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hello , grammarians . i 'd like to bring up the idea of the difference between a common and a proper noun . so the difference between a common and a proper noun is simply the difference between something with a name and a more generic version of that thing . i 'll give you a couple of examples right off the bat . so speaking generally , i am from a city . the specific city that i 'm from is chicago . i could talk about a frog generally , but if i were speaking of a specific frog , i would say kermit . the difference between a common and a proper noun is merely the difference between a general thing , so this side is more general , and a specific thing . it 's a continuum . so if you are speaking of , let 's see , a river , any old river , that 's a common noun , but if you 're talking about a specific river , and it 's a named river here , that would be the nile , say . you could talk about a mountain , and that would be a common noun , because there are many mountains , but if you wanted to talk about a specific mountain , say mount kilimanjaro , in tanzania , that 's a proper noun . so here are the properties of proper nouns . proper nouns are always capitalized . and that means that instead of using a little letter a like that , you would instead use a big letter a like that . common nouns are only capitalized if you find them at the beginning of sentences . so you might say , `` mountains are my favorite . '' but you would also say , `` kilimanjaro `` is my favorite `` mountain . '' and that is a lowercase , non-capitalized m , as opposed to this one , which is uppercase . so that 's the difference between common and proper nouns . if you 're talking about something general , it 's a common noun . if you 're talking about something specific , it 's a proper noun , and the difference between them is that you capitalize a proper noun . you can learn anything . david out .
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the difference between a common and a proper noun is merely the difference between a general thing , so this side is more general , and a specific thing . it 's a continuum . so if you are speaking of , let 's see , a river , any old river , that 's a common noun , but if you 're talking about a specific river , and it 's a named river here , that would be the nile , say .
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in what is a continuum ?
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hello , grammarians . i 'd like to bring up the idea of the difference between a common and a proper noun . so the difference between a common and a proper noun is simply the difference between something with a name and a more generic version of that thing . i 'll give you a couple of examples right off the bat . so speaking generally , i am from a city . the specific city that i 'm from is chicago . i could talk about a frog generally , but if i were speaking of a specific frog , i would say kermit . the difference between a common and a proper noun is merely the difference between a general thing , so this side is more general , and a specific thing . it 's a continuum . so if you are speaking of , let 's see , a river , any old river , that 's a common noun , but if you 're talking about a specific river , and it 's a named river here , that would be the nile , say . you could talk about a mountain , and that would be a common noun , because there are many mountains , but if you wanted to talk about a specific mountain , say mount kilimanjaro , in tanzania , that 's a proper noun . so here are the properties of proper nouns . proper nouns are always capitalized . and that means that instead of using a little letter a like that , you would instead use a big letter a like that . common nouns are only capitalized if you find them at the beginning of sentences . so you might say , `` mountains are my favorite . '' but you would also say , `` kilimanjaro `` is my favorite `` mountain . '' and that is a lowercase , non-capitalized m , as opposed to this one , which is uppercase . so that 's the difference between common and proper nouns . if you 're talking about something general , it 's a common noun . if you 're talking about something specific , it 's a proper noun , and the difference between them is that you capitalize a proper noun . you can learn anything . david out .
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and that is a lowercase , non-capitalized m , as opposed to this one , which is uppercase . so that 's the difference between common and proper nouns . if you 're talking about something general , it 's a common noun .
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are species of animals such as a `` hylidae '' frog common or proper ?
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hello , grammarians . i 'd like to bring up the idea of the difference between a common and a proper noun . so the difference between a common and a proper noun is simply the difference between something with a name and a more generic version of that thing . i 'll give you a couple of examples right off the bat . so speaking generally , i am from a city . the specific city that i 'm from is chicago . i could talk about a frog generally , but if i were speaking of a specific frog , i would say kermit . the difference between a common and a proper noun is merely the difference between a general thing , so this side is more general , and a specific thing . it 's a continuum . so if you are speaking of , let 's see , a river , any old river , that 's a common noun , but if you 're talking about a specific river , and it 's a named river here , that would be the nile , say . you could talk about a mountain , and that would be a common noun , because there are many mountains , but if you wanted to talk about a specific mountain , say mount kilimanjaro , in tanzania , that 's a proper noun . so here are the properties of proper nouns . proper nouns are always capitalized . and that means that instead of using a little letter a like that , you would instead use a big letter a like that . common nouns are only capitalized if you find them at the beginning of sentences . so you might say , `` mountains are my favorite . '' but you would also say , `` kilimanjaro `` is my favorite `` mountain . '' and that is a lowercase , non-capitalized m , as opposed to this one , which is uppercase . so that 's the difference between common and proper nouns . if you 're talking about something general , it 's a common noun . if you 're talking about something specific , it 's a proper noun , and the difference between them is that you capitalize a proper noun . you can learn anything . david out .
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so here are the properties of proper nouns . proper nouns are always capitalized . and that means that instead of using a little letter a like that , you would instead use a big letter a like that .
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does a proper noun always have to have a capital letter ?
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hello , grammarians . i 'd like to bring up the idea of the difference between a common and a proper noun . so the difference between a common and a proper noun is simply the difference between something with a name and a more generic version of that thing . i 'll give you a couple of examples right off the bat . so speaking generally , i am from a city . the specific city that i 'm from is chicago . i could talk about a frog generally , but if i were speaking of a specific frog , i would say kermit . the difference between a common and a proper noun is merely the difference between a general thing , so this side is more general , and a specific thing . it 's a continuum . so if you are speaking of , let 's see , a river , any old river , that 's a common noun , but if you 're talking about a specific river , and it 's a named river here , that would be the nile , say . you could talk about a mountain , and that would be a common noun , because there are many mountains , but if you wanted to talk about a specific mountain , say mount kilimanjaro , in tanzania , that 's a proper noun . so here are the properties of proper nouns . proper nouns are always capitalized . and that means that instead of using a little letter a like that , you would instead use a big letter a like that . common nouns are only capitalized if you find them at the beginning of sentences . so you might say , `` mountains are my favorite . '' but you would also say , `` kilimanjaro `` is my favorite `` mountain . '' and that is a lowercase , non-capitalized m , as opposed to this one , which is uppercase . so that 's the difference between common and proper nouns . if you 're talking about something general , it 's a common noun . if you 're talking about something specific , it 's a proper noun , and the difference between them is that you capitalize a proper noun . you can learn anything . david out .
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and that is a lowercase , non-capitalized m , as opposed to this one , which is uppercase . so that 's the difference between common and proper nouns . if you 're talking about something general , it 's a common noun . if you 're talking about something specific , it 's a proper noun , and the difference between them is that you capitalize a proper noun . you can learn anything .
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is there a proper noun for every common noun ?
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hello , grammarians . i 'd like to bring up the idea of the difference between a common and a proper noun . so the difference between a common and a proper noun is simply the difference between something with a name and a more generic version of that thing . i 'll give you a couple of examples right off the bat . so speaking generally , i am from a city . the specific city that i 'm from is chicago . i could talk about a frog generally , but if i were speaking of a specific frog , i would say kermit . the difference between a common and a proper noun is merely the difference between a general thing , so this side is more general , and a specific thing . it 's a continuum . so if you are speaking of , let 's see , a river , any old river , that 's a common noun , but if you 're talking about a specific river , and it 's a named river here , that would be the nile , say . you could talk about a mountain , and that would be a common noun , because there are many mountains , but if you wanted to talk about a specific mountain , say mount kilimanjaro , in tanzania , that 's a proper noun . so here are the properties of proper nouns . proper nouns are always capitalized . and that means that instead of using a little letter a like that , you would instead use a big letter a like that . common nouns are only capitalized if you find them at the beginning of sentences . so you might say , `` mountains are my favorite . '' but you would also say , `` kilimanjaro `` is my favorite `` mountain . '' and that is a lowercase , non-capitalized m , as opposed to this one , which is uppercase . so that 's the difference between common and proper nouns . if you 're talking about something general , it 's a common noun . if you 're talking about something specific , it 's a proper noun , and the difference between them is that you capitalize a proper noun . you can learn anything . david out .
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if you 're talking about something general , it 's a common noun . if you 're talking about something specific , it 's a proper noun , and the difference between them is that you capitalize a proper noun . you can learn anything .
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would there be a proper noun for paper clip ?
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hello , grammarians . i 'd like to bring up the idea of the difference between a common and a proper noun . so the difference between a common and a proper noun is simply the difference between something with a name and a more generic version of that thing . i 'll give you a couple of examples right off the bat . so speaking generally , i am from a city . the specific city that i 'm from is chicago . i could talk about a frog generally , but if i were speaking of a specific frog , i would say kermit . the difference between a common and a proper noun is merely the difference between a general thing , so this side is more general , and a specific thing . it 's a continuum . so if you are speaking of , let 's see , a river , any old river , that 's a common noun , but if you 're talking about a specific river , and it 's a named river here , that would be the nile , say . you could talk about a mountain , and that would be a common noun , because there are many mountains , but if you wanted to talk about a specific mountain , say mount kilimanjaro , in tanzania , that 's a proper noun . so here are the properties of proper nouns . proper nouns are always capitalized . and that means that instead of using a little letter a like that , you would instead use a big letter a like that . common nouns are only capitalized if you find them at the beginning of sentences . so you might say , `` mountains are my favorite . '' but you would also say , `` kilimanjaro `` is my favorite `` mountain . '' and that is a lowercase , non-capitalized m , as opposed to this one , which is uppercase . so that 's the difference between common and proper nouns . if you 're talking about something general , it 's a common noun . if you 're talking about something specific , it 's a proper noun , and the difference between them is that you capitalize a proper noun . you can learn anything . david out .
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if you 're talking about something general , it 's a common noun . if you 're talking about something specific , it 's a proper noun , and the difference between them is that you capitalize a proper noun . you can learn anything .
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is south asia a proper noun ?
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hello , grammarians . i 'd like to bring up the idea of the difference between a common and a proper noun . so the difference between a common and a proper noun is simply the difference between something with a name and a more generic version of that thing . i 'll give you a couple of examples right off the bat . so speaking generally , i am from a city . the specific city that i 'm from is chicago . i could talk about a frog generally , but if i were speaking of a specific frog , i would say kermit . the difference between a common and a proper noun is merely the difference between a general thing , so this side is more general , and a specific thing . it 's a continuum . so if you are speaking of , let 's see , a river , any old river , that 's a common noun , but if you 're talking about a specific river , and it 's a named river here , that would be the nile , say . you could talk about a mountain , and that would be a common noun , because there are many mountains , but if you wanted to talk about a specific mountain , say mount kilimanjaro , in tanzania , that 's a proper noun . so here are the properties of proper nouns . proper nouns are always capitalized . and that means that instead of using a little letter a like that , you would instead use a big letter a like that . common nouns are only capitalized if you find them at the beginning of sentences . so you might say , `` mountains are my favorite . '' but you would also say , `` kilimanjaro `` is my favorite `` mountain . '' and that is a lowercase , non-capitalized m , as opposed to this one , which is uppercase . so that 's the difference between common and proper nouns . if you 're talking about something general , it 's a common noun . if you 're talking about something specific , it 's a proper noun , and the difference between them is that you capitalize a proper noun . you can learn anything . david out .
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the specific city that i 'm from is chicago . i could talk about a frog generally , but if i were speaking of a specific frog , i would say kermit . the difference between a common and a proper noun is merely the difference between a general thing , so this side is more general , and a specific thing .
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going back to the frog example , out of classification would i capitalize a specific breed of frog ?
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hello , grammarians . i 'd like to bring up the idea of the difference between a common and a proper noun . so the difference between a common and a proper noun is simply the difference between something with a name and a more generic version of that thing . i 'll give you a couple of examples right off the bat . so speaking generally , i am from a city . the specific city that i 'm from is chicago . i could talk about a frog generally , but if i were speaking of a specific frog , i would say kermit . the difference between a common and a proper noun is merely the difference between a general thing , so this side is more general , and a specific thing . it 's a continuum . so if you are speaking of , let 's see , a river , any old river , that 's a common noun , but if you 're talking about a specific river , and it 's a named river here , that would be the nile , say . you could talk about a mountain , and that would be a common noun , because there are many mountains , but if you wanted to talk about a specific mountain , say mount kilimanjaro , in tanzania , that 's a proper noun . so here are the properties of proper nouns . proper nouns are always capitalized . and that means that instead of using a little letter a like that , you would instead use a big letter a like that . common nouns are only capitalized if you find them at the beginning of sentences . so you might say , `` mountains are my favorite . '' but you would also say , `` kilimanjaro `` is my favorite `` mountain . '' and that is a lowercase , non-capitalized m , as opposed to this one , which is uppercase . so that 's the difference between common and proper nouns . if you 're talking about something general , it 's a common noun . if you 're talking about something specific , it 's a proper noun , and the difference between them is that you capitalize a proper noun . you can learn anything . david out .
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you could talk about a mountain , and that would be a common noun , because there are many mountains , but if you wanted to talk about a specific mountain , say mount kilimanjaro , in tanzania , that 's a proper noun . so here are the properties of proper nouns . proper nouns are always capitalized . and that means that instead of using a little letter a like that , you would instead use a big letter a like that .
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are abstract nouns similar to proper nouns ?
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hello , grammarians . i 'd like to bring up the idea of the difference between a common and a proper noun . so the difference between a common and a proper noun is simply the difference between something with a name and a more generic version of that thing . i 'll give you a couple of examples right off the bat . so speaking generally , i am from a city . the specific city that i 'm from is chicago . i could talk about a frog generally , but if i were speaking of a specific frog , i would say kermit . the difference between a common and a proper noun is merely the difference between a general thing , so this side is more general , and a specific thing . it 's a continuum . so if you are speaking of , let 's see , a river , any old river , that 's a common noun , but if you 're talking about a specific river , and it 's a named river here , that would be the nile , say . you could talk about a mountain , and that would be a common noun , because there are many mountains , but if you wanted to talk about a specific mountain , say mount kilimanjaro , in tanzania , that 's a proper noun . so here are the properties of proper nouns . proper nouns are always capitalized . and that means that instead of using a little letter a like that , you would instead use a big letter a like that . common nouns are only capitalized if you find them at the beginning of sentences . so you might say , `` mountains are my favorite . '' but you would also say , `` kilimanjaro `` is my favorite `` mountain . '' and that is a lowercase , non-capitalized m , as opposed to this one , which is uppercase . so that 's the difference between common and proper nouns . if you 're talking about something general , it 's a common noun . if you 're talking about something specific , it 's a proper noun , and the difference between them is that you capitalize a proper noun . you can learn anything . david out .
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hello , grammarians . i 'd like to bring up the idea of the difference between a common and a proper noun .
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when is the next chapter noragami being released ?
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hello , grammarians . i 'd like to bring up the idea of the difference between a common and a proper noun . so the difference between a common and a proper noun is simply the difference between something with a name and a more generic version of that thing . i 'll give you a couple of examples right off the bat . so speaking generally , i am from a city . the specific city that i 'm from is chicago . i could talk about a frog generally , but if i were speaking of a specific frog , i would say kermit . the difference between a common and a proper noun is merely the difference between a general thing , so this side is more general , and a specific thing . it 's a continuum . so if you are speaking of , let 's see , a river , any old river , that 's a common noun , but if you 're talking about a specific river , and it 's a named river here , that would be the nile , say . you could talk about a mountain , and that would be a common noun , because there are many mountains , but if you wanted to talk about a specific mountain , say mount kilimanjaro , in tanzania , that 's a proper noun . so here are the properties of proper nouns . proper nouns are always capitalized . and that means that instead of using a little letter a like that , you would instead use a big letter a like that . common nouns are only capitalized if you find them at the beginning of sentences . so you might say , `` mountains are my favorite . '' but you would also say , `` kilimanjaro `` is my favorite `` mountain . '' and that is a lowercase , non-capitalized m , as opposed to this one , which is uppercase . so that 's the difference between common and proper nouns . if you 're talking about something general , it 's a common noun . if you 're talking about something specific , it 's a proper noun , and the difference between them is that you capitalize a proper noun . you can learn anything . david out .
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so here are the properties of proper nouns . proper nouns are always capitalized . and that means that instead of using a little letter a like that , you would instead use a big letter a like that .
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are the titles of the books always written with uppercase ?
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hello , grammarians . i 'd like to bring up the idea of the difference between a common and a proper noun . so the difference between a common and a proper noun is simply the difference between something with a name and a more generic version of that thing . i 'll give you a couple of examples right off the bat . so speaking generally , i am from a city . the specific city that i 'm from is chicago . i could talk about a frog generally , but if i were speaking of a specific frog , i would say kermit . the difference between a common and a proper noun is merely the difference between a general thing , so this side is more general , and a specific thing . it 's a continuum . so if you are speaking of , let 's see , a river , any old river , that 's a common noun , but if you 're talking about a specific river , and it 's a named river here , that would be the nile , say . you could talk about a mountain , and that would be a common noun , because there are many mountains , but if you wanted to talk about a specific mountain , say mount kilimanjaro , in tanzania , that 's a proper noun . so here are the properties of proper nouns . proper nouns are always capitalized . and that means that instead of using a little letter a like that , you would instead use a big letter a like that . common nouns are only capitalized if you find them at the beginning of sentences . so you might say , `` mountains are my favorite . '' but you would also say , `` kilimanjaro `` is my favorite `` mountain . '' and that is a lowercase , non-capitalized m , as opposed to this one , which is uppercase . so that 's the difference between common and proper nouns . if you 're talking about something general , it 's a common noun . if you 're talking about something specific , it 's a proper noun , and the difference between them is that you capitalize a proper noun . you can learn anything . david out .
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if you 're talking about something general , it 's a common noun . if you 're talking about something specific , it 's a proper noun , and the difference between them is that you capitalize a proper noun . you can learn anything .
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so , if i where to write a sentence about a math equation , 2x+5 for example , and i where to reference one of the terms , would that term be a proper noun ?
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hello , grammarians . i 'd like to bring up the idea of the difference between a common and a proper noun . so the difference between a common and a proper noun is simply the difference between something with a name and a more generic version of that thing . i 'll give you a couple of examples right off the bat . so speaking generally , i am from a city . the specific city that i 'm from is chicago . i could talk about a frog generally , but if i were speaking of a specific frog , i would say kermit . the difference between a common and a proper noun is merely the difference between a general thing , so this side is more general , and a specific thing . it 's a continuum . so if you are speaking of , let 's see , a river , any old river , that 's a common noun , but if you 're talking about a specific river , and it 's a named river here , that would be the nile , say . you could talk about a mountain , and that would be a common noun , because there are many mountains , but if you wanted to talk about a specific mountain , say mount kilimanjaro , in tanzania , that 's a proper noun . so here are the properties of proper nouns . proper nouns are always capitalized . and that means that instead of using a little letter a like that , you would instead use a big letter a like that . common nouns are only capitalized if you find them at the beginning of sentences . so you might say , `` mountains are my favorite . '' but you would also say , `` kilimanjaro `` is my favorite `` mountain . '' and that is a lowercase , non-capitalized m , as opposed to this one , which is uppercase . so that 's the difference between common and proper nouns . if you 're talking about something general , it 's a common noun . if you 're talking about something specific , it 's a proper noun , and the difference between them is that you capitalize a proper noun . you can learn anything . david out .
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you could talk about a mountain , and that would be a common noun , because there are many mountains , but if you wanted to talk about a specific mountain , say mount kilimanjaro , in tanzania , that 's a proper noun . so here are the properties of proper nouns . proper nouns are always capitalized . and that means that instead of using a little letter a like that , you would instead use a big letter a like that .
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if you can have a proper noun , what are improper nouns ?
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hello , grammarians . i 'd like to bring up the idea of the difference between a common and a proper noun . so the difference between a common and a proper noun is simply the difference between something with a name and a more generic version of that thing . i 'll give you a couple of examples right off the bat . so speaking generally , i am from a city . the specific city that i 'm from is chicago . i could talk about a frog generally , but if i were speaking of a specific frog , i would say kermit . the difference between a common and a proper noun is merely the difference between a general thing , so this side is more general , and a specific thing . it 's a continuum . so if you are speaking of , let 's see , a river , any old river , that 's a common noun , but if you 're talking about a specific river , and it 's a named river here , that would be the nile , say . you could talk about a mountain , and that would be a common noun , because there are many mountains , but if you wanted to talk about a specific mountain , say mount kilimanjaro , in tanzania , that 's a proper noun . so here are the properties of proper nouns . proper nouns are always capitalized . and that means that instead of using a little letter a like that , you would instead use a big letter a like that . common nouns are only capitalized if you find them at the beginning of sentences . so you might say , `` mountains are my favorite . '' but you would also say , `` kilimanjaro `` is my favorite `` mountain . '' and that is a lowercase , non-capitalized m , as opposed to this one , which is uppercase . so that 's the difference between common and proper nouns . if you 're talking about something general , it 's a common noun . if you 're talking about something specific , it 's a proper noun , and the difference between them is that you capitalize a proper noun . you can learn anything . david out .
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you could talk about a mountain , and that would be a common noun , because there are many mountains , but if you wanted to talk about a specific mountain , say mount kilimanjaro , in tanzania , that 's a proper noun . so here are the properties of proper nouns . proper nouns are always capitalized . and that means that instead of using a little letter a like that , you would instead use a big letter a like that .
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might be a stupid question but does every language in existence have nouns ?
