitch .
|
let me write this . 1 cycle per second . the velocity of the wave is 10 meters per second . the velocity of the wave is 10 meters per second minus the velocity of the source is 5 meters per second .
|
a listener moves froma to b with a constant velocity u. if the speed of sound is 330m/s , what must be the value of u so that he hears 8 beats per second ?
|
in the last video , we figured out the formulas for the observed period and frequency for an observer sitting in the path of the source . so the source is moving towards the observer . so this is the example where the train is moving towards you , and you perceive the train 's horn as having a higher pitch or a higher frequency . and we were able to do that by doing a thought experiment . saying , ok , my object starts here . after one period -- a period is just a measure of time , but it 's the measure of time over which the source emits a cycle , so it emits a cycle every period . but after one period , we said , ok , where is that first wave front , or that first pulse , or that first crest ? and where is the source ? because exactly one period has passed by , and the source will be ready to emit another crest or another cycle . so the distance between where the source is and that front of the crest , or that first crest , that is going to be the wavelength . because this next thing that emits is going to be traveling at the exact same velocity , and it 's going to be separated by that distance , which we saw is this expression . we said how long will it take it to travel that distance ? well , it 's traveling at a speed of v sub w. that 'll tell you what the observed period would be for this dude over here . we calculated it right here , and then the observed frequency is just the inverse of that . now let 's think about the situation where the observer is over here . so these equations , or these formulas that we came up with right here , this is observer -- or let me say source traveling in direction of observer . now let 's think about the opposite case , where the source is traveling away from the observer , and in this case , the observer is that guy over there . maybe i 'll do it in a different color . he 'll be blue . so this is the observer . so when we started off , our source was right here . after exactly one period from the source 's point of view , that first crest emitted has traveled radially outward that far . this is the distance . the velocity of the wave times the amount of time that passed -- velocity times time is going to give you distance -- and where the source of the wave will have traveled to the right exactly this distance . it 's velocity times the amount of time that 's gone by . now , in the last video , we said , ok , that wave is just passing this guy . how long will it take for that pulse that 's being emitted right then to also reach him ? and then that tells us the period between two pulses or between two crests . now let 's think about that exact same situation here . that first crest is just passing this guy . and a period , or the t sub s , which is a period of the emitted wave has just passed by . so this guy is just about to emit another wave , so ? that other wave is going to be right here . so what is the distance between the crest , or the cycle , or however you want to think about it , the pulse that is passing him by right now , what is the distance between that and the pulse , the front edge of that pulse that is being emitted right at that moment ? right at that moment . well , it 's going to be this radius , which is this value . it 'll be v sub w times the period . that is that distance plus the amount of distance that our source has traveled away from this guy . so plus v sub s times the period . so that 's what this distance is , and this is how far apart this wave pulse is going to be from that one or this crest is going to be from that one . so if he 's seeing this first crest right now , right at this moment , how long will it take him to see the crest that 's being emitted right now , that 's this far away from him ? well , it 's that far away from him . so let me write this down . so the amount of time it takes for him to see the next crest or the same point in the next cycle , that 's the period . that 's the observed period . that 's going to be equal to this distance . the velocity of the wave times the period from the perspective of the source plus the velocity of the source , because the source has gotten that much further away from him , velocity of the source times the period of the source , so that 's how far the next crest is . and then you divide it by the speed of the wave , by the speed of each of the crests , which is just the velocity of the wave . and we can just factor out the t sub s 's here and say this is t sub s times v sub w , the velocity of the wave plus the velocity of the source divided by the velocity the wave . so this will be a larger observed period than if this guy was stationary and especially if the observer was in the path of the guy . that make sense , because every time this guy issues a cycle , he is moving a little bit further away . so every crest the same point in the cycle is going to be further and further apart , so you 're going to have longer wavelengths , longer periods . and if you want the observed frequency for that guy over there -- i 'll do it in the same color . the observed frequency for a guy where the source is traveling away from him is just the inverse of that . so one over the period , one over the period , same argument we did there , one over the period from the point of view of the source is the frequency of the source . let me color code it . so 1 over t sub s is equal to the frequency of the source . this is the inverse of that . so i 'm just taking one over everything here . so 1 over t sub s is the frequency of the source , and then we take the inverse of this over here : the velocity of the wave divided by the velocity of the wave plus the velocity of the source . so we 're done , at least for the simple cases . obviously , it becomes a little more interesting when someone is n't exactly in the direction of the source or exactly being moved away from , but these are kind of the two extreme cases . so this is the situation when it is moving away from you . now , just to check our math and maybe make it a little bit concrete in relation to the video we did where we introduced to the idea of the doppler effect , let 's actually apply those numbers . so on that video , two videos ago , we had a situation where the velocity of our source was 5 meters per second to the right , and the velocity of the wave was 10 meters per second radially outward , and the period of our wave -- i 'll call it the period of our -- let me do it in another color -- from the point of view of the source was 1 second per cycle , and the frequency was just the inverse of that . so 1 cycle per second , or 1 hertz , which is a cycle per second . so using those numbers , let 's see if we get to the exact same answer we got in that first video where we first learned about the doppler effect . so let 's look at the frequency from the point of view of this gentleman right here . so the frequency of the source is going to be 1 cycle per second , 1 hertz . the velocity of the wave is 10 meters per second . let me write this . 1 cycle per second . the velocity of the wave is 10 meters per second . the velocity of the wave is 10 meters per second minus the velocity of the source is 5 meters per second . so what 's this going to be equal to ? the observed frequency for this guy right there , is going to be 1 cycle per second times 10 over 10 minus 5 , and the meters per second cancel out . meters per second in the numerator , meters per second in the denominator . so 10 divided by 10 minus 5 , or 10 divided by 5 , is going to be 2 . so it 's going to be 2 cycles per second . and if you want the observed period for this guy , its going to be the inverse of that , or it 's going to be 1/2 second per cycle . and this is exactly what we got in the previous video , or actually , it was two videos ago . now what about the guy who this guy 's running away from ? well , we 'll do the exact same thing . you have 1 cycle per second , or 1 hertz . that 's the emitted frequency from the point of view of the source times the velocity of the wave divided by the velocity of the wave plus the velocity of the source , because it 's moving away from him . so it 's 10 over 10 plus 5 . that 's 10 over 15 . that is 2/3 , so this is equal to 2/3 . the units over here all cancel out . this was cycles per second . so 2/3 cycle per second , which confirms the numbers we got in that first video , so that should make us feel good and it also makes a lot of sense . this guy is going to see the wave crest more frequently . he 's going to observe a higher frequency . if this is a sound , a higher pitch . this guy , since each crest or the cycles are getting spread out , he 's going to see them less frequently , and if this is sound , he 's going to observe a lower pitch .
|
so the amount of time it takes for him to see the next crest or the same point in the next cycle , that 's the period . that 's the observed period . that 's going to be equal to this distance .
|
what if we wanted the period/frequency if the source had traveled for another period after travelling a period ?
|
in the last video , we figured out the formulas for the observed period and frequency for an observer sitting in the path of the source . so the source is moving towards the observer . so this is the example where the train is moving towards you , and you perceive the train 's horn as having a higher pitch or a higher frequency . and we were able to do that by doing a thought experiment . saying , ok , my object starts here . after one period -- a period is just a measure of time , but it 's the measure of time over which the source emits a cycle , so it emits a cycle every period . but after one period , we said , ok , where is that first wave front , or that first pulse , or that first crest ? and where is the source ? because exactly one period has passed by , and the source will be ready to emit another crest or another cycle . so the distance between where the source is and that front of the crest , or that first crest , that is going to be the wavelength . because this next thing that emits is going to be traveling at the exact same velocity , and it 's going to be separated by that distance , which we saw is this expression . we said how long will it take it to travel that distance ? well , it 's traveling at a speed of v sub w. that 'll tell you what the observed period would be for this dude over here . we calculated it right here , and then the observed frequency is just the inverse of that . now let 's think about the situation where the observer is over here . so these equations , or these formulas that we came up with right here , this is observer -- or let me say source traveling in direction of observer . now let 's think about the opposite case , where the source is traveling away from the observer , and in this case , the observer is that guy over there . maybe i 'll do it in a different color . he 'll be blue . so this is the observer . so when we started off , our source was right here . after exactly one period from the source 's point of view , that first crest emitted has traveled radially outward that far . this is the distance . the velocity of the wave times the amount of time that passed -- velocity times time is going to give you distance -- and where the source of the wave will have traveled to the right exactly this distance . it 's velocity times the amount of time that 's gone by . now , in the last video , we said , ok , that wave is just passing this guy . how long will it take for that pulse that 's being emitted right then to also reach him ? and then that tells us the period between two pulses or between two crests . now let 's think about that exact same situation here . that first crest is just passing this guy . and a period , or the t sub s , which is a period of the emitted wave has just passed by . so this guy is just about to emit another wave , so ? that other wave is going to be right here . so what is the distance between the crest , or the cycle , or however you want to think about it , the pulse that is passing him by right now , what is the distance between that and the pulse , the front edge of that pulse that is being emitted right at that moment ? right at that moment . well , it 's going to be this radius , which is this value . it 'll be v sub w times the period . that is that distance plus the amount of distance that our source has traveled away from this guy . so plus v sub s times the period . so that 's what this distance is , and this is how far apart this wave pulse is going to be from that one or this crest is going to be from that one . so if he 's seeing this first crest right now , right at this moment , how long will it take him to see the crest that 's being emitted right now , that 's this far away from him ? well , it 's that far away from him . so let me write this down . so the amount of time it takes for him to see the next crest or the same point in the next cycle , that 's the period . that 's the observed period . that 's going to be equal to this distance . the velocity of the wave times the period from the perspective of the source plus the velocity of the source , because the source has gotten that much further away from him , velocity of the source times the period of the source , so that 's how far the next crest is . and then you divide it by the speed of the wave , by the speed of each of the crests , which is just the velocity of the wave . and we can just factor out the t sub s 's here and say this is t sub s times v sub w , the velocity of the wave plus the velocity of the source divided by the velocity the wave . so this will be a larger observed period than if this guy was stationary and especially if the observer was in the path of the guy . that make sense , because every time this guy issues a cycle , he is moving a little bit further away . so every crest the same point in the cycle is going to be further and further apart , so you 're going to have longer wavelengths , longer periods . and if you want the observed frequency for that guy over there -- i 'll do it in the same color . the observed frequency for a guy where the source is traveling away from him is just the inverse of that . so one over the period , one over the period , same argument we did there , one over the period from the point of view of the source is the frequency of the source . let me color code it . so 1 over t sub s is equal to the frequency of the source . this is the inverse of that . so i 'm just taking one over everything here . so 1 over t sub s is the frequency of the source , and then we take the inverse of this over here : the velocity of the wave divided by the velocity of the wave plus the velocity of the source . so we 're done , at least for the simple cases . obviously , it becomes a little more interesting when someone is n't exactly in the direction of the source or exactly being moved away from , but these are kind of the two extreme cases . so this is the situation when it is moving away from you . now , just to check our math and maybe make it a little bit concrete in relation to the video we did where we introduced to the idea of the doppler effect , let 's actually apply those numbers . so on that video , two videos ago , we had a situation where the velocity of our source was 5 meters per second to the right , and the velocity of the wave was 10 meters per second radially outward , and the period of our wave -- i 'll call it the period of our -- let me do it in another color -- from the point of view of the source was 1 second per cycle , and the frequency was just the inverse of that . so 1 cycle per second , or 1 hertz , which is a cycle per second . so using those numbers , let 's see if we get to the exact same answer we got in that first video where we first learned about the doppler effect . so let 's look at the frequency from the point of view of this gentleman right here . so the frequency of the source is going to be 1 cycle per second , 1 hertz . the velocity of the wave is 10 meters per second . let me write this . 1 cycle per second . the velocity of the wave is 10 meters per second . the velocity of the wave is 10 meters per second minus the velocity of the source is 5 meters per second . so what 's this going to be equal to ? the observed frequency for this guy right there , is going to be 1 cycle per second times 10 over 10 minus 5 , and the meters per second cancel out . meters per second in the numerator , meters per second in the denominator . so 10 divided by 10 minus 5 , or 10 divided by 5 , is going to be 2 . so it 's going to be 2 cycles per second . and if you want the observed period for this guy , its going to be the inverse of that , or it 's going to be 1/2 second per cycle . and this is exactly what we got in the previous video , or actually , it was two videos ago . now what about the guy who this guy 's running away from ? well , we 'll do the exact same thing . you have 1 cycle per second , or 1 hertz . that 's the emitted frequency from the point of view of the source times the velocity of the wave divided by the velocity of the wave plus the velocity of the source , because it 's moving away from him . so it 's 10 over 10 plus 5 . that 's 10 over 15 . that is 2/3 , so this is equal to 2/3 . the units over here all cancel out . this was cycles per second . so 2/3 cycle per second , which confirms the numbers we got in that first video , so that should make us feel good and it also makes a lot of sense . this guy is going to see the wave crest more frequently . he 's going to observe a higher frequency . if this is a sound , a higher pitch . this guy , since each crest or the cycles are getting spread out , he 's going to see them less frequently , and if this is sound , he 's going to observe a lower pitch .
|
and we can just factor out the t sub s 's here and say this is t sub s times v sub w , the velocity of the wave plus the velocity of the source divided by the velocity the wave . so this will be a larger observed period than if this guy was stationary and especially if the observer was in the path of the guy . that make sense , because every time this guy issues a cycle , he is moving a little bit further away .
|
what are stationary or standing waves ?
