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let 's figure out the shape of the methane molecule using vsepr theory . so the first thing that you do is draw a dot structure to show it the valence electrons . so for methane , carbon is in group four . so 4 valence electrons . hydrogen is in group one , and i have four of them , so 1 times 4 is 4 , plus 4 is 8 vale... | 2 , 4 , six , and 8 . so carbon has an octet and we are done . the next thing we need to do is count the number of electron clouds that surround our central atom . | does carbon follow the octet rule ? |
let 's figure out the shape of the methane molecule using vsepr theory . so the first thing that you do is draw a dot structure to show it the valence electrons . so for methane , carbon is in group four . so 4 valence electrons . hydrogen is in group one , and i have four of them , so 1 times 4 is 4 , plus 4 is 8 vale... | carbon goes in the center and carbon is bonded to 4 hydrogens , so i can go ahead and put my hydrogens in there like that . and this is a very simple dot structure . we 've already shown all 8 of our valence electrons . | hello , when drawing dot structure , should n't the less electronegative atom sit in the centre ? |
let 's figure out the shape of the methane molecule using vsepr theory . so the first thing that you do is draw a dot structure to show it the valence electrons . so for methane , carbon is in group four . so 4 valence electrons . hydrogen is in group one , and i have four of them , so 1 times 4 is 4 , plus 4 is 8 vale... | and when you have four electron clouds , the electron clouds are farthest away from each other if they point towards the course of a tetrahedron , which is a four sided figure . so let me go ahead and draw the molecule here , draw the methane molecule . i 'm going to attempt to show it in a tetrahedral geometry . | or , is there a rule that says the only atom in a molecule sits in the centre ? |
let 's figure out the shape of the methane molecule using vsepr theory . so the first thing that you do is draw a dot structure to show it the valence electrons . so for methane , carbon is in group four . so 4 valence electrons . hydrogen is in group one , and i have four of them , so 1 times 4 is 4 , plus 4 is 8 vale... | so 5 valence electrons . hydrogen in group one , and i have three of them . so 1 times 3 plus 5 is 8 . | and also , when drawing ammonia , can one of the hydrogen be drawn above the nitrogen ? |
let 's figure out the shape of the methane molecule using vsepr theory . so the first thing that you do is draw a dot structure to show it the valence electrons . so for methane , carbon is in group four . so 4 valence electrons . hydrogen is in group one , and i have four of them , so 1 times 4 is 4 , plus 4 is 8 vale... | that 's 2 , that 's 4 . so 8 minus 4 is 4 valence electrons left . we first think about putting them on our terminal atoms , but those are our hydrogens , so they 're already happy with two electrons . | how can i identify whether a molecule with 4 electron pairs has tetrahedral or square planar structure ? |
let 's figure out the shape of the methane molecule using vsepr theory . so the first thing that you do is draw a dot structure to show it the valence electrons . so for methane , carbon is in group four . so 4 valence electrons . hydrogen is in group one , and i have four of them , so 1 times 4 is 4 , plus 4 is 8 vale... | and that 's going to make our bond angle even smaller than before . so it 's going to be even smaller than 107 degrees . and so this bond angle right here , you 'll see it listed as approximately 104.5 degrees , or some textbooks will say 105 degrees . so that 's approximately what it is for this . | why is n't the angle 90 degrees ? |
let 's figure out the shape of the methane molecule using vsepr theory . so the first thing that you do is draw a dot structure to show it the valence electrons . so for methane , carbon is in group four . so 4 valence electrons . hydrogen is in group one , and i have four of them , so 1 times 4 is 4 , plus 4 is 8 vale... | so we can think about these bonding electrons here as being electron clouds . so that 's one electron cloud . here 's another one down here . | i thought hydrogen could only take one valence electron ? |
let 's figure out the shape of the methane molecule using vsepr theory . so the first thing that you do is draw a dot structure to show it the valence electrons . so for methane , carbon is in group four . so 4 valence electrons . hydrogen is in group one , and i have four of them , so 1 times 4 is 4 , plus 4 is 8 vale... | so we go ahead and put those four valence electrons on our central atom , which is our oxygen . and four valence electrons means two lone pairs of electrons now . and so now we 've represented all eight valence electrons for water . | the electrons are being pushed further away/out from each other ? |
let 's figure out the shape of the methane molecule using vsepr theory . so the first thing that you do is draw a dot structure to show it the valence electrons . so for methane , carbon is in group four . so 4 valence electrons . hydrogen is in group one , and i have four of them , so 1 times 4 is 4 , plus 4 is 8 vale... | so we can say that methane is a tetrahedral molecule like that . all right , in terms of bond angles . so our goal now is to figure out what the bond angles are in a tetrahedral molecule . | how do we determine the bond angles of the different shapes ? |
let 's figure out the shape of the methane molecule using vsepr theory . so the first thing that you do is draw a dot structure to show it the valence electrons . so for methane , carbon is in group four . so 4 valence electrons . hydrogen is in group one , and i have four of them , so 1 times 4 is 4 , plus 4 is 8 vale... | so the geometry of the electron clouds are attempting to be , once again , a tetrahedral fashion , but the geometry of the molecule is different because you ignore lone pairs of electrons . and so when you look at the shape , if you look at the shape of this -- i 'll go ahead and draw the shape over here . if you 're i... | why is xef4 not a tetrahedral shape ? |
let 's figure out the shape of the methane molecule using vsepr theory . so the first thing that you do is draw a dot structure to show it the valence electrons . so for methane , carbon is in group four . so 4 valence electrons . hydrogen is in group one , and i have four of them , so 1 times 4 is 4 , plus 4 is 8 vale... | so we go ahead and put those four valence electrons on our central atom , which is our oxygen . and four valence electrons means two lone pairs of electrons now . and so now we 've represented all eight valence electrons for water . | why are lone pairs of electrons ignored ? |
let 's figure out the shape of the methane molecule using vsepr theory . so the first thing that you do is draw a dot structure to show it the valence electrons . so for methane , carbon is in group four . so 4 valence electrons . hydrogen is in group one , and i have four of them , so 1 times 4 is 4 , plus 4 is 8 vale... | here 's another one . so that 's two . here 's another one . | what is the angle between the plane containing two c-h bonds in ch4 , and the plane containing the other two c-h bonds in the same molecule ? |
let 's figure out the shape of the methane molecule using vsepr theory . so the first thing that you do is draw a dot structure to show it the valence electrons . so for methane , carbon is in group four . so 4 valence electrons . hydrogen is in group one , and i have four of them , so 1 times 4 is 4 , plus 4 is 8 vale... | so that 's three . and this lone pair of electrons , this non-bonding pair of electrons is also going to be counted as an electron cloud . it 's a region of electron density too . | if you ignore the lone electron pair or if they did n't have effect on the geometry , should n't ammonia be a flat triangle ? |