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hello , grammarians . i 'd like to bring up the idea of the difference between a common and a proper noun . so the difference between a common and a proper noun is simply the difference between something with a name and a more generic version of that thing . i 'll give you a couple of examples right off the bat . so speaking generally , i am from a city . the specific city that i 'm from is chicago . i could talk about a frog generally , but if i were speaking of a specific frog , i would say kermit . the difference between a common and a proper noun is merely the difference between a general thing , so this side is more general , and a specific thing . it 's a continuum . so if you are speaking of , let 's see , a river , any old river , that 's a common noun , but if you 're talking about a specific river , and it 's a named river here , that would be the nile , say . you could talk about a mountain , and that would be a common noun , because there are many mountains , but if you wanted to talk about a specific mountain , say mount kilimanjaro , in tanzania , that 's a proper noun . so here are the properties of proper nouns . proper nouns are always capitalized . and that means that instead of using a little letter a like that , you would instead use a big letter a like that . common nouns are only capitalized if you find them at the beginning of sentences . so you might say , `` mountains are my favorite . '' but you would also say , `` kilimanjaro `` is my favorite `` mountain . '' and that is a lowercase , non-capitalized m , as opposed to this one , which is uppercase . so that 's the difference between common and proper nouns . if you 're talking about something general , it 's a common noun . if you 're talking about something specific , it 's a proper noun , and the difference between them is that you capitalize a proper noun . you can learn anything . david out .
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and that is a lowercase , non-capitalized m , as opposed to this one , which is uppercase . so that 's the difference between common and proper nouns . if you 're talking about something general , it 's a common noun . if you 're talking about something specific , it 's a proper noun , and the difference between them is that you capitalize a proper noun .
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will boy be a propernoun or a common noun ?
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hello , grammarians . i 'd like to bring up the idea of the difference between a common and a proper noun . so the difference between a common and a proper noun is simply the difference between something with a name and a more generic version of that thing . i 'll give you a couple of examples right off the bat . so speaking generally , i am from a city . the specific city that i 'm from is chicago . i could talk about a frog generally , but if i were speaking of a specific frog , i would say kermit . the difference between a common and a proper noun is merely the difference between a general thing , so this side is more general , and a specific thing . it 's a continuum . so if you are speaking of , let 's see , a river , any old river , that 's a common noun , but if you 're talking about a specific river , and it 's a named river here , that would be the nile , say . you could talk about a mountain , and that would be a common noun , because there are many mountains , but if you wanted to talk about a specific mountain , say mount kilimanjaro , in tanzania , that 's a proper noun . so here are the properties of proper nouns . proper nouns are always capitalized . and that means that instead of using a little letter a like that , you would instead use a big letter a like that . common nouns are only capitalized if you find them at the beginning of sentences . so you might say , `` mountains are my favorite . '' but you would also say , `` kilimanjaro `` is my favorite `` mountain . '' and that is a lowercase , non-capitalized m , as opposed to this one , which is uppercase . so that 's the difference between common and proper nouns . if you 're talking about something general , it 's a common noun . if you 're talking about something specific , it 's a proper noun , and the difference between them is that you capitalize a proper noun . you can learn anything . david out .
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you could talk about a mountain , and that would be a common noun , because there are many mountains , but if you wanted to talk about a specific mountain , say mount kilimanjaro , in tanzania , that 's a proper noun . so here are the properties of proper nouns . proper nouns are always capitalized . and that means that instead of using a little letter a like that , you would instead use a big letter a like that .
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is there any circumstances where you do n't capitalize proper nouns ?
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( lighthearted music ) : you 're looking at one of ed reinhardt 's black paintings but actually there is no black on this painting . what may first appear as an all over black square , actually is a grid . a three by three grid of well , nine squares and each square contains an intensely deep shade of either red , green or blue . one could call this chromatic blacks or colored blacks because at the four corners of the painting , we see actually a deep shade of red . across the center of the painting we see a very deep green . in midway , along the top and bottom edges we find a very deep blue . now if you do n't see this at first , there 's a reason for it because even in front of the painting in the gallery perceiving this painting is a function of the rods and cones adjusting in your eye . it 's actually the same experience you have when waking up in the middle of the night . when at first everything is black and then gradually as your rods and cones adjusts , color forms slowly , gradually emerge . perceiving this composition takes time , it takes patience and it takes attention . ad reinhardt was actually very , very interested in exactly those qualities for he 's not to purify art and the experience of it . reinhardt wanted to keep art and business totally separate . he relished the fact that these paintings are almost impossible to reproduce in photography . reinhardt was in the abstract expressionist circle . however the paint qualities that you associate with that movement are totally lacking in reinhardt , and there 's a reason for that . reinhardt was an oppositional figure . if he knew that he were lumped in with the abstract expressionists , he would cringe at the thought . ( lighthearted music ) ad reinhardt 's painting process was a very individualistic one , a very unique one and he never made a mystery of his technique like so many other new york school painters did . corey : the first step in preparing this exquisite matte quality paint is actually involving these jars here . interestingly his materials , despite the fact that his paintings look so odd and unique , his materials are straight up classical . nothing more than oil paint out of the tube and turpentine , the typical solvent for all oil painting . what he would typically do is to use quite a bit of mars black paint . to that entire quantity he would add just a little bit of one of the three colors he painted with : red , green and blue . next a generous dose of turpentine . what i 'm doing now is making sure that that oil paint is dissolving into that turpentine very , very thoroughly . reinhardt would then leave this jars on his shelves in his studio for a week , for two weeks , for perhaps a month . the reason for that waiting period is that the dense part of the paint , in other words the pigment , would settle to the bottom . meanwhile the light part of that mixture would rise to the top . what is the light part ? well it 's the turpentine that he just added . now the oil , the binder from that tube of paint now extracted from that pigment , lifted up to the top of this jar . what he would do next is this , he would open the jar and then pour off all of that solvent phase if you will or all that light part of the paint mixture . leaving behind only that sludge of paint . ( lighthearted music ) : because reinhardt has withdrawn so much medium from his paint , the resulting paint surfaces are almost free of any trace of brush work . in addition , they are the most matte paint surfaces you will probably ever see . because there 's no gloss , because there 's no reflection on that surface there 's no other light hitting us in the eye . in other words , we have the opportunity to perceive color directly . reinhardt was by far in a way the most subtle colorist of the abstract expressionist painters . his use of color was so subtle , in fact , that it 's on the very threshold of perception . to see these painting we quite literally have to slow down the pace of every day life . his paintings demand our patience and close looking . ( lighthearted music )
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he relished the fact that these paintings are almost impossible to reproduce in photography . reinhardt was in the abstract expressionist circle . however the paint qualities that you associate with that movement are totally lacking in reinhardt , and there 's a reason for that .
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if reinhardt did n't want to be clumped in with abstract expressionists , why was he ?
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( lighthearted music ) : you 're looking at one of ed reinhardt 's black paintings but actually there is no black on this painting . what may first appear as an all over black square , actually is a grid . a three by three grid of well , nine squares and each square contains an intensely deep shade of either red , green or blue . one could call this chromatic blacks or colored blacks because at the four corners of the painting , we see actually a deep shade of red . across the center of the painting we see a very deep green . in midway , along the top and bottom edges we find a very deep blue . now if you do n't see this at first , there 's a reason for it because even in front of the painting in the gallery perceiving this painting is a function of the rods and cones adjusting in your eye . it 's actually the same experience you have when waking up in the middle of the night . when at first everything is black and then gradually as your rods and cones adjusts , color forms slowly , gradually emerge . perceiving this composition takes time , it takes patience and it takes attention . ad reinhardt was actually very , very interested in exactly those qualities for he 's not to purify art and the experience of it . reinhardt wanted to keep art and business totally separate . he relished the fact that these paintings are almost impossible to reproduce in photography . reinhardt was in the abstract expressionist circle . however the paint qualities that you associate with that movement are totally lacking in reinhardt , and there 's a reason for that . reinhardt was an oppositional figure . if he knew that he were lumped in with the abstract expressionists , he would cringe at the thought . ( lighthearted music ) ad reinhardt 's painting process was a very individualistic one , a very unique one and he never made a mystery of his technique like so many other new york school painters did . corey : the first step in preparing this exquisite matte quality paint is actually involving these jars here . interestingly his materials , despite the fact that his paintings look so odd and unique , his materials are straight up classical . nothing more than oil paint out of the tube and turpentine , the typical solvent for all oil painting . what he would typically do is to use quite a bit of mars black paint . to that entire quantity he would add just a little bit of one of the three colors he painted with : red , green and blue . next a generous dose of turpentine . what i 'm doing now is making sure that that oil paint is dissolving into that turpentine very , very thoroughly . reinhardt would then leave this jars on his shelves in his studio for a week , for two weeks , for perhaps a month . the reason for that waiting period is that the dense part of the paint , in other words the pigment , would settle to the bottom . meanwhile the light part of that mixture would rise to the top . what is the light part ? well it 's the turpentine that he just added . now the oil , the binder from that tube of paint now extracted from that pigment , lifted up to the top of this jar . what he would do next is this , he would open the jar and then pour off all of that solvent phase if you will or all that light part of the paint mixture . leaving behind only that sludge of paint . ( lighthearted music ) : because reinhardt has withdrawn so much medium from his paint , the resulting paint surfaces are almost free of any trace of brush work . in addition , they are the most matte paint surfaces you will probably ever see . because there 's no gloss , because there 's no reflection on that surface there 's no other light hitting us in the eye . in other words , we have the opportunity to perceive color directly . reinhardt was by far in a way the most subtle colorist of the abstract expressionist painters . his use of color was so subtle , in fact , that it 's on the very threshold of perception . to see these painting we quite literally have to slow down the pace of every day life . his paintings demand our patience and close looking . ( lighthearted music )
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perceiving this composition takes time , it takes patience and it takes attention . ad reinhardt was actually very , very interested in exactly those qualities for he 's not to purify art and the experience of it . reinhardt wanted to keep art and business totally separate . he relished the fact that these paintings are almost impossible to reproduce in photography .
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~- `` purify art '' - what was reinhardt 's view of `` purified '' art ?
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( lighthearted music ) : you 're looking at one of ed reinhardt 's black paintings but actually there is no black on this painting . what may first appear as an all over black square , actually is a grid . a three by three grid of well , nine squares and each square contains an intensely deep shade of either red , green or blue . one could call this chromatic blacks or colored blacks because at the four corners of the painting , we see actually a deep shade of red . across the center of the painting we see a very deep green . in midway , along the top and bottom edges we find a very deep blue . now if you do n't see this at first , there 's a reason for it because even in front of the painting in the gallery perceiving this painting is a function of the rods and cones adjusting in your eye . it 's actually the same experience you have when waking up in the middle of the night . when at first everything is black and then gradually as your rods and cones adjusts , color forms slowly , gradually emerge . perceiving this composition takes time , it takes patience and it takes attention . ad reinhardt was actually very , very interested in exactly those qualities for he 's not to purify art and the experience of it . reinhardt wanted to keep art and business totally separate . he relished the fact that these paintings are almost impossible to reproduce in photography . reinhardt was in the abstract expressionist circle . however the paint qualities that you associate with that movement are totally lacking in reinhardt , and there 's a reason for that . reinhardt was an oppositional figure . if he knew that he were lumped in with the abstract expressionists , he would cringe at the thought . ( lighthearted music ) ad reinhardt 's painting process was a very individualistic one , a very unique one and he never made a mystery of his technique like so many other new york school painters did . corey : the first step in preparing this exquisite matte quality paint is actually involving these jars here . interestingly his materials , despite the fact that his paintings look so odd and unique , his materials are straight up classical . nothing more than oil paint out of the tube and turpentine , the typical solvent for all oil painting . what he would typically do is to use quite a bit of mars black paint . to that entire quantity he would add just a little bit of one of the three colors he painted with : red , green and blue . next a generous dose of turpentine . what i 'm doing now is making sure that that oil paint is dissolving into that turpentine very , very thoroughly . reinhardt would then leave this jars on his shelves in his studio for a week , for two weeks , for perhaps a month . the reason for that waiting period is that the dense part of the paint , in other words the pigment , would settle to the bottom . meanwhile the light part of that mixture would rise to the top . what is the light part ? well it 's the turpentine that he just added . now the oil , the binder from that tube of paint now extracted from that pigment , lifted up to the top of this jar . what he would do next is this , he would open the jar and then pour off all of that solvent phase if you will or all that light part of the paint mixture . leaving behind only that sludge of paint . ( lighthearted music ) : because reinhardt has withdrawn so much medium from his paint , the resulting paint surfaces are almost free of any trace of brush work . in addition , they are the most matte paint surfaces you will probably ever see . because there 's no gloss , because there 's no reflection on that surface there 's no other light hitting us in the eye . in other words , we have the opportunity to perceive color directly . reinhardt was by far in a way the most subtle colorist of the abstract expressionist painters . his use of color was so subtle , in fact , that it 's on the very threshold of perception . to see these painting we quite literally have to slow down the pace of every day life . his paintings demand our patience and close looking . ( lighthearted music )
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however the paint qualities that you associate with that movement are totally lacking in reinhardt , and there 's a reason for that . reinhardt was an oppositional figure . if he knew that he were lumped in with the abstract expressionists , he would cringe at the thought . ( lighthearted music ) ad reinhardt 's painting process was a very individualistic one , a very unique one and he never made a mystery of his technique like so many other new york school painters did .
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~- why would reinhardt cringe at the idea of being associated with others of his genre ?
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( lighthearted music ) : you 're looking at one of ed reinhardt 's black paintings but actually there is no black on this painting . what may first appear as an all over black square , actually is a grid . a three by three grid of well , nine squares and each square contains an intensely deep shade of either red , green or blue . one could call this chromatic blacks or colored blacks because at the four corners of the painting , we see actually a deep shade of red . across the center of the painting we see a very deep green . in midway , along the top and bottom edges we find a very deep blue . now if you do n't see this at first , there 's a reason for it because even in front of the painting in the gallery perceiving this painting is a function of the rods and cones adjusting in your eye . it 's actually the same experience you have when waking up in the middle of the night . when at first everything is black and then gradually as your rods and cones adjusts , color forms slowly , gradually emerge . perceiving this composition takes time , it takes patience and it takes attention . ad reinhardt was actually very , very interested in exactly those qualities for he 's not to purify art and the experience of it . reinhardt wanted to keep art and business totally separate . he relished the fact that these paintings are almost impossible to reproduce in photography . reinhardt was in the abstract expressionist circle . however the paint qualities that you associate with that movement are totally lacking in reinhardt , and there 's a reason for that . reinhardt was an oppositional figure . if he knew that he were lumped in with the abstract expressionists , he would cringe at the thought . ( lighthearted music ) ad reinhardt 's painting process was a very individualistic one , a very unique one and he never made a mystery of his technique like so many other new york school painters did . corey : the first step in preparing this exquisite matte quality paint is actually involving these jars here . interestingly his materials , despite the fact that his paintings look so odd and unique , his materials are straight up classical . nothing more than oil paint out of the tube and turpentine , the typical solvent for all oil painting . what he would typically do is to use quite a bit of mars black paint . to that entire quantity he would add just a little bit of one of the three colors he painted with : red , green and blue . next a generous dose of turpentine . what i 'm doing now is making sure that that oil paint is dissolving into that turpentine very , very thoroughly . reinhardt would then leave this jars on his shelves in his studio for a week , for two weeks , for perhaps a month . the reason for that waiting period is that the dense part of the paint , in other words the pigment , would settle to the bottom . meanwhile the light part of that mixture would rise to the top . what is the light part ? well it 's the turpentine that he just added . now the oil , the binder from that tube of paint now extracted from that pigment , lifted up to the top of this jar . what he would do next is this , he would open the jar and then pour off all of that solvent phase if you will or all that light part of the paint mixture . leaving behind only that sludge of paint . ( lighthearted music ) : because reinhardt has withdrawn so much medium from his paint , the resulting paint surfaces are almost free of any trace of brush work . in addition , they are the most matte paint surfaces you will probably ever see . because there 's no gloss , because there 's no reflection on that surface there 's no other light hitting us in the eye . in other words , we have the opportunity to perceive color directly . reinhardt was by far in a way the most subtle colorist of the abstract expressionist painters . his use of color was so subtle , in fact , that it 's on the very threshold of perception . to see these painting we quite literally have to slow down the pace of every day life . his paintings demand our patience and close looking . ( lighthearted music )
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what he would do next is this , he would open the jar and then pour off all of that solvent phase if you will or all that light part of the paint mixture . leaving behind only that sludge of paint . ( lighthearted music ) : because reinhardt has withdrawn so much medium from his paint , the resulting paint surfaces are almost free of any trace of brush work .
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~- acquiring the `` sludge of paint '' - would it not be possible to acquire this in a more expeditious manner ?
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( lighthearted music ) : you 're looking at one of ed reinhardt 's black paintings but actually there is no black on this painting . what may first appear as an all over black square , actually is a grid . a three by three grid of well , nine squares and each square contains an intensely deep shade of either red , green or blue . one could call this chromatic blacks or colored blacks because at the four corners of the painting , we see actually a deep shade of red . across the center of the painting we see a very deep green . in midway , along the top and bottom edges we find a very deep blue . now if you do n't see this at first , there 's a reason for it because even in front of the painting in the gallery perceiving this painting is a function of the rods and cones adjusting in your eye . it 's actually the same experience you have when waking up in the middle of the night . when at first everything is black and then gradually as your rods and cones adjusts , color forms slowly , gradually emerge . perceiving this composition takes time , it takes patience and it takes attention . ad reinhardt was actually very , very interested in exactly those qualities for he 's not to purify art and the experience of it . reinhardt wanted to keep art and business totally separate . he relished the fact that these paintings are almost impossible to reproduce in photography . reinhardt was in the abstract expressionist circle . however the paint qualities that you associate with that movement are totally lacking in reinhardt , and there 's a reason for that . reinhardt was an oppositional figure . if he knew that he were lumped in with the abstract expressionists , he would cringe at the thought . ( lighthearted music ) ad reinhardt 's painting process was a very individualistic one , a very unique one and he never made a mystery of his technique like so many other new york school painters did . corey : the first step in preparing this exquisite matte quality paint is actually involving these jars here . interestingly his materials , despite the fact that his paintings look so odd and unique , his materials are straight up classical . nothing more than oil paint out of the tube and turpentine , the typical solvent for all oil painting . what he would typically do is to use quite a bit of mars black paint . to that entire quantity he would add just a little bit of one of the three colors he painted with : red , green and blue . next a generous dose of turpentine . what i 'm doing now is making sure that that oil paint is dissolving into that turpentine very , very thoroughly . reinhardt would then leave this jars on his shelves in his studio for a week , for two weeks , for perhaps a month . the reason for that waiting period is that the dense part of the paint , in other words the pigment , would settle to the bottom . meanwhile the light part of that mixture would rise to the top . what is the light part ? well it 's the turpentine that he just added . now the oil , the binder from that tube of paint now extracted from that pigment , lifted up to the top of this jar . what he would do next is this , he would open the jar and then pour off all of that solvent phase if you will or all that light part of the paint mixture . leaving behind only that sludge of paint . ( lighthearted music ) : because reinhardt has withdrawn so much medium from his paint , the resulting paint surfaces are almost free of any trace of brush work . in addition , they are the most matte paint surfaces you will probably ever see . because there 's no gloss , because there 's no reflection on that surface there 's no other light hitting us in the eye . in other words , we have the opportunity to perceive color directly . reinhardt was by far in a way the most subtle colorist of the abstract expressionist painters . his use of color was so subtle , in fact , that it 's on the very threshold of perception . to see these painting we quite literally have to slow down the pace of every day life . his paintings demand our patience and close looking . ( lighthearted music )
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he relished the fact that these paintings are almost impossible to reproduce in photography . reinhardt was in the abstract expressionist circle . however the paint qualities that you associate with that movement are totally lacking in reinhardt , and there 's a reason for that .
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does anyone know if there are any images of any these artist pre-abstract expressionism ?
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let k equal six , so that f of x is equal to one over x-squared minus six x . find the partial fraction decomposition for the function f. find the integral of f of x d-x . and so let 's first thinl about the partial fraction decomposition for the function f. so , f of x , i could rewrite it where i factor the denominator . if i factor out an x , i get x times x minus six . so , i can rewrite this as ... and this is where i 'm going to decompose it into partial fractions . a over x plus b over x minus six . if i actually had to add these two , i would try to find a common denominator . the best common denominator is just to take the product of these two expressions . so i could multiply the numerator and denomenator of this first time by x minus six . x minus six . and the numerator and denominator of the second term , i can multiply it by the denominator of the other one . so times x and times x . so that would give us , let 's see if i distribute the a , it would give us a x minus six a plus b x over x times x minus six . if this looks completely foreign to you , i encourage you to watch the videos on kahn academy on partial fraction decomposition . so let 's see if we can , so what we wan na do is solve for the a 's and the b 's . and so let 's see , if we see that these have to add up to one over x times x minus six . so these have to add up . i 'm just going back to this , these have to add up . the numerator . so it has to add up to one over x times x minus six . so this numerator has to add up to one . so what we can see is is that the x terms right over here must cancel out since we have no x terms here . we have a x plus b x must be equal to zero . or you could say , well that means a plus b is equal to zero . so we took care of that term and that term . and then we know , that this must be the constant term that adds up to one or that is equal to one . and so we also know that negative six a is equal to one . or a is equal to , divide both sides by negative six , negative one sixth . and if a is negative one sixth , well b is going to be the negative of that . and negative one sixth plus b. i 'm just substituting a back into that equation . need to be equal to zero . and so add one sixth to both sides , you get b is equal to one sixth . so i can decompose f of x. i can decompose f of x as being equal to a over x so that 's negative one sixth over x. i just write it that way . i could write it as negative one over six x or something like that . but i 'll just write it like this , just be clear that this was our a . plus b , which is one sixth over x minus six . so that right over there , that 's the partial fraction decomposition for our function f. and if i want to evaluate the integral . so the integral of f of x , the indefinite integral . well , that 's where this partial fractions decomposition is going to be valuable . that 's going to be the indefinite integral of negative one sixth over x plus positive one sixth over x minus six , and then we have d x . well what 's the anti-derivative of this right over here ? well the anti-derivative of one over x is the natural log of the absolute value of x . and so we can just this is going to be negative one sixth times the natural log of the absolute value of x. anti-derivative of this part . then plus one sixth . you could do u-substitution but you could just say `` hey , look the derivative of this bottom part , x minus six . '' that 's just one . so you could say that i have that derivative laying around . so i can just take the anti-derivative with respect to that . so that 's going to be one sixth times the natural log of the absolute value of x minus six , and then i have plus c. do n't forget this is an indefinite integral over here . and then you 're done . and you see that partial fraction decomposition was actually quite useful . so they were helping us how to figure out this . you did n't just have to have that insight that , okay how do i evaluate this anti-derivative . well , the they 're telling us to use partial fraction decomposition .