|
you know , it 's not unheard of for a guy to have a uti or urinary tract infection , but still , it 's pretty reasonable to say that utis are way more common in women . and in fact , there 's a statistic that i once read that said that one in five women will have a uti at some point in their life . so , yeah , i guess you could say it is pretty common in women . and to understand why this infection is so common in women , you have to understand the female anatomy . and i know that 's not an easy thing to do . female urinary anatomy is a bit complicated . so let 's simplify it . so , okay , if this is a woman 's body , what we 're going to do is kind of slice it down the middle and take a look down the midline . so if we do that , towards the back , towards the back of the body , we have the digestive system , so we have the rectum . and that 's what i 'm drawing here , the rectum . the rectum and with the anus , so this is the anus right here . and almost immediately in front of the anus is the vagina . okay , so the vagina . and the vagina is this canal that leads up to the uterus . so this is the uterus , the uterus , and this canal is the vagina . okay . and then immediately in front of the vagina is the urethra . now , urethra , urethra is the end or the outlet for the bladder . so this is the bladder , the bladder , and this is the urethra . and the bladder stores urine , right ? we know that . and that urine is made up in the kidney , which is then carried through the ureters and into the bladder . so this is the kidney , all right , the kidney . and these are the ureters or a ureter , leading up to the bladder . so it 's really apparent from this diagram that there 's a lot cramped into a pretty small space . okay . so back to urinary tract infection . so the term urinary tract really implies anything from the kidney down to the urethra . but to be a little bit more specific , infection of the urethra or this part right here , this part , and i 'm coloring it a different color , is called urethritis . so the infection of this part right here , the urethra , is called the urethritis . infection of the bladder , right , so this portion right here , this portion , is called cystitis . so the infection of the bladder is called cystitis . and then finally , infection of the kidneys , right here , so infection of the kidney is called pyelonephritis . it 's called pyelonephritis . and the fact is when most people talk about utis , they 're actually talking about cystitis or infection of the bladder . and the word `` infection '' implies some sort of pathogen , some bacteria of sorts . the bacteria in this case is usually e-coli , which comes from contamination from feces from back here in the anus . and that actually explains why utis are so common in women . the anus and the urethra are so close together in women that it makes contamination a lot more likely . and the urethra in women is also a lot shorter than it is in men , so the bacteria have a much shorter climb on their way to the bladder , also contributing to the fact that utis are more common in women . and that brings me to kind of a random but important point . so like i said , most utis , something like 80 % of them , are caused by e-coli . and one of the most important reasons for that is that e-coli have these things on them called pili that actually enable them to stick to the urinary tract and to climb up the urinary tract . so the pili are kind of like those suction cup things that ninjas and spies use in those movies to climb up the outside of buildings . that 's how i like to think of them . and pili are important because when urine comes rushing down the urinary tract , the e-coli are washed away like other bacteria without pili would be . so then , why are we talking about utis in pregnancy ? well , it turns out that they 're even more common in pregnant women . and a lot of that has to do with the uterus becoming larger throughout the course of the pregnancy , with the growing baby . so the large uterus does a couple of different things . firstly , it makes it pretty difficult to maintain hygiene in the area , making contamination more likely . and secondly , the uterus pushes on the bladder , making it hard to empty the bladder completely . so that means that each time the woman urinates , some urine is left behind . and that urine sitting around for a long time serves as a good medium for bacteria to grow and to thrive in . and another thing that happens in pregnancy is that the gfr or the glomerular filtration rate , or the rate at which blood is filtered through the kidneys , increases , so the gfr increases in pregnancy . and so what that means is that things like glucose are filtered a lot more out of blood and into the urine . and more glucose in the urine means more food for bacteria to grow , right ? and finally , pregnancy , especially the very end of pregnancy , is a sort of state of immune suppression , right ? and so that predisposes the mom to all sorts of infections , infections including utis . so then , let 's talk about some of the symptoms of cystitis , right , so some of the symptoms that you get when you have an infection of the bladder . burning and discomfort with urination , right , that 's called dysuria , occurs pretty commonly with cystitis . also a symptom called urinary frequency or feeling like you have to urinate really frequently , occurs pretty commonly . and urinary urgency , or feeling like you have to urinate immediately , is another common symptom . and occasionally , you can also have something called hematuria . and hematuria refers to blood in the urine . and this consolidation of symptoms or this group of symptoms , they 're really important to know because a person has to have these symptoms in order to be treated for the uti . that is to say that they ca n't be treated if they have only evidence of bacteria in their urine , but they do n't have any symptoms of the uti . but one of the important exceptions to that rule is pregnancy . so a pregnant woman who has bacteria in her urine without any symptoms of uti , which is called asymptomatic , it 's called asymptomatic , so without symptoms , bacteriuria , so when you have bacteria in the urine but do n't have any symptoms , right ? in a pregnant woman , that needs to be treated . and the reason for that is something like 25 % of pregnant women with asymptomatic bacteriuria end up with pyelonephritis . and that 's dangerous , very dangerous . in fact , because pyelonephritis in a pregnant woman can lead to sepsis . it can lead to kidney dysfunction , premature labor , and even death of the fetus . so that 's why we always , always collect the sample of urine in the very first trimester , even if a woman has absolutely no symptoms of uti , to look for bacteria in the urine . and if there 's evidence of bacteria in the urine , again , even if she has no symptoms , we go ahead and treat . and that brings me to what it takes to diagnose a uti . so when we 're trying to diagnose a uti , we collect a sample of urine , and what we 're looking for is a urine culture in which we have more than 100,000 colony-forming units , so more than 100,000 colonies of one type of bacteria that 's known to cause a uti . now , i want to mention one other important difference in utis in non-pregnant versus pregnant women , and that is treatment . so in a non-pregnant woman , or even a man , for that matter , your choice of treatment is pretty diverse . it could be something like trimethoprim , trimethoprim , sulfamethoxazole , right ? i know it 's a mouthful . and that 's a very popular treatment . another popular treatment is nitrofurantoin , right ? and a third popular choice is ciprofloxacin . ciprofloxacin , okay . and there are other less commonly used antibiotics also available for your choice . but your treatment choice in a pregnant woman is very different from that in a non-pregnant woman , because a lot of these choices are dangerous to the fetus . trimethoprim , for example , is a folate antagonist , so it blocks the use of folate by the baby . and we know that folate is really important for neural tube development . so for that reason , it 's usually avoided in the first trimester . and sulfamethoxazole can increase bilirubin levels in the baby , which can be pretty dangerous . so for that reason , it 's avoided . and ciprofloxacin is also avoided in pregnancy due to problems with bone development . so it leads to problems with bone development . so if you ca n't have those options available to you , some of the options that are available in pregnant women for the treatment of cystitis include amoxicillin , right ? you may have heard of that one before . amoxicillin is used for a bunch of different things . you can also use nitrofurantoin . so nitrofurantoin is one of those drugs that you can use safely in both pregnant and non-pregnant women . another popular option is cephalexin . cephalexin . and also fosfomycin . so those are some of the options that you have available to you for the use in pregnant women . so there you have it . that 's quite a bit of information on a very common disorder , urinary tract infection .
|
but one of the important exceptions to that rule is pregnancy . so a pregnant woman who has bacteria in her urine without any symptoms of uti , which is called asymptomatic , it 's called asymptomatic , so without symptoms , bacteriuria , so when you have bacteria in the urine but do n't have any symptoms , right ? in a pregnant woman , that needs to be treated . and the reason for that is something like 25 % of pregnant women with asymptomatic bacteriuria end up with pyelonephritis .
|
if a pregnant woman is diagnosed with a uti caused by bacteria ( and not a fungus or a virus ) then since cranberry juice not only prevents more bacteria from getting in like blueberry juice does but also acts as a natural antibiotic could n't the pregnant woman drink cranberry juice for about a week or 2 to naturally cure her uti ?
|
you know , it 's not unheard of for a guy to have a uti or urinary tract infection , but still , it 's pretty reasonable to say that utis are way more common in women . and in fact , there 's a statistic that i once read that said that one in five women will have a uti at some point in their life . so , yeah , i guess you could say it is pretty common in women . and to understand why this infection is so common in women , you have to understand the female anatomy . and i know that 's not an easy thing to do . female urinary anatomy is a bit complicated . so let 's simplify it . so , okay , if this is a woman 's body , what we 're going to do is kind of slice it down the middle and take a look down the midline . so if we do that , towards the back , towards the back of the body , we have the digestive system , so we have the rectum . and that 's what i 'm drawing here , the rectum . the rectum and with the anus , so this is the anus right here . and almost immediately in front of the anus is the vagina . okay , so the vagina . and the vagina is this canal that leads up to the uterus . so this is the uterus , the uterus , and this canal is the vagina . okay . and then immediately in front of the vagina is the urethra . now , urethra , urethra is the end or the outlet for the bladder . so this is the bladder , the bladder , and this is the urethra . and the bladder stores urine , right ? we know that . and that urine is made up in the kidney , which is then carried through the ureters and into the bladder . so this is the kidney , all right , the kidney . and these are the ureters or a ureter , leading up to the bladder . so it 's really apparent from this diagram that there 's a lot cramped into a pretty small space . okay . so back to urinary tract infection . so the term urinary tract really implies anything from the kidney down to the urethra . but to be a little bit more specific , infection of the urethra or this part right here , this part , and i 'm coloring it a different color , is called urethritis . so the infection of this part right here , the urethra , is called the urethritis . infection of the bladder , right , so this portion right here , this portion , is called cystitis . so the infection of the bladder is called cystitis . and then finally , infection of the kidneys , right here , so infection of the kidney is called pyelonephritis . it 's called pyelonephritis . and the fact is when most people talk about utis , they 're actually talking about cystitis or infection of the bladder . and the word `` infection '' implies some sort of pathogen , some bacteria of sorts . the bacteria in this case is usually e-coli , which comes from contamination from feces from back here in the anus . and that actually explains why utis are so common in women . the anus and the urethra are so close together in women that it makes contamination a lot more likely . and the urethra in women is also a lot shorter than it is in men , so the bacteria have a much shorter climb on their way to the bladder , also contributing to the fact that utis are more common in women . and that brings me to kind of a random but important point . so like i said , most utis , something like 80 % of them , are caused by e-coli . and one of the most important reasons for that is that e-coli have these things on them called pili that actually enable them to stick to the urinary tract and to climb up the urinary tract . so the pili are kind of like those suction cup things that ninjas and spies use in those movies to climb up the outside of buildings . that 's how i like to think of them . and pili are important because when urine comes rushing down the urinary tract , the e-coli are washed away like other bacteria without pili would be . so then , why are we talking about utis in pregnancy ? well , it turns out that they 're even more common in pregnant women . and a lot of that has to do with the uterus becoming larger throughout the course of the pregnancy , with the growing baby . so the large uterus does a couple of different things . firstly , it makes it pretty difficult to maintain hygiene in the area , making contamination more likely . and secondly , the uterus pushes on the bladder , making it hard to empty the bladder completely . so that means that each time the woman urinates , some urine is left behind . and that urine sitting around for a long time serves as a good medium for bacteria to grow and to thrive in . and another thing that happens in pregnancy is that the gfr or the glomerular filtration rate , or the rate at which blood is filtered through the kidneys , increases , so the gfr increases in pregnancy . and so what that means is that things like glucose are filtered a lot more out of blood and into the urine . and more glucose in the urine means more food for bacteria to grow , right ? and finally , pregnancy , especially the very end of pregnancy , is a sort of state of immune suppression , right ? and so that predisposes the mom to all sorts of infections , infections including utis . so then , let 's talk about some of the symptoms of cystitis , right , so some of the symptoms that you get when you have an infection of the bladder . burning and discomfort with urination , right , that 's called dysuria , occurs pretty commonly with cystitis . also a symptom called urinary frequency or feeling like you have to urinate really frequently , occurs pretty commonly . and urinary urgency , or feeling like you have to urinate immediately , is another common symptom . and occasionally , you can also have something called hematuria . and hematuria refers to blood in the urine . and this consolidation of symptoms or this group of symptoms , they 're really important to know because a person has to have these symptoms in order to be treated for the uti . that is to say that they ca n't be treated if they have only evidence of bacteria in their urine , but they do n't have any symptoms of the uti . but one of the important exceptions to that rule is pregnancy . so a pregnant woman who has bacteria in her urine without any symptoms of uti , which is called asymptomatic , it 's called asymptomatic , so without symptoms , bacteriuria , so when you have bacteria in the urine but do n't have any symptoms , right ? in a pregnant woman , that needs to be treated . and the reason for that is something like 25 % of pregnant women with asymptomatic bacteriuria end up with pyelonephritis . and that 's dangerous , very dangerous . in fact , because pyelonephritis in a pregnant woman can lead to sepsis . it can lead to kidney dysfunction , premature labor , and even death of the fetus . so that 's why we always , always collect the sample of urine in the very first trimester , even if a woman has absolutely no symptoms of uti , to look for bacteria in the urine . and if there 's evidence of bacteria in the urine , again , even if she has no symptoms , we go ahead and treat . and that brings me to what it takes to diagnose a uti . so when we 're trying to diagnose a uti , we collect a sample of urine , and what we 're looking for is a urine culture in which we have more than 100,000 colony-forming units , so more than 100,000 colonies of one type of bacteria that 's known to cause a uti . now , i want to mention one other important difference in utis in non-pregnant versus pregnant women , and that is treatment . so in a non-pregnant woman , or even a man , for that matter , your choice of treatment is pretty diverse . it could be something like trimethoprim , trimethoprim , sulfamethoxazole , right ? i know it 's a mouthful . and that 's a very popular treatment . another popular treatment is nitrofurantoin , right ? and a third popular choice is ciprofloxacin . ciprofloxacin , okay . and there are other less commonly used antibiotics also available for your choice . but your treatment choice in a pregnant woman is very different from that in a non-pregnant woman , because a lot of these choices are dangerous to the fetus . trimethoprim , for example , is a folate antagonist , so it blocks the use of folate by the baby . and we know that folate is really important for neural tube development . so for that reason , it 's usually avoided in the first trimester . and sulfamethoxazole can increase bilirubin levels in the baby , which can be pretty dangerous . so for that reason , it 's avoided . and ciprofloxacin is also avoided in pregnancy due to problems with bone development . so it leads to problems with bone development . so if you ca n't have those options available to you , some of the options that are available in pregnant women for the treatment of cystitis include amoxicillin , right ? you may have heard of that one before . amoxicillin is used for a bunch of different things . you can also use nitrofurantoin . so nitrofurantoin is one of those drugs that you can use safely in both pregnant and non-pregnant women . another popular option is cephalexin . cephalexin . and also fosfomycin . so those are some of the options that you have available to you for the use in pregnant women . so there you have it . that 's quite a bit of information on a very common disorder , urinary tract infection .
|
so a pregnant woman who has bacteria in her urine without any symptoms of uti , which is called asymptomatic , it 's called asymptomatic , so without symptoms , bacteriuria , so when you have bacteria in the urine but do n't have any symptoms , right ? in a pregnant woman , that needs to be treated . and the reason for that is something like 25 % of pregnant women with asymptomatic bacteriuria end up with pyelonephritis .
|
how would you treat a pregnant woman who has a uti because of a fungus or a virus ( which is less common but could potentially be more serious than a bacterial uti ) ?