let 's figure out the shape of the methane molecule using vsepr theory . so the first thing that you do is draw a dot structure to show it the valence electrons . so for methane , carbon is in group four . so 4 valence electrons . hydrogen is in group one , and i have four of them , so 1 times 4 is 4 , plus 4 is 8 vale... | and so when you look at the shape , if you look at the shape of this -- i 'll go ahead and draw the shape over here . if you 're ignoring lone pairs of electrons , it looks like that . and we 've seen that shape before . | the molecules of ammonia and water have lone pairs , and they are drawn like a balloon ; are n't they supposed to be dumbbell shaped ? |
let 's figure out the shape of the methane molecule using vsepr theory . so the first thing that you do is draw a dot structure to show it the valence electrons . so for methane , carbon is in group four . so 4 valence electrons . hydrogen is in group one , and i have four of them , so 1 times 4 is 4 , plus 4 is 8 vale... | so in step three , we predicted the geometry electron clouds are going to attempt to be in a tetrahedron shape around our central atom . but when we 're actually talking about the geometry or shape of the molecule , we 're going to ignore any lone pairs when we predict the geometry of the molecule . so when we look at ... | why do we ignore the lone pairs while predicting the geometry of the molecule ? |
let 's figure out the shape of the methane molecule using vsepr theory . so the first thing that you do is draw a dot structure to show it the valence electrons . so for methane , carbon is in group four . so 4 valence electrons . hydrogen is in group one , and i have four of them , so 1 times 4 is 4 , plus 4 is 8 vale... | so carbon has an octet and we are done . the next thing we need to do is count the number of electron clouds that surround our central atom . so remember , electron clouds are regions of electron density , all right ? so we can think about these bonding electrons here as being electron clouds . | so why do n't they need more space , when the electron clouds have a trigonal planar shape ? |
let 's figure out the shape of the methane molecule using vsepr theory . so the first thing that you do is draw a dot structure to show it the valence electrons . so for methane , carbon is in group four . so 4 valence electrons . hydrogen is in group one , and i have four of them , so 1 times 4 is 4 , plus 4 is 8 vale... | so we go ahead and put those four valence electrons on our central atom , which is our oxygen . and four valence electrons means two lone pairs of electrons now . and so now we 've represented all eight valence electrons for water . | 8 ; 20 if you count the lone pairs as two diferents clouds does taht mean those valence electrons are gon na take 2 exclusive zones for them different of where they been before hydrogens appear ? |
let 's figure out the shape of the methane molecule using vsepr theory . so the first thing that you do is draw a dot structure to show it the valence electrons . so for methane , carbon is in group four . so 4 valence electrons . hydrogen is in group one , and i have four of them , so 1 times 4 is 4 , plus 4 is 8 vale... | so 8 minus 6 is 2 valence electrons left . we ca n't put them on our terminal atoms , because the hydrogens are already surrounded by two electrons . so we go ahead and put those two valence electrons on our central atom , which is our nitrogen like that . | at 5.10 , ca n't we ignore the lone pair above n like sulphur dioxide molecule ? |
let 's figure out the shape of the methane molecule using vsepr theory . so the first thing that you do is draw a dot structure to show it the valence electrons . so for methane , carbon is in group four . so 4 valence electrons . hydrogen is in group one , and i have four of them , so 1 times 4 is 4 , plus 4 is 8 vale... | so we can think about these bonding electrons here as being electron clouds . so that 's one electron cloud . here 's another one down here . | he says that the bon angle is 109.5 ' and if the electron cloud repel each other equally then every electron cloud will be equally far from each other , so ... .. should n't the bond angle be 90 ' ? |
let 's figure out the shape of the methane molecule using vsepr theory . so the first thing that you do is draw a dot structure to show it the valence electrons . so for methane , carbon is in group four . so 4 valence electrons . hydrogen is in group one , and i have four of them , so 1 times 4 is 4 , plus 4 is 8 vale... | so that 's three . and this lone pair of electrons , this non-bonding pair of electrons is also going to be counted as an electron cloud . it 's a region of electron density too . | in case of water we have lone electrons onn oxygen why do we pair it ? |
let 's figure out the shape of the methane molecule using vsepr theory . so the first thing that you do is draw a dot structure to show it the valence electrons . so for methane , carbon is in group four . so 4 valence electrons . hydrogen is in group one , and i have four of them , so 1 times 4 is 4 , plus 4 is 8 vale... | and then finally , here 's another one . so we have four electron clouds surrounding our central atom . the next step is to predict the geometry of your electron clouds around your central atom . | to draw dot structure we should take less electronegative atom as the central atom.in water molecular structure why is oxygen taken as central atom eventhough it is highly electronegative than hydrogen why so ? |
let 's figure out the shape of the methane molecule using vsepr theory . so the first thing that you do is draw a dot structure to show it the valence electrons . so for methane , carbon is in group four . so 4 valence electrons . hydrogen is in group one , and i have four of them , so 1 times 4 is 4 , plus 4 is 8 vale... | and we have our water molecule and we have our lone pairs of electrons like that . and in this case we have two lone pairs of electrons . remember , lone pairs or non-bonding electrons take up a little bit more space than bonding electrons . and therefore , they 're going to repel these electrons right in here a little... | why does lone pair of electrons take up more space than bonded electrons ? |
let 's figure out the shape of the methane molecule using vsepr theory . so the first thing that you do is draw a dot structure to show it the valence electrons . so for methane , carbon is in group four . so 4 valence electrons . hydrogen is in group one , and i have four of them , so 1 times 4 is 4 , plus 4 is 8 vale... | so we go ahead and put those four valence electrons on our central atom , which is our oxygen . and four valence electrons means two lone pairs of electrons now . and so now we 've represented all eight valence electrons for water . | how important are the different types of electrons and what are there purposes ? |
let 's figure out the shape of the methane molecule using vsepr theory . so the first thing that you do is draw a dot structure to show it the valence electrons . so for methane , carbon is in group four . so 4 valence electrons . hydrogen is in group one , and i have four of them , so 1 times 4 is 4 , plus 4 is 8 vale... | the next thing we need to do is count the number of electron clouds that surround our central atom . so remember , electron clouds are regions of electron density , all right ? so we can think about these bonding electrons here as being electron clouds . so that 's one electron cloud . here 's another one down here . | what is the importance of vsepr and how does it relate to electron clouds ? |
let 's figure out the shape of the methane molecule using vsepr theory . so the first thing that you do is draw a dot structure to show it the valence electrons . so for methane , carbon is in group four . so 4 valence electrons . hydrogen is in group one , and i have four of them , so 1 times 4 is 4 , plus 4 is 8 vale... | so we have nh3 . first thing we need to do is draw the dot structure . so we start by finding our valence electrons , nitrogen in group five . | how do you draw lewis structure and show vsepr classification , 3-d structure + bond angles , molecular shape , and electron group geometry ? |