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and so let 's first thinl about the partial fraction decomposition for the function f. so , f of x , i could rewrite it where i factor the denominator . if i factor out an x , i get x times x minus six . so , i can rewrite this as ... and this is where i 'm going to decompose it into partial fractions .
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could n't you use the natural log properties to simplify it to be 1/6ln abs ( x-6/x ) + c ?
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what i wan na do in this video is a few examples that test our intuition of the derivative as a rate of change or the steepness of a curve or the slope of a curve or the slope of a tangent line of a curve depending on how you actually want to think about it . so here it says f prime of five so this notation , prime this is another way of saying well what 's the derivative let 's estimate the derivative of our function at five . and when we say f prime of five this is the slope slope of tangent line tangent line at five or you could view it as the you could view it as the rate of change of y with respect to x which is really how we define slope respect to x of our function f. so let 's think about that a little bit . we see they put the point the point five comma f of five right over here and so if we want to estimate the slope of the tangent line if we want to estimate the steepness of this curve we could try to draw a line that is tangent right at that point . so let me see if i can do that . so if i were to draw a line starting there if i just wanted to make a tangent it looks like it would do something like that . right at that point that looks to be about how steep that curve is now what makes this an interesting thing in non-linear is that it 's constantly changing the steepness it 's very low here and it gets steeper and steeper and steeper as we move to the right for larger and larger x values . but if we look at the point in question when x is equal to five remember f prime of five would be if you were estimating it this would be the slope of this line here . and the slope of this line it looks like for every time we move one in the x direction we 're moving two in the y direction . delta y is equal to two when delta x is equal to one . so our change in y with respect to x at least for this tangent line here which would represent our change in y with respect to x right at that point is going to be equal to two over one , or two . and it 's almost estimated , but all of these are way off . having a negative two derivative would mean that as we increase our x our y is decreasing . so if our curve looks something like this we would have a slope of negative two . if having slopes in this a positive of point one that would be very flat something down here we might have a slope closer to point one . negative point one that might be closer on this side now we 're sloping but very close to flat . a slope of zero , that would be right over here at the bottom where right at that moment as we change x y is not increasing or decreasing the slope of the tangent line right at that bottom point would have a slope of zero . so i feel really good about that response . let 's do one more of these . so alright , so they 're telling us to compare the derivative of g at four to the derivative of g at six and which of these is greater and like always , pause the video and see if you can figure this out . well this is just an exercise let 's see if we were to if we were to make a line that indicates the slope there you can do this as a tangent line let me try to do that . so now that would n't , that does n't do a good job so right over here at that looks like a i think i can do a better job than that no that 's too shallow to see not shallow 's not the word , that 's too flat . so let me try to really okay , that looks pretty good . so that line that i just drew seems to be indicative of the rate of change of y with respect to x or the slope of that curve or that line you can view it as a tangent line so we could think about what its slope is going to be and then if we go further down over here this one is , it looks like it is steeper but in the negative direction so it looks like it is steeper for sure but it 's in the negative direction . as we increase , think of it this way as we increase x one here it looks like we are decreasing y by about one . so it looks like g prime of four g prime of four , the derivative when x is equal to four is approximately , i 'm estimating it negative one while the derivative here when we increase x if we increase x by if we increase x by one it looks like we 're decreasing y by close to three so g prime of six looks like it 's closer to negative three . so which one of these is larger ? well , this one is less negative so it 's going to be greater than the other one and you could have done this intuitively if you just look at the curve this is some type of a sinusoid here you have right over here the curve is flat you have right at that moment you have no change in y with respect to x then it starts to decrease then it decreases at an even faster rate then it decreases at a faster rate then it starts , it 's still decreasing but it 's decreasing at slower and slower rates decreasing at slower rates and right at that moment you have your slope of your tangent line is zero then it starts to increase , increase , so on and so forth and it just keeps happening over and over again . so you can also think about this in a more intuitive way .
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delta y is equal to two when delta x is equal to one . so our change in y with respect to x at least for this tangent line here which would represent our change in y with respect to x right at that point is going to be equal to two over one , or two . and it 's almost estimated , but all of these are way off .
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why ca n't we differentiate acceleration again with respect to time ?
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what i wan na do in this video is a few examples that test our intuition of the derivative as a rate of change or the steepness of a curve or the slope of a curve or the slope of a tangent line of a curve depending on how you actually want to think about it . so here it says f prime of five so this notation , prime this is another way of saying well what 's the derivative let 's estimate the derivative of our function at five . and when we say f prime of five this is the slope slope of tangent line tangent line at five or you could view it as the you could view it as the rate of change of y with respect to x which is really how we define slope respect to x of our function f. so let 's think about that a little bit . we see they put the point the point five comma f of five right over here and so if we want to estimate the slope of the tangent line if we want to estimate the steepness of this curve we could try to draw a line that is tangent right at that point . so let me see if i can do that . so if i were to draw a line starting there if i just wanted to make a tangent it looks like it would do something like that . right at that point that looks to be about how steep that curve is now what makes this an interesting thing in non-linear is that it 's constantly changing the steepness it 's very low here and it gets steeper and steeper and steeper as we move to the right for larger and larger x values . but if we look at the point in question when x is equal to five remember f prime of five would be if you were estimating it this would be the slope of this line here . and the slope of this line it looks like for every time we move one in the x direction we 're moving two in the y direction . delta y is equal to two when delta x is equal to one . so our change in y with respect to x at least for this tangent line here which would represent our change in y with respect to x right at that point is going to be equal to two over one , or two . and it 's almost estimated , but all of these are way off . having a negative two derivative would mean that as we increase our x our y is decreasing . so if our curve looks something like this we would have a slope of negative two . if having slopes in this a positive of point one that would be very flat something down here we might have a slope closer to point one . negative point one that might be closer on this side now we 're sloping but very close to flat . a slope of zero , that would be right over here at the bottom where right at that moment as we change x y is not increasing or decreasing the slope of the tangent line right at that bottom point would have a slope of zero . so i feel really good about that response . let 's do one more of these . so alright , so they 're telling us to compare the derivative of g at four to the derivative of g at six and which of these is greater and like always , pause the video and see if you can figure this out . well this is just an exercise let 's see if we were to if we were to make a line that indicates the slope there you can do this as a tangent line let me try to do that . so now that would n't , that does n't do a good job so right over here at that looks like a i think i can do a better job than that no that 's too shallow to see not shallow 's not the word , that 's too flat . so let me try to really okay , that looks pretty good . so that line that i just drew seems to be indicative of the rate of change of y with respect to x or the slope of that curve or that line you can view it as a tangent line so we could think about what its slope is going to be and then if we go further down over here this one is , it looks like it is steeper but in the negative direction so it looks like it is steeper for sure but it 's in the negative direction . as we increase , think of it this way as we increase x one here it looks like we are decreasing y by about one . so it looks like g prime of four g prime of four , the derivative when x is equal to four is approximately , i 'm estimating it negative one while the derivative here when we increase x if we increase x by if we increase x by one it looks like we 're decreasing y by close to three so g prime of six looks like it 's closer to negative three . so which one of these is larger ? well , this one is less negative so it 's going to be greater than the other one and you could have done this intuitively if you just look at the curve this is some type of a sinusoid here you have right over here the curve is flat you have right at that moment you have no change in y with respect to x then it starts to decrease then it decreases at an even faster rate then it decreases at a faster rate then it starts , it 's still decreasing but it 's decreasing at slower and slower rates decreasing at slower rates and right at that moment you have your slope of your tangent line is zero then it starts to increase , increase , so on and so forth and it just keeps happening over and over again . so you can also think about this in a more intuitive way .
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what i wan na do in this video is a few examples that test our intuition of the derivative as a rate of change or the steepness of a curve or the slope of a curve or the slope of a tangent line of a curve depending on how you actually want to think about it . so here it says f prime of five so this notation , prime this is another way of saying well what 's the derivative let 's estimate the derivative of our function at five . and when we say f prime of five this is the slope slope of tangent line tangent line at five or you could view it as the you could view it as the rate of change of y with respect to x which is really how we define slope respect to x of our function f. so let 's think about that a little bit .
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what exactly is a derivative ?
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what i wan na do in this video is a few examples that test our intuition of the derivative as a rate of change or the steepness of a curve or the slope of a curve or the slope of a tangent line of a curve depending on how you actually want to think about it . so here it says f prime of five so this notation , prime this is another way of saying well what 's the derivative let 's estimate the derivative of our function at five . and when we say f prime of five this is the slope slope of tangent line tangent line at five or you could view it as the you could view it as the rate of change of y with respect to x which is really how we define slope respect to x of our function f. so let 's think about that a little bit . we see they put the point the point five comma f of five right over here and so if we want to estimate the slope of the tangent line if we want to estimate the steepness of this curve we could try to draw a line that is tangent right at that point . so let me see if i can do that . so if i were to draw a line starting there if i just wanted to make a tangent it looks like it would do something like that . right at that point that looks to be about how steep that curve is now what makes this an interesting thing in non-linear is that it 's constantly changing the steepness it 's very low here and it gets steeper and steeper and steeper as we move to the right for larger and larger x values . but if we look at the point in question when x is equal to five remember f prime of five would be if you were estimating it this would be the slope of this line here . and the slope of this line it looks like for every time we move one in the x direction we 're moving two in the y direction . delta y is equal to two when delta x is equal to one . so our change in y with respect to x at least for this tangent line here which would represent our change in y with respect to x right at that point is going to be equal to two over one , or two . and it 's almost estimated , but all of these are way off . having a negative two derivative would mean that as we increase our x our y is decreasing . so if our curve looks something like this we would have a slope of negative two . if having slopes in this a positive of point one that would be very flat something down here we might have a slope closer to point one . negative point one that might be closer on this side now we 're sloping but very close to flat . a slope of zero , that would be right over here at the bottom where right at that moment as we change x y is not increasing or decreasing the slope of the tangent line right at that bottom point would have a slope of zero . so i feel really good about that response . let 's do one more of these . so alright , so they 're telling us to compare the derivative of g at four to the derivative of g at six and which of these is greater and like always , pause the video and see if you can figure this out . well this is just an exercise let 's see if we were to if we were to make a line that indicates the slope there you can do this as a tangent line let me try to do that . so now that would n't , that does n't do a good job so right over here at that looks like a i think i can do a better job than that no that 's too shallow to see not shallow 's not the word , that 's too flat . so let me try to really okay , that looks pretty good . so that line that i just drew seems to be indicative of the rate of change of y with respect to x or the slope of that curve or that line you can view it as a tangent line so we could think about what its slope is going to be and then if we go further down over here this one is , it looks like it is steeper but in the negative direction so it looks like it is steeper for sure but it 's in the negative direction . as we increase , think of it this way as we increase x one here it looks like we are decreasing y by about one . so it looks like g prime of four g prime of four , the derivative when x is equal to four is approximately , i 'm estimating it negative one while the derivative here when we increase x if we increase x by if we increase x by one it looks like we 're decreasing y by close to three so g prime of six looks like it 's closer to negative three . so which one of these is larger ? well , this one is less negative so it 's going to be greater than the other one and you could have done this intuitively if you just look at the curve this is some type of a sinusoid here you have right over here the curve is flat you have right at that moment you have no change in y with respect to x then it starts to decrease then it decreases at an even faster rate then it decreases at a faster rate then it starts , it 's still decreasing but it 's decreasing at slower and slower rates decreasing at slower rates and right at that moment you have your slope of your tangent line is zero then it starts to increase , increase , so on and so forth and it just keeps happening over and over again . so you can also think about this in a more intuitive way .
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as we increase , think of it this way as we increase x one here it looks like we are decreasing y by about one . so it looks like g prime of four g prime of four , the derivative when x is equal to four is approximately , i 'm estimating it negative one while the derivative here when we increase x if we increase x by if we increase x by one it looks like we 're decreasing y by close to three so g prime of six looks like it 's closer to negative three . so which one of these is larger ?
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lim as x approaches toward zero , x * sin ( 1/x ) = 0 , how ?
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what i wan na do in this video is a few examples that test our intuition of the derivative as a rate of change or the steepness of a curve or the slope of a curve or the slope of a tangent line of a curve depending on how you actually want to think about it . so here it says f prime of five so this notation , prime this is another way of saying well what 's the derivative let 's estimate the derivative of our function at five . and when we say f prime of five this is the slope slope of tangent line tangent line at five or you could view it as the you could view it as the rate of change of y with respect to x which is really how we define slope respect to x of our function f. so let 's think about that a little bit . we see they put the point the point five comma f of five right over here and so if we want to estimate the slope of the tangent line if we want to estimate the steepness of this curve we could try to draw a line that is tangent right at that point . so let me see if i can do that . so if i were to draw a line starting there if i just wanted to make a tangent it looks like it would do something like that . right at that point that looks to be about how steep that curve is now what makes this an interesting thing in non-linear is that it 's constantly changing the steepness it 's very low here and it gets steeper and steeper and steeper as we move to the right for larger and larger x values . but if we look at the point in question when x is equal to five remember f prime of five would be if you were estimating it this would be the slope of this line here . and the slope of this line it looks like for every time we move one in the x direction we 're moving two in the y direction . delta y is equal to two when delta x is equal to one . so our change in y with respect to x at least for this tangent line here which would represent our change in y with respect to x right at that point is going to be equal to two over one , or two . and it 's almost estimated , but all of these are way off . having a negative two derivative would mean that as we increase our x our y is decreasing . so if our curve looks something like this we would have a slope of negative two . if having slopes in this a positive of point one that would be very flat something down here we might have a slope closer to point one . negative point one that might be closer on this side now we 're sloping but very close to flat . a slope of zero , that would be right over here at the bottom where right at that moment as we change x y is not increasing or decreasing the slope of the tangent line right at that bottom point would have a slope of zero . so i feel really good about that response . let 's do one more of these . so alright , so they 're telling us to compare the derivative of g at four to the derivative of g at six and which of these is greater and like always , pause the video and see if you can figure this out . well this is just an exercise let 's see if we were to if we were to make a line that indicates the slope there you can do this as a tangent line let me try to do that . so now that would n't , that does n't do a good job so right over here at that looks like a i think i can do a better job than that no that 's too shallow to see not shallow 's not the word , that 's too flat . so let me try to really okay , that looks pretty good . so that line that i just drew seems to be indicative of the rate of change of y with respect to x or the slope of that curve or that line you can view it as a tangent line so we could think about what its slope is going to be and then if we go further down over here this one is , it looks like it is steeper but in the negative direction so it looks like it is steeper for sure but it 's in the negative direction . as we increase , think of it this way as we increase x one here it looks like we are decreasing y by about one . so it looks like g prime of four g prime of four , the derivative when x is equal to four is approximately , i 'm estimating it negative one while the derivative here when we increase x if we increase x by if we increase x by one it looks like we 're decreasing y by close to three so g prime of six looks like it 's closer to negative three . so which one of these is larger ? well , this one is less negative so it 's going to be greater than the other one and you could have done this intuitively if you just look at the curve this is some type of a sinusoid here you have right over here the curve is flat you have right at that moment you have no change in y with respect to x then it starts to decrease then it decreases at an even faster rate then it decreases at a faster rate then it starts , it 's still decreasing but it 's decreasing at slower and slower rates decreasing at slower rates and right at that moment you have your slope of your tangent line is zero then it starts to increase , increase , so on and so forth and it just keeps happening over and over again . so you can also think about this in a more intuitive way .
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what i wan na do in this video is a few examples that test our intuition of the derivative as a rate of change or the steepness of a curve or the slope of a curve or the slope of a tangent line of a curve depending on how you actually want to think about it . so here it says f prime of five so this notation , prime this is another way of saying well what 's the derivative let 's estimate the derivative of our function at five . and when we say f prime of five this is the slope slope of tangent line tangent line at five or you could view it as the you could view it as the rate of change of y with respect to x which is really how we define slope respect to x of our function f. so let 's think about that a little bit .
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is f ' ( 5 ) derivative of 5 or function of 5 ?
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what i wan na do in this video is a few examples that test our intuition of the derivative as a rate of change or the steepness of a curve or the slope of a curve or the slope of a tangent line of a curve depending on how you actually want to think about it . so here it says f prime of five so this notation , prime this is another way of saying well what 's the derivative let 's estimate the derivative of our function at five . and when we say f prime of five this is the slope slope of tangent line tangent line at five or you could view it as the you could view it as the rate of change of y with respect to x which is really how we define slope respect to x of our function f. so let 's think about that a little bit . we see they put the point the point five comma f of five right over here and so if we want to estimate the slope of the tangent line if we want to estimate the steepness of this curve we could try to draw a line that is tangent right at that point . so let me see if i can do that . so if i were to draw a line starting there if i just wanted to make a tangent it looks like it would do something like that . right at that point that looks to be about how steep that curve is now what makes this an interesting thing in non-linear is that it 's constantly changing the steepness it 's very low here and it gets steeper and steeper and steeper as we move to the right for larger and larger x values . but if we look at the point in question when x is equal to five remember f prime of five would be if you were estimating it this would be the slope of this line here . and the slope of this line it looks like for every time we move one in the x direction we 're moving two in the y direction . delta y is equal to two when delta x is equal to one . so our change in y with respect to x at least for this tangent line here which would represent our change in y with respect to x right at that point is going to be equal to two over one , or two . and it 's almost estimated , but all of these are way off . having a negative two derivative would mean that as we increase our x our y is decreasing . so if our curve looks something like this we would have a slope of negative two . if having slopes in this a positive of point one that would be very flat something down here we might have a slope closer to point one . negative point one that might be closer on this side now we 're sloping but very close to flat . a slope of zero , that would be right over here at the bottom where right at that moment as we change x y is not increasing or decreasing the slope of the tangent line right at that bottom point would have a slope of zero . so i feel really good about that response . let 's do one more of these . so alright , so they 're telling us to compare the derivative of g at four to the derivative of g at six and which of these is greater and like always , pause the video and see if you can figure this out . well this is just an exercise let 's see if we were to if we were to make a line that indicates the slope there you can do this as a tangent line let me try to do that . so now that would n't , that does n't do a good job so right over here at that looks like a i think i can do a better job than that no that 's too shallow to see not shallow 's not the word , that 's too flat . so let me try to really okay , that looks pretty good . so that line that i just drew seems to be indicative of the rate of change of y with respect to x or the slope of that curve or that line you can view it as a tangent line so we could think about what its slope is going to be and then if we go further down over here this one is , it looks like it is steeper but in the negative direction so it looks like it is steeper for sure but it 's in the negative direction . as we increase , think of it this way as we increase x one here it looks like we are decreasing y by about one . so it looks like g prime of four g prime of four , the derivative when x is equal to four is approximately , i 'm estimating it negative one while the derivative here when we increase x if we increase x by if we increase x by one it looks like we 're decreasing y by close to three so g prime of six looks like it 's closer to negative three . so which one of these is larger ? well , this one is less negative so it 's going to be greater than the other one and you could have done this intuitively if you just look at the curve this is some type of a sinusoid here you have right over here the curve is flat you have right at that moment you have no change in y with respect to x then it starts to decrease then it decreases at an even faster rate then it decreases at a faster rate then it starts , it 's still decreasing but it 's decreasing at slower and slower rates decreasing at slower rates and right at that moment you have your slope of your tangent line is zero then it starts to increase , increase , so on and so forth and it just keeps happening over and over again . so you can also think about this in a more intuitive way .
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let 's do one more of these . so alright , so they 're telling us to compare the derivative of g at four to the derivative of g at six and which of these is greater and like always , pause the video and see if you can figure this out . well this is just an exercise let 's see if we were to if we were to make a line that indicates the slope there you can do this as a tangent line let me try to do that .
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still unclear on this topic do we always have to take deltax as 1 ?