|
you know , it 's not unheard of for a guy to have a uti or urinary tract infection , but still , it 's pretty reasonable to say that utis are way more common in women . and in fact , there 's a statistic that i once read that said that one in five women will have a uti at some point in their life . so , yeah , i guess you could say it is pretty common in women . and to understand why this infection is so common in women , you have to understand the female anatomy . and i know that 's not an easy thing to do . female urinary anatomy is a bit complicated . so let 's simplify it . so , okay , if this is a woman 's body , what we 're going to do is kind of slice it down the middle and take a look down the midline . so if we do that , towards the back , towards the back of the body , we have the digestive system , so we have the rectum . and that 's what i 'm drawing here , the rectum . the rectum and with the anus , so this is the anus right here . and almost immediately in front of the anus is the vagina . okay , so the vagina . and the vagina is this canal that leads up to the uterus . so this is the uterus , the uterus , and this canal is the vagina . okay . and then immediately in front of the vagina is the urethra . now , urethra , urethra is the end or the outlet for the bladder . so this is the bladder , the bladder , and this is the urethra . and the bladder stores urine , right ? we know that . and that urine is made up in the kidney , which is then carried through the ureters and into the bladder . so this is the kidney , all right , the kidney . and these are the ureters or a ureter , leading up to the bladder . so it 's really apparent from this diagram that there 's a lot cramped into a pretty small space . okay . so back to urinary tract infection . so the term urinary tract really implies anything from the kidney down to the urethra . but to be a little bit more specific , infection of the urethra or this part right here , this part , and i 'm coloring it a different color , is called urethritis . so the infection of this part right here , the urethra , is called the urethritis . infection of the bladder , right , so this portion right here , this portion , is called cystitis . so the infection of the bladder is called cystitis . and then finally , infection of the kidneys , right here , so infection of the kidney is called pyelonephritis . it 's called pyelonephritis . and the fact is when most people talk about utis , they 're actually talking about cystitis or infection of the bladder . and the word `` infection '' implies some sort of pathogen , some bacteria of sorts . the bacteria in this case is usually e-coli , which comes from contamination from feces from back here in the anus . and that actually explains why utis are so common in women . the anus and the urethra are so close together in women that it makes contamination a lot more likely . and the urethra in women is also a lot shorter than it is in men , so the bacteria have a much shorter climb on their way to the bladder , also contributing to the fact that utis are more common in women . and that brings me to kind of a random but important point . so like i said , most utis , something like 80 % of them , are caused by e-coli . and one of the most important reasons for that is that e-coli have these things on them called pili that actually enable them to stick to the urinary tract and to climb up the urinary tract . so the pili are kind of like those suction cup things that ninjas and spies use in those movies to climb up the outside of buildings . that 's how i like to think of them . and pili are important because when urine comes rushing down the urinary tract , the e-coli are washed away like other bacteria without pili would be . so then , why are we talking about utis in pregnancy ? well , it turns out that they 're even more common in pregnant women . and a lot of that has to do with the uterus becoming larger throughout the course of the pregnancy , with the growing baby . so the large uterus does a couple of different things . firstly , it makes it pretty difficult to maintain hygiene in the area , making contamination more likely . and secondly , the uterus pushes on the bladder , making it hard to empty the bladder completely . so that means that each time the woman urinates , some urine is left behind . and that urine sitting around for a long time serves as a good medium for bacteria to grow and to thrive in . and another thing that happens in pregnancy is that the gfr or the glomerular filtration rate , or the rate at which blood is filtered through the kidneys , increases , so the gfr increases in pregnancy . and so what that means is that things like glucose are filtered a lot more out of blood and into the urine . and more glucose in the urine means more food for bacteria to grow , right ? and finally , pregnancy , especially the very end of pregnancy , is a sort of state of immune suppression , right ? and so that predisposes the mom to all sorts of infections , infections including utis . so then , let 's talk about some of the symptoms of cystitis , right , so some of the symptoms that you get when you have an infection of the bladder . burning and discomfort with urination , right , that 's called dysuria , occurs pretty commonly with cystitis . also a symptom called urinary frequency or feeling like you have to urinate really frequently , occurs pretty commonly . and urinary urgency , or feeling like you have to urinate immediately , is another common symptom . and occasionally , you can also have something called hematuria . and hematuria refers to blood in the urine . and this consolidation of symptoms or this group of symptoms , they 're really important to know because a person has to have these symptoms in order to be treated for the uti . that is to say that they ca n't be treated if they have only evidence of bacteria in their urine , but they do n't have any symptoms of the uti . but one of the important exceptions to that rule is pregnancy . so a pregnant woman who has bacteria in her urine without any symptoms of uti , which is called asymptomatic , it 's called asymptomatic , so without symptoms , bacteriuria , so when you have bacteria in the urine but do n't have any symptoms , right ? in a pregnant woman , that needs to be treated . and the reason for that is something like 25 % of pregnant women with asymptomatic bacteriuria end up with pyelonephritis . and that 's dangerous , very dangerous . in fact , because pyelonephritis in a pregnant woman can lead to sepsis . it can lead to kidney dysfunction , premature labor , and even death of the fetus . so that 's why we always , always collect the sample of urine in the very first trimester , even if a woman has absolutely no symptoms of uti , to look for bacteria in the urine . and if there 's evidence of bacteria in the urine , again , even if she has no symptoms , we go ahead and treat . and that brings me to what it takes to diagnose a uti . so when we 're trying to diagnose a uti , we collect a sample of urine , and what we 're looking for is a urine culture in which we have more than 100,000 colony-forming units , so more than 100,000 colonies of one type of bacteria that 's known to cause a uti . now , i want to mention one other important difference in utis in non-pregnant versus pregnant women , and that is treatment . so in a non-pregnant woman , or even a man , for that matter , your choice of treatment is pretty diverse . it could be something like trimethoprim , trimethoprim , sulfamethoxazole , right ? i know it 's a mouthful . and that 's a very popular treatment . another popular treatment is nitrofurantoin , right ? and a third popular choice is ciprofloxacin . ciprofloxacin , okay . and there are other less commonly used antibiotics also available for your choice . but your treatment choice in a pregnant woman is very different from that in a non-pregnant woman , because a lot of these choices are dangerous to the fetus . trimethoprim , for example , is a folate antagonist , so it blocks the use of folate by the baby . and we know that folate is really important for neural tube development . so for that reason , it 's usually avoided in the first trimester . and sulfamethoxazole can increase bilirubin levels in the baby , which can be pretty dangerous . so for that reason , it 's avoided . and ciprofloxacin is also avoided in pregnancy due to problems with bone development . so it leads to problems with bone development . so if you ca n't have those options available to you , some of the options that are available in pregnant women for the treatment of cystitis include amoxicillin , right ? you may have heard of that one before . amoxicillin is used for a bunch of different things . you can also use nitrofurantoin . so nitrofurantoin is one of those drugs that you can use safely in both pregnant and non-pregnant women . another popular option is cephalexin . cephalexin . and also fosfomycin . so those are some of the options that you have available to you for the use in pregnant women . so there you have it . that 's quite a bit of information on a very common disorder , urinary tract infection .
|
and we know that folate is really important for neural tube development . so for that reason , it 's usually avoided in the first trimester . and sulfamethoxazole can increase bilirubin levels in the baby , which can be pretty dangerous .
|
is recommended screening for asymptomatic bacteriuria limited to just once during the first trimester as mentioned ?
|
you know , it 's not unheard of for a guy to have a uti or urinary tract infection , but still , it 's pretty reasonable to say that utis are way more common in women . and in fact , there 's a statistic that i once read that said that one in five women will have a uti at some point in their life . so , yeah , i guess you could say it is pretty common in women . and to understand why this infection is so common in women , you have to understand the female anatomy . and i know that 's not an easy thing to do . female urinary anatomy is a bit complicated . so let 's simplify it . so , okay , if this is a woman 's body , what we 're going to do is kind of slice it down the middle and take a look down the midline . so if we do that , towards the back , towards the back of the body , we have the digestive system , so we have the rectum . and that 's what i 'm drawing here , the rectum . the rectum and with the anus , so this is the anus right here . and almost immediately in front of the anus is the vagina . okay , so the vagina . and the vagina is this canal that leads up to the uterus . so this is the uterus , the uterus , and this canal is the vagina . okay . and then immediately in front of the vagina is the urethra . now , urethra , urethra is the end or the outlet for the bladder . so this is the bladder , the bladder , and this is the urethra . and the bladder stores urine , right ? we know that . and that urine is made up in the kidney , which is then carried through the ureters and into the bladder . so this is the kidney , all right , the kidney . and these are the ureters or a ureter , leading up to the bladder . so it 's really apparent from this diagram that there 's a lot cramped into a pretty small space . okay . so back to urinary tract infection . so the term urinary tract really implies anything from the kidney down to the urethra . but to be a little bit more specific , infection of the urethra or this part right here , this part , and i 'm coloring it a different color , is called urethritis . so the infection of this part right here , the urethra , is called the urethritis . infection of the bladder , right , so this portion right here , this portion , is called cystitis . so the infection of the bladder is called cystitis . and then finally , infection of the kidneys , right here , so infection of the kidney is called pyelonephritis . it 's called pyelonephritis . and the fact is when most people talk about utis , they 're actually talking about cystitis or infection of the bladder . and the word `` infection '' implies some sort of pathogen , some bacteria of sorts . the bacteria in this case is usually e-coli , which comes from contamination from feces from back here in the anus . and that actually explains why utis are so common in women . the anus and the urethra are so close together in women that it makes contamination a lot more likely . and the urethra in women is also a lot shorter than it is in men , so the bacteria have a much shorter climb on their way to the bladder , also contributing to the fact that utis are more common in women . and that brings me to kind of a random but important point . so like i said , most utis , something like 80 % of them , are caused by e-coli . and one of the most important reasons for that is that e-coli have these things on them called pili that actually enable them to stick to the urinary tract and to climb up the urinary tract . so the pili are kind of like those suction cup things that ninjas and spies use in those movies to climb up the outside of buildings . that 's how i like to think of them . and pili are important because when urine comes rushing down the urinary tract , the e-coli are washed away like other bacteria without pili would be . so then , why are we talking about utis in pregnancy ? well , it turns out that they 're even more common in pregnant women . and a lot of that has to do with the uterus becoming larger throughout the course of the pregnancy , with the growing baby . so the large uterus does a couple of different things . firstly , it makes it pretty difficult to maintain hygiene in the area , making contamination more likely . and secondly , the uterus pushes on the bladder , making it hard to empty the bladder completely . so that means that each time the woman urinates , some urine is left behind . and that urine sitting around for a long time serves as a good medium for bacteria to grow and to thrive in . and another thing that happens in pregnancy is that the gfr or the glomerular filtration rate , or the rate at which blood is filtered through the kidneys , increases , so the gfr increases in pregnancy . and so what that means is that things like glucose are filtered a lot more out of blood and into the urine . and more glucose in the urine means more food for bacteria to grow , right ? and finally , pregnancy , especially the very end of pregnancy , is a sort of state of immune suppression , right ? and so that predisposes the mom to all sorts of infections , infections including utis . so then , let 's talk about some of the symptoms of cystitis , right , so some of the symptoms that you get when you have an infection of the bladder . burning and discomfort with urination , right , that 's called dysuria , occurs pretty commonly with cystitis . also a symptom called urinary frequency or feeling like you have to urinate really frequently , occurs pretty commonly . and urinary urgency , or feeling like you have to urinate immediately , is another common symptom . and occasionally , you can also have something called hematuria . and hematuria refers to blood in the urine . and this consolidation of symptoms or this group of symptoms , they 're really important to know because a person has to have these symptoms in order to be treated for the uti . that is to say that they ca n't be treated if they have only evidence of bacteria in their urine , but they do n't have any symptoms of the uti . but one of the important exceptions to that rule is pregnancy . so a pregnant woman who has bacteria in her urine without any symptoms of uti , which is called asymptomatic , it 's called asymptomatic , so without symptoms , bacteriuria , so when you have bacteria in the urine but do n't have any symptoms , right ? in a pregnant woman , that needs to be treated . and the reason for that is something like 25 % of pregnant women with asymptomatic bacteriuria end up with pyelonephritis . and that 's dangerous , very dangerous . in fact , because pyelonephritis in a pregnant woman can lead to sepsis . it can lead to kidney dysfunction , premature labor , and even death of the fetus . so that 's why we always , always collect the sample of urine in the very first trimester , even if a woman has absolutely no symptoms of uti , to look for bacteria in the urine . and if there 's evidence of bacteria in the urine , again , even if she has no symptoms , we go ahead and treat . and that brings me to what it takes to diagnose a uti . so when we 're trying to diagnose a uti , we collect a sample of urine , and what we 're looking for is a urine culture in which we have more than 100,000 colony-forming units , so more than 100,000 colonies of one type of bacteria that 's known to cause a uti . now , i want to mention one other important difference in utis in non-pregnant versus pregnant women , and that is treatment . so in a non-pregnant woman , or even a man , for that matter , your choice of treatment is pretty diverse . it could be something like trimethoprim , trimethoprim , sulfamethoxazole , right ? i know it 's a mouthful . and that 's a very popular treatment . another popular treatment is nitrofurantoin , right ? and a third popular choice is ciprofloxacin . ciprofloxacin , okay . and there are other less commonly used antibiotics also available for your choice . but your treatment choice in a pregnant woman is very different from that in a non-pregnant woman , because a lot of these choices are dangerous to the fetus . trimethoprim , for example , is a folate antagonist , so it blocks the use of folate by the baby . and we know that folate is really important for neural tube development . so for that reason , it 's usually avoided in the first trimester . and sulfamethoxazole can increase bilirubin levels in the baby , which can be pretty dangerous . so for that reason , it 's avoided . and ciprofloxacin is also avoided in pregnancy due to problems with bone development . so it leads to problems with bone development . so if you ca n't have those options available to you , some of the options that are available in pregnant women for the treatment of cystitis include amoxicillin , right ? you may have heard of that one before . amoxicillin is used for a bunch of different things . you can also use nitrofurantoin . so nitrofurantoin is one of those drugs that you can use safely in both pregnant and non-pregnant women . another popular option is cephalexin . cephalexin . and also fosfomycin . so those are some of the options that you have available to you for the use in pregnant women . so there you have it . that 's quite a bit of information on a very common disorder , urinary tract infection .