let 's figure out the shape of the methane molecule using vsepr theory . so the first thing that you do is draw a dot structure to show it the valence electrons . so for methane , carbon is in group four . so 4 valence electrons . hydrogen is in group one , and i have four of them , so 1 times 4 is 4 , plus 4 is 8 vale... | let 's go ahead and do the water molecule . all right , so we have h2o . and to follow our steps , we know that hydrogen 's in group one . | for example , electronegativity can determine the boiling point , right ? |
let 's figure out the shape of the methane molecule using vsepr theory . so the first thing that you do is draw a dot structure to show it the valence electrons . so for methane , carbon is in group four . so 4 valence electrons . hydrogen is in group one , and i have four of them , so 1 times 4 is 4 , plus 4 is 8 vale... | so we can think about these bonding electrons here as being electron clouds . so that 's one electron cloud . here 's another one down here . | why is step # 3 not first honored by drawing the h2o molecule as a tetrahedral ( with one electron cloud as the vertical top piece and one electron cloud as one of the legs ) before then going to step # 4 and showing it as a bent shape ? |
let 's figure out the shape of the methane molecule using vsepr theory . so the first thing that you do is draw a dot structure to show it the valence electrons . so for methane , carbon is in group four . so 4 valence electrons . hydrogen is in group one , and i have four of them , so 1 times 4 is 4 , plus 4 is 8 vale... | so we go ahead and put those four valence electrons on our central atom , which is our oxygen . and four valence electrons means two lone pairs of electrons now . and so now we 've represented all eight valence electrons for water . | could you please tell me why it is important to ignore the lone pairs of electrons whilst determining the final geometry of the molecule ? |
let 's figure out the shape of the methane molecule using vsepr theory . so the first thing that you do is draw a dot structure to show it the valence electrons . so for methane , carbon is in group four . so 4 valence electrons . hydrogen is in group one , and i have four of them , so 1 times 4 is 4 , plus 4 is 8 vale... | that lone pair of electrons up there at the top , it 's going to look something like that for the shape . and we call this trigonal pyramidal . so this is a trigonal pyramidal shape . | why are both diagrams for a tetrahedral and trigonal pyramidal the same ? |
let 's figure out the shape of the methane molecule using vsepr theory . so the first thing that you do is draw a dot structure to show it the valence electrons . so for methane , carbon is in group four . so 4 valence electrons . hydrogen is in group one , and i have four of them , so 1 times 4 is 4 , plus 4 is 8 vale... | and we call this trigonal pyramidal . so this is a trigonal pyramidal shape . so even though the electron clouds are attempting to be in a tetrahedron fashion , the shape is more trigonal pyramidal because we ignore any lone pairs of electrons . in terms of a bond angle , this lone pair of electrons on the nitrogen act... | in nh3 compound ... why do we represent its shape as a trigonal pyramidal , ,i think that its shape should be trigonal planar because we are agreed to ignore the lone pair , ,so in this compound there are just 3 electron clouds to form the shape..there is no electron cloud in the top to form the trigonal shape ..do you... |
let 's figure out the shape of the methane molecule using vsepr theory . so the first thing that you do is draw a dot structure to show it the valence electrons . so for methane , carbon is in group four . so 4 valence electrons . hydrogen is in group one , and i have four of them , so 1 times 4 is 4 , plus 4 is 8 vale... | that lone pair of electrons up there at the top , it 's going to look something like that for the shape . and we call this trigonal pyramidal . so this is a trigonal pyramidal shape . | what is the difference between tetrahedral and trigonal pyramid ? |
let 's figure out the shape of the methane molecule using vsepr theory . so the first thing that you do is draw a dot structure to show it the valence electrons . so for methane , carbon is in group four . so 4 valence electrons . hydrogen is in group one , and i have four of them , so 1 times 4 is 4 , plus 4 is 8 vale... | so 6 plus 2 is once again 8 valence electrons to represent for our dot structure . and we put oxygen in the center . oxygen is bonded to two hydrogens , so we go ahead and draw those in there . | why did n't we put the electron pairs on oxygen 's up and down ? |
let 's figure out the shape of the methane molecule using vsepr theory . so the first thing that you do is draw a dot structure to show it the valence electrons . so for methane , carbon is in group four . so 4 valence electrons . hydrogen is in group one , and i have four of them , so 1 times 4 is 4 , plus 4 is 8 vale... | so 6 plus 2 is once again 8 valence electrons to represent for our dot structure . and we put oxygen in the center . oxygen is bonded to two hydrogens , so we go ahead and draw those in there . | why did n't we put the electron pairs on oxygen 's up and down ? |
let 's figure out the shape of the methane molecule using vsepr theory . so the first thing that you do is draw a dot structure to show it the valence electrons . so for methane , carbon is in group four . so 4 valence electrons . hydrogen is in group one , and i have four of them , so 1 times 4 is 4 , plus 4 is 8 vale... | that 's 2 , that 's 4 . so 8 minus 4 is 4 valence electrons left . we first think about putting them on our terminal atoms , but those are our hydrogens , so they 're already happy with two electrons . | how can we know that nh3 molecule can also be written in the manner in which the instructor has written in the video , time ? |
let 's figure out the shape of the methane molecule using vsepr theory . so the first thing that you do is draw a dot structure to show it the valence electrons . so for methane , carbon is in group four . so 4 valence electrons . hydrogen is in group one , and i have four of them , so 1 times 4 is 4 , plus 4 is 8 vale... | so we say that the geometry of the water molecule is bent or angular with an approximately 104.5 degree bond angle . so those are a couple examples of four electron clouds and how to figure out the geometry while also thinking about the bond angles . | at 3 why ca n't all the electron clouds arrange in a square planar instead of a tetrahedral shape ? |
let 's figure out the shape of the methane molecule using vsepr theory . so the first thing that you do is draw a dot structure to show it the valence electrons . so for methane , carbon is in group four . so 4 valence electrons . hydrogen is in group one , and i have four of them , so 1 times 4 is 4 , plus 4 is 8 vale... | that 's 2 , that 's 4 . so 8 minus 4 is 4 valence electrons left . we first think about putting them on our terminal atoms , but those are our hydrogens , so they 're already happy with two electrons . | why do n't we consider all 4 lone pair of electrons as one whole single cloud on oxygen ? |
let 's figure out the shape of the methane molecule using vsepr theory . so the first thing that you do is draw a dot structure to show it the valence electrons . so for methane , carbon is in group four . so 4 valence electrons . hydrogen is in group one , and i have four of them , so 1 times 4 is 4 , plus 4 is 8 vale... | and in step four , we ignore any lone pairs around our central atom , which we have none this time . and so therefore , the geometry the molecule is the same as the geometry of our electron pairs . so we can say that methane is a tetrahedral molecule like that . | what is the electron pair geometry and molecular geometry of c2h2 ? |
let 's figure out the shape of the methane molecule using vsepr theory . so the first thing that you do is draw a dot structure to show it the valence electrons . so for methane , carbon is in group four . so 4 valence electrons . hydrogen is in group one , and i have four of them , so 1 times 4 is 4 , plus 4 is 8 vale... | and then we 're going to go ahead and put our lone pair of electrons right up here . and so again , it 's an attempt to show the electron clouds in a tetrahedral geometry . let 's go back up here and look at our steps again . | is tetrahedral for electron pair geometry ? |