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what i wan na do in this video is a few examples that test our intuition of the derivative as a rate of change or the steepness of a curve or the slope of a curve or the slope of a tangent line of a curve depending on how you actually want to think about it . so here it says f prime of five so this notation , prime this is another way of saying well what 's the derivative let 's estimate the derivative of our function at five . and when we say f prime of five this is the slope slope of tangent line tangent line at five or you could view it as the you could view it as the rate of change of y with respect to x which is really how we define slope respect to x of our function f. so let 's think about that a little bit . we see they put the point the point five comma f of five right over here and so if we want to estimate the slope of the tangent line if we want to estimate the steepness of this curve we could try to draw a line that is tangent right at that point . so let me see if i can do that . so if i were to draw a line starting there if i just wanted to make a tangent it looks like it would do something like that . right at that point that looks to be about how steep that curve is now what makes this an interesting thing in non-linear is that it 's constantly changing the steepness it 's very low here and it gets steeper and steeper and steeper as we move to the right for larger and larger x values . but if we look at the point in question when x is equal to five remember f prime of five would be if you were estimating it this would be the slope of this line here . and the slope of this line it looks like for every time we move one in the x direction we 're moving two in the y direction . delta y is equal to two when delta x is equal to one . so our change in y with respect to x at least for this tangent line here which would represent our change in y with respect to x right at that point is going to be equal to two over one , or two . and it 's almost estimated , but all of these are way off . having a negative two derivative would mean that as we increase our x our y is decreasing . so if our curve looks something like this we would have a slope of negative two . if having slopes in this a positive of point one that would be very flat something down here we might have a slope closer to point one . negative point one that might be closer on this side now we 're sloping but very close to flat . a slope of zero , that would be right over here at the bottom where right at that moment as we change x y is not increasing or decreasing the slope of the tangent line right at that bottom point would have a slope of zero . so i feel really good about that response . let 's do one more of these . so alright , so they 're telling us to compare the derivative of g at four to the derivative of g at six and which of these is greater and like always , pause the video and see if you can figure this out . well this is just an exercise let 's see if we were to if we were to make a line that indicates the slope there you can do this as a tangent line let me try to do that . so now that would n't , that does n't do a good job so right over here at that looks like a i think i can do a better job than that no that 's too shallow to see not shallow 's not the word , that 's too flat . so let me try to really okay , that looks pretty good . so that line that i just drew seems to be indicative of the rate of change of y with respect to x or the slope of that curve or that line you can view it as a tangent line so we could think about what its slope is going to be and then if we go further down over here this one is , it looks like it is steeper but in the negative direction so it looks like it is steeper for sure but it 's in the negative direction . as we increase , think of it this way as we increase x one here it looks like we are decreasing y by about one . so it looks like g prime of four g prime of four , the derivative when x is equal to four is approximately , i 'm estimating it negative one while the derivative here when we increase x if we increase x by if we increase x by one it looks like we 're decreasing y by close to three so g prime of six looks like it 's closer to negative three . so which one of these is larger ? well , this one is less negative so it 's going to be greater than the other one and you could have done this intuitively if you just look at the curve this is some type of a sinusoid here you have right over here the curve is flat you have right at that moment you have no change in y with respect to x then it starts to decrease then it decreases at an even faster rate then it decreases at a faster rate then it starts , it 's still decreasing but it 's decreasing at slower and slower rates decreasing at slower rates and right at that moment you have your slope of your tangent line is zero then it starts to increase , increase , so on and so forth and it just keeps happening over and over again . so you can also think about this in a more intuitive way .
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what i wan na do in this video is a few examples that test our intuition of the derivative as a rate of change or the steepness of a curve or the slope of a curve or the slope of a tangent line of a curve depending on how you actually want to think about it . so here it says f prime of five so this notation , prime this is another way of saying well what 's the derivative let 's estimate the derivative of our function at five . and when we say f prime of five this is the slope slope of tangent line tangent line at five or you could view it as the you could view it as the rate of change of y with respect to x which is really how we define slope respect to x of our function f. so let 's think about that a little bit .
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why the derivative of a horizontal line equals to 0 ?
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what i wan na do in this video is a few examples that test our intuition of the derivative as a rate of change or the steepness of a curve or the slope of a curve or the slope of a tangent line of a curve depending on how you actually want to think about it . so here it says f prime of five so this notation , prime this is another way of saying well what 's the derivative let 's estimate the derivative of our function at five . and when we say f prime of five this is the slope slope of tangent line tangent line at five or you could view it as the you could view it as the rate of change of y with respect to x which is really how we define slope respect to x of our function f. so let 's think about that a little bit . we see they put the point the point five comma f of five right over here and so if we want to estimate the slope of the tangent line if we want to estimate the steepness of this curve we could try to draw a line that is tangent right at that point . so let me see if i can do that . so if i were to draw a line starting there if i just wanted to make a tangent it looks like it would do something like that . right at that point that looks to be about how steep that curve is now what makes this an interesting thing in non-linear is that it 's constantly changing the steepness it 's very low here and it gets steeper and steeper and steeper as we move to the right for larger and larger x values . but if we look at the point in question when x is equal to five remember f prime of five would be if you were estimating it this would be the slope of this line here . and the slope of this line it looks like for every time we move one in the x direction we 're moving two in the y direction . delta y is equal to two when delta x is equal to one . so our change in y with respect to x at least for this tangent line here which would represent our change in y with respect to x right at that point is going to be equal to two over one , or two . and it 's almost estimated , but all of these are way off . having a negative two derivative would mean that as we increase our x our y is decreasing . so if our curve looks something like this we would have a slope of negative two . if having slopes in this a positive of point one that would be very flat something down here we might have a slope closer to point one . negative point one that might be closer on this side now we 're sloping but very close to flat . a slope of zero , that would be right over here at the bottom where right at that moment as we change x y is not increasing or decreasing the slope of the tangent line right at that bottom point would have a slope of zero . so i feel really good about that response . let 's do one more of these . so alright , so they 're telling us to compare the derivative of g at four to the derivative of g at six and which of these is greater and like always , pause the video and see if you can figure this out . well this is just an exercise let 's see if we were to if we were to make a line that indicates the slope there you can do this as a tangent line let me try to do that . so now that would n't , that does n't do a good job so right over here at that looks like a i think i can do a better job than that no that 's too shallow to see not shallow 's not the word , that 's too flat . so let me try to really okay , that looks pretty good . so that line that i just drew seems to be indicative of the rate of change of y with respect to x or the slope of that curve or that line you can view it as a tangent line so we could think about what its slope is going to be and then if we go further down over here this one is , it looks like it is steeper but in the negative direction so it looks like it is steeper for sure but it 's in the negative direction . as we increase , think of it this way as we increase x one here it looks like we are decreasing y by about one . so it looks like g prime of four g prime of four , the derivative when x is equal to four is approximately , i 'm estimating it negative one while the derivative here when we increase x if we increase x by if we increase x by one it looks like we 're decreasing y by close to three so g prime of six looks like it 's closer to negative three . so which one of these is larger ? well , this one is less negative so it 's going to be greater than the other one and you could have done this intuitively if you just look at the curve this is some type of a sinusoid here you have right over here the curve is flat you have right at that moment you have no change in y with respect to x then it starts to decrease then it decreases at an even faster rate then it decreases at a faster rate then it starts , it 's still decreasing but it 's decreasing at slower and slower rates decreasing at slower rates and right at that moment you have your slope of your tangent line is zero then it starts to increase , increase , so on and so forth and it just keeps happening over and over again . so you can also think about this in a more intuitive way .
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so let me see if i can do that . so if i were to draw a line starting there if i just wanted to make a tangent it looks like it would do something like that . right at that point that looks to be about how steep that curve is now what makes this an interesting thing in non-linear is that it 's constantly changing the steepness it 's very low here and it gets steeper and steeper and steeper as we move to the right for larger and larger x values .
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so is it okay to draw any arbitrary tangent line ?
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what i wan na do in this video is a few examples that test our intuition of the derivative as a rate of change or the steepness of a curve or the slope of a curve or the slope of a tangent line of a curve depending on how you actually want to think about it . so here it says f prime of five so this notation , prime this is another way of saying well what 's the derivative let 's estimate the derivative of our function at five . and when we say f prime of five this is the slope slope of tangent line tangent line at five or you could view it as the you could view it as the rate of change of y with respect to x which is really how we define slope respect to x of our function f. so let 's think about that a little bit . we see they put the point the point five comma f of five right over here and so if we want to estimate the slope of the tangent line if we want to estimate the steepness of this curve we could try to draw a line that is tangent right at that point . so let me see if i can do that . so if i were to draw a line starting there if i just wanted to make a tangent it looks like it would do something like that . right at that point that looks to be about how steep that curve is now what makes this an interesting thing in non-linear is that it 's constantly changing the steepness it 's very low here and it gets steeper and steeper and steeper as we move to the right for larger and larger x values . but if we look at the point in question when x is equal to five remember f prime of five would be if you were estimating it this would be the slope of this line here . and the slope of this line it looks like for every time we move one in the x direction we 're moving two in the y direction . delta y is equal to two when delta x is equal to one . so our change in y with respect to x at least for this tangent line here which would represent our change in y with respect to x right at that point is going to be equal to two over one , or two . and it 's almost estimated , but all of these are way off . having a negative two derivative would mean that as we increase our x our y is decreasing . so if our curve looks something like this we would have a slope of negative two . if having slopes in this a positive of point one that would be very flat something down here we might have a slope closer to point one . negative point one that might be closer on this side now we 're sloping but very close to flat . a slope of zero , that would be right over here at the bottom where right at that moment as we change x y is not increasing or decreasing the slope of the tangent line right at that bottom point would have a slope of zero . so i feel really good about that response . let 's do one more of these . so alright , so they 're telling us to compare the derivative of g at four to the derivative of g at six and which of these is greater and like always , pause the video and see if you can figure this out . well this is just an exercise let 's see if we were to if we were to make a line that indicates the slope there you can do this as a tangent line let me try to do that . so now that would n't , that does n't do a good job so right over here at that looks like a i think i can do a better job than that no that 's too shallow to see not shallow 's not the word , that 's too flat . so let me try to really okay , that looks pretty good . so that line that i just drew seems to be indicative of the rate of change of y with respect to x or the slope of that curve or that line you can view it as a tangent line so we could think about what its slope is going to be and then if we go further down over here this one is , it looks like it is steeper but in the negative direction so it looks like it is steeper for sure but it 's in the negative direction . as we increase , think of it this way as we increase x one here it looks like we are decreasing y by about one . so it looks like g prime of four g prime of four , the derivative when x is equal to four is approximately , i 'm estimating it negative one while the derivative here when we increase x if we increase x by if we increase x by one it looks like we 're decreasing y by close to three so g prime of six looks like it 's closer to negative three . so which one of these is larger ? well , this one is less negative so it 's going to be greater than the other one and you could have done this intuitively if you just look at the curve this is some type of a sinusoid here you have right over here the curve is flat you have right at that moment you have no change in y with respect to x then it starts to decrease then it decreases at an even faster rate then it decreases at a faster rate then it starts , it 's still decreasing but it 's decreasing at slower and slower rates decreasing at slower rates and right at that moment you have your slope of your tangent line is zero then it starts to increase , increase , so on and so forth and it just keeps happening over and over again . so you can also think about this in a more intuitive way .
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so alright , so they 're telling us to compare the derivative of g at four to the derivative of g at six and which of these is greater and like always , pause the video and see if you can figure this out . well this is just an exercise let 's see if we were to if we were to make a line that indicates the slope there you can do this as a tangent line let me try to do that . so now that would n't , that does n't do a good job so right over here at that looks like a i think i can do a better job than that no that 's too shallow to see not shallow 's not the word , that 's too flat .
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is there any way to find the equation for a tangent line ?
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what i wan na do in this video is a few examples that test our intuition of the derivative as a rate of change or the steepness of a curve or the slope of a curve or the slope of a tangent line of a curve depending on how you actually want to think about it . so here it says f prime of five so this notation , prime this is another way of saying well what 's the derivative let 's estimate the derivative of our function at five . and when we say f prime of five this is the slope slope of tangent line tangent line at five or you could view it as the you could view it as the rate of change of y with respect to x which is really how we define slope respect to x of our function f. so let 's think about that a little bit . we see they put the point the point five comma f of five right over here and so if we want to estimate the slope of the tangent line if we want to estimate the steepness of this curve we could try to draw a line that is tangent right at that point . so let me see if i can do that . so if i were to draw a line starting there if i just wanted to make a tangent it looks like it would do something like that . right at that point that looks to be about how steep that curve is now what makes this an interesting thing in non-linear is that it 's constantly changing the steepness it 's very low here and it gets steeper and steeper and steeper as we move to the right for larger and larger x values . but if we look at the point in question when x is equal to five remember f prime of five would be if you were estimating it this would be the slope of this line here . and the slope of this line it looks like for every time we move one in the x direction we 're moving two in the y direction . delta y is equal to two when delta x is equal to one . so our change in y with respect to x at least for this tangent line here which would represent our change in y with respect to x right at that point is going to be equal to two over one , or two . and it 's almost estimated , but all of these are way off . having a negative two derivative would mean that as we increase our x our y is decreasing . so if our curve looks something like this we would have a slope of negative two . if having slopes in this a positive of point one that would be very flat something down here we might have a slope closer to point one . negative point one that might be closer on this side now we 're sloping but very close to flat . a slope of zero , that would be right over here at the bottom where right at that moment as we change x y is not increasing or decreasing the slope of the tangent line right at that bottom point would have a slope of zero . so i feel really good about that response . let 's do one more of these . so alright , so they 're telling us to compare the derivative of g at four to the derivative of g at six and which of these is greater and like always , pause the video and see if you can figure this out . well this is just an exercise let 's see if we were to if we were to make a line that indicates the slope there you can do this as a tangent line let me try to do that . so now that would n't , that does n't do a good job so right over here at that looks like a i think i can do a better job than that no that 's too shallow to see not shallow 's not the word , that 's too flat . so let me try to really okay , that looks pretty good . so that line that i just drew seems to be indicative of the rate of change of y with respect to x or the slope of that curve or that line you can view it as a tangent line so we could think about what its slope is going to be and then if we go further down over here this one is , it looks like it is steeper but in the negative direction so it looks like it is steeper for sure but it 's in the negative direction . as we increase , think of it this way as we increase x one here it looks like we are decreasing y by about one . so it looks like g prime of four g prime of four , the derivative when x is equal to four is approximately , i 'm estimating it negative one while the derivative here when we increase x if we increase x by if we increase x by one it looks like we 're decreasing y by close to three so g prime of six looks like it 's closer to negative three . so which one of these is larger ? well , this one is less negative so it 's going to be greater than the other one and you could have done this intuitively if you just look at the curve this is some type of a sinusoid here you have right over here the curve is flat you have right at that moment you have no change in y with respect to x then it starts to decrease then it decreases at an even faster rate then it decreases at a faster rate then it starts , it 's still decreasing but it 's decreasing at slower and slower rates decreasing at slower rates and right at that moment you have your slope of your tangent line is zero then it starts to increase , increase , so on and so forth and it just keeps happening over and over again . so you can also think about this in a more intuitive way .
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and it 's almost estimated , but all of these are way off . having a negative two derivative would mean that as we increase our x our y is decreasing . so if our curve looks something like this we would have a slope of negative two .
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what does derivative practically and literally mean ?
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what i wan na do in this video is a few examples that test our intuition of the derivative as a rate of change or the steepness of a curve or the slope of a curve or the slope of a tangent line of a curve depending on how you actually want to think about it . so here it says f prime of five so this notation , prime this is another way of saying well what 's the derivative let 's estimate the derivative of our function at five . and when we say f prime of five this is the slope slope of tangent line tangent line at five or you could view it as the you could view it as the rate of change of y with respect to x which is really how we define slope respect to x of our function f. so let 's think about that a little bit . we see they put the point the point five comma f of five right over here and so if we want to estimate the slope of the tangent line if we want to estimate the steepness of this curve we could try to draw a line that is tangent right at that point . so let me see if i can do that . so if i were to draw a line starting there if i just wanted to make a tangent it looks like it would do something like that . right at that point that looks to be about how steep that curve is now what makes this an interesting thing in non-linear is that it 's constantly changing the steepness it 's very low here and it gets steeper and steeper and steeper as we move to the right for larger and larger x values . but if we look at the point in question when x is equal to five remember f prime of five would be if you were estimating it this would be the slope of this line here . and the slope of this line it looks like for every time we move one in the x direction we 're moving two in the y direction . delta y is equal to two when delta x is equal to one . so our change in y with respect to x at least for this tangent line here which would represent our change in y with respect to x right at that point is going to be equal to two over one , or two . and it 's almost estimated , but all of these are way off . having a negative two derivative would mean that as we increase our x our y is decreasing . so if our curve looks something like this we would have a slope of negative two . if having slopes in this a positive of point one that would be very flat something down here we might have a slope closer to point one . negative point one that might be closer on this side now we 're sloping but very close to flat . a slope of zero , that would be right over here at the bottom where right at that moment as we change x y is not increasing or decreasing the slope of the tangent line right at that bottom point would have a slope of zero . so i feel really good about that response . let 's do one more of these . so alright , so they 're telling us to compare the derivative of g at four to the derivative of g at six and which of these is greater and like always , pause the video and see if you can figure this out . well this is just an exercise let 's see if we were to if we were to make a line that indicates the slope there you can do this as a tangent line let me try to do that . so now that would n't , that does n't do a good job so right over here at that looks like a i think i can do a better job than that no that 's too shallow to see not shallow 's not the word , that 's too flat . so let me try to really okay , that looks pretty good . so that line that i just drew seems to be indicative of the rate of change of y with respect to x or the slope of that curve or that line you can view it as a tangent line so we could think about what its slope is going to be and then if we go further down over here this one is , it looks like it is steeper but in the negative direction so it looks like it is steeper for sure but it 's in the negative direction . as we increase , think of it this way as we increase x one here it looks like we are decreasing y by about one . so it looks like g prime of four g prime of four , the derivative when x is equal to four is approximately , i 'm estimating it negative one while the derivative here when we increase x if we increase x by if we increase x by one it looks like we 're decreasing y by close to three so g prime of six looks like it 's closer to negative three . so which one of these is larger ? well , this one is less negative so it 's going to be greater than the other one and you could have done this intuitively if you just look at the curve this is some type of a sinusoid here you have right over here the curve is flat you have right at that moment you have no change in y with respect to x then it starts to decrease then it decreases at an even faster rate then it decreases at a faster rate then it starts , it 's still decreasing but it 's decreasing at slower and slower rates decreasing at slower rates and right at that moment you have your slope of your tangent line is zero then it starts to increase , increase , so on and so forth and it just keeps happening over and over again . so you can also think about this in a more intuitive way .
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having a negative two derivative would mean that as we increase our x our y is decreasing . so if our curve looks something like this we would have a slope of negative two . if having slopes in this a positive of point one that would be very flat something down here we might have a slope closer to point one .
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how do you know if the slope is positive or negative ?
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what i wan na do in this video is a few examples that test our intuition of the derivative as a rate of change or the steepness of a curve or the slope of a curve or the slope of a tangent line of a curve depending on how you actually want to think about it . so here it says f prime of five so this notation , prime this is another way of saying well what 's the derivative let 's estimate the derivative of our function at five . and when we say f prime of five this is the slope slope of tangent line tangent line at five or you could view it as the you could view it as the rate of change of y with respect to x which is really how we define slope respect to x of our function f. so let 's think about that a little bit . we see they put the point the point five comma f of five right over here and so if we want to estimate the slope of the tangent line if we want to estimate the steepness of this curve we could try to draw a line that is tangent right at that point . so let me see if i can do that . so if i were to draw a line starting there if i just wanted to make a tangent it looks like it would do something like that . right at that point that looks to be about how steep that curve is now what makes this an interesting thing in non-linear is that it 's constantly changing the steepness it 's very low here and it gets steeper and steeper and steeper as we move to the right for larger and larger x values . but if we look at the point in question when x is equal to five remember f prime of five would be if you were estimating it this would be the slope of this line here . and the slope of this line it looks like for every time we move one in the x direction we 're moving two in the y direction . delta y is equal to two when delta x is equal to one . so our change in y with respect to x at least for this tangent line here which would represent our change in y with respect to x right at that point is going to be equal to two over one , or two . and it 's almost estimated , but all of these are way off . having a negative two derivative would mean that as we increase our x our y is decreasing . so if our curve looks something like this we would have a slope of negative two . if having slopes in this a positive of point one that would be very flat something down here we might have a slope closer to point one . negative point one that might be closer on this side now we 're sloping but very close to flat . a slope of zero , that would be right over here at the bottom where right at that moment as we change x y is not increasing or decreasing the slope of the tangent line right at that bottom point would have a slope of zero . so i feel really good about that response . let 's do one more of these . so alright , so they 're telling us to compare the derivative of g at four to the derivative of g at six and which of these is greater and like always , pause the video and see if you can figure this out . well this is just an exercise let 's see if we were to if we were to make a line that indicates the slope there you can do this as a tangent line let me try to do that . so now that would n't , that does n't do a good job so right over here at that looks like a i think i can do a better job than that no that 's too shallow to see not shallow 's not the word , that 's too flat . so let me try to really okay , that looks pretty good . so that line that i just drew seems to be indicative of the rate of change of y with respect to x or the slope of that curve or that line you can view it as a tangent line so we could think about what its slope is going to be and then if we go further down over here this one is , it looks like it is steeper but in the negative direction so it looks like it is steeper for sure but it 's in the negative direction . as we increase , think of it this way as we increase x one here it looks like we are decreasing y by about one . so it looks like g prime of four g prime of four , the derivative when x is equal to four is approximately , i 'm estimating it negative one while the derivative here when we increase x if we increase x by if we increase x by one it looks like we 're decreasing y by close to three so g prime of six looks like it 's closer to negative three . so which one of these is larger ? well , this one is less negative so it 's going to be greater than the other one and you could have done this intuitively if you just look at the curve this is some type of a sinusoid here you have right over here the curve is flat you have right at that moment you have no change in y with respect to x then it starts to decrease then it decreases at an even faster rate then it decreases at a faster rate then it starts , it 's still decreasing but it 's decreasing at slower and slower rates decreasing at slower rates and right at that moment you have your slope of your tangent line is zero then it starts to increase , increase , so on and so forth and it just keeps happening over and over again . so you can also think about this in a more intuitive way .
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delta y is equal to two when delta x is equal to one . so our change in y with respect to x at least for this tangent line here which would represent our change in y with respect to x right at that point is going to be equal to two over one , or two . and it 's almost estimated , but all of these are way off .
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simply , when estimating with the tangent line the x is more important to find first , because we 're trying to estimate the rate of change to where the graph intersects the next x unit ?