|
it can lead to kidney dysfunction , premature labor , and even death of the fetus . so that 's why we always , always collect the sample of urine in the very first trimester , even if a woman has absolutely no symptoms of uti , to look for bacteria in the urine . and if there 's evidence of bacteria in the urine , again , even if she has no symptoms , we go ahead and treat .
|
is there any strip available for uti detection which can detect the whole bacteria cell in urine sample ?
|
you know , it 's not unheard of for a guy to have a uti or urinary tract infection , but still , it 's pretty reasonable to say that utis are way more common in women . and in fact , there 's a statistic that i once read that said that one in five women will have a uti at some point in their life . so , yeah , i guess you could say it is pretty common in women . and to understand why this infection is so common in women , you have to understand the female anatomy . and i know that 's not an easy thing to do . female urinary anatomy is a bit complicated . so let 's simplify it . so , okay , if this is a woman 's body , what we 're going to do is kind of slice it down the middle and take a look down the midline . so if we do that , towards the back , towards the back of the body , we have the digestive system , so we have the rectum . and that 's what i 'm drawing here , the rectum . the rectum and with the anus , so this is the anus right here . and almost immediately in front of the anus is the vagina . okay , so the vagina . and the vagina is this canal that leads up to the uterus . so this is the uterus , the uterus , and this canal is the vagina . okay . and then immediately in front of the vagina is the urethra . now , urethra , urethra is the end or the outlet for the bladder . so this is the bladder , the bladder , and this is the urethra . and the bladder stores urine , right ? we know that . and that urine is made up in the kidney , which is then carried through the ureters and into the bladder . so this is the kidney , all right , the kidney . and these are the ureters or a ureter , leading up to the bladder . so it 's really apparent from this diagram that there 's a lot cramped into a pretty small space . okay . so back to urinary tract infection . so the term urinary tract really implies anything from the kidney down to the urethra . but to be a little bit more specific , infection of the urethra or this part right here , this part , and i 'm coloring it a different color , is called urethritis . so the infection of this part right here , the urethra , is called the urethritis . infection of the bladder , right , so this portion right here , this portion , is called cystitis . so the infection of the bladder is called cystitis . and then finally , infection of the kidneys , right here , so infection of the kidney is called pyelonephritis . it 's called pyelonephritis . and the fact is when most people talk about utis , they 're actually talking about cystitis or infection of the bladder . and the word `` infection '' implies some sort of pathogen , some bacteria of sorts . the bacteria in this case is usually e-coli , which comes from contamination from feces from back here in the anus . and that actually explains why utis are so common in women . the anus and the urethra are so close together in women that it makes contamination a lot more likely . and the urethra in women is also a lot shorter than it is in men , so the bacteria have a much shorter climb on their way to the bladder , also contributing to the fact that utis are more common in women . and that brings me to kind of a random but important point . so like i said , most utis , something like 80 % of them , are caused by e-coli . and one of the most important reasons for that is that e-coli have these things on them called pili that actually enable them to stick to the urinary tract and to climb up the urinary tract . so the pili are kind of like those suction cup things that ninjas and spies use in those movies to climb up the outside of buildings . that 's how i like to think of them . and pili are important because when urine comes rushing down the urinary tract , the e-coli are washed away like other bacteria without pili would be . so then , why are we talking about utis in pregnancy ? well , it turns out that they 're even more common in pregnant women . and a lot of that has to do with the uterus becoming larger throughout the course of the pregnancy , with the growing baby . so the large uterus does a couple of different things . firstly , it makes it pretty difficult to maintain hygiene in the area , making contamination more likely . and secondly , the uterus pushes on the bladder , making it hard to empty the bladder completely . so that means that each time the woman urinates , some urine is left behind . and that urine sitting around for a long time serves as a good medium for bacteria to grow and to thrive in . and another thing that happens in pregnancy is that the gfr or the glomerular filtration rate , or the rate at which blood is filtered through the kidneys , increases , so the gfr increases in pregnancy . and so what that means is that things like glucose are filtered a lot more out of blood and into the urine . and more glucose in the urine means more food for bacteria to grow , right ? and finally , pregnancy , especially the very end of pregnancy , is a sort of state of immune suppression , right ? and so that predisposes the mom to all sorts of infections , infections including utis . so then , let 's talk about some of the symptoms of cystitis , right , so some of the symptoms that you get when you have an infection of the bladder . burning and discomfort with urination , right , that 's called dysuria , occurs pretty commonly with cystitis . also a symptom called urinary frequency or feeling like you have to urinate really frequently , occurs pretty commonly . and urinary urgency , or feeling like you have to urinate immediately , is another common symptom . and occasionally , you can also have something called hematuria . and hematuria refers to blood in the urine . and this consolidation of symptoms or this group of symptoms , they 're really important to know because a person has to have these symptoms in order to be treated for the uti . that is to say that they ca n't be treated if they have only evidence of bacteria in their urine , but they do n't have any symptoms of the uti . but one of the important exceptions to that rule is pregnancy . so a pregnant woman who has bacteria in her urine without any symptoms of uti , which is called asymptomatic , it 's called asymptomatic , so without symptoms , bacteriuria , so when you have bacteria in the urine but do n't have any symptoms , right ? in a pregnant woman , that needs to be treated . and the reason for that is something like 25 % of pregnant women with asymptomatic bacteriuria end up with pyelonephritis . and that 's dangerous , very dangerous . in fact , because pyelonephritis in a pregnant woman can lead to sepsis . it can lead to kidney dysfunction , premature labor , and even death of the fetus . so that 's why we always , always collect the sample of urine in the very first trimester , even if a woman has absolutely no symptoms of uti , to look for bacteria in the urine . and if there 's evidence of bacteria in the urine , again , even if she has no symptoms , we go ahead and treat . and that brings me to what it takes to diagnose a uti . so when we 're trying to diagnose a uti , we collect a sample of urine , and what we 're looking for is a urine culture in which we have more than 100,000 colony-forming units , so more than 100,000 colonies of one type of bacteria that 's known to cause a uti . now , i want to mention one other important difference in utis in non-pregnant versus pregnant women , and that is treatment . so in a non-pregnant woman , or even a man , for that matter , your choice of treatment is pretty diverse . it could be something like trimethoprim , trimethoprim , sulfamethoxazole , right ? i know it 's a mouthful . and that 's a very popular treatment . another popular treatment is nitrofurantoin , right ? and a third popular choice is ciprofloxacin . ciprofloxacin , okay . and there are other less commonly used antibiotics also available for your choice . but your treatment choice in a pregnant woman is very different from that in a non-pregnant woman , because a lot of these choices are dangerous to the fetus . trimethoprim , for example , is a folate antagonist , so it blocks the use of folate by the baby . and we know that folate is really important for neural tube development . so for that reason , it 's usually avoided in the first trimester . and sulfamethoxazole can increase bilirubin levels in the baby , which can be pretty dangerous . so for that reason , it 's avoided . and ciprofloxacin is also avoided in pregnancy due to problems with bone development . so it leads to problems with bone development . so if you ca n't have those options available to you , some of the options that are available in pregnant women for the treatment of cystitis include amoxicillin , right ? you may have heard of that one before . amoxicillin is used for a bunch of different things . you can also use nitrofurantoin . so nitrofurantoin is one of those drugs that you can use safely in both pregnant and non-pregnant women . another popular option is cephalexin . cephalexin . and also fosfomycin . so those are some of the options that you have available to you for the use in pregnant women . so there you have it . that 's quite a bit of information on a very common disorder , urinary tract infection .
|
i know it 's a mouthful . and that 's a very popular treatment . another popular treatment is nitrofurantoin , right ? and a third popular choice is ciprofloxacin .
|
what are the differences in treatment of urethritis and cystitis ?
|
you know , it 's not unheard of for a guy to have a uti or urinary tract infection , but still , it 's pretty reasonable to say that utis are way more common in women . and in fact , there 's a statistic that i once read that said that one in five women will have a uti at some point in their life . so , yeah , i guess you could say it is pretty common in women . and to understand why this infection is so common in women , you have to understand the female anatomy . and i know that 's not an easy thing to do . female urinary anatomy is a bit complicated . so let 's simplify it . so , okay , if this is a woman 's body , what we 're going to do is kind of slice it down the middle and take a look down the midline . so if we do that , towards the back , towards the back of the body , we have the digestive system , so we have the rectum . and that 's what i 'm drawing here , the rectum . the rectum and with the anus , so this is the anus right here . and almost immediately in front of the anus is the vagina . okay , so the vagina . and the vagina is this canal that leads up to the uterus . so this is the uterus , the uterus , and this canal is the vagina . okay . and then immediately in front of the vagina is the urethra . now , urethra , urethra is the end or the outlet for the bladder . so this is the bladder , the bladder , and this is the urethra . and the bladder stores urine , right ? we know that . and that urine is made up in the kidney , which is then carried through the ureters and into the bladder . so this is the kidney , all right , the kidney . and these are the ureters or a ureter , leading up to the bladder . so it 's really apparent from this diagram that there 's a lot cramped into a pretty small space . okay . so back to urinary tract infection . so the term urinary tract really implies anything from the kidney down to the urethra . but to be a little bit more specific , infection of the urethra or this part right here , this part , and i 'm coloring it a different color , is called urethritis . so the infection of this part right here , the urethra , is called the urethritis . infection of the bladder , right , so this portion right here , this portion , is called cystitis . so the infection of the bladder is called cystitis . and then finally , infection of the kidneys , right here , so infection of the kidney is called pyelonephritis . it 's called pyelonephritis . and the fact is when most people talk about utis , they 're actually talking about cystitis or infection of the bladder . and the word `` infection '' implies some sort of pathogen , some bacteria of sorts . the bacteria in this case is usually e-coli , which comes from contamination from feces from back here in the anus . and that actually explains why utis are so common in women . the anus and the urethra are so close together in women that it makes contamination a lot more likely . and the urethra in women is also a lot shorter than it is in men , so the bacteria have a much shorter climb on their way to the bladder , also contributing to the fact that utis are more common in women . and that brings me to kind of a random but important point . so like i said , most utis , something like 80 % of them , are caused by e-coli . and one of the most important reasons for that is that e-coli have these things on them called pili that actually enable them to stick to the urinary tract and to climb up the urinary tract . so the pili are kind of like those suction cup things that ninjas and spies use in those movies to climb up the outside of buildings . that 's how i like to think of them . and pili are important because when urine comes rushing down the urinary tract , the e-coli are washed away like other bacteria without pili would be . so then , why are we talking about utis in pregnancy ? well , it turns out that they 're even more common in pregnant women . and a lot of that has to do with the uterus becoming larger throughout the course of the pregnancy , with the growing baby . so the large uterus does a couple of different things . firstly , it makes it pretty difficult to maintain hygiene in the area , making contamination more likely . and secondly , the uterus pushes on the bladder , making it hard to empty the bladder completely . so that means that each time the woman urinates , some urine is left behind . and that urine sitting around for a long time serves as a good medium for bacteria to grow and to thrive in . and another thing that happens in pregnancy is that the gfr or the glomerular filtration rate , or the rate at which blood is filtered through the kidneys , increases , so the gfr increases in pregnancy . and so what that means is that things like glucose are filtered a lot more out of blood and into the urine . and more glucose in the urine means more food for bacteria to grow , right ? and finally , pregnancy , especially the very end of pregnancy , is a sort of state of immune suppression , right ? and so that predisposes the mom to all sorts of infections , infections including utis . so then , let 's talk about some of the symptoms of cystitis , right , so some of the symptoms that you get when you have an infection of the bladder . burning and discomfort with urination , right , that 's called dysuria , occurs pretty commonly with cystitis . also a symptom called urinary frequency or feeling like you have to urinate really frequently , occurs pretty commonly . and urinary urgency , or feeling like you have to urinate immediately , is another common symptom . and occasionally , you can also have something called hematuria . and hematuria refers to blood in the urine . and this consolidation of symptoms or this group of symptoms , they 're really important to know because a person has to have these symptoms in order to be treated for the uti . that is to say that they ca n't be treated if they have only evidence of bacteria in their urine , but they do n't have any symptoms of the uti . but one of the important exceptions to that rule is pregnancy . so a pregnant woman who has bacteria in her urine without any symptoms of uti , which is called asymptomatic , it 's called asymptomatic , so without symptoms , bacteriuria , so when you have bacteria in the urine but do n't have any symptoms , right ? in a pregnant woman , that needs to be treated . and the reason for that is something like 25 % of pregnant women with asymptomatic bacteriuria end up with pyelonephritis . and that 's dangerous , very dangerous . in fact , because pyelonephritis in a pregnant woman can lead to sepsis . it can lead to kidney dysfunction , premature labor , and even death of the fetus . so that 's why we always , always collect the sample of urine in the very first trimester , even if a woman has absolutely no symptoms of uti , to look for bacteria in the urine . and if there 's evidence of bacteria in the urine , again , even if she has no symptoms , we go ahead and treat . and that brings me to what it takes to diagnose a uti . so when we 're trying to diagnose a uti , we collect a sample of urine , and what we 're looking for is a urine culture in which we have more than 100,000 colony-forming units , so more than 100,000 colonies of one type of bacteria that 's known to cause a uti . now , i want to mention one other important difference in utis in non-pregnant versus pregnant women , and that is treatment . so in a non-pregnant woman , or even a man , for that matter , your choice of treatment is pretty diverse . it could be something like trimethoprim , trimethoprim , sulfamethoxazole , right ? i know it 's a mouthful . and that 's a very popular treatment . another popular treatment is nitrofurantoin , right ? and a third popular choice is ciprofloxacin . ciprofloxacin , okay . and there are other less commonly used antibiotics also available for your choice . but your treatment choice in a pregnant woman is very different from that in a non-pregnant woman , because a lot of these choices are dangerous to the fetus . trimethoprim , for example , is a folate antagonist , so it blocks the use of folate by the baby . and we know that folate is really important for neural tube development . so for that reason , it 's usually avoided in the first trimester . and sulfamethoxazole can increase bilirubin levels in the baby , which can be pretty dangerous . so for that reason , it 's avoided . and ciprofloxacin is also avoided in pregnancy due to problems with bone development . so it leads to problems with bone development . so if you ca n't have those options available to you , some of the options that are available in pregnant women for the treatment of cystitis include amoxicillin , right ? you may have heard of that one before . amoxicillin is used for a bunch of different things . you can also use nitrofurantoin . so nitrofurantoin is one of those drugs that you can use safely in both pregnant and non-pregnant women . another popular option is cephalexin . cephalexin . and also fosfomycin . so those are some of the options that you have available to you for the use in pregnant women . so there you have it . that 's quite a bit of information on a very common disorder , urinary tract infection .