let 's figure out the shape of the methane molecule using vsepr theory . so the first thing that you do is draw a dot structure to show it the valence electrons . so for methane , carbon is in group four . so 4 valence electrons . hydrogen is in group one , and i have four of them , so 1 times 4 is 4 , plus 4 is 8 vale... | so let 's go ahead . and we have our water molecule and we have our lone pairs of electrons like that . and in this case we have two lone pairs of electrons . remember , lone pairs or non-bonding electrons take up a little bit more space than bonding electrons . | can the water molecule be linear ... ..having two lone pairs on oxygen ( up and down ) ? |
let 's figure out the shape of the methane molecule using vsepr theory . so the first thing that you do is draw a dot structure to show it the valence electrons . so for methane , carbon is in group four . so 4 valence electrons . hydrogen is in group one , and i have four of them , so 1 times 4 is 4 , plus 4 is 8 vale... | so we go ahead and put those four valence electrons on our central atom , which is our oxygen . and four valence electrons means two lone pairs of electrons now . and so now we 've represented all eight valence electrons for water . | in the structure of water , why does n't oxygen use its lone pairs of electrons to form a double bond with the two hydrogens ? |
let 's figure out the shape of the methane molecule using vsepr theory . so the first thing that you do is draw a dot structure to show it the valence electrons . so for methane , carbon is in group four . so 4 valence electrons . hydrogen is in group one , and i have four of them , so 1 times 4 is 4 , plus 4 is 8 vale... | so we can say that methane is a tetrahedral molecule like that . all right , in terms of bond angles . so our goal now is to figure out what the bond angles are in a tetrahedral molecule . | why is the measurement not from the same to the same when comparing bond angles ? |
vera rents bicycles to tourists . she recorded the height , in centimeters , of each customer and the frame size , in centimeters , of the bicycle that customer rented . after plotting her results , vera noticed that the relationship between the two variables was fairly linear , so she used the data to calculate the f... | well , the residual is going to be the difference between what they actually produce and what the line , what our regression line would have predicted . so we could say residual , let me write it this way , residual is going to be actual , actual minus predicted . so if predicted is larger than actual , this is actuall... | how do we find the residual when there are two y values for one x value ? |
so we have got a very interesting scenario here . i have this conical thimble-like cup that is 4 centimeters high . and also , the diameter of the top of the cup is also 4 centimeters . and i 'm pouring water into this cup right now . and i 'm pouring the water at a rate of 1 cubic centimeter . 1 cubic centimeter per s... | so we 're taking the derivative of everything with respect to time . so that 's why the chain rule came into play when we were taking the derivative of h , or the derivative of h of t , because we 're assuming that h is a function of time . now what does this thing right over here get us ? | why ca n't you make the volume formula h = v/.3 ( area of base ) and then take the derivative of that ? |
so we have got a very interesting scenario here . i have this conical thimble-like cup that is 4 centimeters high . and also , the diameter of the top of the cup is also 4 centimeters . and i 'm pouring water into this cup right now . and i 'm pouring the water at a rate of 1 cubic centimeter . 1 cubic centimeter per s... | and we get our drum roll now . the rate at which our height is changing with respect to time as we 're putting 1 cubic centimeter of water per second in . and right when our height is 2 centimeters high , the rate at which this height is changing with respect to time is 1 over pi . | how would we derive a function to determine the rate at which the increase of the height of the water is decreasing with respect to time ? |
so we have got a very interesting scenario here . i have this conical thimble-like cup that is 4 centimeters high . and also , the diameter of the top of the cup is also 4 centimeters . and i 'm pouring water into this cup right now . and i 'm pouring the water at a rate of 1 cubic centimeter . 1 cubic centimeter per s... | because we 're pouring more and more water here . so instead of just writing h to the third power , which i could write over here , let me write h of t to the third power . just to make it clear that this is a function of t. h of t to the third power . | at 0: why can we write that h is dependent on t ? |
so we have got a very interesting scenario here . i have this conical thimble-like cup that is 4 centimeters high . and also , the diameter of the top of the cup is also 4 centimeters . and i 'm pouring water into this cup right now . and i 'm pouring the water at a rate of 1 cubic centimeter . 1 cubic centimeter per s... | h over 2 squared . that 's the area of the surface of the water . and of course we still have the 1/3 out here . | the area of the triangle is ? |
so we have got a very interesting scenario here . i have this conical thimble-like cup that is 4 centimeters high . and also , the diameter of the top of the cup is also 4 centimeters . and i 'm pouring water into this cup right now . and i 'm pouring the water at a rate of 1 cubic centimeter . 1 cubic centimeter per s... | so let me see if i can simplify this . so this gives us 1/3 times pi h squared over 4 times another h , which is equal to -- we have pi , h to the third power over 12 . so that is our volume . | i am a bit confused sal said that the ratios between the height and the diameter are the same then when h =2 then r=1 and since ( v=1/3x pix r^2x h ) by taking the derivative we can say that the r is not a function of time ( correct me if i am mistaken ) then by all these results we have h'=3/pi so i still do n't get i... |
so we have got a very interesting scenario here . i have this conical thimble-like cup that is 4 centimeters high . and also , the diameter of the top of the cup is also 4 centimeters . and i 'm pouring water into this cup right now . and i 'm pouring the water at a rate of 1 cubic centimeter . 1 cubic centimeter per s... | so i just move this over to the right . and so now we can take the derivative with respect to time of both sides of this business . so the derivative with respect to time of our volume and the derivative with respect to time of this business . | how do you know it needs to , is it impossible to take the derivative of pi ? |
so we have got a very interesting scenario here . i have this conical thimble-like cup that is 4 centimeters high . and also , the diameter of the top of the cup is also 4 centimeters . and i 'm pouring water into this cup right now . and i 'm pouring the water at a rate of 1 cubic centimeter . 1 cubic centimeter per s... | let me rewrite everything i 've done . so we 've got dv , the rate at which our volume is changing with respect to time . the rate at which our volume is changing with respect to time is equal to pi over 12 times 3 h of t squared , or i could just write that as 3h squared , times the rate at which the height is changin... | i am thinking doing rate of change as my ib math ia , is it possible to find the most cost effective way to fill up the pool taking the salary of staff , fuel of the machine , volume of water ( pool is not perfect square ) with respect to time and make a relationship between these ? |
so we have got a very interesting scenario here . i have this conical thimble-like cup that is 4 centimeters high . and also , the diameter of the top of the cup is also 4 centimeters . and i 'm pouring water into this cup right now . and i 'm pouring the water at a rate of 1 cubic centimeter . 1 cubic centimeter per s... | because we 're pouring more and more water here . so instead of just writing h to the third power , which i could write over here , let me write h of t to the third power . just to make it clear that this is a function of t. h of t to the third power . | starting why does sal write h ( t ) ^3 over and over again just to take the derivative of it later on ? |