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what i wan na do in this video is a few examples that test our intuition of the derivative as a rate of change or the steepness of a curve or the slope of a curve or the slope of a tangent line of a curve depending on how you actually want to think about it . so here it says f prime of five so this notation , prime this is another way of saying well what 's the derivative let 's estimate the derivative of our function at five . and when we say f prime of five this is the slope slope of tangent line tangent line at five or you could view it as the you could view it as the rate of change of y with respect to x which is really how we define slope respect to x of our function f. so let 's think about that a little bit . we see they put the point the point five comma f of five right over here and so if we want to estimate the slope of the tangent line if we want to estimate the steepness of this curve we could try to draw a line that is tangent right at that point . so let me see if i can do that . so if i were to draw a line starting there if i just wanted to make a tangent it looks like it would do something like that . right at that point that looks to be about how steep that curve is now what makes this an interesting thing in non-linear is that it 's constantly changing the steepness it 's very low here and it gets steeper and steeper and steeper as we move to the right for larger and larger x values . but if we look at the point in question when x is equal to five remember f prime of five would be if you were estimating it this would be the slope of this line here . and the slope of this line it looks like for every time we move one in the x direction we 're moving two in the y direction . delta y is equal to two when delta x is equal to one . so our change in y with respect to x at least for this tangent line here which would represent our change in y with respect to x right at that point is going to be equal to two over one , or two . and it 's almost estimated , but all of these are way off . having a negative two derivative would mean that as we increase our x our y is decreasing . so if our curve looks something like this we would have a slope of negative two . if having slopes in this a positive of point one that would be very flat something down here we might have a slope closer to point one . negative point one that might be closer on this side now we 're sloping but very close to flat . a slope of zero , that would be right over here at the bottom where right at that moment as we change x y is not increasing or decreasing the slope of the tangent line right at that bottom point would have a slope of zero . so i feel really good about that response . let 's do one more of these . so alright , so they 're telling us to compare the derivative of g at four to the derivative of g at six and which of these is greater and like always , pause the video and see if you can figure this out . well this is just an exercise let 's see if we were to if we were to make a line that indicates the slope there you can do this as a tangent line let me try to do that . so now that would n't , that does n't do a good job so right over here at that looks like a i think i can do a better job than that no that 's too shallow to see not shallow 's not the word , that 's too flat . so let me try to really okay , that looks pretty good . so that line that i just drew seems to be indicative of the rate of change of y with respect to x or the slope of that curve or that line you can view it as a tangent line so we could think about what its slope is going to be and then if we go further down over here this one is , it looks like it is steeper but in the negative direction so it looks like it is steeper for sure but it 's in the negative direction . as we increase , think of it this way as we increase x one here it looks like we are decreasing y by about one . so it looks like g prime of four g prime of four , the derivative when x is equal to four is approximately , i 'm estimating it negative one while the derivative here when we increase x if we increase x by if we increase x by one it looks like we 're decreasing y by close to three so g prime of six looks like it 's closer to negative three . so which one of these is larger ? well , this one is less negative so it 's going to be greater than the other one and you could have done this intuitively if you just look at the curve this is some type of a sinusoid here you have right over here the curve is flat you have right at that moment you have no change in y with respect to x then it starts to decrease then it decreases at an even faster rate then it decreases at a faster rate then it starts , it 's still decreasing but it 's decreasing at slower and slower rates decreasing at slower rates and right at that moment you have your slope of your tangent line is zero then it starts to increase , increase , so on and so forth and it just keeps happening over and over again . so you can also think about this in a more intuitive way .
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negative point one that might be closer on this side now we 're sloping but very close to flat . a slope of zero , that would be right over here at the bottom where right at that moment as we change x y is not increasing or decreasing the slope of the tangent line right at that bottom point would have a slope of zero . so i feel really good about that response .
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is the slope of the tangent line the derivative ( there 's only one tangent right ) ?
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: here 's a general structure for an acyl chloride , also called an acid chloride , and it 's a carboxylic acid derivative , so we can form them from carboxylic acid , so if we start with a carboxylic acid and add thionyl chloride , we can form our acyl chlorides , and we would also form a sulfur dioxide and hcl in this reaction . let 's look at the structure of thionyl chloride . here we have the dot structure right here and we could draw a resonance structure for this , so we could show these electrons in here moving off onto the oxygen , so let 's go ahead and draw what we would form from that . our oxygen would now have three loan pairs of electrons on it , giving it a negative formal charge . our sulfur would still be barred to these chlorines here . it would solve a loan pair of electrons and it would get a plus one formal charge like that . this is a major contributor to the overall structure . oxygen is more electronegative than sulfur , and if you think about this pie bond in here , there 's ineffective overlap of those p orbitals , and that 's because sulfur and oxygen are in different periods on the periodic table , so sulfur is in the third period , so it has a larger p orbital than oxygen . oxygen 's in the second period . you get ineffective overlap of these orbitals here , and so that 's another reason why this is going to contribute to the overall structure . plus , you have these chlorines here , withdrawing some electron density from the sulfur , so chlorine is more electronegative than the sulfur , so that the end result of all this is going to make this sulfur very electrophilic right here , and so therefore , our carboxylic acid is able to act as a nucleophile , and so if these electrons move into here , these electrons can attack our sulfur so the nucleophile attacks our electrophile , and then these electrons kick off onto the oxygen , so let 's go ahead and show that . we would have our r group bonded to our carbon , and then that 's bonded to an oxygen . the oxygen has two loan pairs of electrons on it . the oxygen formed a bond with the sulfur , and now the sulfur 's bonded to this oxygen , which gets a negative one formal charge . this sulfur is bonded to two chlorines , so we draw in our chlorines here with all the loan pairs of electrons , and there 's still a loan pair of electrons on our sulfur . we now have a double bond between the carbon and this oxygen , and one loan pair of electrons on this oxygen gives it a plus one formal charge . following some of our electrons , these electrons in magenta move in here , they 'd form our double bond , and then we can think about these electrons in blue , forming the bonds between the oxygen and the sulfur . finally , we could think about these electrons in here in green moving off onto our oxygen like that . so for the next step , we could think about these electrons moving in here to form our double bond between oxygen and sulfur . that would kick off the chloride anion as a leaving group , and we know the chloride anion is an excellent leaving group because it 's stable on it 's own . when we draw the results of that , we would have our group . we would have our carbon , we would have that carbon double-bonded to our oxygen here with a loan pair of electrons plus one formal charge , and then we would have our oxygen with two loan pairs of electrons , and our sulfur is now double-bonded to this oxygen . all right , there 's still a loan pair of electrons on that sulfur . only one chlorine bonded to this sulfur now , so we lost the chloride anion . let 's go ahead and show some of those electrons . let 's say these electrons in red here move in to reform the dull bond between oxygen and sulfur , and then we had some electrons kick off onto the chlorine , so we have the chloride anion that forms . let 's go ahead and show that as well . these electrons in here come off onto chlorine , then we have the chloride anions . let 's draw in the chloride anion . let 's get some more space down here . we also have the chloride anion that forms , so we draw that in here like that . a negative , more informal charge . at this stage , we need to consider whether the chloride anion is going to function as a base or a nucleophile , and it could do both , so let 's first think about the chloride anion functioning as a base . if it functions as a base , it could take this proton , leaving these electrons behind on the oxygen , so let 's go ahead and draw the lewd form . lewd form are carbonile with two loan pairs of electrons on the oxygen , and then we would have this oxygen here with two loan pairs , and then this sulfur , right when it would still would be double-bonded , one loan pair , and then the chlorine , like that . we 're saying that in this acid-based reaction , these electrons in here move back onto here to form the carbonile . you can consider this to be the intermediate for this mechanism . you 'll see some versions of this mechanism take this intermediate and continue on to form your product . i 'm going to show the chloride anion functioning as a nucleophile over here for this . it 's an acid-based reaction , so it 's possible to , once again , protonate your carbonile . that 's going to activate it . this carbon right here becomes more electrophilic , and that means that the chloride anion can function as a nucleophile and attack our electrophile since the chloride anion is n't a great nucleophile on it 's own , so if the chloride anion attacks here , that would push these electrons off onto the oxygen , and we could go ahead and draw what we would form , so we would have our r group , we would have our carbon , we would have this oxygen with two loan pairs of electrons bonded to hydrogen here , so let 's show those electrons in blue . if these electrons in blue kick off onto the oxygen , we could say that those are these electrons . then we could also say that these electrons here in green , on the chloride anion are going to form a bond between the chlorine and the carbon , so we can go ahead and draw in the bond between the chlorine and the carbon like that . we still have our oxygen right here bonded to our sulfur , double-bonded to this oxygen , and then we have our loan pair of electrons here . then we have our chlorine . what all this does is make a much better leaving group than what we started off with so you could think about this as our leaving group . this is a leaving group , and then the next step of the mechanism , and if we go back up here , that 's a much better leaving group than the oh that we started out with . if that 's going to be our leaving group , we could show these electrons in here moving into form our double bond which would kick these electrons off onto our oxygen . let 's go ahead and show what we would make from that . we would now form our r group , our carbon would be double-bonded to an oxygen like this , loan pair of electrons plus one formal charge , and then we 'd have our chlorine over here with all of its electrons too . let 's go ahead and show where those electrons came from so if these electrons move in here , that would reform our carbonile , so i can go ahead and draw that in . then we still have this bond in green , here , just to clarify . then we would have some electrons , and now let 's make a magenta , so these electrons in here are going to kick off onto the oxygen , so we could go ahead and draw that . we would have our oxygen . so we 'd have three loan pairs of electrons . one of them would be the magenta ones , so we could go ahead and do that . that gives our oxygen a negative one formal charge because it 's bonded to the sulfur here , the sulfur 's still double-bonded to this oxygen , and once again , still a loan pair of electrons and a chlorine over here like that . the reason why this is a good leaving group is because we can push these electrons into here , push these electrons off onto here , and we can form sulfur dioxide as a gas . if we go ahead and draw the result of that , we would now have this oxygen double-bonded to this sulfur , double-bonded to this oxygen , so draw in loan pairs of electrons here , and we still have a loan pair of electrons on our sulfur . if these electrons in red here move in to form this bond , we 've now formed sulfur dioxide , and we also formed a chloride anion , so it once again follows some electrons . if these electrons move off onto chlorine , we have the chloride anion , so now we have our chloride anion which could function as a base and in the last step of our mechanism , deprotonate , so take this proton , leave these electrons behind and finally , we formed our acyl chloride . we have our carbon double-bonded to an oxygen with two loan pairs of electrons , and we have our chlorine over here , like that . if we follow those final electrons , these electrons in blue , move off onto your carbonile . you would also form hcl by this , so you form hcl , which is also a gas , so the formation of these gases , the formation of sulfur dioxide and hcl drives the reaction to completion here . that was a long mechanism . once again , you 'll see some slight variations on this , so if i go back up to here , you could show this step , the nucleophile attacking the electrophile , and directly form here if you wanted to , so as your nucleophile attacks your sulfur , these electrons could kick off onto chlorine to form the chloride anion at this step , and at this step , and then you would get this at this point . once again , i talked about the possibility of using this as your intermediate . depending on which textbook you look in , you might see some slight variations on this mechanism . it is a little bit of a long one here . all right , let 's look at two other ways to make an acyl chloride , so starting with a carboxylic acid , you could add phosphorus , a pentachloride , or a phosphorus trichloride , and both of those will give you an acyl chloride as well . the mechanism is pretty similar and also we could think about , instead of phosphorus trichloride , we could add a pbr3 , so a phosphorus tribromide instead of putting a chlorine here . that would put a bromine , and so we could form an acyl bromide this way , and we 'll see a use for this reaction in a later video .
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if we follow those final electrons , these electrons in blue , move off onto your carbonile . you would also form hcl by this , so you form hcl , which is also a gas , so the formation of these gases , the formation of sulfur dioxide and hcl drives the reaction to completion here . that was a long mechanism .
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why is hcl a gas ?
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: here 's a general structure for an acyl chloride , also called an acid chloride , and it 's a carboxylic acid derivative , so we can form them from carboxylic acid , so if we start with a carboxylic acid and add thionyl chloride , we can form our acyl chlorides , and we would also form a sulfur dioxide and hcl in this reaction . let 's look at the structure of thionyl chloride . here we have the dot structure right here and we could draw a resonance structure for this , so we could show these electrons in here moving off onto the oxygen , so let 's go ahead and draw what we would form from that . our oxygen would now have three loan pairs of electrons on it , giving it a negative formal charge . our sulfur would still be barred to these chlorines here . it would solve a loan pair of electrons and it would get a plus one formal charge like that . this is a major contributor to the overall structure . oxygen is more electronegative than sulfur , and if you think about this pie bond in here , there 's ineffective overlap of those p orbitals , and that 's because sulfur and oxygen are in different periods on the periodic table , so sulfur is in the third period , so it has a larger p orbital than oxygen . oxygen 's in the second period . you get ineffective overlap of these orbitals here , and so that 's another reason why this is going to contribute to the overall structure . plus , you have these chlorines here , withdrawing some electron density from the sulfur , so chlorine is more electronegative than the sulfur , so that the end result of all this is going to make this sulfur very electrophilic right here , and so therefore , our carboxylic acid is able to act as a nucleophile , and so if these electrons move into here , these electrons can attack our sulfur so the nucleophile attacks our electrophile , and then these electrons kick off onto the oxygen , so let 's go ahead and show that . we would have our r group bonded to our carbon , and then that 's bonded to an oxygen . the oxygen has two loan pairs of electrons on it . the oxygen formed a bond with the sulfur , and now the sulfur 's bonded to this oxygen , which gets a negative one formal charge . this sulfur is bonded to two chlorines , so we draw in our chlorines here with all the loan pairs of electrons , and there 's still a loan pair of electrons on our sulfur . we now have a double bond between the carbon and this oxygen , and one loan pair of electrons on this oxygen gives it a plus one formal charge . following some of our electrons , these electrons in magenta move in here , they 'd form our double bond , and then we can think about these electrons in blue , forming the bonds between the oxygen and the sulfur . finally , we could think about these electrons in here in green moving off onto our oxygen like that . so for the next step , we could think about these electrons moving in here to form our double bond between oxygen and sulfur . that would kick off the chloride anion as a leaving group , and we know the chloride anion is an excellent leaving group because it 's stable on it 's own . when we draw the results of that , we would have our group . we would have our carbon , we would have that carbon double-bonded to our oxygen here with a loan pair of electrons plus one formal charge , and then we would have our oxygen with two loan pairs of electrons , and our sulfur is now double-bonded to this oxygen . all right , there 's still a loan pair of electrons on that sulfur . only one chlorine bonded to this sulfur now , so we lost the chloride anion . let 's go ahead and show some of those electrons . let 's say these electrons in red here move in to reform the dull bond between oxygen and sulfur , and then we had some electrons kick off onto the chlorine , so we have the chloride anion that forms . let 's go ahead and show that as well . these electrons in here come off onto chlorine , then we have the chloride anions . let 's draw in the chloride anion . let 's get some more space down here . we also have the chloride anion that forms , so we draw that in here like that . a negative , more informal charge . at this stage , we need to consider whether the chloride anion is going to function as a base or a nucleophile , and it could do both , so let 's first think about the chloride anion functioning as a base . if it functions as a base , it could take this proton , leaving these electrons behind on the oxygen , so let 's go ahead and draw the lewd form . lewd form are carbonile with two loan pairs of electrons on the oxygen , and then we would have this oxygen here with two loan pairs , and then this sulfur , right when it would still would be double-bonded , one loan pair , and then the chlorine , like that . we 're saying that in this acid-based reaction , these electrons in here move back onto here to form the carbonile . you can consider this to be the intermediate for this mechanism . you 'll see some versions of this mechanism take this intermediate and continue on to form your product . i 'm going to show the chloride anion functioning as a nucleophile over here for this . it 's an acid-based reaction , so it 's possible to , once again , protonate your carbonile . that 's going to activate it . this carbon right here becomes more electrophilic , and that means that the chloride anion can function as a nucleophile and attack our electrophile since the chloride anion is n't a great nucleophile on it 's own , so if the chloride anion attacks here , that would push these electrons off onto the oxygen , and we could go ahead and draw what we would form , so we would have our r group , we would have our carbon , we would have this oxygen with two loan pairs of electrons bonded to hydrogen here , so let 's show those electrons in blue . if these electrons in blue kick off onto the oxygen , we could say that those are these electrons . then we could also say that these electrons here in green , on the chloride anion are going to form a bond between the chlorine and the carbon , so we can go ahead and draw in the bond between the chlorine and the carbon like that . we still have our oxygen right here bonded to our sulfur , double-bonded to this oxygen , and then we have our loan pair of electrons here . then we have our chlorine . what all this does is make a much better leaving group than what we started off with so you could think about this as our leaving group . this is a leaving group , and then the next step of the mechanism , and if we go back up here , that 's a much better leaving group than the oh that we started out with . if that 's going to be our leaving group , we could show these electrons in here moving into form our double bond which would kick these electrons off onto our oxygen . let 's go ahead and show what we would make from that . we would now form our r group , our carbon would be double-bonded to an oxygen like this , loan pair of electrons plus one formal charge , and then we 'd have our chlorine over here with all of its electrons too . let 's go ahead and show where those electrons came from so if these electrons move in here , that would reform our carbonile , so i can go ahead and draw that in . then we still have this bond in green , here , just to clarify . then we would have some electrons , and now let 's make a magenta , so these electrons in here are going to kick off onto the oxygen , so we could go ahead and draw that . we would have our oxygen . so we 'd have three loan pairs of electrons . one of them would be the magenta ones , so we could go ahead and do that . that gives our oxygen a negative one formal charge because it 's bonded to the sulfur here , the sulfur 's still double-bonded to this oxygen , and once again , still a loan pair of electrons and a chlorine over here like that . the reason why this is a good leaving group is because we can push these electrons into here , push these electrons off onto here , and we can form sulfur dioxide as a gas . if we go ahead and draw the result of that , we would now have this oxygen double-bonded to this sulfur , double-bonded to this oxygen , so draw in loan pairs of electrons here , and we still have a loan pair of electrons on our sulfur . if these electrons in red here move in to form this bond , we 've now formed sulfur dioxide , and we also formed a chloride anion , so it once again follows some electrons . if these electrons move off onto chlorine , we have the chloride anion , so now we have our chloride anion which could function as a base and in the last step of our mechanism , deprotonate , so take this proton , leave these electrons behind and finally , we formed our acyl chloride . we have our carbon double-bonded to an oxygen with two loan pairs of electrons , and we have our chlorine over here , like that . if we follow those final electrons , these electrons in blue , move off onto your carbonile . you would also form hcl by this , so you form hcl , which is also a gas , so the formation of these gases , the formation of sulfur dioxide and hcl drives the reaction to completion here . that was a long mechanism . once again , you 'll see some slight variations on this , so if i go back up to here , you could show this step , the nucleophile attacking the electrophile , and directly form here if you wanted to , so as your nucleophile attacks your sulfur , these electrons could kick off onto chlorine to form the chloride anion at this step , and at this step , and then you would get this at this point . once again , i talked about the possibility of using this as your intermediate . depending on which textbook you look in , you might see some slight variations on this mechanism . it is a little bit of a long one here . all right , let 's look at two other ways to make an acyl chloride , so starting with a carboxylic acid , you could add phosphorus , a pentachloride , or a phosphorus trichloride , and both of those will give you an acyl chloride as well . the mechanism is pretty similar and also we could think about , instead of phosphorus trichloride , we could add a pbr3 , so a phosphorus tribromide instead of putting a chlorine here . that would put a bromine , and so we could form an acyl bromide this way , and we 'll see a use for this reaction in a later video .
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i 'm going to show the chloride anion functioning as a nucleophile over here for this . it 's an acid-based reaction , so it 's possible to , once again , protonate your carbonile . that 's going to activate it .
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if there is a nitrogen atom directly attached with the carbonyl carbon of the acid , whether this reaction will occour in that same mechanism ?