|
it can lead to kidney dysfunction , premature labor , and even death of the fetus . so that 's why we always , always collect the sample of urine in the very first trimester , even if a woman has absolutely no symptoms of uti , to look for bacteria in the urine . and if there 's evidence of bacteria in the urine , again , even if she has no symptoms , we go ahead and treat .
|
is there any colorimetric strip which can detect whole bacterial cell in urine sample in case of uti ?
|
two videos ago we learned about half-lives and we saw that they 're good if we are trying to figure out how much of a compound we have left after one half-life or two half-life or three half-lives . we can just take half of the compound at every period . but it 's not as useful if we 're trying to figure out how much of a compound we have after one half of a half-life or after one day or 10 seconds or 10 billion years . and to solve to address that issue in the last video , i proved that it involved a little bit of sophisticated math . and if you have n't taken calculus , you can really just skip that video . you do n't have to watch it for an intro math class but if you 're curious , that 's where we proved the following formula that at any given point of time , if you have some decaying atom , some element , it can be described as the amount of element you have at any period of time is equal to the amount you started off with times `` e '' to some constant . in the last video i used lambda . i could use `` k '' this time , minus `` k '' times `` t '' . and then for a particular element with a particular half-life you could just solve for the `` k '' and then apply it to your problem . so let 's do that in this video just so that all of these variables can become a little bit more concrete . so let 's figure out the general formula for carbon . carbon-14 that 's the one that we addressed in the half-life . we saw that carbon-14 has a half-life of 5,730 years . so let 's see if we can somehow take this information and apply it to this equation . so this tells us that after one half-life , so `` t '' is equal to 5,730 . `` n '' of 5,730 is equal to the amount we start off with so we 're starting off with `` n '' sub zero times `` e '' to the minus wherever you see the `` t '' , you put the 5,730 so minus `` k '' times 5,730 . that 's how many years have gone by . and half-life tells us that after 5,730 years , we 'll have half of our initial sample left . so we 'll have half of our initial sample left . so if we try to solve this equation for `` k '' what do we get ? divide both sides by `` n '' naught , get rid of that variable and we 're left with `` e '' to the minus 5,730 `` k '' , i 'm just switching these two around , is equal to one half . take the natural log of both sides . what do we get ? we get the natural log of `` e '' to anything , the natural log of `` e '' to the `` a '' is just `` a '' . the natural log of this is minus 5,730 `` k '' is equal to the natural log of one half . i just took the natural log of both sides . natural log and natural log of both sides of that . and so to solve for `` k '' we could just say `` k '' is equal to the natural log of one half over minus 5,730 so it equals 1.2 times 10 to the minus four . so now we have the general formula for carbon-14 given its half-life , at any given point in time after our starting point . so this is for let 's call this for carbon-14 or c-14 . the amount of carbon-14 we 're going to have left is going to be the amount that we started with times `` e '' to the minus `` k '' , `` k '' we just solved for-1.2 times 10 to the minus four , times the amount of time that has passed by . this is our formula for carbon , if we were doing it for carbon-14 . if we were doing this for some other element , we would use that element 's half-life to figure out how much we 're going to have at any given period of time , to figure out the `` k '' value . so let 's use this to solve a problem . let 's say that i start off with ... let 's say that i start off with 300 grams of carbon-14 , and i want to know how much do i have after 2,000 years . how much do i have ? well , i just plug into the formula . `` n '' of 2,000 is equal to the amount that i started off with , 300 grams , times `` e '' to the minus 1.2 times 10 to the minus four times `` t '' , times 2,000 . so what is that ? so this is equal to 236 grams . so just like that using this exponential decay formula i was able to figure out how much of the carbon i have after kind of an unusual period of time , a non-half life period of time . let 's do another one like this . let 's say let 's go the other way around . let 's say i 'm trying to figure out , let 's say i start off with 400 grams of c-14 and i wan na know how long so i wan na know a certain amount of time , does it take for me to get to 350 grams of c-14 ? so you just say that 350 grams is how much i 'm ending up with it 's equal to the amount that i started off with , 400 grams times `` e '' to the minus `` k '' , that 's minus 1.2 times 10 to the minus four , times `` time '' and now we solve for `` time '' . so you get .875 is equal to `` e '' to the minus 1.2 times 10 to the minus four `` t '' . take the natural log of both sides . you get the natural of .875 is equal to , the natural log of e to anything is just the anything , so it 's equal to minus 1.2 times 10 to the minus four `` t '' . so `` t '' is equal to this divided by 1.2 times 10 to the minus four . so the natural log of .875 divided by minus 1.2 times 10 to the minus four is equal to the amount of time it would take us to get from 400 grams to 350 . my cell phone is ringing let me turn that off . to 350 so let me do the math . so this is equal to 1,112 years to get from 400 to 350 grams of my substance . this might seem a little complicated , but you just have to . if there 's one thing you just have to do , is you just have to remember this formula .
|
the amount of carbon-14 we 're going to have left is going to be the amount that we started with times `` e '' to the minus `` k '' , `` k '' we just solved for-1.2 times 10 to the minus four , times the amount of time that has passed by . this is our formula for carbon , if we were doing it for carbon-14 . if we were doing this for some other element , we would use that element 's half-life to figure out how much we 're going to have at any given period of time , to figure out the `` k '' value .
|
what is the mathematical process of using carbon-14 to calculate the age of very old objects ?
|
steven zucker : as with many triptychs , viewers could see the exterior of the closed triptych during the weekdays . and on feast days or on the weekends , the painting would be opened up . you would move from the rather prosaic expressions of our world to a brilliantly colored scene of the horrors of limbo and the horrors of hell . beth harris : we see a saint on each wing . and these are painted in grisaille , in tones of gray . so it really would have been amazing when it opened to this colorful vision . steven zucker : the idea of painting the exterior in grisaille was meant to mimic the exterior , that is , the stone , of the church . but here the artist has moved far beyond that earlier tradition . and he 's actually not painting niches and sculptures , but actual people , a city and the landscape . beth harris : so here in grisaille one side , we see st. bavo , who was associated with the northern city of ghent . he 's shown distributing alms to the sick and the poor through a doorway , a view of a flemish city . steven zucker : showing the wonderful detail of that cityscape . let 's walk around to the other side and take a look at st. james . you can see he 's traveled past all kinds of expressions of wickedness . his faith , however , has kept him safe . and he 's very much associated with pilgrimage . it is his pilgrimage that so many medieval faithful would follow . beth harris : these wings could have given us clues to the patron of this very large triptych , but unfortunately , the coat of arms is blank . steven zucker : we do n't know why , but some art historians have suggested that perhaps the donor may have died before the work was finished . let 's look inside . the image , in a sense , unfolds as a kind of story , beginning in the left wing . we see god in heaven . he is in majesty , in a kind of brilliant mandorla surrounded by clouds . but when you look more closely , you see that there are hundreds , perhaps thousands , of angels that seem to be battling each other . this is the fall of the rebel angels . beth harris : this is a rebellion of angels led by lucifer , the devil . steven zucker : and they will be expelled from heaven and , of course , will go to reside in hell . and down at the very foot of the panel , we can see god extracting eve from adam 's side . that is the last part of the creation of adam and eve . just above that , we have the temptation . there is a sense of peace in the foreground . but this act of defiance against god 's law is this important breaking point because you can see that beyond that original sin , you have one animal eating another instead of living in harmony . and then you have an avenging angel who is expelling adam and eve from paradise and leading them into to the world that we know . beth harris : we have an unfolding of events at the top , beginning with god and the fall of the rebel angels , the event that happens first . then we jump down to the bottom and the creation of eve , then just above that the temptation , and above that the expulsion of adam and even from the garden of eden . and of course , this is the origin of original sin . and after the expulsion of adam and eve from the garden of eden , mankind knows sin and temptation and death . steven zucker : these stories echo each other . you have lucifer disobeying god . you have adam and eve disobeying god . you have lucifer being expelled from heaven . and you have man being expelled from the garden . so there is this parallel of the heavenly and the earthly . beth harris : and what bosch is really concerned with are the wages of sin . this is what bosch was famous , for even in his time . steven zucker : this deeply pessimistic philosophy , this questioning , is there any possibility of redemption given the sins of the world ? beth harris : it certainly does n't seem that way . steven zucker : well , let 's take a look at the evidence that he offers . ok , so we 're moving to the central panel . at the top , we see christ , functioning as judge . we see angels with long golden trumpets who are announcing the end of time . beth harris : and below it , taking up most of the central panel , is limbo , or the edges of hell . and this is a scene that bosch has combined with images of the seven deadly sins , the sins that cause mankind to spend eternity in hell . steven zucker : this is a painting whose bottom 2/3 is filled with torture and the terrible crimes that people inflict upon each other , but here enacted by devils and composite creatures that are incredibly fantastic . beth harris : the punishments that we see here are punishments for specific crimes . and the punishments are related to the crimes . steven zucker : let 's take a look at a few specifics . beth harris : on the left side , we see something that resembles an inn . on the roof , a figure who seems remarkably oblivious to everything that 's going on . she walks as though she 's on a fashion runway . but surrounding her and biting her is a hideous insect . and she 's led by a hideous dragon . steven zucker : led to a kind of hellish brothel . beth harris : and all accompanied by a lute , played by another demon . steven zucker : as well as a horn played by a demon in the back , where the horn actually looks as if it 's an extension of this nose . bosch uses music as one vehicle for sinfulness . beth harris : a kind of sign of indulgence in pleasure . steven zucker : below the representation of pride or vanity , you have the sins of gluttony . you see a rather overweight man who 's having liquid forcibly poured in to him as he 's restrained by devils . beth harris : and it 's not a very nice liquid . steven zucker : no . if you look a little bit above that barrel , you can see that there 's a siphon that 's receiving the excrement of a devil whose backside just be seen through the gated window . below that , you see one large demonic fish devouring another , which seems to be a reference to a northern proverb , the big fish eats the little . beth harris : that we take advantage of those who are smaller and weaker and less powerful than we are . steven zucker : to the right of that , we can see just inside the inn a series of hanging figures , and below that , a large cauldron with a series of figures that seem to be boiling . and we know that they 're boiling in molten metal , the metal that had been melted from their money . beth harris : so this is the sin of avarice or greed . steven zucker : there are endless representations of pain and suffering . you see men being roasted or fried by demonic frogs . you see , in one case , a frying pan with pieces of a body . this frog-like figure seems ready to take her two eggs that sit beside her and crack those into the pan as well . beth harris : and make a yummy omelet . steven zucker : yeah . beth harris : i think that what 's so disturbing here is the everydayness of the devilish figures who torture the human beings . they 're just going about their roasting and cooking and frying and torturing as thought it were a normal , everyday activity . and it reminds us that hell is eternity . steven zucker : in the middle of the large panel , you can see the sin of anger . and it 's represented by three knights who are particularly awful . there 's one knight in the middle who has upon his helmet a severed , blinded head . below that , you see images of corruption . and scattered throughout the foreground , you see images of bodies that have been mutilated , that have been shot with arrows . bodies have been cut and wounded and devoured in various ways . and all of this , of course , is a lead-in to the right panel , to hell itself . beth harris : when we think about the triptych as a whole , we have god in the upper left and satan diagonally across on the lower right . steven zucker : lucifer here sits in a kind of mock judgement of the souls that have been found to have been sinful . and here he is meting out the terrible punishments according to their crimes in life . beth harris : and you can see in the doorway behind him images of toads , which often torture figures in images of the last judgment , and then above , on the roof , all of the damned in hell who 've recognized where they 're spending eternity , who are wailing and crying and flailing their arms . steven zucker : and will populate the city of hell , which we see rising above this image . it is a place of fire and brimstone . it is a place of ruined cities , of absolute neglect . it is an apocalyptic scene most horrible . beth harris : and if this did n't make you want to live a virtuous life , i do n't know what would .
|
the image , in a sense , unfolds as a kind of story , beginning in the left wing . we see god in heaven . he is in majesty , in a kind of brilliant mandorla surrounded by clouds .
|
what word is used to describe god 's setting ?