so we have got a very interesting scenario here . i have this conical thimble-like cup that is 4 centimeters high . and also , the diameter of the top of the cup is also 4 centimeters . and i 'm pouring water into this cup right now . and i 'm pouring the water at a rate of 1 cubic centimeter . 1 cubic centimeter per s... | so at any given point , the diameter across the surface of the water -- if the depth is h , the diameter across the surface of the water is also going to be h. and so from that we can figure out what are the radius is going to be . the radius is going to be h over 2 . and so the area of the water surface is going to be... | also , what do i do if the radius does n't happen to be as nice as r = h/2 with say arbitrarily a max height of 20 cm and water is at 6 cm ? |
so we have got a very interesting scenario here . i have this conical thimble-like cup that is 4 centimeters high . and also , the diameter of the top of the cup is also 4 centimeters . and i 'm pouring water into this cup right now . and i 'm pouring the water at a rate of 1 cubic centimeter . 1 cubic centimeter per s... | so let me see if i can simplify this . so this gives us 1/3 times pi h squared over 4 times another h , which is equal to -- we have pi , h to the third power over 12 . so that is our volume . | = 0.29 however , 1/pi = 0.31 can you please tell me where i went wrong ? |
so we have got a very interesting scenario here . i have this conical thimble-like cup that is 4 centimeters high . and also , the diameter of the top of the cup is also 4 centimeters . and i 'm pouring water into this cup right now . and i 'm pouring the water at a rate of 1 cubic centimeter . 1 cubic centimeter per s... | and so the area of the water surface is going to be pi r squared , pi times the radius squared . h over 2 squared . that 's the area of the surface of the water . | is it cm/sec in the answer because we squared ( 2cm ) ^2 and that becomes ( 1cm^3 sec ) / ( pi cm^2 ) ? |
so we have got a very interesting scenario here . i have this conical thimble-like cup that is 4 centimeters high . and also , the diameter of the top of the cup is also 4 centimeters . and i 'm pouring water into this cup right now . and i 'm pouring the water at a rate of 1 cubic centimeter . 1 cubic centimeter per s... | so we 're assuming -- we did express volume as a function of height -- but we 're saying that height itself is a function of time . so we 're taking the derivative of everything with respect to time . so that 's why the chain rule came into play when we were taking the derivative of h , or the derivative of h of t , be... | how do you know when to make a function like h with respect to time and use the chain rule ? |
so we have got a very interesting scenario here . i have this conical thimble-like cup that is 4 centimeters high . and also , the diameter of the top of the cup is also 4 centimeters . and i 'm pouring water into this cup right now . and i 'm pouring the water at a rate of 1 cubic centimeter . 1 cubic centimeter per s... | you might have been tempted to take the derivative over here with respect to h. but remember , we 're thinking about how things are changing with respect to time . so we 're assuming -- we did express volume as a function of height -- but we 're saying that height itself is a function of time . so we 're taking the der... | is n't v also a function of time or height ? |
so we have got a very interesting scenario here . i have this conical thimble-like cup that is 4 centimeters high . and also , the diameter of the top of the cup is also 4 centimeters . and i 'm pouring water into this cup right now . and i 'm pouring the water at a rate of 1 cubic centimeter . 1 cubic centimeter per s... | and so the area of the water surface is going to be pi r squared , pi times the radius squared . h over 2 squared . that 's the area of the surface of the water . | i have a quick question concerning the product rule : f ( x ) = [ u ( x ) x v ( x ) ] f ' ( x ) = u ( v ' ( x ) ) + v ( u ' ( x ) ) when you are supposed to find the derivative of the function ( x ) ( 1-x^2 ) ^1/2 it should be separated into ( x ) [ 1/2 ( 1-x^2 ) ^-1/2 ] + ( 1-x^2 ) ^1/2 right ? |
so we have got a very interesting scenario here . i have this conical thimble-like cup that is 4 centimeters high . and also , the diameter of the top of the cup is also 4 centimeters . and i 'm pouring water into this cup right now . and i 'm pouring the water at a rate of 1 cubic centimeter . 1 cubic centimeter per s... | my question to you is right at this moment , right when we are filling our cup at 1 cubic centimeter per second . and we have exactly 2 centimeters of water in the cup , 2 centimeters deep of water in the cup , what is the rate at which the height of the water is changing ? what is the rate at which this height right o... | for example the base of the cone has a radius of 10 and the total height is 18 , while the water level/height is 10 deep and you need to find the radius of the water level in terms of h. how would you exactly do that ? |
so we have got a very interesting scenario here . i have this conical thimble-like cup that is 4 centimeters high . and also , the diameter of the top of the cup is also 4 centimeters . and i 'm pouring water into this cup right now . and i 'm pouring the water at a rate of 1 cubic centimeter . 1 cubic centimeter per s... | we get 1 is equal to , well 3 times 4 is 12 , cancels out with that 12 . we get one is equal to pi times dh dt . to solve for dh dt divide both sides by pi . | if i was trying to determine the rate of change of an angle ( dtheta/dt ) in rad/s and i 'm given dx/dt in km/h and x and y in meters ... is it necessary to first convert the units of dx/dt to some standardized unit ? |
so we have got a very interesting scenario here . i have this conical thimble-like cup that is 4 centimeters high . and also , the diameter of the top of the cup is also 4 centimeters . and i 'm pouring water into this cup right now . and i 'm pouring the water at a rate of 1 cubic centimeter . 1 cubic centimeter per s... | we get 1 is equal to , well 3 times 4 is 12 , cancels out with that 12 . we get one is equal to pi times dh dt . to solve for dh dt divide both sides by pi . | dx/dt = 50km/h should correspond to a much different dtheta/dt than if dx/dt = 50cm/s , correct ? |
so we have got a very interesting scenario here . i have this conical thimble-like cup that is 4 centimeters high . and also , the diameter of the top of the cup is also 4 centimeters . and i 'm pouring water into this cup right now . and i 'm pouring the water at a rate of 1 cubic centimeter . 1 cubic centimeter per s... | we were told it is 2 centimeters . so the only unknown we have over here is the rate at which our height is changing with respect to time . which is exactly what we needed to figure out in the first place . | does dh/dt going to be the same thing until the cup is full ? |
so we have got a very interesting scenario here . i have this conical thimble-like cup that is 4 centimeters high . and also , the diameter of the top of the cup is also 4 centimeters . and i 'm pouring water into this cup right now . and i 'm pouring the water at a rate of 1 cubic centimeter . 1 cubic centimeter per s... | and so the area of the water surface is going to be pi r squared , pi times the radius squared . h over 2 squared . that 's the area of the surface of the water . | would n't the process be much more simplified finding that r= 1/2 h and plugging that into the function for volume , then taking a derivative to find dh/dt ? |
so we have got a very interesting scenario here . i have this conical thimble-like cup that is 4 centimeters high . and also , the diameter of the top of the cup is also 4 centimeters . and i 'm pouring water into this cup right now . and i 'm pouring the water at a rate of 1 cubic centimeter . 1 cubic centimeter per s... | and so that ratio is going to be true of any -- at any depth of water . it 's always going to have the same ratio between the diameter across the top and the height . because these are lines right over here . | what do you do when the height and diameter are n't the same ? |