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: here 's a general structure for an acyl chloride , also called an acid chloride , and it 's a carboxylic acid derivative , so we can form them from carboxylic acid , so if we start with a carboxylic acid and add thionyl chloride , we can form our acyl chlorides , and we would also form a sulfur dioxide and hcl in this reaction . let 's look at the structure of thionyl chloride . here we have the dot structure right here and we could draw a resonance structure for this , so we could show these electrons in here moving off onto the oxygen , so let 's go ahead and draw what we would form from that . our oxygen would now have three loan pairs of electrons on it , giving it a negative formal charge . our sulfur would still be barred to these chlorines here . it would solve a loan pair of electrons and it would get a plus one formal charge like that . this is a major contributor to the overall structure . oxygen is more electronegative than sulfur , and if you think about this pie bond in here , there 's ineffective overlap of those p orbitals , and that 's because sulfur and oxygen are in different periods on the periodic table , so sulfur is in the third period , so it has a larger p orbital than oxygen . oxygen 's in the second period . you get ineffective overlap of these orbitals here , and so that 's another reason why this is going to contribute to the overall structure . plus , you have these chlorines here , withdrawing some electron density from the sulfur , so chlorine is more electronegative than the sulfur , so that the end result of all this is going to make this sulfur very electrophilic right here , and so therefore , our carboxylic acid is able to act as a nucleophile , and so if these electrons move into here , these electrons can attack our sulfur so the nucleophile attacks our electrophile , and then these electrons kick off onto the oxygen , so let 's go ahead and show that . we would have our r group bonded to our carbon , and then that 's bonded to an oxygen . the oxygen has two loan pairs of electrons on it . the oxygen formed a bond with the sulfur , and now the sulfur 's bonded to this oxygen , which gets a negative one formal charge . this sulfur is bonded to two chlorines , so we draw in our chlorines here with all the loan pairs of electrons , and there 's still a loan pair of electrons on our sulfur . we now have a double bond between the carbon and this oxygen , and one loan pair of electrons on this oxygen gives it a plus one formal charge . following some of our electrons , these electrons in magenta move in here , they 'd form our double bond , and then we can think about these electrons in blue , forming the bonds between the oxygen and the sulfur . finally , we could think about these electrons in here in green moving off onto our oxygen like that . so for the next step , we could think about these electrons moving in here to form our double bond between oxygen and sulfur . that would kick off the chloride anion as a leaving group , and we know the chloride anion is an excellent leaving group because it 's stable on it 's own . when we draw the results of that , we would have our group . we would have our carbon , we would have that carbon double-bonded to our oxygen here with a loan pair of electrons plus one formal charge , and then we would have our oxygen with two loan pairs of electrons , and our sulfur is now double-bonded to this oxygen . all right , there 's still a loan pair of electrons on that sulfur . only one chlorine bonded to this sulfur now , so we lost the chloride anion . let 's go ahead and show some of those electrons . let 's say these electrons in red here move in to reform the dull bond between oxygen and sulfur , and then we had some electrons kick off onto the chlorine , so we have the chloride anion that forms . let 's go ahead and show that as well . these electrons in here come off onto chlorine , then we have the chloride anions . let 's draw in the chloride anion . let 's get some more space down here . we also have the chloride anion that forms , so we draw that in here like that . a negative , more informal charge . at this stage , we need to consider whether the chloride anion is going to function as a base or a nucleophile , and it could do both , so let 's first think about the chloride anion functioning as a base . if it functions as a base , it could take this proton , leaving these electrons behind on the oxygen , so let 's go ahead and draw the lewd form . lewd form are carbonile with two loan pairs of electrons on the oxygen , and then we would have this oxygen here with two loan pairs , and then this sulfur , right when it would still would be double-bonded , one loan pair , and then the chlorine , like that . we 're saying that in this acid-based reaction , these electrons in here move back onto here to form the carbonile . you can consider this to be the intermediate for this mechanism . you 'll see some versions of this mechanism take this intermediate and continue on to form your product . i 'm going to show the chloride anion functioning as a nucleophile over here for this . it 's an acid-based reaction , so it 's possible to , once again , protonate your carbonile . that 's going to activate it . this carbon right here becomes more electrophilic , and that means that the chloride anion can function as a nucleophile and attack our electrophile since the chloride anion is n't a great nucleophile on it 's own , so if the chloride anion attacks here , that would push these electrons off onto the oxygen , and we could go ahead and draw what we would form , so we would have our r group , we would have our carbon , we would have this oxygen with two loan pairs of electrons bonded to hydrogen here , so let 's show those electrons in blue . if these electrons in blue kick off onto the oxygen , we could say that those are these electrons . then we could also say that these electrons here in green , on the chloride anion are going to form a bond between the chlorine and the carbon , so we can go ahead and draw in the bond between the chlorine and the carbon like that . we still have our oxygen right here bonded to our sulfur , double-bonded to this oxygen , and then we have our loan pair of electrons here . then we have our chlorine . what all this does is make a much better leaving group than what we started off with so you could think about this as our leaving group . this is a leaving group , and then the next step of the mechanism , and if we go back up here , that 's a much better leaving group than the oh that we started out with . if that 's going to be our leaving group , we could show these electrons in here moving into form our double bond which would kick these electrons off onto our oxygen . let 's go ahead and show what we would make from that . we would now form our r group , our carbon would be double-bonded to an oxygen like this , loan pair of electrons plus one formal charge , and then we 'd have our chlorine over here with all of its electrons too . let 's go ahead and show where those electrons came from so if these electrons move in here , that would reform our carbonile , so i can go ahead and draw that in . then we still have this bond in green , here , just to clarify . then we would have some electrons , and now let 's make a magenta , so these electrons in here are going to kick off onto the oxygen , so we could go ahead and draw that . we would have our oxygen . so we 'd have three loan pairs of electrons . one of them would be the magenta ones , so we could go ahead and do that . that gives our oxygen a negative one formal charge because it 's bonded to the sulfur here , the sulfur 's still double-bonded to this oxygen , and once again , still a loan pair of electrons and a chlorine over here like that . the reason why this is a good leaving group is because we can push these electrons into here , push these electrons off onto here , and we can form sulfur dioxide as a gas . if we go ahead and draw the result of that , we would now have this oxygen double-bonded to this sulfur , double-bonded to this oxygen , so draw in loan pairs of electrons here , and we still have a loan pair of electrons on our sulfur . if these electrons in red here move in to form this bond , we 've now formed sulfur dioxide , and we also formed a chloride anion , so it once again follows some electrons . if these electrons move off onto chlorine , we have the chloride anion , so now we have our chloride anion which could function as a base and in the last step of our mechanism , deprotonate , so take this proton , leave these electrons behind and finally , we formed our acyl chloride . we have our carbon double-bonded to an oxygen with two loan pairs of electrons , and we have our chlorine over here , like that . if we follow those final electrons , these electrons in blue , move off onto your carbonile . you would also form hcl by this , so you form hcl , which is also a gas , so the formation of these gases , the formation of sulfur dioxide and hcl drives the reaction to completion here . that was a long mechanism . once again , you 'll see some slight variations on this , so if i go back up to here , you could show this step , the nucleophile attacking the electrophile , and directly form here if you wanted to , so as your nucleophile attacks your sulfur , these electrons could kick off onto chlorine to form the chloride anion at this step , and at this step , and then you would get this at this point . once again , i talked about the possibility of using this as your intermediate . depending on which textbook you look in , you might see some slight variations on this mechanism . it is a little bit of a long one here . all right , let 's look at two other ways to make an acyl chloride , so starting with a carboxylic acid , you could add phosphorus , a pentachloride , or a phosphorus trichloride , and both of those will give you an acyl chloride as well . the mechanism is pretty similar and also we could think about , instead of phosphorus trichloride , we could add a pbr3 , so a phosphorus tribromide instead of putting a chlorine here . that would put a bromine , and so we could form an acyl bromide this way , and we 'll see a use for this reaction in a later video .
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: here 's a general structure for an acyl chloride , also called an acid chloride , and it 's a carboxylic acid derivative , so we can form them from carboxylic acid , so if we start with a carboxylic acid and add thionyl chloride , we can form our acyl chlorides , and we would also form a sulfur dioxide and hcl in this reaction . let 's look at the structure of thionyl chloride .
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why are acyl chlorides called acid chlorides ?
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: here 's a general structure for an acyl chloride , also called an acid chloride , and it 's a carboxylic acid derivative , so we can form them from carboxylic acid , so if we start with a carboxylic acid and add thionyl chloride , we can form our acyl chlorides , and we would also form a sulfur dioxide and hcl in this reaction . let 's look at the structure of thionyl chloride . here we have the dot structure right here and we could draw a resonance structure for this , so we could show these electrons in here moving off onto the oxygen , so let 's go ahead and draw what we would form from that . our oxygen would now have three loan pairs of electrons on it , giving it a negative formal charge . our sulfur would still be barred to these chlorines here . it would solve a loan pair of electrons and it would get a plus one formal charge like that . this is a major contributor to the overall structure . oxygen is more electronegative than sulfur , and if you think about this pie bond in here , there 's ineffective overlap of those p orbitals , and that 's because sulfur and oxygen are in different periods on the periodic table , so sulfur is in the third period , so it has a larger p orbital than oxygen . oxygen 's in the second period . you get ineffective overlap of these orbitals here , and so that 's another reason why this is going to contribute to the overall structure . plus , you have these chlorines here , withdrawing some electron density from the sulfur , so chlorine is more electronegative than the sulfur , so that the end result of all this is going to make this sulfur very electrophilic right here , and so therefore , our carboxylic acid is able to act as a nucleophile , and so if these electrons move into here , these electrons can attack our sulfur so the nucleophile attacks our electrophile , and then these electrons kick off onto the oxygen , so let 's go ahead and show that . we would have our r group bonded to our carbon , and then that 's bonded to an oxygen . the oxygen has two loan pairs of electrons on it . the oxygen formed a bond with the sulfur , and now the sulfur 's bonded to this oxygen , which gets a negative one formal charge . this sulfur is bonded to two chlorines , so we draw in our chlorines here with all the loan pairs of electrons , and there 's still a loan pair of electrons on our sulfur . we now have a double bond between the carbon and this oxygen , and one loan pair of electrons on this oxygen gives it a plus one formal charge . following some of our electrons , these electrons in magenta move in here , they 'd form our double bond , and then we can think about these electrons in blue , forming the bonds between the oxygen and the sulfur . finally , we could think about these electrons in here in green moving off onto our oxygen like that . so for the next step , we could think about these electrons moving in here to form our double bond between oxygen and sulfur . that would kick off the chloride anion as a leaving group , and we know the chloride anion is an excellent leaving group because it 's stable on it 's own . when we draw the results of that , we would have our group . we would have our carbon , we would have that carbon double-bonded to our oxygen here with a loan pair of electrons plus one formal charge , and then we would have our oxygen with two loan pairs of electrons , and our sulfur is now double-bonded to this oxygen . all right , there 's still a loan pair of electrons on that sulfur . only one chlorine bonded to this sulfur now , so we lost the chloride anion . let 's go ahead and show some of those electrons . let 's say these electrons in red here move in to reform the dull bond between oxygen and sulfur , and then we had some electrons kick off onto the chlorine , so we have the chloride anion that forms . let 's go ahead and show that as well . these electrons in here come off onto chlorine , then we have the chloride anions . let 's draw in the chloride anion . let 's get some more space down here . we also have the chloride anion that forms , so we draw that in here like that . a negative , more informal charge . at this stage , we need to consider whether the chloride anion is going to function as a base or a nucleophile , and it could do both , so let 's first think about the chloride anion functioning as a base . if it functions as a base , it could take this proton , leaving these electrons behind on the oxygen , so let 's go ahead and draw the lewd form . lewd form are carbonile with two loan pairs of electrons on the oxygen , and then we would have this oxygen here with two loan pairs , and then this sulfur , right when it would still would be double-bonded , one loan pair , and then the chlorine , like that . we 're saying that in this acid-based reaction , these electrons in here move back onto here to form the carbonile . you can consider this to be the intermediate for this mechanism . you 'll see some versions of this mechanism take this intermediate and continue on to form your product . i 'm going to show the chloride anion functioning as a nucleophile over here for this . it 's an acid-based reaction , so it 's possible to , once again , protonate your carbonile . that 's going to activate it . this carbon right here becomes more electrophilic , and that means that the chloride anion can function as a nucleophile and attack our electrophile since the chloride anion is n't a great nucleophile on it 's own , so if the chloride anion attacks here , that would push these electrons off onto the oxygen , and we could go ahead and draw what we would form , so we would have our r group , we would have our carbon , we would have this oxygen with two loan pairs of electrons bonded to hydrogen here , so let 's show those electrons in blue . if these electrons in blue kick off onto the oxygen , we could say that those are these electrons . then we could also say that these electrons here in green , on the chloride anion are going to form a bond between the chlorine and the carbon , so we can go ahead and draw in the bond between the chlorine and the carbon like that . we still have our oxygen right here bonded to our sulfur , double-bonded to this oxygen , and then we have our loan pair of electrons here . then we have our chlorine . what all this does is make a much better leaving group than what we started off with so you could think about this as our leaving group . this is a leaving group , and then the next step of the mechanism , and if we go back up here , that 's a much better leaving group than the oh that we started out with . if that 's going to be our leaving group , we could show these electrons in here moving into form our double bond which would kick these electrons off onto our oxygen . let 's go ahead and show what we would make from that . we would now form our r group , our carbon would be double-bonded to an oxygen like this , loan pair of electrons plus one formal charge , and then we 'd have our chlorine over here with all of its electrons too . let 's go ahead and show where those electrons came from so if these electrons move in here , that would reform our carbonile , so i can go ahead and draw that in . then we still have this bond in green , here , just to clarify . then we would have some electrons , and now let 's make a magenta , so these electrons in here are going to kick off onto the oxygen , so we could go ahead and draw that . we would have our oxygen . so we 'd have three loan pairs of electrons . one of them would be the magenta ones , so we could go ahead and do that . that gives our oxygen a negative one formal charge because it 's bonded to the sulfur here , the sulfur 's still double-bonded to this oxygen , and once again , still a loan pair of electrons and a chlorine over here like that . the reason why this is a good leaving group is because we can push these electrons into here , push these electrons off onto here , and we can form sulfur dioxide as a gas . if we go ahead and draw the result of that , we would now have this oxygen double-bonded to this sulfur , double-bonded to this oxygen , so draw in loan pairs of electrons here , and we still have a loan pair of electrons on our sulfur . if these electrons in red here move in to form this bond , we 've now formed sulfur dioxide , and we also formed a chloride anion , so it once again follows some electrons . if these electrons move off onto chlorine , we have the chloride anion , so now we have our chloride anion which could function as a base and in the last step of our mechanism , deprotonate , so take this proton , leave these electrons behind and finally , we formed our acyl chloride . we have our carbon double-bonded to an oxygen with two loan pairs of electrons , and we have our chlorine over here , like that . if we follow those final electrons , these electrons in blue , move off onto your carbonile . you would also form hcl by this , so you form hcl , which is also a gas , so the formation of these gases , the formation of sulfur dioxide and hcl drives the reaction to completion here . that was a long mechanism . once again , you 'll see some slight variations on this , so if i go back up to here , you could show this step , the nucleophile attacking the electrophile , and directly form here if you wanted to , so as your nucleophile attacks your sulfur , these electrons could kick off onto chlorine to form the chloride anion at this step , and at this step , and then you would get this at this point . once again , i talked about the possibility of using this as your intermediate . depending on which textbook you look in , you might see some slight variations on this mechanism . it is a little bit of a long one here . all right , let 's look at two other ways to make an acyl chloride , so starting with a carboxylic acid , you could add phosphorus , a pentachloride , or a phosphorus trichloride , and both of those will give you an acyl chloride as well . the mechanism is pretty similar and also we could think about , instead of phosphorus trichloride , we could add a pbr3 , so a phosphorus tribromide instead of putting a chlorine here . that would put a bromine , and so we could form an acyl bromide this way , and we 'll see a use for this reaction in a later video .
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what all this does is make a much better leaving group than what we started off with so you could think about this as our leaving group . this is a leaving group , and then the next step of the mechanism , and if we go back up here , that 's a much better leaving group than the oh that we started out with . if that 's going to be our leaving group , we could show these electrons in here moving into form our double bond which would kick these electrons off onto our oxygen . let 's go ahead and show what we would make from that .
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if the molecule is stable at that point , with the alcohol group as well as the oxygen-sulfur group , then why would the electrons from the -oh move back in to kick off the -os group and form a positively-charged oxygen ?
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: here 's a general structure for an acyl chloride , also called an acid chloride , and it 's a carboxylic acid derivative , so we can form them from carboxylic acid , so if we start with a carboxylic acid and add thionyl chloride , we can form our acyl chlorides , and we would also form a sulfur dioxide and hcl in this reaction . let 's look at the structure of thionyl chloride . here we have the dot structure right here and we could draw a resonance structure for this , so we could show these electrons in here moving off onto the oxygen , so let 's go ahead and draw what we would form from that . our oxygen would now have three loan pairs of electrons on it , giving it a negative formal charge . our sulfur would still be barred to these chlorines here . it would solve a loan pair of electrons and it would get a plus one formal charge like that . this is a major contributor to the overall structure . oxygen is more electronegative than sulfur , and if you think about this pie bond in here , there 's ineffective overlap of those p orbitals , and that 's because sulfur and oxygen are in different periods on the periodic table , so sulfur is in the third period , so it has a larger p orbital than oxygen . oxygen 's in the second period . you get ineffective overlap of these orbitals here , and so that 's another reason why this is going to contribute to the overall structure . plus , you have these chlorines here , withdrawing some electron density from the sulfur , so chlorine is more electronegative than the sulfur , so that the end result of all this is going to make this sulfur very electrophilic right here , and so therefore , our carboxylic acid is able to act as a nucleophile , and so if these electrons move into here , these electrons can attack our sulfur so the nucleophile attacks our electrophile , and then these electrons kick off onto the oxygen , so let 's go ahead and show that . we would have our r group bonded to our carbon , and then that 's bonded to an oxygen . the oxygen has two loan pairs of electrons on it . the oxygen formed a bond with the sulfur , and now the sulfur 's bonded to this oxygen , which gets a negative one formal charge . this sulfur is bonded to two chlorines , so we draw in our chlorines here with all the loan pairs of electrons , and there 's still a loan pair of electrons on our sulfur . we now have a double bond between the carbon and this oxygen , and one loan pair of electrons on this oxygen gives it a plus one formal charge . following some of our electrons , these electrons in magenta move in here , they 'd form our double bond , and then we can think about these electrons in blue , forming the bonds between the oxygen and the sulfur . finally , we could think about these electrons in here in green moving off onto our oxygen like that . so for the next step , we could think about these electrons moving in here to form our double bond between oxygen and sulfur . that would kick off the chloride anion as a leaving group , and we know the chloride anion is an excellent leaving group because it 's stable on it 's own . when we draw the results of that , we would have our group . we would have our carbon , we would have that carbon double-bonded to our oxygen here with a loan pair of electrons plus one formal charge , and then we would have our oxygen with two loan pairs of electrons , and our sulfur is now double-bonded to this oxygen . all right , there 's still a loan pair of electrons on that sulfur . only one chlorine bonded to this sulfur now , so we lost the chloride anion . let 's go ahead and show some of those electrons . let 's say these electrons in red here move in to reform the dull bond between oxygen and sulfur , and then we had some electrons kick off onto the chlorine , so we have the chloride anion that forms . let 's go ahead and show that as well . these electrons in here come off onto chlorine , then we have the chloride anions . let 's draw in the chloride anion . let 's get some more space down here . we also have the chloride anion that forms , so we draw that in here like that . a negative , more informal charge . at this stage , we need to consider whether the chloride anion is going to function as a base or a nucleophile , and it could do both , so let 's first think about the chloride anion functioning as a base . if it functions as a base , it could take this proton , leaving these electrons behind on the oxygen , so let 's go ahead and draw the lewd form . lewd form are carbonile with two loan pairs of electrons on the oxygen , and then we would have this oxygen here with two loan pairs , and then this sulfur , right when it would still would be double-bonded , one loan pair , and then the chlorine , like that . we 're saying that in this acid-based reaction , these electrons in here move back onto here to form the carbonile . you can consider this to be the intermediate for this mechanism . you 'll see some versions of this mechanism take this intermediate and continue on to form your product . i 'm going to show the chloride anion functioning as a nucleophile over here for this . it 's an acid-based reaction , so it 's possible to , once again , protonate your carbonile . that 's going to activate it . this carbon right here becomes more electrophilic , and that means that the chloride anion can function as a nucleophile and attack our electrophile since the chloride anion is n't a great nucleophile on it 's own , so if the chloride anion attacks here , that would push these electrons off onto the oxygen , and we could go ahead and draw what we would form , so we would have our r group , we would have our carbon , we would have this oxygen with two loan pairs of electrons bonded to hydrogen here , so let 's show those electrons in blue . if these electrons in blue kick off onto the oxygen , we could say that those are these electrons . then we could also say that these electrons here in green , on the chloride anion are going to form a bond between the chlorine and the carbon , so we can go ahead and draw in the bond between the chlorine and the carbon like that . we still have our oxygen right here bonded to our sulfur , double-bonded to this oxygen , and then we have our loan pair of electrons here . then we have our chlorine . what all this does is make a much better leaving group than what we started off with so you could think about this as our leaving group . this is a leaving group , and then the next step of the mechanism , and if we go back up here , that 's a much better leaving group than the oh that we started out with . if that 's going to be our leaving group , we could show these electrons in here moving into form our double bond which would kick these electrons off onto our oxygen . let 's go ahead and show what we would make from that . we would now form our r group , our carbon would be double-bonded to an oxygen like this , loan pair of electrons plus one formal charge , and then we 'd have our chlorine over here with all of its electrons too . let 's go ahead and show where those electrons came from so if these electrons move in here , that would reform our carbonile , so i can go ahead and draw that in . then we still have this bond in green , here , just to clarify . then we would have some electrons , and now let 's make a magenta , so these electrons in here are going to kick off onto the oxygen , so we could go ahead and draw that . we would have our oxygen . so we 'd have three loan pairs of electrons . one of them would be the magenta ones , so we could go ahead and do that . that gives our oxygen a negative one formal charge because it 's bonded to the sulfur here , the sulfur 's still double-bonded to this oxygen , and once again , still a loan pair of electrons and a chlorine over here like that . the reason why this is a good leaving group is because we can push these electrons into here , push these electrons off onto here , and we can form sulfur dioxide as a gas . if we go ahead and draw the result of that , we would now have this oxygen double-bonded to this sulfur , double-bonded to this oxygen , so draw in loan pairs of electrons here , and we still have a loan pair of electrons on our sulfur . if these electrons in red here move in to form this bond , we 've now formed sulfur dioxide , and we also formed a chloride anion , so it once again follows some electrons . if these electrons move off onto chlorine , we have the chloride anion , so now we have our chloride anion which could function as a base and in the last step of our mechanism , deprotonate , so take this proton , leave these electrons behind and finally , we formed our acyl chloride . we have our carbon double-bonded to an oxygen with two loan pairs of electrons , and we have our chlorine over here , like that . if we follow those final electrons , these electrons in blue , move off onto your carbonile . you would also form hcl by this , so you form hcl , which is also a gas , so the formation of these gases , the formation of sulfur dioxide and hcl drives the reaction to completion here . that was a long mechanism . once again , you 'll see some slight variations on this , so if i go back up to here , you could show this step , the nucleophile attacking the electrophile , and directly form here if you wanted to , so as your nucleophile attacks your sulfur , these electrons could kick off onto chlorine to form the chloride anion at this step , and at this step , and then you would get this at this point . once again , i talked about the possibility of using this as your intermediate . depending on which textbook you look in , you might see some slight variations on this mechanism . it is a little bit of a long one here . all right , let 's look at two other ways to make an acyl chloride , so starting with a carboxylic acid , you could add phosphorus , a pentachloride , or a phosphorus trichloride , and both of those will give you an acyl chloride as well . the mechanism is pretty similar and also we could think about , instead of phosphorus trichloride , we could add a pbr3 , so a phosphorus tribromide instead of putting a chlorine here . that would put a bromine , and so we could form an acyl bromide this way , and we 'll see a use for this reaction in a later video .