|
steven zucker : as with many triptychs , viewers could see the exterior of the closed triptych during the weekdays . and on feast days or on the weekends , the painting would be opened up . you would move from the rather prosaic expressions of our world to a brilliantly colored scene of the horrors of limbo and the horrors of hell . beth harris : we see a saint on each wing . and these are painted in grisaille , in tones of gray . so it really would have been amazing when it opened to this colorful vision . steven zucker : the idea of painting the exterior in grisaille was meant to mimic the exterior , that is , the stone , of the church . but here the artist has moved far beyond that earlier tradition . and he 's actually not painting niches and sculptures , but actual people , a city and the landscape . beth harris : so here in grisaille one side , we see st. bavo , who was associated with the northern city of ghent . he 's shown distributing alms to the sick and the poor through a doorway , a view of a flemish city . steven zucker : showing the wonderful detail of that cityscape . let 's walk around to the other side and take a look at st. james . you can see he 's traveled past all kinds of expressions of wickedness . his faith , however , has kept him safe . and he 's very much associated with pilgrimage . it is his pilgrimage that so many medieval faithful would follow . beth harris : these wings could have given us clues to the patron of this very large triptych , but unfortunately , the coat of arms is blank . steven zucker : we do n't know why , but some art historians have suggested that perhaps the donor may have died before the work was finished . let 's look inside . the image , in a sense , unfolds as a kind of story , beginning in the left wing . we see god in heaven . he is in majesty , in a kind of brilliant mandorla surrounded by clouds . but when you look more closely , you see that there are hundreds , perhaps thousands , of angels that seem to be battling each other . this is the fall of the rebel angels . beth harris : this is a rebellion of angels led by lucifer , the devil . steven zucker : and they will be expelled from heaven and , of course , will go to reside in hell . and down at the very foot of the panel , we can see god extracting eve from adam 's side . that is the last part of the creation of adam and eve . just above that , we have the temptation . there is a sense of peace in the foreground . but this act of defiance against god 's law is this important breaking point because you can see that beyond that original sin , you have one animal eating another instead of living in harmony . and then you have an avenging angel who is expelling adam and eve from paradise and leading them into to the world that we know . beth harris : we have an unfolding of events at the top , beginning with god and the fall of the rebel angels , the event that happens first . then we jump down to the bottom and the creation of eve , then just above that the temptation , and above that the expulsion of adam and even from the garden of eden . and of course , this is the origin of original sin . and after the expulsion of adam and eve from the garden of eden , mankind knows sin and temptation and death . steven zucker : these stories echo each other . you have lucifer disobeying god . you have adam and eve disobeying god . you have lucifer being expelled from heaven . and you have man being expelled from the garden . so there is this parallel of the heavenly and the earthly . beth harris : and what bosch is really concerned with are the wages of sin . this is what bosch was famous , for even in his time . steven zucker : this deeply pessimistic philosophy , this questioning , is there any possibility of redemption given the sins of the world ? beth harris : it certainly does n't seem that way . steven zucker : well , let 's take a look at the evidence that he offers . ok , so we 're moving to the central panel . at the top , we see christ , functioning as judge . we see angels with long golden trumpets who are announcing the end of time . beth harris : and below it , taking up most of the central panel , is limbo , or the edges of hell . and this is a scene that bosch has combined with images of the seven deadly sins , the sins that cause mankind to spend eternity in hell . steven zucker : this is a painting whose bottom 2/3 is filled with torture and the terrible crimes that people inflict upon each other , but here enacted by devils and composite creatures that are incredibly fantastic . beth harris : the punishments that we see here are punishments for specific crimes . and the punishments are related to the crimes . steven zucker : let 's take a look at a few specifics . beth harris : on the left side , we see something that resembles an inn . on the roof , a figure who seems remarkably oblivious to everything that 's going on . she walks as though she 's on a fashion runway . but surrounding her and biting her is a hideous insect . and she 's led by a hideous dragon . steven zucker : led to a kind of hellish brothel . beth harris : and all accompanied by a lute , played by another demon . steven zucker : as well as a horn played by a demon in the back , where the horn actually looks as if it 's an extension of this nose . bosch uses music as one vehicle for sinfulness . beth harris : a kind of sign of indulgence in pleasure . steven zucker : below the representation of pride or vanity , you have the sins of gluttony . you see a rather overweight man who 's having liquid forcibly poured in to him as he 's restrained by devils . beth harris : and it 's not a very nice liquid . steven zucker : no . if you look a little bit above that barrel , you can see that there 's a siphon that 's receiving the excrement of a devil whose backside just be seen through the gated window . below that , you see one large demonic fish devouring another , which seems to be a reference to a northern proverb , the big fish eats the little . beth harris : that we take advantage of those who are smaller and weaker and less powerful than we are . steven zucker : to the right of that , we can see just inside the inn a series of hanging figures , and below that , a large cauldron with a series of figures that seem to be boiling . and we know that they 're boiling in molten metal , the metal that had been melted from their money . beth harris : so this is the sin of avarice or greed . steven zucker : there are endless representations of pain and suffering . you see men being roasted or fried by demonic frogs . you see , in one case , a frying pan with pieces of a body . this frog-like figure seems ready to take her two eggs that sit beside her and crack those into the pan as well . beth harris : and make a yummy omelet . steven zucker : yeah . beth harris : i think that what 's so disturbing here is the everydayness of the devilish figures who torture the human beings . they 're just going about their roasting and cooking and frying and torturing as thought it were a normal , everyday activity . and it reminds us that hell is eternity . steven zucker : in the middle of the large panel , you can see the sin of anger . and it 's represented by three knights who are particularly awful . there 's one knight in the middle who has upon his helmet a severed , blinded head . below that , you see images of corruption . and scattered throughout the foreground , you see images of bodies that have been mutilated , that have been shot with arrows . bodies have been cut and wounded and devoured in various ways . and all of this , of course , is a lead-in to the right panel , to hell itself . beth harris : when we think about the triptych as a whole , we have god in the upper left and satan diagonally across on the lower right . steven zucker : lucifer here sits in a kind of mock judgement of the souls that have been found to have been sinful . and here he is meting out the terrible punishments according to their crimes in life . beth harris : and you can see in the doorway behind him images of toads , which often torture figures in images of the last judgment , and then above , on the roof , all of the damned in hell who 've recognized where they 're spending eternity , who are wailing and crying and flailing their arms . steven zucker : and will populate the city of hell , which we see rising above this image . it is a place of fire and brimstone . it is a place of ruined cities , of absolute neglect . it is an apocalyptic scene most horrible . beth harris : and if this did n't make you want to live a virtuous life , i do n't know what would .
|
steven zucker : as well as a horn played by a demon in the back , where the horn actually looks as if it 's an extension of this nose . bosch uses music as one vehicle for sinfulness . beth harris : a kind of sign of indulgence in pleasure .
|
is there any known history regarding the life of the artist , bosch ?
|
steven zucker : as with many triptychs , viewers could see the exterior of the closed triptych during the weekdays . and on feast days or on the weekends , the painting would be opened up . you would move from the rather prosaic expressions of our world to a brilliantly colored scene of the horrors of limbo and the horrors of hell . beth harris : we see a saint on each wing . and these are painted in grisaille , in tones of gray . so it really would have been amazing when it opened to this colorful vision . steven zucker : the idea of painting the exterior in grisaille was meant to mimic the exterior , that is , the stone , of the church . but here the artist has moved far beyond that earlier tradition . and he 's actually not painting niches and sculptures , but actual people , a city and the landscape . beth harris : so here in grisaille one side , we see st. bavo , who was associated with the northern city of ghent . he 's shown distributing alms to the sick and the poor through a doorway , a view of a flemish city . steven zucker : showing the wonderful detail of that cityscape . let 's walk around to the other side and take a look at st. james . you can see he 's traveled past all kinds of expressions of wickedness . his faith , however , has kept him safe . and he 's very much associated with pilgrimage . it is his pilgrimage that so many medieval faithful would follow . beth harris : these wings could have given us clues to the patron of this very large triptych , but unfortunately , the coat of arms is blank . steven zucker : we do n't know why , but some art historians have suggested that perhaps the donor may have died before the work was finished . let 's look inside . the image , in a sense , unfolds as a kind of story , beginning in the left wing . we see god in heaven . he is in majesty , in a kind of brilliant mandorla surrounded by clouds . but when you look more closely , you see that there are hundreds , perhaps thousands , of angels that seem to be battling each other . this is the fall of the rebel angels . beth harris : this is a rebellion of angels led by lucifer , the devil . steven zucker : and they will be expelled from heaven and , of course , will go to reside in hell . and down at the very foot of the panel , we can see god extracting eve from adam 's side . that is the last part of the creation of adam and eve . just above that , we have the temptation . there is a sense of peace in the foreground . but this act of defiance against god 's law is this important breaking point because you can see that beyond that original sin , you have one animal eating another instead of living in harmony . and then you have an avenging angel who is expelling adam and eve from paradise and leading them into to the world that we know . beth harris : we have an unfolding of events at the top , beginning with god and the fall of the rebel angels , the event that happens first . then we jump down to the bottom and the creation of eve , then just above that the temptation , and above that the expulsion of adam and even from the garden of eden . and of course , this is the origin of original sin . and after the expulsion of adam and eve from the garden of eden , mankind knows sin and temptation and death . steven zucker : these stories echo each other . you have lucifer disobeying god . you have adam and eve disobeying god . you have lucifer being expelled from heaven . and you have man being expelled from the garden . so there is this parallel of the heavenly and the earthly . beth harris : and what bosch is really concerned with are the wages of sin . this is what bosch was famous , for even in his time . steven zucker : this deeply pessimistic philosophy , this questioning , is there any possibility of redemption given the sins of the world ? beth harris : it certainly does n't seem that way . steven zucker : well , let 's take a look at the evidence that he offers . ok , so we 're moving to the central panel . at the top , we see christ , functioning as judge . we see angels with long golden trumpets who are announcing the end of time . beth harris : and below it , taking up most of the central panel , is limbo , or the edges of hell . and this is a scene that bosch has combined with images of the seven deadly sins , the sins that cause mankind to spend eternity in hell . steven zucker : this is a painting whose bottom 2/3 is filled with torture and the terrible crimes that people inflict upon each other , but here enacted by devils and composite creatures that are incredibly fantastic . beth harris : the punishments that we see here are punishments for specific crimes . and the punishments are related to the crimes . steven zucker : let 's take a look at a few specifics . beth harris : on the left side , we see something that resembles an inn . on the roof , a figure who seems remarkably oblivious to everything that 's going on . she walks as though she 's on a fashion runway . but surrounding her and biting her is a hideous insect . and she 's led by a hideous dragon . steven zucker : led to a kind of hellish brothel . beth harris : and all accompanied by a lute , played by another demon . steven zucker : as well as a horn played by a demon in the back , where the horn actually looks as if it 's an extension of this nose . bosch uses music as one vehicle for sinfulness . beth harris : a kind of sign of indulgence in pleasure . steven zucker : below the representation of pride or vanity , you have the sins of gluttony . you see a rather overweight man who 's having liquid forcibly poured in to him as he 's restrained by devils . beth harris : and it 's not a very nice liquid . steven zucker : no . if you look a little bit above that barrel , you can see that there 's a siphon that 's receiving the excrement of a devil whose backside just be seen through the gated window . below that , you see one large demonic fish devouring another , which seems to be a reference to a northern proverb , the big fish eats the little . beth harris : that we take advantage of those who are smaller and weaker and less powerful than we are . steven zucker : to the right of that , we can see just inside the inn a series of hanging figures , and below that , a large cauldron with a series of figures that seem to be boiling . and we know that they 're boiling in molten metal , the metal that had been melted from their money . beth harris : so this is the sin of avarice or greed . steven zucker : there are endless representations of pain and suffering . you see men being roasted or fried by demonic frogs . you see , in one case , a frying pan with pieces of a body . this frog-like figure seems ready to take her two eggs that sit beside her and crack those into the pan as well . beth harris : and make a yummy omelet . steven zucker : yeah . beth harris : i think that what 's so disturbing here is the everydayness of the devilish figures who torture the human beings . they 're just going about their roasting and cooking and frying and torturing as thought it were a normal , everyday activity . and it reminds us that hell is eternity . steven zucker : in the middle of the large panel , you can see the sin of anger . and it 's represented by three knights who are particularly awful . there 's one knight in the middle who has upon his helmet a severed , blinded head . below that , you see images of corruption . and scattered throughout the foreground , you see images of bodies that have been mutilated , that have been shot with arrows . bodies have been cut and wounded and devoured in various ways . and all of this , of course , is a lead-in to the right panel , to hell itself . beth harris : when we think about the triptych as a whole , we have god in the upper left and satan diagonally across on the lower right . steven zucker : lucifer here sits in a kind of mock judgement of the souls that have been found to have been sinful . and here he is meting out the terrible punishments according to their crimes in life . beth harris : and you can see in the doorway behind him images of toads , which often torture figures in images of the last judgment , and then above , on the roof , all of the damned in hell who 've recognized where they 're spending eternity , who are wailing and crying and flailing their arms . steven zucker : and will populate the city of hell , which we see rising above this image . it is a place of fire and brimstone . it is a place of ruined cities , of absolute neglect . it is an apocalyptic scene most horrible . beth harris : and if this did n't make you want to live a virtuous life , i do n't know what would .
|
and he 's very much associated with pilgrimage . it is his pilgrimage that so many medieval faithful would follow . beth harris : these wings could have given us clues to the patron of this very large triptych , but unfortunately , the coat of arms is blank .
|
would a christian of this era not have viewed nature as just a collection of gods creations ?