so we have got a very interesting scenario here . i have this conical thimble-like cup that is 4 centimeters high . and also , the diameter of the top of the cup is also 4 centimeters . and i 'm pouring water into this cup right now . and i 'm pouring the water at a rate of 1 cubic centimeter . 1 cubic centimeter per s... | and so the area of the water surface is going to be pi r squared , pi times the radius squared . h over 2 squared . that 's the area of the surface of the water . | is it just an indicator that h is a function of t or is it a separate term that is multiplied to x ? |
so we have got a very interesting scenario here . i have this conical thimble-like cup that is 4 centimeters high . and also , the diameter of the top of the cup is also 4 centimeters . and i 'm pouring water into this cup right now . and i 'm pouring the water at a rate of 1 cubic centimeter . 1 cubic centimeter per s... | all right let me be careful , that was n't pi over 2 , that was pi over 12 . this is a pi over 12 right over here . so you get pi over 12 , times 3 times 2 squared , times dh dt . all of this is equal to 1 . | can i write dv/dt = pi/12 * d/dt ( h^3 ) instead ? |
so we have got a very interesting scenario here . i have this conical thimble-like cup that is 4 centimeters high . and also , the diameter of the top of the cup is also 4 centimeters . and i 'm pouring water into this cup right now . and i 'm pouring the water at a rate of 1 cubic centimeter . 1 cubic centimeter per s... | so 3 times 4 . all right let me be careful , that was n't pi over 2 , that was pi over 12 . this is a pi over 12 right over here . | if r was pi/2 could this problem still be solved ? |
so we have got a very interesting scenario here . i have this conical thimble-like cup that is 4 centimeters high . and also , the diameter of the top of the cup is also 4 centimeters . and i 'm pouring water into this cup right now . and i 'm pouring the water at a rate of 1 cubic centimeter . 1 cubic centimeter per s... | and of course we still have the 1/3 out here . and we 're still multiplying by this h over here . so let me see if i can simplify this . | could you still put r in terms of h ? |
so we have got a very interesting scenario here . i have this conical thimble-like cup that is 4 centimeters high . and also , the diameter of the top of the cup is also 4 centimeters . and i 'm pouring water into this cup right now . and i 'm pouring the water at a rate of 1 cubic centimeter . 1 cubic centimeter per s... | the radius is going to be h over 2 . and so the area of the water surface is going to be pi r squared , pi times the radius squared . h over 2 squared . | or would i just solve and have my area be pi*r^2 and have 144pi ? |
so we have got a very interesting scenario here . i have this conical thimble-like cup that is 4 centimeters high . and also , the diameter of the top of the cup is also 4 centimeters . and i 'm pouring water into this cup right now . and i 'm pouring the water at a rate of 1 cubic centimeter . 1 cubic centimeter per s... | h is 2 so you 're going to get 4 squared centimeters if we kept the units . so 3 times 4 . all right let me be careful , that was n't pi over 2 , that was pi over 12 . | sal talk about the ratio of the diameter and height as a raito , because they are equal i do n't know how to translate that characteristic to cones with differing heights and diameters i have one cone with r=4 , and h=9 is the ratio 4/9 ? |
so we have got a very interesting scenario here . i have this conical thimble-like cup that is 4 centimeters high . and also , the diameter of the top of the cup is also 4 centimeters . and i 'm pouring water into this cup right now . and i 'm pouring the water at a rate of 1 cubic centimeter . 1 cubic centimeter per s... | is equal to pi over 2 , times 3 , times h squared . h is 2 so you 're going to get 4 squared centimeters if we kept the units . so 3 times 4 . all right let me be careful , that was n't pi over 2 , that was pi over 12 . this is a pi over 12 right over here . | if the similar triangles are in similar proportions , would n't you set it up to be 4/4 : r/2 , then solving to get r=2 ? |
so we have got a very interesting scenario here . i have this conical thimble-like cup that is 4 centimeters high . and also , the diameter of the top of the cup is also 4 centimeters . and i 'm pouring water into this cup right now . and i 'm pouring the water at a rate of 1 cubic centimeter . 1 cubic centimeter per s... | so let me see if i can simplify this . so this gives us 1/3 times pi h squared over 4 times another h , which is equal to -- we have pi , h to the third power over 12 . so that is our volume . | would n't he have to take the derivative of h^3/12 rather than just h^3 itself ? |
so we have got a very interesting scenario here . i have this conical thimble-like cup that is 4 centimeters high . and also , the diameter of the top of the cup is also 4 centimeters . and i 'm pouring water into this cup right now . and i 'm pouring the water at a rate of 1 cubic centimeter . 1 cubic centimeter per s... | 1 cubic centimeter per second . and right at this moment , there is a height of 2 centimeters of water in the cup right now . so the height right now from the bottom of the cup to this point right over here is 2 centimeters . | so if this problem was asking for the rate of change of the water 's radius ( instead of the height ) , the answer would be 2pi/9 cm/sec right ? |
so we have got a very interesting scenario here . i have this conical thimble-like cup that is 4 centimeters high . and also , the diameter of the top of the cup is also 4 centimeters . and i 'm pouring water into this cup right now . and i 'm pouring the water at a rate of 1 cubic centimeter . 1 cubic centimeter per s... | in fact , it 's definitely a function of time . as time goes on , the height will change . because we 're pouring more and more water here . | so , what if we want to determine the rate of change of height when we start to fill the thing up , when the cone is empty ( h=0 ) ? |
so we have got a very interesting scenario here . i have this conical thimble-like cup that is 4 centimeters high . and also , the diameter of the top of the cup is also 4 centimeters . and i 'm pouring water into this cup right now . and i 'm pouring the water at a rate of 1 cubic centimeter . 1 cubic centimeter per s... | and so the area of the water surface is going to be pi r squared , pi times the radius squared . h over 2 squared . that 's the area of the surface of the water . | our formula seems to break down if we do n't have h > 0 ? |
so we have got a very interesting scenario here . i have this conical thimble-like cup that is 4 centimeters high . and also , the diameter of the top of the cup is also 4 centimeters . and i 'm pouring water into this cup right now . and i 'm pouring the water at a rate of 1 cubic centimeter . 1 cubic centimeter per s... | in fact , it 's definitely a function of time . as time goes on , the height will change . because we 're pouring more and more water here . | so , what if we want to determine the rate of change of height when we start to fill the thing up , when the cone is empty ( h=0 ) ? |
so we have got a very interesting scenario here . i have this conical thimble-like cup that is 4 centimeters high . and also , the diameter of the top of the cup is also 4 centimeters . and i 'm pouring water into this cup right now . and i 'm pouring the water at a rate of 1 cubic centimeter . 1 cubic centimeter per s... | and so the area of the water surface is going to be pi r squared , pi times the radius squared . h over 2 squared . that 's the area of the surface of the water . | our formula seems to break down if we do n't have h > 0 ? |