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i 'm going to show the chloride anion functioning as a nucleophile over here for this . it 's an acid-based reaction , so it 's possible to , once again , protonate your carbonile . that 's going to activate it .
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if there is a nitrogen atom directly attached with the carbonyl carbon of the acid whether the reaction will occur in the same mechanism ?
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: here 's a general structure for an acyl chloride , also called an acid chloride , and it 's a carboxylic acid derivative , so we can form them from carboxylic acid , so if we start with a carboxylic acid and add thionyl chloride , we can form our acyl chlorides , and we would also form a sulfur dioxide and hcl in this reaction . let 's look at the structure of thionyl chloride . here we have the dot structure right here and we could draw a resonance structure for this , so we could show these electrons in here moving off onto the oxygen , so let 's go ahead and draw what we would form from that . our oxygen would now have three loan pairs of electrons on it , giving it a negative formal charge . our sulfur would still be barred to these chlorines here . it would solve a loan pair of electrons and it would get a plus one formal charge like that . this is a major contributor to the overall structure . oxygen is more electronegative than sulfur , and if you think about this pie bond in here , there 's ineffective overlap of those p orbitals , and that 's because sulfur and oxygen are in different periods on the periodic table , so sulfur is in the third period , so it has a larger p orbital than oxygen . oxygen 's in the second period . you get ineffective overlap of these orbitals here , and so that 's another reason why this is going to contribute to the overall structure . plus , you have these chlorines here , withdrawing some electron density from the sulfur , so chlorine is more electronegative than the sulfur , so that the end result of all this is going to make this sulfur very electrophilic right here , and so therefore , our carboxylic acid is able to act as a nucleophile , and so if these electrons move into here , these electrons can attack our sulfur so the nucleophile attacks our electrophile , and then these electrons kick off onto the oxygen , so let 's go ahead and show that . we would have our r group bonded to our carbon , and then that 's bonded to an oxygen . the oxygen has two loan pairs of electrons on it . the oxygen formed a bond with the sulfur , and now the sulfur 's bonded to this oxygen , which gets a negative one formal charge . this sulfur is bonded to two chlorines , so we draw in our chlorines here with all the loan pairs of electrons , and there 's still a loan pair of electrons on our sulfur . we now have a double bond between the carbon and this oxygen , and one loan pair of electrons on this oxygen gives it a plus one formal charge . following some of our electrons , these electrons in magenta move in here , they 'd form our double bond , and then we can think about these electrons in blue , forming the bonds between the oxygen and the sulfur . finally , we could think about these electrons in here in green moving off onto our oxygen like that . so for the next step , we could think about these electrons moving in here to form our double bond between oxygen and sulfur . that would kick off the chloride anion as a leaving group , and we know the chloride anion is an excellent leaving group because it 's stable on it 's own . when we draw the results of that , we would have our group . we would have our carbon , we would have that carbon double-bonded to our oxygen here with a loan pair of electrons plus one formal charge , and then we would have our oxygen with two loan pairs of electrons , and our sulfur is now double-bonded to this oxygen . all right , there 's still a loan pair of electrons on that sulfur . only one chlorine bonded to this sulfur now , so we lost the chloride anion . let 's go ahead and show some of those electrons . let 's say these electrons in red here move in to reform the dull bond between oxygen and sulfur , and then we had some electrons kick off onto the chlorine , so we have the chloride anion that forms . let 's go ahead and show that as well . these electrons in here come off onto chlorine , then we have the chloride anions . let 's draw in the chloride anion . let 's get some more space down here . we also have the chloride anion that forms , so we draw that in here like that . a negative , more informal charge . at this stage , we need to consider whether the chloride anion is going to function as a base or a nucleophile , and it could do both , so let 's first think about the chloride anion functioning as a base . if it functions as a base , it could take this proton , leaving these electrons behind on the oxygen , so let 's go ahead and draw the lewd form . lewd form are carbonile with two loan pairs of electrons on the oxygen , and then we would have this oxygen here with two loan pairs , and then this sulfur , right when it would still would be double-bonded , one loan pair , and then the chlorine , like that . we 're saying that in this acid-based reaction , these electrons in here move back onto here to form the carbonile . you can consider this to be the intermediate for this mechanism . you 'll see some versions of this mechanism take this intermediate and continue on to form your product . i 'm going to show the chloride anion functioning as a nucleophile over here for this . it 's an acid-based reaction , so it 's possible to , once again , protonate your carbonile . that 's going to activate it . this carbon right here becomes more electrophilic , and that means that the chloride anion can function as a nucleophile and attack our electrophile since the chloride anion is n't a great nucleophile on it 's own , so if the chloride anion attacks here , that would push these electrons off onto the oxygen , and we could go ahead and draw what we would form , so we would have our r group , we would have our carbon , we would have this oxygen with two loan pairs of electrons bonded to hydrogen here , so let 's show those electrons in blue . if these electrons in blue kick off onto the oxygen , we could say that those are these electrons . then we could also say that these electrons here in green , on the chloride anion are going to form a bond between the chlorine and the carbon , so we can go ahead and draw in the bond between the chlorine and the carbon like that . we still have our oxygen right here bonded to our sulfur , double-bonded to this oxygen , and then we have our loan pair of electrons here . then we have our chlorine . what all this does is make a much better leaving group than what we started off with so you could think about this as our leaving group . this is a leaving group , and then the next step of the mechanism , and if we go back up here , that 's a much better leaving group than the oh that we started out with . if that 's going to be our leaving group , we could show these electrons in here moving into form our double bond which would kick these electrons off onto our oxygen . let 's go ahead and show what we would make from that . we would now form our r group , our carbon would be double-bonded to an oxygen like this , loan pair of electrons plus one formal charge , and then we 'd have our chlorine over here with all of its electrons too . let 's go ahead and show where those electrons came from so if these electrons move in here , that would reform our carbonile , so i can go ahead and draw that in . then we still have this bond in green , here , just to clarify . then we would have some electrons , and now let 's make a magenta , so these electrons in here are going to kick off onto the oxygen , so we could go ahead and draw that . we would have our oxygen . so we 'd have three loan pairs of electrons . one of them would be the magenta ones , so we could go ahead and do that . that gives our oxygen a negative one formal charge because it 's bonded to the sulfur here , the sulfur 's still double-bonded to this oxygen , and once again , still a loan pair of electrons and a chlorine over here like that . the reason why this is a good leaving group is because we can push these electrons into here , push these electrons off onto here , and we can form sulfur dioxide as a gas . if we go ahead and draw the result of that , we would now have this oxygen double-bonded to this sulfur , double-bonded to this oxygen , so draw in loan pairs of electrons here , and we still have a loan pair of electrons on our sulfur . if these electrons in red here move in to form this bond , we 've now formed sulfur dioxide , and we also formed a chloride anion , so it once again follows some electrons . if these electrons move off onto chlorine , we have the chloride anion , so now we have our chloride anion which could function as a base and in the last step of our mechanism , deprotonate , so take this proton , leave these electrons behind and finally , we formed our acyl chloride . we have our carbon double-bonded to an oxygen with two loan pairs of electrons , and we have our chlorine over here , like that . if we follow those final electrons , these electrons in blue , move off onto your carbonile . you would also form hcl by this , so you form hcl , which is also a gas , so the formation of these gases , the formation of sulfur dioxide and hcl drives the reaction to completion here . that was a long mechanism . once again , you 'll see some slight variations on this , so if i go back up to here , you could show this step , the nucleophile attacking the electrophile , and directly form here if you wanted to , so as your nucleophile attacks your sulfur , these electrons could kick off onto chlorine to form the chloride anion at this step , and at this step , and then you would get this at this point . once again , i talked about the possibility of using this as your intermediate . depending on which textbook you look in , you might see some slight variations on this mechanism . it is a little bit of a long one here . all right , let 's look at two other ways to make an acyl chloride , so starting with a carboxylic acid , you could add phosphorus , a pentachloride , or a phosphorus trichloride , and both of those will give you an acyl chloride as well . the mechanism is pretty similar and also we could think about , instead of phosphorus trichloride , we could add a pbr3 , so a phosphorus tribromide instead of putting a chlorine here . that would put a bromine , and so we could form an acyl bromide this way , and we 'll see a use for this reaction in a later video .
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you get ineffective overlap of these orbitals here , and so that 's another reason why this is going to contribute to the overall structure . plus , you have these chlorines here , withdrawing some electron density from the sulfur , so chlorine is more electronegative than the sulfur , so that the end result of all this is going to make this sulfur very electrophilic right here , and so therefore , our carboxylic acid is able to act as a nucleophile , and so if these electrons move into here , these electrons can attack our sulfur so the nucleophile attacks our electrophile , and then these electrons kick off onto the oxygen , so let 's go ahead and show that . we would have our r group bonded to our carbon , and then that 's bonded to an oxygen .
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will the mcat 2016 require extensive knowledge of these mechanisms or just end products ?
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: here 's a general structure for an acyl chloride , also called an acid chloride , and it 's a carboxylic acid derivative , so we can form them from carboxylic acid , so if we start with a carboxylic acid and add thionyl chloride , we can form our acyl chlorides , and we would also form a sulfur dioxide and hcl in this reaction . let 's look at the structure of thionyl chloride . here we have the dot structure right here and we could draw a resonance structure for this , so we could show these electrons in here moving off onto the oxygen , so let 's go ahead and draw what we would form from that . our oxygen would now have three loan pairs of electrons on it , giving it a negative formal charge . our sulfur would still be barred to these chlorines here . it would solve a loan pair of electrons and it would get a plus one formal charge like that . this is a major contributor to the overall structure . oxygen is more electronegative than sulfur , and if you think about this pie bond in here , there 's ineffective overlap of those p orbitals , and that 's because sulfur and oxygen are in different periods on the periodic table , so sulfur is in the third period , so it has a larger p orbital than oxygen . oxygen 's in the second period . you get ineffective overlap of these orbitals here , and so that 's another reason why this is going to contribute to the overall structure . plus , you have these chlorines here , withdrawing some electron density from the sulfur , so chlorine is more electronegative than the sulfur , so that the end result of all this is going to make this sulfur very electrophilic right here , and so therefore , our carboxylic acid is able to act as a nucleophile , and so if these electrons move into here , these electrons can attack our sulfur so the nucleophile attacks our electrophile , and then these electrons kick off onto the oxygen , so let 's go ahead and show that . we would have our r group bonded to our carbon , and then that 's bonded to an oxygen . the oxygen has two loan pairs of electrons on it . the oxygen formed a bond with the sulfur , and now the sulfur 's bonded to this oxygen , which gets a negative one formal charge . this sulfur is bonded to two chlorines , so we draw in our chlorines here with all the loan pairs of electrons , and there 's still a loan pair of electrons on our sulfur . we now have a double bond between the carbon and this oxygen , and one loan pair of electrons on this oxygen gives it a plus one formal charge . following some of our electrons , these electrons in magenta move in here , they 'd form our double bond , and then we can think about these electrons in blue , forming the bonds between the oxygen and the sulfur . finally , we could think about these electrons in here in green moving off onto our oxygen like that . so for the next step , we could think about these electrons moving in here to form our double bond between oxygen and sulfur . that would kick off the chloride anion as a leaving group , and we know the chloride anion is an excellent leaving group because it 's stable on it 's own . when we draw the results of that , we would have our group . we would have our carbon , we would have that carbon double-bonded to our oxygen here with a loan pair of electrons plus one formal charge , and then we would have our oxygen with two loan pairs of electrons , and our sulfur is now double-bonded to this oxygen . all right , there 's still a loan pair of electrons on that sulfur . only one chlorine bonded to this sulfur now , so we lost the chloride anion . let 's go ahead and show some of those electrons . let 's say these electrons in red here move in to reform the dull bond between oxygen and sulfur , and then we had some electrons kick off onto the chlorine , so we have the chloride anion that forms . let 's go ahead and show that as well . these electrons in here come off onto chlorine , then we have the chloride anions . let 's draw in the chloride anion . let 's get some more space down here . we also have the chloride anion that forms , so we draw that in here like that . a negative , more informal charge . at this stage , we need to consider whether the chloride anion is going to function as a base or a nucleophile , and it could do both , so let 's first think about the chloride anion functioning as a base . if it functions as a base , it could take this proton , leaving these electrons behind on the oxygen , so let 's go ahead and draw the lewd form . lewd form are carbonile with two loan pairs of electrons on the oxygen , and then we would have this oxygen here with two loan pairs , and then this sulfur , right when it would still would be double-bonded , one loan pair , and then the chlorine , like that . we 're saying that in this acid-based reaction , these electrons in here move back onto here to form the carbonile . you can consider this to be the intermediate for this mechanism . you 'll see some versions of this mechanism take this intermediate and continue on to form your product . i 'm going to show the chloride anion functioning as a nucleophile over here for this . it 's an acid-based reaction , so it 's possible to , once again , protonate your carbonile . that 's going to activate it . this carbon right here becomes more electrophilic , and that means that the chloride anion can function as a nucleophile and attack our electrophile since the chloride anion is n't a great nucleophile on it 's own , so if the chloride anion attacks here , that would push these electrons off onto the oxygen , and we could go ahead and draw what we would form , so we would have our r group , we would have our carbon , we would have this oxygen with two loan pairs of electrons bonded to hydrogen here , so let 's show those electrons in blue . if these electrons in blue kick off onto the oxygen , we could say that those are these electrons . then we could also say that these electrons here in green , on the chloride anion are going to form a bond between the chlorine and the carbon , so we can go ahead and draw in the bond between the chlorine and the carbon like that . we still have our oxygen right here bonded to our sulfur , double-bonded to this oxygen , and then we have our loan pair of electrons here . then we have our chlorine . what all this does is make a much better leaving group than what we started off with so you could think about this as our leaving group . this is a leaving group , and then the next step of the mechanism , and if we go back up here , that 's a much better leaving group than the oh that we started out with . if that 's going to be our leaving group , we could show these electrons in here moving into form our double bond which would kick these electrons off onto our oxygen . let 's go ahead and show what we would make from that . we would now form our r group , our carbon would be double-bonded to an oxygen like this , loan pair of electrons plus one formal charge , and then we 'd have our chlorine over here with all of its electrons too . let 's go ahead and show where those electrons came from so if these electrons move in here , that would reform our carbonile , so i can go ahead and draw that in . then we still have this bond in green , here , just to clarify . then we would have some electrons , and now let 's make a magenta , so these electrons in here are going to kick off onto the oxygen , so we could go ahead and draw that . we would have our oxygen . so we 'd have three loan pairs of electrons . one of them would be the magenta ones , so we could go ahead and do that . that gives our oxygen a negative one formal charge because it 's bonded to the sulfur here , the sulfur 's still double-bonded to this oxygen , and once again , still a loan pair of electrons and a chlorine over here like that . the reason why this is a good leaving group is because we can push these electrons into here , push these electrons off onto here , and we can form sulfur dioxide as a gas . if we go ahead and draw the result of that , we would now have this oxygen double-bonded to this sulfur , double-bonded to this oxygen , so draw in loan pairs of electrons here , and we still have a loan pair of electrons on our sulfur . if these electrons in red here move in to form this bond , we 've now formed sulfur dioxide , and we also formed a chloride anion , so it once again follows some electrons . if these electrons move off onto chlorine , we have the chloride anion , so now we have our chloride anion which could function as a base and in the last step of our mechanism , deprotonate , so take this proton , leave these electrons behind and finally , we formed our acyl chloride . we have our carbon double-bonded to an oxygen with two loan pairs of electrons , and we have our chlorine over here , like that . if we follow those final electrons , these electrons in blue , move off onto your carbonile . you would also form hcl by this , so you form hcl , which is also a gas , so the formation of these gases , the formation of sulfur dioxide and hcl drives the reaction to completion here . that was a long mechanism . once again , you 'll see some slight variations on this , so if i go back up to here , you could show this step , the nucleophile attacking the electrophile , and directly form here if you wanted to , so as your nucleophile attacks your sulfur , these electrons could kick off onto chlorine to form the chloride anion at this step , and at this step , and then you would get this at this point . once again , i talked about the possibility of using this as your intermediate . depending on which textbook you look in , you might see some slight variations on this mechanism . it is a little bit of a long one here . all right , let 's look at two other ways to make an acyl chloride , so starting with a carboxylic acid , you could add phosphorus , a pentachloride , or a phosphorus trichloride , and both of those will give you an acyl chloride as well . the mechanism is pretty similar and also we could think about , instead of phosphorus trichloride , we could add a pbr3 , so a phosphorus tribromide instead of putting a chlorine here . that would put a bromine , and so we could form an acyl bromide this way , and we 'll see a use for this reaction in a later video .
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: here 's a general structure for an acyl chloride , also called an acid chloride , and it 's a carboxylic acid derivative , so we can form them from carboxylic acid , so if we start with a carboxylic acid and add thionyl chloride , we can form our acyl chlorides , and we would also form a sulfur dioxide and hcl in this reaction . let 's look at the structure of thionyl chloride .
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why does this video begin the mechanism from the double bond on the carboxylic acid , while in the previous video titled `` acid chloride formation '' the electrons from the oh acted to attack the socl2 ?