|
steven zucker : as with many triptychs , viewers could see the exterior of the closed triptych during the weekdays . and on feast days or on the weekends , the painting would be opened up . you would move from the rather prosaic expressions of our world to a brilliantly colored scene of the horrors of limbo and the horrors of hell . beth harris : we see a saint on each wing . and these are painted in grisaille , in tones of gray . so it really would have been amazing when it opened to this colorful vision . steven zucker : the idea of painting the exterior in grisaille was meant to mimic the exterior , that is , the stone , of the church . but here the artist has moved far beyond that earlier tradition . and he 's actually not painting niches and sculptures , but actual people , a city and the landscape . beth harris : so here in grisaille one side , we see st. bavo , who was associated with the northern city of ghent . he 's shown distributing alms to the sick and the poor through a doorway , a view of a flemish city . steven zucker : showing the wonderful detail of that cityscape . let 's walk around to the other side and take a look at st. james . you can see he 's traveled past all kinds of expressions of wickedness . his faith , however , has kept him safe . and he 's very much associated with pilgrimage . it is his pilgrimage that so many medieval faithful would follow . beth harris : these wings could have given us clues to the patron of this very large triptych , but unfortunately , the coat of arms is blank . steven zucker : we do n't know why , but some art historians have suggested that perhaps the donor may have died before the work was finished . let 's look inside . the image , in a sense , unfolds as a kind of story , beginning in the left wing . we see god in heaven . he is in majesty , in a kind of brilliant mandorla surrounded by clouds . but when you look more closely , you see that there are hundreds , perhaps thousands , of angels that seem to be battling each other . this is the fall of the rebel angels . beth harris : this is a rebellion of angels led by lucifer , the devil . steven zucker : and they will be expelled from heaven and , of course , will go to reside in hell . and down at the very foot of the panel , we can see god extracting eve from adam 's side . that is the last part of the creation of adam and eve . just above that , we have the temptation . there is a sense of peace in the foreground . but this act of defiance against god 's law is this important breaking point because you can see that beyond that original sin , you have one animal eating another instead of living in harmony . and then you have an avenging angel who is expelling adam and eve from paradise and leading them into to the world that we know . beth harris : we have an unfolding of events at the top , beginning with god and the fall of the rebel angels , the event that happens first . then we jump down to the bottom and the creation of eve , then just above that the temptation , and above that the expulsion of adam and even from the garden of eden . and of course , this is the origin of original sin . and after the expulsion of adam and eve from the garden of eden , mankind knows sin and temptation and death . steven zucker : these stories echo each other . you have lucifer disobeying god . you have adam and eve disobeying god . you have lucifer being expelled from heaven . and you have man being expelled from the garden . so there is this parallel of the heavenly and the earthly . beth harris : and what bosch is really concerned with are the wages of sin . this is what bosch was famous , for even in his time . steven zucker : this deeply pessimistic philosophy , this questioning , is there any possibility of redemption given the sins of the world ? beth harris : it certainly does n't seem that way . steven zucker : well , let 's take a look at the evidence that he offers . ok , so we 're moving to the central panel . at the top , we see christ , functioning as judge . we see angels with long golden trumpets who are announcing the end of time . beth harris : and below it , taking up most of the central panel , is limbo , or the edges of hell . and this is a scene that bosch has combined with images of the seven deadly sins , the sins that cause mankind to spend eternity in hell . steven zucker : this is a painting whose bottom 2/3 is filled with torture and the terrible crimes that people inflict upon each other , but here enacted by devils and composite creatures that are incredibly fantastic . beth harris : the punishments that we see here are punishments for specific crimes . and the punishments are related to the crimes . steven zucker : let 's take a look at a few specifics . beth harris : on the left side , we see something that resembles an inn . on the roof , a figure who seems remarkably oblivious to everything that 's going on . she walks as though she 's on a fashion runway . but surrounding her and biting her is a hideous insect . and she 's led by a hideous dragon . steven zucker : led to a kind of hellish brothel . beth harris : and all accompanied by a lute , played by another demon . steven zucker : as well as a horn played by a demon in the back , where the horn actually looks as if it 's an extension of this nose . bosch uses music as one vehicle for sinfulness . beth harris : a kind of sign of indulgence in pleasure . steven zucker : below the representation of pride or vanity , you have the sins of gluttony . you see a rather overweight man who 's having liquid forcibly poured in to him as he 's restrained by devils . beth harris : and it 's not a very nice liquid . steven zucker : no . if you look a little bit above that barrel , you can see that there 's a siphon that 's receiving the excrement of a devil whose backside just be seen through the gated window . below that , you see one large demonic fish devouring another , which seems to be a reference to a northern proverb , the big fish eats the little . beth harris : that we take advantage of those who are smaller and weaker and less powerful than we are . steven zucker : to the right of that , we can see just inside the inn a series of hanging figures , and below that , a large cauldron with a series of figures that seem to be boiling . and we know that they 're boiling in molten metal , the metal that had been melted from their money . beth harris : so this is the sin of avarice or greed . steven zucker : there are endless representations of pain and suffering . you see men being roasted or fried by demonic frogs . you see , in one case , a frying pan with pieces of a body . this frog-like figure seems ready to take her two eggs that sit beside her and crack those into the pan as well . beth harris : and make a yummy omelet . steven zucker : yeah . beth harris : i think that what 's so disturbing here is the everydayness of the devilish figures who torture the human beings . they 're just going about their roasting and cooking and frying and torturing as thought it were a normal , everyday activity . and it reminds us that hell is eternity . steven zucker : in the middle of the large panel , you can see the sin of anger . and it 's represented by three knights who are particularly awful . there 's one knight in the middle who has upon his helmet a severed , blinded head . below that , you see images of corruption . and scattered throughout the foreground , you see images of bodies that have been mutilated , that have been shot with arrows . bodies have been cut and wounded and devoured in various ways . and all of this , of course , is a lead-in to the right panel , to hell itself . beth harris : when we think about the triptych as a whole , we have god in the upper left and satan diagonally across on the lower right . steven zucker : lucifer here sits in a kind of mock judgement of the souls that have been found to have been sinful . and here he is meting out the terrible punishments according to their crimes in life . beth harris : and you can see in the doorway behind him images of toads , which often torture figures in images of the last judgment , and then above , on the roof , all of the damned in hell who 've recognized where they 're spending eternity , who are wailing and crying and flailing their arms . steven zucker : and will populate the city of hell , which we see rising above this image . it is a place of fire and brimstone . it is a place of ruined cities , of absolute neglect . it is an apocalyptic scene most horrible . beth harris : and if this did n't make you want to live a virtuous life , i do n't know what would .
|
and he 's very much associated with pilgrimage . it is his pilgrimage that so many medieval faithful would follow . beth harris : these wings could have given us clues to the patron of this very large triptych , but unfortunately , the coat of arms is blank .
|
why are so many art pieces fold-able panels ?
|
steven zucker : as with many triptychs , viewers could see the exterior of the closed triptych during the weekdays . and on feast days or on the weekends , the painting would be opened up . you would move from the rather prosaic expressions of our world to a brilliantly colored scene of the horrors of limbo and the horrors of hell . beth harris : we see a saint on each wing . and these are painted in grisaille , in tones of gray . so it really would have been amazing when it opened to this colorful vision . steven zucker : the idea of painting the exterior in grisaille was meant to mimic the exterior , that is , the stone , of the church . but here the artist has moved far beyond that earlier tradition . and he 's actually not painting niches and sculptures , but actual people , a city and the landscape . beth harris : so here in grisaille one side , we see st. bavo , who was associated with the northern city of ghent . he 's shown distributing alms to the sick and the poor through a doorway , a view of a flemish city . steven zucker : showing the wonderful detail of that cityscape . let 's walk around to the other side and take a look at st. james . you can see he 's traveled past all kinds of expressions of wickedness . his faith , however , has kept him safe . and he 's very much associated with pilgrimage . it is his pilgrimage that so many medieval faithful would follow . beth harris : these wings could have given us clues to the patron of this very large triptych , but unfortunately , the coat of arms is blank . steven zucker : we do n't know why , but some art historians have suggested that perhaps the donor may have died before the work was finished . let 's look inside . the image , in a sense , unfolds as a kind of story , beginning in the left wing . we see god in heaven . he is in majesty , in a kind of brilliant mandorla surrounded by clouds . but when you look more closely , you see that there are hundreds , perhaps thousands , of angels that seem to be battling each other . this is the fall of the rebel angels . beth harris : this is a rebellion of angels led by lucifer , the devil . steven zucker : and they will be expelled from heaven and , of course , will go to reside in hell . and down at the very foot of the panel , we can see god extracting eve from adam 's side . that is the last part of the creation of adam and eve . just above that , we have the temptation . there is a sense of peace in the foreground . but this act of defiance against god 's law is this important breaking point because you can see that beyond that original sin , you have one animal eating another instead of living in harmony . and then you have an avenging angel who is expelling adam and eve from paradise and leading them into to the world that we know . beth harris : we have an unfolding of events at the top , beginning with god and the fall of the rebel angels , the event that happens first . then we jump down to the bottom and the creation of eve , then just above that the temptation , and above that the expulsion of adam and even from the garden of eden . and of course , this is the origin of original sin . and after the expulsion of adam and eve from the garden of eden , mankind knows sin and temptation and death . steven zucker : these stories echo each other . you have lucifer disobeying god . you have adam and eve disobeying god . you have lucifer being expelled from heaven . and you have man being expelled from the garden . so there is this parallel of the heavenly and the earthly . beth harris : and what bosch is really concerned with are the wages of sin . this is what bosch was famous , for even in his time . steven zucker : this deeply pessimistic philosophy , this questioning , is there any possibility of redemption given the sins of the world ? beth harris : it certainly does n't seem that way . steven zucker : well , let 's take a look at the evidence that he offers . ok , so we 're moving to the central panel . at the top , we see christ , functioning as judge . we see angels with long golden trumpets who are announcing the end of time . beth harris : and below it , taking up most of the central panel , is limbo , or the edges of hell . and this is a scene that bosch has combined with images of the seven deadly sins , the sins that cause mankind to spend eternity in hell . steven zucker : this is a painting whose bottom 2/3 is filled with torture and the terrible crimes that people inflict upon each other , but here enacted by devils and composite creatures that are incredibly fantastic . beth harris : the punishments that we see here are punishments for specific crimes . and the punishments are related to the crimes . steven zucker : let 's take a look at a few specifics . beth harris : on the left side , we see something that resembles an inn . on the roof , a figure who seems remarkably oblivious to everything that 's going on . she walks as though she 's on a fashion runway . but surrounding her and biting her is a hideous insect . and she 's led by a hideous dragon . steven zucker : led to a kind of hellish brothel . beth harris : and all accompanied by a lute , played by another demon . steven zucker : as well as a horn played by a demon in the back , where the horn actually looks as if it 's an extension of this nose . bosch uses music as one vehicle for sinfulness . beth harris : a kind of sign of indulgence in pleasure . steven zucker : below the representation of pride or vanity , you have the sins of gluttony . you see a rather overweight man who 's having liquid forcibly poured in to him as he 's restrained by devils . beth harris : and it 's not a very nice liquid . steven zucker : no . if you look a little bit above that barrel , you can see that there 's a siphon that 's receiving the excrement of a devil whose backside just be seen through the gated window . below that , you see one large demonic fish devouring another , which seems to be a reference to a northern proverb , the big fish eats the little . beth harris : that we take advantage of those who are smaller and weaker and less powerful than we are . steven zucker : to the right of that , we can see just inside the inn a series of hanging figures , and below that , a large cauldron with a series of figures that seem to be boiling . and we know that they 're boiling in molten metal , the metal that had been melted from their money . beth harris : so this is the sin of avarice or greed . steven zucker : there are endless representations of pain and suffering . you see men being roasted or fried by demonic frogs . you see , in one case , a frying pan with pieces of a body . this frog-like figure seems ready to take her two eggs that sit beside her and crack those into the pan as well . beth harris : and make a yummy omelet . steven zucker : yeah . beth harris : i think that what 's so disturbing here is the everydayness of the devilish figures who torture the human beings . they 're just going about their roasting and cooking and frying and torturing as thought it were a normal , everyday activity . and it reminds us that hell is eternity . steven zucker : in the middle of the large panel , you can see the sin of anger . and it 's represented by three knights who are particularly awful . there 's one knight in the middle who has upon his helmet a severed , blinded head . below that , you see images of corruption . and scattered throughout the foreground , you see images of bodies that have been mutilated , that have been shot with arrows . bodies have been cut and wounded and devoured in various ways . and all of this , of course , is a lead-in to the right panel , to hell itself . beth harris : when we think about the triptych as a whole , we have god in the upper left and satan diagonally across on the lower right . steven zucker : lucifer here sits in a kind of mock judgement of the souls that have been found to have been sinful . and here he is meting out the terrible punishments according to their crimes in life . beth harris : and you can see in the doorway behind him images of toads , which often torture figures in images of the last judgment , and then above , on the roof , all of the damned in hell who 've recognized where they 're spending eternity , who are wailing and crying and flailing their arms . steven zucker : and will populate the city of hell , which we see rising above this image . it is a place of fire and brimstone . it is a place of ruined cities , of absolute neglect . it is an apocalyptic scene most horrible . beth harris : and if this did n't make you want to live a virtuous life , i do n't know what would .
|
then we jump down to the bottom and the creation of eve , then just above that the temptation , and above that the expulsion of adam and even from the garden of eden . and of course , this is the origin of original sin . and after the expulsion of adam and eve from the garden of eden , mankind knows sin and temptation and death .
|
how original is this painting ?
|
steven zucker : as with many triptychs , viewers could see the exterior of the closed triptych during the weekdays . and on feast days or on the weekends , the painting would be opened up . you would move from the rather prosaic expressions of our world to a brilliantly colored scene of the horrors of limbo and the horrors of hell . beth harris : we see a saint on each wing . and these are painted in grisaille , in tones of gray . so it really would have been amazing when it opened to this colorful vision . steven zucker : the idea of painting the exterior in grisaille was meant to mimic the exterior , that is , the stone , of the church . but here the artist has moved far beyond that earlier tradition . and he 's actually not painting niches and sculptures , but actual people , a city and the landscape . beth harris : so here in grisaille one side , we see st. bavo , who was associated with the northern city of ghent . he 's shown distributing alms to the sick and the poor through a doorway , a view of a flemish city . steven zucker : showing the wonderful detail of that cityscape . let 's walk around to the other side and take a look at st. james . you can see he 's traveled past all kinds of expressions of wickedness . his faith , however , has kept him safe . and he 's very much associated with pilgrimage . it is his pilgrimage that so many medieval faithful would follow . beth harris : these wings could have given us clues to the patron of this very large triptych , but unfortunately , the coat of arms is blank . steven zucker : we do n't know why , but some art historians have suggested that perhaps the donor may have died before the work was finished . let 's look inside . the image , in a sense , unfolds as a kind of story , beginning in the left wing . we see god in heaven . he is in majesty , in a kind of brilliant mandorla surrounded by clouds . but when you look more closely , you see that there are hundreds , perhaps thousands , of angels that seem to be battling each other . this is the fall of the rebel angels . beth harris : this is a rebellion of angels led by lucifer , the devil . steven zucker : and they will be expelled from heaven and , of course , will go to reside in hell . and down at the very foot of the panel , we can see god extracting eve from adam 's side . that is the last part of the creation of adam and eve . just above that , we have the temptation . there is a sense of peace in the foreground . but this act of defiance against god 's law is this important breaking point because you can see that beyond that original sin , you have one animal eating another instead of living in harmony . and then you have an avenging angel who is expelling adam and eve from paradise and leading them into to the world that we know . beth harris : we have an unfolding of events at the top , beginning with god and the fall of the rebel angels , the event that happens first . then we jump down to the bottom and the creation of eve , then just above that the temptation , and above that the expulsion of adam and even from the garden of eden . and of course , this is the origin of original sin . and after the expulsion of adam and eve from the garden of eden , mankind knows sin and temptation and death . steven zucker : these stories echo each other . you have lucifer disobeying god . you have adam and eve disobeying god . you have lucifer being expelled from heaven . and you have man being expelled from the garden . so there is this parallel of the heavenly and the earthly . beth harris : and what bosch is really concerned with are the wages of sin . this is what bosch was famous , for even in his time . steven zucker : this deeply pessimistic philosophy , this questioning , is there any possibility of redemption given the sins of the world ? beth harris : it certainly does n't seem that way . steven zucker : well , let 's take a look at the evidence that he offers . ok , so we 're moving to the central panel . at the top , we see christ , functioning as judge . we see angels with long golden trumpets who are announcing the end of time . beth harris : and below it , taking up most of the central panel , is limbo , or the edges of hell . and this is a scene that bosch has combined with images of the seven deadly sins , the sins that cause mankind to spend eternity in hell . steven zucker : this is a painting whose bottom 2/3 is filled with torture and the terrible crimes that people inflict upon each other , but here enacted by devils and composite creatures that are incredibly fantastic . beth harris : the punishments that we see here are punishments for specific crimes . and the punishments are related to the crimes . steven zucker : let 's take a look at a few specifics . beth harris : on the left side , we see something that resembles an inn . on the roof , a figure who seems remarkably oblivious to everything that 's going on . she walks as though she 's on a fashion runway . but surrounding her and biting her is a hideous insect . and she 's led by a hideous dragon . steven zucker : led to a kind of hellish brothel . beth harris : and all accompanied by a lute , played by another demon . steven zucker : as well as a horn played by a demon in the back , where the horn actually looks as if it 's an extension of this nose . bosch uses music as one vehicle for sinfulness . beth harris : a kind of sign of indulgence in pleasure . steven zucker : below the representation of pride or vanity , you have the sins of gluttony . you see a rather overweight man who 's having liquid forcibly poured in to him as he 's restrained by devils . beth harris : and it 's not a very nice liquid . steven zucker : no . if you look a little bit above that barrel , you can see that there 's a siphon that 's receiving the excrement of a devil whose backside just be seen through the gated window . below that , you see one large demonic fish devouring another , which seems to be a reference to a northern proverb , the big fish eats the little . beth harris : that we take advantage of those who are smaller and weaker and less powerful than we are . steven zucker : to the right of that , we can see just inside the inn a series of hanging figures , and below that , a large cauldron with a series of figures that seem to be boiling . and we know that they 're boiling in molten metal , the metal that had been melted from their money . beth harris : so this is the sin of avarice or greed . steven zucker : there are endless representations of pain and suffering . you see men being roasted or fried by demonic frogs . you see , in one case , a frying pan with pieces of a body . this frog-like figure seems ready to take her two eggs that sit beside her and crack those into the pan as well . beth harris : and make a yummy omelet . steven zucker : yeah . beth harris : i think that what 's so disturbing here is the everydayness of the devilish figures who torture the human beings . they 're just going about their roasting and cooking and frying and torturing as thought it were a normal , everyday activity . and it reminds us that hell is eternity . steven zucker : in the middle of the large panel , you can see the sin of anger . and it 's represented by three knights who are particularly awful . there 's one knight in the middle who has upon his helmet a severed , blinded head . below that , you see images of corruption . and scattered throughout the foreground , you see images of bodies that have been mutilated , that have been shot with arrows . bodies have been cut and wounded and devoured in various ways . and all of this , of course , is a lead-in to the right panel , to hell itself . beth harris : when we think about the triptych as a whole , we have god in the upper left and satan diagonally across on the lower right . steven zucker : lucifer here sits in a kind of mock judgement of the souls that have been found to have been sinful . and here he is meting out the terrible punishments according to their crimes in life . beth harris : and you can see in the doorway behind him images of toads , which often torture figures in images of the last judgment , and then above , on the roof , all of the damned in hell who 've recognized where they 're spending eternity , who are wailing and crying and flailing their arms . steven zucker : and will populate the city of hell , which we see rising above this image . it is a place of fire and brimstone . it is a place of ruined cities , of absolute neglect . it is an apocalyptic scene most horrible . beth harris : and if this did n't make you want to live a virtuous life , i do n't know what would .