so we have got a very interesting scenario here . i have this conical thimble-like cup that is 4 centimeters high . and also , the diameter of the top of the cup is also 4 centimeters . and i 'm pouring water into this cup right now . and i 'm pouring the water at a rate of 1 cubic centimeter . 1 cubic centimeter per s... | in fact , it 's definitely a function of time . as time goes on , the height will change . because we 're pouring more and more water here . | so , what if we want to determine the rate of change of height when we start to fill the thing up , when the cone is empty ( h=0 ) ? |
so we have got a very interesting scenario here . i have this conical thimble-like cup that is 4 centimeters high . and also , the diameter of the top of the cup is also 4 centimeters . and i 'm pouring water into this cup right now . and i 'm pouring the water at a rate of 1 cubic centimeter . 1 cubic centimeter per s... | and so the area of the water surface is going to be pi r squared , pi times the radius squared . h over 2 squared . that 's the area of the surface of the water . | our formula seems to break down if we do n't have h > 0 ? |
so we have got a very interesting scenario here . i have this conical thimble-like cup that is 4 centimeters high . and also , the diameter of the top of the cup is also 4 centimeters . and i 'm pouring water into this cup right now . and i 'm pouring the water at a rate of 1 cubic centimeter . 1 cubic centimeter per s... | and so the area of the water surface is going to be pi r squared , pi times the radius squared . h over 2 squared . that 's the area of the surface of the water . | i understand how to get h/2 , but why ca n't i immediately sub the height of the water which is 2 , to the h ? |
so we have got a very interesting scenario here . i have this conical thimble-like cup that is 4 centimeters high . and also , the diameter of the top of the cup is also 4 centimeters . and i 'm pouring water into this cup right now . and i 'm pouring the water at a rate of 1 cubic centimeter . 1 cubic centimeter per s... | so we have got a very interesting scenario here . i have this conical thimble-like cup that is 4 centimeters high . | what is the use of such problems in the real world ? |
so we have got a very interesting scenario here . i have this conical thimble-like cup that is 4 centimeters high . and also , the diameter of the top of the cup is also 4 centimeters . and i 'm pouring water into this cup right now . and i 'm pouring the water at a rate of 1 cubic centimeter . 1 cubic centimeter per s... | so the height right now from the bottom of the cup to this point right over here is 2 centimeters . so my question to you is , at what rate -- we know the rate at which the water is flowing into the cup , we 're being given a volume per time . my question to you is right at this moment , right when we are filling our c... | what will we do by knowing the instantaneous rate of change of depth or the volume of water ? |
so we have got a very interesting scenario here . i have this conical thimble-like cup that is 4 centimeters high . and also , the diameter of the top of the cup is also 4 centimeters . and i 'm pouring water into this cup right now . and i 'm pouring the water at a rate of 1 cubic centimeter . 1 cubic centimeter per s... | h is 2 so you 're going to get 4 squared centimeters if we kept the units . so 3 times 4 . all right let me be careful , that was n't pi over 2 , that was pi over 12 . | why did sal raise to the power of 3 ? |
so we have got a very interesting scenario here . i have this conical thimble-like cup that is 4 centimeters high . and also , the diameter of the top of the cup is also 4 centimeters . and i 'm pouring water into this cup right now . and i 'm pouring the water at a rate of 1 cubic centimeter . 1 cubic centimeter per s... | well we could say that volume -- and i 'll do it in this blue color -- the volume of water is what we really care about . the volume of water is going to be equal to 1/3 times the area of the surface of the water -- area of water surface -- times our height of the water . so times h. so how can we figure out the area o... | how would you find the surface area of the water if the ratio between the height and diameter of the cone was not ? |
so we have got a very interesting scenario here . i have this conical thimble-like cup that is 4 centimeters high . and also , the diameter of the top of the cup is also 4 centimeters . and i 'm pouring water into this cup right now . and i 'm pouring the water at a rate of 1 cubic centimeter . 1 cubic centimeter per s... | and i 'm pouring the water at a rate of 1 cubic centimeter . 1 cubic centimeter per second . and right at this moment , there is a height of 2 centimeters of water in the cup right now . | a six tall person is walking away from a 14 foot tall lamp post at 3 feet per second.when is 10 feet from the lamp post , how fast is the length of the person 's shadow changing ? |
so we have got a very interesting scenario here . i have this conical thimble-like cup that is 4 centimeters high . and also , the diameter of the top of the cup is also 4 centimeters . and i 'm pouring water into this cup right now . and i 'm pouring the water at a rate of 1 cubic centimeter . 1 cubic centimeter per s... | this orange term right over here -- and i 'm just using the chain rule -- this is the derivative of h of t to the third power with respect to h of t. and then we 're going to multiply that times the derivative of h of t with respect to t. and then that gives us the derivative of this entire thing , h of t to the third ... | and how fast is the tip of the shadow moving away from the lamp post ? |
so we have got a very interesting scenario here . i have this conical thimble-like cup that is 4 centimeters high . and also , the diameter of the top of the cup is also 4 centimeters . and i 'm pouring water into this cup right now . and i 'm pouring the water at a rate of 1 cubic centimeter . 1 cubic centimeter per s... | but we 'll just take it on faith right now , that this is how we can figure out the volume of a cone . so given this can we figure out volume -- can we figure out an expression that relates volume to the height of the cone ? well we could say that volume -- and i 'll do it in this blue color -- the volume of water is w... | and hey everybody , i have a question regarding related rates , how would you go about solving a related rates problem involving a cone , but they ask for the change n height when the volume is given ? |
so we have got a very interesting scenario here . i have this conical thimble-like cup that is 4 centimeters high . and also , the diameter of the top of the cup is also 4 centimeters . and i 'm pouring water into this cup right now . and i 'm pouring the water at a rate of 1 cubic centimeter . 1 cubic centimeter per s... | so let me see if i can simplify this . so this gives us 1/3 times pi h squared over 4 times another h , which is equal to -- we have pi , h to the third power over 12 . so that is our volume . | at d/dt [ v ] =d/dt [ pi*h^3/12 ] , that h we know is 2 , so i got 1=d/dt [ pi*8/12 ] =0 , but 1=0 is wrong , so where is the mistake ? |
so we have got a very interesting scenario here . i have this conical thimble-like cup that is 4 centimeters high . and also , the diameter of the top of the cup is also 4 centimeters . and i 'm pouring water into this cup right now . and i 'm pouring the water at a rate of 1 cubic centimeter . 1 cubic centimeter per s... | so let me see if i can simplify this . so this gives us 1/3 times pi h squared over 4 times another h , which is equal to -- we have pi , h to the third power over 12 . so that is our volume . | would n't it behoove you ( for the sake of time ) to split the equation for volume into pi/12 * h^3 , and then just use the power rule to differentiate h^3 ? |
so we have got a very interesting scenario here . i have this conical thimble-like cup that is 4 centimeters high . and also , the diameter of the top of the cup is also 4 centimeters . and i 'm pouring water into this cup right now . and i 'm pouring the water at a rate of 1 cubic centimeter . 1 cubic centimeter per s... | and so that ratio is going to be true of any -- at any depth of water . it 's always going to have the same ratio between the diameter across the top and the height . because these are lines right over here . | what do you do when the ratio between diameter and height do n't work out so perfectly ? |
so we have got a very interesting scenario here . i have this conical thimble-like cup that is 4 centimeters high . and also , the diameter of the top of the cup is also 4 centimeters . and i 'm pouring water into this cup right now . and i 'm pouring the water at a rate of 1 cubic centimeter . 1 cubic centimeter per s... | well we could say that volume -- and i 'll do it in this blue color -- the volume of water is what we really care about . the volume of water is going to be equal to 1/3 times the area of the surface of the water -- area of water surface -- times our height of the water . so times h. so how can we figure out the area o... | so if water was draining out of the cone , the rates would just become negative ? |
so we have got a very interesting scenario here . i have this conical thimble-like cup that is 4 centimeters high . and also , the diameter of the top of the cup is also 4 centimeters . and i 'm pouring water into this cup right now . and i 'm pouring the water at a rate of 1 cubic centimeter . 1 cubic centimeter per s... | this orange term right over here -- and i 'm just using the chain rule -- this is the derivative of h of t to the third power with respect to h of t. and then we 're going to multiply that times the derivative of h of t with respect to t. and then that gives us the derivative of this entire thing , h of t to the third ... | would you get negative 1cm/sec for how fast the water would be `` decreasing '' instead of increasing ? |
so we have got a very interesting scenario here . i have this conical thimble-like cup that is 4 centimeters high . and also , the diameter of the top of the cup is also 4 centimeters . and i 'm pouring water into this cup right now . and i 'm pouring the water at a rate of 1 cubic centimeter . 1 cubic centimeter per s... | and so now we can take the derivative with respect to time of both sides of this business . so the derivative with respect to time of our volume and the derivative with respect to time of this business . well the derivative with respect to time of our volume , we could just rewrite that as dv dt , this thing right over... | if you are still left with r in the volume equation and when you take the derivative you have dr/dt , does it become zero ? |
so we have got a very interesting scenario here . i have this conical thimble-like cup that is 4 centimeters high . and also , the diameter of the top of the cup is also 4 centimeters . and i 'm pouring water into this cup right now . and i 'm pouring the water at a rate of 1 cubic centimeter . 1 cubic centimeter per s... | and so the area of the water surface is going to be pi r squared , pi times the radius squared . h over 2 squared . that 's the area of the surface of the water . | why is n't the derivative of d/dt ( h ( t ) ) ^3 ) equal to d/dt ( h ( t ) ^2 ) x 3 h ( t ) ^2 ? |
so we have got a very interesting scenario here . i have this conical thimble-like cup that is 4 centimeters high . and also , the diameter of the top of the cup is also 4 centimeters . and i 'm pouring water into this cup right now . and i 'm pouring the water at a rate of 1 cubic centimeter . 1 cubic centimeter per s... | so we 've got dv , the rate at which our volume is changing with respect to time . the rate at which our volume is changing with respect to time is equal to pi over 12 times 3 h of t squared , or i could just write that as 3h squared , times the rate at which the height is changing with respect to time , times dh dt . ... | at 0: why can you rewrite dv/dt as pi/2*3h^2*dh/dt ? |
to understand the pathophysiology of shock , we first need to remind ourselves what shock is . shock at its basic level means cells are not getting oxygen they need . essentially the o2 delivery , the amount of oxygen that the body can deliver to its cells becomes less than the amount of oxygen that is required by the... | less oxygen is getting back to the heart . two types of shock that are an example of this increased extraction are cardiogenic shock and hypovolemic shock . if you think about it , hypovolemic and cardiogenic shock , blood is not getting pushed forward fast enough . | so is shock essentially severe ischemia ? |
to understand the pathophysiology of shock , we first need to remind ourselves what shock is . shock at its basic level means cells are not getting oxygen they need . essentially the o2 delivery , the amount of oxygen that the body can deliver to its cells becomes less than the amount of oxygen that is required by the... | just a smaller version than this big one up here . now remember , in shock the issue is tissue perfusion . tissues and cells are not getting enough oxygen , they 're not getting enough blood that they need for oxygenation , and without this oxygen they ca n't create the energy necessary to sustain life . | so have that covered the next part of the question is would the same number of atp be produced if the four molecules of glucose were metabolized by muscle tissue ? |
we 're asked to find the square root of 100 . let me write this down bigger . so the square root is this big check-looking thing . the square root of 100 . when you see it like this , this means the positive square root . if you 're familiar with negative numbers , you know that there 's also a negative square root , b... | so the square root is this big check-looking thing . the square root of 100 . when you see it like this , this means the positive square root . if you 're familiar with negative numbers , you know that there 's also a negative square root , but when you just see this symbol , that means the positive square root . so le... | what does the negetive square root symbol look like ? |
we 're asked to find the square root of 100 . let me write this down bigger . so the square root is this big check-looking thing . the square root of 100 . when you see it like this , this means the positive square root . if you 're familiar with negative numbers , you know that there 's also a negative square root , b... | we 're asked to find the square root of 100 . let me write this down bigger . | is there a methed to find the square root of anay number like there is for other problems an example is long divison ? |
we 're asked to find the square root of 100 . let me write this down bigger . so the square root is this big check-looking thing . the square root of 100 . when you see it like this , this means the positive square root . if you 're familiar with negative numbers , you know that there 's also a negative square root , b... | so the square root is this big check-looking thing . the square root of 100 . when you see it like this , this means the positive square root . if you 're familiar with negative numbers , you know that there 's also a negative square root , but when you just see this symbol , that means the positive square root . | why do people say that square roots are always positive ? |
we 're asked to find the square root of 100 . let me write this down bigger . so the square root is this big check-looking thing . the square root of 100 . when you see it like this , this means the positive square root . if you 're familiar with negative numbers , you know that there 's also a negative square root , b... | so the square root is this big check-looking thing . the square root of 100 . when you see it like this , this means the positive square root . | what does one do if they are adding two square roots , such as the square root of 13 + the square root of 13 ? |
we 're asked to find the square root of 100 . let me write this down bigger . so the square root is this big check-looking thing . the square root of 100 . when you see it like this , this means the positive square root . if you 're familiar with negative numbers , you know that there 's also a negative square root , b... | so let 's think about what this is saying . this is asking us find the number , the positive number , that when i multiply that number by itself , i get 100 . so what number when i multiply it by itself do i get 100 ? | so the square root of a number is the number that when you multiply it by itself is equal to that number ? |
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