|
steven zucker : as with many triptychs , viewers could see the exterior of the closed triptych during the weekdays . and on feast days or on the weekends , the painting would be opened up . you would move from the rather prosaic expressions of our world to a brilliantly colored scene of the horrors of limbo and the horrors of hell .
|
i can vaguely recall maybe some cruel egyptian hieroglyphs , and some minotaurs , but these torturing figures and excrement seem like a big leap forward to me in terms of painting from the imagination ... who or what influenced bosch to paint this painting full of torture and gross things ?
|
here 's the second problem from ck12.org 's ap statistics flexbook . it 's an open source textbook , essentially . i 'm using it essentially to get some practice on some statistics problems . so here , number 2 . the grades on a statistics midterm for a high school are normally distributed with a mean of 81 and a standard deviation of 6.3 . all right . calculate the z-scores for each of the following exam grades . draw and label a sketch for each example . we can probably do it all on the same example . but the first thing we 'd have to do is just remember what is a z-score . what is a z-score ? a z-score is literally just measuring how many standard deviations away from the mean ? just like that . so we literally just have to calculate how many standard deviations each of these guys are from the mean , and that 's their z-scores . so let me do part a . so we have 65 . so first we can just figure out how far is 65 from the mean . let me just draw one chart here that we can use the entire time . so it 's just our distribution . let 's see . we have a mean of 81 . that 's our mean . and then a standard deviation of 6.3 . so our distribution , they 're telling us that it 's normally distributed . so i can draw a nice bell curve here . they 're saying it 's normally distributed , so that 's as good of a bell curve as i 'm capable of drawing . this is the mean right there at 81 . and the standard deviation is 6.3 . so one standard deviation above and below is going to be 6.3 away from that mean . so if we go 6.3 in the positive direction , that value right there is going to be 87.3 . if we go 6.3 in the negative direction , where does that get us ? what , 74.7 ? right , if we add 6 , it 'll get us to 80.7 , and then 0.3 will get us to 81 . so that 's one standard deviation below and above the mean , and then you 'd add another 6.3 to go 2 standard deviations , so on and so forth . so that 's a drawing of the distribution itself . so let 's figure out the z-scores for each of these grades . 65 is how far ? 65 is maybe going to be here someplace . so we first want to say , well how far is it just from our mean ? so the distance is , you just want to positive number here . well actually , you want a negative number . because you want your z-score to be positive or negative . negative would mean to the left of the mean and positive would mean to the right of the mean . so we say 65 minus 81 . so that 's literally how far away we are . but we want that in terms of standard deviations . so we divide that by the length or the magnitude of our standard deviation . so 65 minus 81 . let 's see , 81 minus 65 is what ? it is 5 plus 11 . it 's 16 . so this is going to be minus 16 over 6.3 . we 'll take our calculator out . and let 's see , if we have minus 16 divided by 6.3 , you get minus 2 point -- oh , it 's like 54 . approximately equal to minus 2.54 . that 's the z-score for a grade of 65 . pretty straightforward . let 's do a couple more . let 's do all of them . 83 . so how is it away from the mean ? well , it 's 83 minus 81 . it 's two grades above the mean . but we want it in terms of standard deviations . how many standard deviations . so this was part a . a was right here . we were 2.5 standard deviations below the mean . so this is part a . 1 , 2 , and then 0.5 . so this was a right there , 65 . and then part b , 83 , 83 is going to be right here . a little bit higher , but right here . and the z-score here , 83 minus 81 divided by 6.3 will get us -- let 's see , clear the calculator . so we have 83 minus 81 is 2 divided by 6.3 . it 's 0.32 , roughly . so here we get 0.32 . so 83 is 0.32 standard deviations above the mean . and so it would be roughly 1/3 third of the standard deviation along the way , right ? because this as one whole standard deviation . so we 're 0.3 of a standard deviation above the mean . choice number c. or not choice , part c , i guess i should call it . 93 . well , we do the same exercise . 93 is how much above the mean ? well , it 's 93 minus 81 is 12 . but we want it in terms of standard deviations . so 12 is how many standard deviations above the mean ? well , it 's going to be almost 2 . let 's take the calculator out . so we get 12 divided by 6.3 . it 's 1.9 standard deviations . its z-score is 1.9 . which means it 's 1.9 standard deviations above the mean . so the mean is 81 , we go one whole standard deviation , and then 0.9 standard deviations , and that 's where a score of 93 would lie , right there . its z-score is 1.9 . and all that means is 1.9 standard deviations above the mean . let 's do the last one . i 'll do it in magenta . d , part d. a score of 100 . we do n't even need the problem anymore . a score of 100 . well , same thing . we figure out how far is 100 above the mean -- remember , the mean was 81 -- and we divide that by the length or the size or the magnitude of our standard deviation . so 100 minus 81 is equal to 19 over 6.3 . so it 's going to be a little over 3 standard deviations . and in the next problem we 'll see what does that imply in terms of the probability of that actually occurring . but if we just want to figure out the z-score , 19 divided by 6.3 is equal to 3.01 . so it 's very close . 3.02 , really , if i were to round . so it 's very close to 3.02 . its z-score is 3.02 , or a grade of 100 is 3.02 standard deviations above the mean . so remember , this was the mean right here at 81 . we go 1 standard deviation above the mean , 2 standard deviations above the mean , the third standard deviation above the mean is right there . so we 're sitting right there on our chart . a little bit above that , 3.02 standard deviations above the mean , that 's where a score of 100 will be . and you can see the probability , the height of this -- that 's what the chart tells us -- it 's actually a very low probability . actually , not just a very low probability of getting something higher than that . because as we learned before , in a probably density function , if this is a continuous , not a discreet , the probability of getting exactly that is 0 , if this was n't discrete . but since this is scores on a test , we know that it 's actually a discrete probability function . but the probability is low of getting higher than that , because you can see where we sit on the bell curve . well anyway , hopefully this at least clarified how to solve for z-scores , which is pretty straightforward mathematically . and in the next video , we 'll interpret z-scores and probabilities a little bit more .
|
but the first thing we 'd have to do is just remember what is a z-score . what is a z-score ? a z-score is literally just measuring how many standard deviations away from the mean ?
|
what is the difference between z score and standard deviation ?
|
here 's the second problem from ck12.org 's ap statistics flexbook . it 's an open source textbook , essentially . i 'm using it essentially to get some practice on some statistics problems . so here , number 2 . the grades on a statistics midterm for a high school are normally distributed with a mean of 81 and a standard deviation of 6.3 . all right . calculate the z-scores for each of the following exam grades . draw and label a sketch for each example . we can probably do it all on the same example . but the first thing we 'd have to do is just remember what is a z-score . what is a z-score ? a z-score is literally just measuring how many standard deviations away from the mean ? just like that . so we literally just have to calculate how many standard deviations each of these guys are from the mean , and that 's their z-scores . so let me do part a . so we have 65 . so first we can just figure out how far is 65 from the mean . let me just draw one chart here that we can use the entire time . so it 's just our distribution . let 's see . we have a mean of 81 . that 's our mean . and then a standard deviation of 6.3 . so our distribution , they 're telling us that it 's normally distributed . so i can draw a nice bell curve here . they 're saying it 's normally distributed , so that 's as good of a bell curve as i 'm capable of drawing . this is the mean right there at 81 . and the standard deviation is 6.3 . so one standard deviation above and below is going to be 6.3 away from that mean . so if we go 6.3 in the positive direction , that value right there is going to be 87.3 . if we go 6.3 in the negative direction , where does that get us ? what , 74.7 ? right , if we add 6 , it 'll get us to 80.7 , and then 0.3 will get us to 81 . so that 's one standard deviation below and above the mean , and then you 'd add another 6.3 to go 2 standard deviations , so on and so forth . so that 's a drawing of the distribution itself . so let 's figure out the z-scores for each of these grades . 65 is how far ? 65 is maybe going to be here someplace . so we first want to say , well how far is it just from our mean ? so the distance is , you just want to positive number here . well actually , you want a negative number . because you want your z-score to be positive or negative . negative would mean to the left of the mean and positive would mean to the right of the mean . so we say 65 minus 81 . so that 's literally how far away we are . but we want that in terms of standard deviations . so we divide that by the length or the magnitude of our standard deviation . so 65 minus 81 . let 's see , 81 minus 65 is what ? it is 5 plus 11 . it 's 16 . so this is going to be minus 16 over 6.3 . we 'll take our calculator out . and let 's see , if we have minus 16 divided by 6.3 , you get minus 2 point -- oh , it 's like 54 . approximately equal to minus 2.54 . that 's the z-score for a grade of 65 . pretty straightforward . let 's do a couple more . let 's do all of them . 83 . so how is it away from the mean ? well , it 's 83 minus 81 . it 's two grades above the mean . but we want it in terms of standard deviations . how many standard deviations . so this was part a . a was right here . we were 2.5 standard deviations below the mean . so this is part a . 1 , 2 , and then 0.5 . so this was a right there , 65 . and then part b , 83 , 83 is going to be right here . a little bit higher , but right here . and the z-score here , 83 minus 81 divided by 6.3 will get us -- let 's see , clear the calculator . so we have 83 minus 81 is 2 divided by 6.3 . it 's 0.32 , roughly . so here we get 0.32 . so 83 is 0.32 standard deviations above the mean . and so it would be roughly 1/3 third of the standard deviation along the way , right ? because this as one whole standard deviation . so we 're 0.3 of a standard deviation above the mean . choice number c. or not choice , part c , i guess i should call it . 93 . well , we do the same exercise . 93 is how much above the mean ? well , it 's 93 minus 81 is 12 . but we want it in terms of standard deviations . so 12 is how many standard deviations above the mean ? well , it 's going to be almost 2 . let 's take the calculator out . so we get 12 divided by 6.3 . it 's 1.9 standard deviations . its z-score is 1.9 . which means it 's 1.9 standard deviations above the mean . so the mean is 81 , we go one whole standard deviation , and then 0.9 standard deviations , and that 's where a score of 93 would lie , right there . its z-score is 1.9 . and all that means is 1.9 standard deviations above the mean . let 's do the last one . i 'll do it in magenta . d , part d. a score of 100 . we do n't even need the problem anymore . a score of 100 . well , same thing . we figure out how far is 100 above the mean -- remember , the mean was 81 -- and we divide that by the length or the size or the magnitude of our standard deviation . so 100 minus 81 is equal to 19 over 6.3 . so it 's going to be a little over 3 standard deviations . and in the next problem we 'll see what does that imply in terms of the probability of that actually occurring . but if we just want to figure out the z-score , 19 divided by 6.3 is equal to 3.01 . so it 's very close . 3.02 , really , if i were to round . so it 's very close to 3.02 . its z-score is 3.02 , or a grade of 100 is 3.02 standard deviations above the mean . so remember , this was the mean right here at 81 . we go 1 standard deviation above the mean , 2 standard deviations above the mean , the third standard deviation above the mean is right there . so we 're sitting right there on our chart . a little bit above that , 3.02 standard deviations above the mean , that 's where a score of 100 will be . and you can see the probability , the height of this -- that 's what the chart tells us -- it 's actually a very low probability . actually , not just a very low probability of getting something higher than that . because as we learned before , in a probably density function , if this is a continuous , not a discreet , the probability of getting exactly that is 0 , if this was n't discrete . but since this is scores on a test , we know that it 's actually a discrete probability function . but the probability is low of getting higher than that , because you can see where we sit on the bell curve . well anyway , hopefully this at least clarified how to solve for z-scores , which is pretty straightforward mathematically . and in the next video , we 'll interpret z-scores and probabilities a little bit more .
|
well , it 's 83 minus 81 . it 's two grades above the mean . but we want it in terms of standard deviations .
|
how do i find an area under a normal curve if only two plots are given ?
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.