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: here 's a general structure for an acyl chloride , also called an acid chloride , and it 's a carboxylic acid derivative , so we can form them from carboxylic acid , so if we start with a carboxylic acid and add thionyl chloride , we can form our acyl chlorides , and we would also form a sulfur dioxide and hcl in this reaction . let 's look at the structure of thionyl chloride . here we have the dot structure right here and we could draw a resonance structure for this , so we could show these electrons in here moving off onto the oxygen , so let 's go ahead and draw what we would form from that . our oxygen would now have three loan pairs of electrons on it , giving it a negative formal charge . our sulfur would still be barred to these chlorines here . it would solve a loan pair of electrons and it would get a plus one formal charge like that . this is a major contributor to the overall structure . oxygen is more electronegative than sulfur , and if you think about this pie bond in here , there 's ineffective overlap of those p orbitals , and that 's because sulfur and oxygen are in different periods on the periodic table , so sulfur is in the third period , so it has a larger p orbital than oxygen . oxygen 's in the second period . you get ineffective overlap of these orbitals here , and so that 's another reason why this is going to contribute to the overall structure . plus , you have these chlorines here , withdrawing some electron density from the sulfur , so chlorine is more electronegative than the sulfur , so that the end result of all this is going to make this sulfur very electrophilic right here , and so therefore , our carboxylic acid is able to act as a nucleophile , and so if these electrons move into here , these electrons can attack our sulfur so the nucleophile attacks our electrophile , and then these electrons kick off onto the oxygen , so let 's go ahead and show that . we would have our r group bonded to our carbon , and then that 's bonded to an oxygen . the oxygen has two loan pairs of electrons on it . the oxygen formed a bond with the sulfur , and now the sulfur 's bonded to this oxygen , which gets a negative one formal charge . this sulfur is bonded to two chlorines , so we draw in our chlorines here with all the loan pairs of electrons , and there 's still a loan pair of electrons on our sulfur . we now have a double bond between the carbon and this oxygen , and one loan pair of electrons on this oxygen gives it a plus one formal charge . following some of our electrons , these electrons in magenta move in here , they 'd form our double bond , and then we can think about these electrons in blue , forming the bonds between the oxygen and the sulfur . finally , we could think about these electrons in here in green moving off onto our oxygen like that . so for the next step , we could think about these electrons moving in here to form our double bond between oxygen and sulfur . that would kick off the chloride anion as a leaving group , and we know the chloride anion is an excellent leaving group because it 's stable on it 's own . when we draw the results of that , we would have our group . we would have our carbon , we would have that carbon double-bonded to our oxygen here with a loan pair of electrons plus one formal charge , and then we would have our oxygen with two loan pairs of electrons , and our sulfur is now double-bonded to this oxygen . all right , there 's still a loan pair of electrons on that sulfur . only one chlorine bonded to this sulfur now , so we lost the chloride anion . let 's go ahead and show some of those electrons . let 's say these electrons in red here move in to reform the dull bond between oxygen and sulfur , and then we had some electrons kick off onto the chlorine , so we have the chloride anion that forms . let 's go ahead and show that as well . these electrons in here come off onto chlorine , then we have the chloride anions . let 's draw in the chloride anion . let 's get some more space down here . we also have the chloride anion that forms , so we draw that in here like that . a negative , more informal charge . at this stage , we need to consider whether the chloride anion is going to function as a base or a nucleophile , and it could do both , so let 's first think about the chloride anion functioning as a base . if it functions as a base , it could take this proton , leaving these electrons behind on the oxygen , so let 's go ahead and draw the lewd form . lewd form are carbonile with two loan pairs of electrons on the oxygen , and then we would have this oxygen here with two loan pairs , and then this sulfur , right when it would still would be double-bonded , one loan pair , and then the chlorine , like that . we 're saying that in this acid-based reaction , these electrons in here move back onto here to form the carbonile . you can consider this to be the intermediate for this mechanism . you 'll see some versions of this mechanism take this intermediate and continue on to form your product . i 'm going to show the chloride anion functioning as a nucleophile over here for this . it 's an acid-based reaction , so it 's possible to , once again , protonate your carbonile . that 's going to activate it . this carbon right here becomes more electrophilic , and that means that the chloride anion can function as a nucleophile and attack our electrophile since the chloride anion is n't a great nucleophile on it 's own , so if the chloride anion attacks here , that would push these electrons off onto the oxygen , and we could go ahead and draw what we would form , so we would have our r group , we would have our carbon , we would have this oxygen with two loan pairs of electrons bonded to hydrogen here , so let 's show those electrons in blue . if these electrons in blue kick off onto the oxygen , we could say that those are these electrons . then we could also say that these electrons here in green , on the chloride anion are going to form a bond between the chlorine and the carbon , so we can go ahead and draw in the bond between the chlorine and the carbon like that . we still have our oxygen right here bonded to our sulfur , double-bonded to this oxygen , and then we have our loan pair of electrons here . then we have our chlorine . what all this does is make a much better leaving group than what we started off with so you could think about this as our leaving group . this is a leaving group , and then the next step of the mechanism , and if we go back up here , that 's a much better leaving group than the oh that we started out with . if that 's going to be our leaving group , we could show these electrons in here moving into form our double bond which would kick these electrons off onto our oxygen . let 's go ahead and show what we would make from that . we would now form our r group , our carbon would be double-bonded to an oxygen like this , loan pair of electrons plus one formal charge , and then we 'd have our chlorine over here with all of its electrons too . let 's go ahead and show where those electrons came from so if these electrons move in here , that would reform our carbonile , so i can go ahead and draw that in . then we still have this bond in green , here , just to clarify . then we would have some electrons , and now let 's make a magenta , so these electrons in here are going to kick off onto the oxygen , so we could go ahead and draw that . we would have our oxygen . so we 'd have three loan pairs of electrons . one of them would be the magenta ones , so we could go ahead and do that . that gives our oxygen a negative one formal charge because it 's bonded to the sulfur here , the sulfur 's still double-bonded to this oxygen , and once again , still a loan pair of electrons and a chlorine over here like that . the reason why this is a good leaving group is because we can push these electrons into here , push these electrons off onto here , and we can form sulfur dioxide as a gas . if we go ahead and draw the result of that , we would now have this oxygen double-bonded to this sulfur , double-bonded to this oxygen , so draw in loan pairs of electrons here , and we still have a loan pair of electrons on our sulfur . if these electrons in red here move in to form this bond , we 've now formed sulfur dioxide , and we also formed a chloride anion , so it once again follows some electrons . if these electrons move off onto chlorine , we have the chloride anion , so now we have our chloride anion which could function as a base and in the last step of our mechanism , deprotonate , so take this proton , leave these electrons behind and finally , we formed our acyl chloride . we have our carbon double-bonded to an oxygen with two loan pairs of electrons , and we have our chlorine over here , like that . if we follow those final electrons , these electrons in blue , move off onto your carbonile . you would also form hcl by this , so you form hcl , which is also a gas , so the formation of these gases , the formation of sulfur dioxide and hcl drives the reaction to completion here . that was a long mechanism . once again , you 'll see some slight variations on this , so if i go back up to here , you could show this step , the nucleophile attacking the electrophile , and directly form here if you wanted to , so as your nucleophile attacks your sulfur , these electrons could kick off onto chlorine to form the chloride anion at this step , and at this step , and then you would get this at this point . once again , i talked about the possibility of using this as your intermediate . depending on which textbook you look in , you might see some slight variations on this mechanism . it is a little bit of a long one here . all right , let 's look at two other ways to make an acyl chloride , so starting with a carboxylic acid , you could add phosphorus , a pentachloride , or a phosphorus trichloride , and both of those will give you an acyl chloride as well . the mechanism is pretty similar and also we could think about , instead of phosphorus trichloride , we could add a pbr3 , so a phosphorus tribromide instead of putting a chlorine here . that would put a bromine , and so we could form an acyl bromide this way , and we 'll see a use for this reaction in a later video .
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it is a little bit of a long one here . all right , let 's look at two other ways to make an acyl chloride , so starting with a carboxylic acid , you could add phosphorus , a pentachloride , or a phosphorus trichloride , and both of those will give you an acyl chloride as well . the mechanism is pretty similar and also we could think about , instead of phosphorus trichloride , we could add a pbr3 , so a phosphorus tribromide instead of putting a chlorine here .
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why cl- ion is not behaving as a nucleophile and rather than making acyl chloride , it make 1,1dichloroalcohol ?
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: here 's a general structure for an acyl chloride , also called an acid chloride , and it 's a carboxylic acid derivative , so we can form them from carboxylic acid , so if we start with a carboxylic acid and add thionyl chloride , we can form our acyl chlorides , and we would also form a sulfur dioxide and hcl in this reaction . let 's look at the structure of thionyl chloride . here we have the dot structure right here and we could draw a resonance structure for this , so we could show these electrons in here moving off onto the oxygen , so let 's go ahead and draw what we would form from that . our oxygen would now have three loan pairs of electrons on it , giving it a negative formal charge . our sulfur would still be barred to these chlorines here . it would solve a loan pair of electrons and it would get a plus one formal charge like that . this is a major contributor to the overall structure . oxygen is more electronegative than sulfur , and if you think about this pie bond in here , there 's ineffective overlap of those p orbitals , and that 's because sulfur and oxygen are in different periods on the periodic table , so sulfur is in the third period , so it has a larger p orbital than oxygen . oxygen 's in the second period . you get ineffective overlap of these orbitals here , and so that 's another reason why this is going to contribute to the overall structure . plus , you have these chlorines here , withdrawing some electron density from the sulfur , so chlorine is more electronegative than the sulfur , so that the end result of all this is going to make this sulfur very electrophilic right here , and so therefore , our carboxylic acid is able to act as a nucleophile , and so if these electrons move into here , these electrons can attack our sulfur so the nucleophile attacks our electrophile , and then these electrons kick off onto the oxygen , so let 's go ahead and show that . we would have our r group bonded to our carbon , and then that 's bonded to an oxygen . the oxygen has two loan pairs of electrons on it . the oxygen formed a bond with the sulfur , and now the sulfur 's bonded to this oxygen , which gets a negative one formal charge . this sulfur is bonded to two chlorines , so we draw in our chlorines here with all the loan pairs of electrons , and there 's still a loan pair of electrons on our sulfur . we now have a double bond between the carbon and this oxygen , and one loan pair of electrons on this oxygen gives it a plus one formal charge . following some of our electrons , these electrons in magenta move in here , they 'd form our double bond , and then we can think about these electrons in blue , forming the bonds between the oxygen and the sulfur . finally , we could think about these electrons in here in green moving off onto our oxygen like that . so for the next step , we could think about these electrons moving in here to form our double bond between oxygen and sulfur . that would kick off the chloride anion as a leaving group , and we know the chloride anion is an excellent leaving group because it 's stable on it 's own . when we draw the results of that , we would have our group . we would have our carbon , we would have that carbon double-bonded to our oxygen here with a loan pair of electrons plus one formal charge , and then we would have our oxygen with two loan pairs of electrons , and our sulfur is now double-bonded to this oxygen . all right , there 's still a loan pair of electrons on that sulfur . only one chlorine bonded to this sulfur now , so we lost the chloride anion . let 's go ahead and show some of those electrons . let 's say these electrons in red here move in to reform the dull bond between oxygen and sulfur , and then we had some electrons kick off onto the chlorine , so we have the chloride anion that forms . let 's go ahead and show that as well . these electrons in here come off onto chlorine , then we have the chloride anions . let 's draw in the chloride anion . let 's get some more space down here . we also have the chloride anion that forms , so we draw that in here like that . a negative , more informal charge . at this stage , we need to consider whether the chloride anion is going to function as a base or a nucleophile , and it could do both , so let 's first think about the chloride anion functioning as a base . if it functions as a base , it could take this proton , leaving these electrons behind on the oxygen , so let 's go ahead and draw the lewd form . lewd form are carbonile with two loan pairs of electrons on the oxygen , and then we would have this oxygen here with two loan pairs , and then this sulfur , right when it would still would be double-bonded , one loan pair , and then the chlorine , like that . we 're saying that in this acid-based reaction , these electrons in here move back onto here to form the carbonile . you can consider this to be the intermediate for this mechanism . you 'll see some versions of this mechanism take this intermediate and continue on to form your product . i 'm going to show the chloride anion functioning as a nucleophile over here for this . it 's an acid-based reaction , so it 's possible to , once again , protonate your carbonile . that 's going to activate it . this carbon right here becomes more electrophilic , and that means that the chloride anion can function as a nucleophile and attack our electrophile since the chloride anion is n't a great nucleophile on it 's own , so if the chloride anion attacks here , that would push these electrons off onto the oxygen , and we could go ahead and draw what we would form , so we would have our r group , we would have our carbon , we would have this oxygen with two loan pairs of electrons bonded to hydrogen here , so let 's show those electrons in blue . if these electrons in blue kick off onto the oxygen , we could say that those are these electrons . then we could also say that these electrons here in green , on the chloride anion are going to form a bond between the chlorine and the carbon , so we can go ahead and draw in the bond between the chlorine and the carbon like that . we still have our oxygen right here bonded to our sulfur , double-bonded to this oxygen , and then we have our loan pair of electrons here . then we have our chlorine . what all this does is make a much better leaving group than what we started off with so you could think about this as our leaving group . this is a leaving group , and then the next step of the mechanism , and if we go back up here , that 's a much better leaving group than the oh that we started out with . if that 's going to be our leaving group , we could show these electrons in here moving into form our double bond which would kick these electrons off onto our oxygen . let 's go ahead and show what we would make from that . we would now form our r group , our carbon would be double-bonded to an oxygen like this , loan pair of electrons plus one formal charge , and then we 'd have our chlorine over here with all of its electrons too . let 's go ahead and show where those electrons came from so if these electrons move in here , that would reform our carbonile , so i can go ahead and draw that in . then we still have this bond in green , here , just to clarify . then we would have some electrons , and now let 's make a magenta , so these electrons in here are going to kick off onto the oxygen , so we could go ahead and draw that . we would have our oxygen . so we 'd have three loan pairs of electrons . one of them would be the magenta ones , so we could go ahead and do that . that gives our oxygen a negative one formal charge because it 's bonded to the sulfur here , the sulfur 's still double-bonded to this oxygen , and once again , still a loan pair of electrons and a chlorine over here like that . the reason why this is a good leaving group is because we can push these electrons into here , push these electrons off onto here , and we can form sulfur dioxide as a gas . if we go ahead and draw the result of that , we would now have this oxygen double-bonded to this sulfur , double-bonded to this oxygen , so draw in loan pairs of electrons here , and we still have a loan pair of electrons on our sulfur . if these electrons in red here move in to form this bond , we 've now formed sulfur dioxide , and we also formed a chloride anion , so it once again follows some electrons . if these electrons move off onto chlorine , we have the chloride anion , so now we have our chloride anion which could function as a base and in the last step of our mechanism , deprotonate , so take this proton , leave these electrons behind and finally , we formed our acyl chloride . we have our carbon double-bonded to an oxygen with two loan pairs of electrons , and we have our chlorine over here , like that . if we follow those final electrons , these electrons in blue , move off onto your carbonile . you would also form hcl by this , so you form hcl , which is also a gas , so the formation of these gases , the formation of sulfur dioxide and hcl drives the reaction to completion here . that was a long mechanism . once again , you 'll see some slight variations on this , so if i go back up to here , you could show this step , the nucleophile attacking the electrophile , and directly form here if you wanted to , so as your nucleophile attacks your sulfur , these electrons could kick off onto chlorine to form the chloride anion at this step , and at this step , and then you would get this at this point . once again , i talked about the possibility of using this as your intermediate . depending on which textbook you look in , you might see some slight variations on this mechanism . it is a little bit of a long one here . all right , let 's look at two other ways to make an acyl chloride , so starting with a carboxylic acid , you could add phosphorus , a pentachloride , or a phosphorus trichloride , and both of those will give you an acyl chloride as well . the mechanism is pretty similar and also we could think about , instead of phosphorus trichloride , we could add a pbr3 , so a phosphorus tribromide instead of putting a chlorine here . that would put a bromine , and so we could form an acyl bromide this way , and we 'll see a use for this reaction in a later video .
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so for the next step , we could think about these electrons moving in here to form our double bond between oxygen and sulfur . that would kick off the chloride anion as a leaving group , and we know the chloride anion is an excellent leaving group because it 's stable on it 's own . when we draw the results of that , we would have our group .
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how do you know which will be the leaving group , or what is going to be included in the leaving group ?
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: here 's a general structure for an acyl chloride , also called an acid chloride , and it 's a carboxylic acid derivative , so we can form them from carboxylic acid , so if we start with a carboxylic acid and add thionyl chloride , we can form our acyl chlorides , and we would also form a sulfur dioxide and hcl in this reaction . let 's look at the structure of thionyl chloride . here we have the dot structure right here and we could draw a resonance structure for this , so we could show these electrons in here moving off onto the oxygen , so let 's go ahead and draw what we would form from that . our oxygen would now have three loan pairs of electrons on it , giving it a negative formal charge . our sulfur would still be barred to these chlorines here . it would solve a loan pair of electrons and it would get a plus one formal charge like that . this is a major contributor to the overall structure . oxygen is more electronegative than sulfur , and if you think about this pie bond in here , there 's ineffective overlap of those p orbitals , and that 's because sulfur and oxygen are in different periods on the periodic table , so sulfur is in the third period , so it has a larger p orbital than oxygen . oxygen 's in the second period . you get ineffective overlap of these orbitals here , and so that 's another reason why this is going to contribute to the overall structure . plus , you have these chlorines here , withdrawing some electron density from the sulfur , so chlorine is more electronegative than the sulfur , so that the end result of all this is going to make this sulfur very electrophilic right here , and so therefore , our carboxylic acid is able to act as a nucleophile , and so if these electrons move into here , these electrons can attack our sulfur so the nucleophile attacks our electrophile , and then these electrons kick off onto the oxygen , so let 's go ahead and show that . we would have our r group bonded to our carbon , and then that 's bonded to an oxygen . the oxygen has two loan pairs of electrons on it . the oxygen formed a bond with the sulfur , and now the sulfur 's bonded to this oxygen , which gets a negative one formal charge . this sulfur is bonded to two chlorines , so we draw in our chlorines here with all the loan pairs of electrons , and there 's still a loan pair of electrons on our sulfur . we now have a double bond between the carbon and this oxygen , and one loan pair of electrons on this oxygen gives it a plus one formal charge . following some of our electrons , these electrons in magenta move in here , they 'd form our double bond , and then we can think about these electrons in blue , forming the bonds between the oxygen and the sulfur . finally , we could think about these electrons in here in green moving off onto our oxygen like that . so for the next step , we could think about these electrons moving in here to form our double bond between oxygen and sulfur . that would kick off the chloride anion as a leaving group , and we know the chloride anion is an excellent leaving group because it 's stable on it 's own . when we draw the results of that , we would have our group . we would have our carbon , we would have that carbon double-bonded to our oxygen here with a loan pair of electrons plus one formal charge , and then we would have our oxygen with two loan pairs of electrons , and our sulfur is now double-bonded to this oxygen . all right , there 's still a loan pair of electrons on that sulfur . only one chlorine bonded to this sulfur now , so we lost the chloride anion . let 's go ahead and show some of those electrons . let 's say these electrons in red here move in to reform the dull bond between oxygen and sulfur , and then we had some electrons kick off onto the chlorine , so we have the chloride anion that forms . let 's go ahead and show that as well . these electrons in here come off onto chlorine , then we have the chloride anions . let 's draw in the chloride anion . let 's get some more space down here . we also have the chloride anion that forms , so we draw that in here like that . a negative , more informal charge . at this stage , we need to consider whether the chloride anion is going to function as a base or a nucleophile , and it could do both , so let 's first think about the chloride anion functioning as a base . if it functions as a base , it could take this proton , leaving these electrons behind on the oxygen , so let 's go ahead and draw the lewd form . lewd form are carbonile with two loan pairs of electrons on the oxygen , and then we would have this oxygen here with two loan pairs , and then this sulfur , right when it would still would be double-bonded , one loan pair , and then the chlorine , like that . we 're saying that in this acid-based reaction , these electrons in here move back onto here to form the carbonile . you can consider this to be the intermediate for this mechanism . you 'll see some versions of this mechanism take this intermediate and continue on to form your product . i 'm going to show the chloride anion functioning as a nucleophile over here for this . it 's an acid-based reaction , so it 's possible to , once again , protonate your carbonile . that 's going to activate it . this carbon right here becomes more electrophilic , and that means that the chloride anion can function as a nucleophile and attack our electrophile since the chloride anion is n't a great nucleophile on it 's own , so if the chloride anion attacks here , that would push these electrons off onto the oxygen , and we could go ahead and draw what we would form , so we would have our r group , we would have our carbon , we would have this oxygen with two loan pairs of electrons bonded to hydrogen here , so let 's show those electrons in blue . if these electrons in blue kick off onto the oxygen , we could say that those are these electrons . then we could also say that these electrons here in green , on the chloride anion are going to form a bond between the chlorine and the carbon , so we can go ahead and draw in the bond between the chlorine and the carbon like that . we still have our oxygen right here bonded to our sulfur , double-bonded to this oxygen , and then we have our loan pair of electrons here . then we have our chlorine . what all this does is make a much better leaving group than what we started off with so you could think about this as our leaving group . this is a leaving group , and then the next step of the mechanism , and if we go back up here , that 's a much better leaving group than the oh that we started out with . if that 's going to be our leaving group , we could show these electrons in here moving into form our double bond which would kick these electrons off onto our oxygen . let 's go ahead and show what we would make from that . we would now form our r group , our carbon would be double-bonded to an oxygen like this , loan pair of electrons plus one formal charge , and then we 'd have our chlorine over here with all of its electrons too . let 's go ahead and show where those electrons came from so if these electrons move in here , that would reform our carbonile , so i can go ahead and draw that in . then we still have this bond in green , here , just to clarify . then we would have some electrons , and now let 's make a magenta , so these electrons in here are going to kick off onto the oxygen , so we could go ahead and draw that . we would have our oxygen . so we 'd have three loan pairs of electrons . one of them would be the magenta ones , so we could go ahead and do that . that gives our oxygen a negative one formal charge because it 's bonded to the sulfur here , the sulfur 's still double-bonded to this oxygen , and once again , still a loan pair of electrons and a chlorine over here like that . the reason why this is a good leaving group is because we can push these electrons into here , push these electrons off onto here , and we can form sulfur dioxide as a gas . if we go ahead and draw the result of that , we would now have this oxygen double-bonded to this sulfur , double-bonded to this oxygen , so draw in loan pairs of electrons here , and we still have a loan pair of electrons on our sulfur . if these electrons in red here move in to form this bond , we 've now formed sulfur dioxide , and we also formed a chloride anion , so it once again follows some electrons . if these electrons move off onto chlorine , we have the chloride anion , so now we have our chloride anion which could function as a base and in the last step of our mechanism , deprotonate , so take this proton , leave these electrons behind and finally , we formed our acyl chloride . we have our carbon double-bonded to an oxygen with two loan pairs of electrons , and we have our chlorine over here , like that . if we follow those final electrons , these electrons in blue , move off onto your carbonile . you would also form hcl by this , so you form hcl , which is also a gas , so the formation of these gases , the formation of sulfur dioxide and hcl drives the reaction to completion here . that was a long mechanism . once again , you 'll see some slight variations on this , so if i go back up to here , you could show this step , the nucleophile attacking the electrophile , and directly form here if you wanted to , so as your nucleophile attacks your sulfur , these electrons could kick off onto chlorine to form the chloride anion at this step , and at this step , and then you would get this at this point . once again , i talked about the possibility of using this as your intermediate . depending on which textbook you look in , you might see some slight variations on this mechanism . it is a little bit of a long one here . all right , let 's look at two other ways to make an acyl chloride , so starting with a carboxylic acid , you could add phosphorus , a pentachloride , or a phosphorus trichloride , and both of those will give you an acyl chloride as well . the mechanism is pretty similar and also we could think about , instead of phosphorus trichloride , we could add a pbr3 , so a phosphorus tribromide instead of putting a chlorine here . that would put a bromine , and so we could form an acyl bromide this way , and we 'll see a use for this reaction in a later video .
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then we would have some electrons , and now let 's make a magenta , so these electrons in here are going to kick off onto the oxygen , so we could go ahead and draw that . we would have our oxygen . so we 'd have three loan pairs of electrons .
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why did n't the hydrogen attached to the positive oxygen leave as a proton and attack the negatively charged hydrogen ?
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: here 's a general structure for an acyl chloride , also called an acid chloride , and it 's a carboxylic acid derivative , so we can form them from carboxylic acid , so if we start with a carboxylic acid and add thionyl chloride , we can form our acyl chlorides , and we would also form a sulfur dioxide and hcl in this reaction . let 's look at the structure of thionyl chloride . here we have the dot structure right here and we could draw a resonance structure for this , so we could show these electrons in here moving off onto the oxygen , so let 's go ahead and draw what we would form from that . our oxygen would now have three loan pairs of electrons on it , giving it a negative formal charge . our sulfur would still be barred to these chlorines here . it would solve a loan pair of electrons and it would get a plus one formal charge like that . this is a major contributor to the overall structure . oxygen is more electronegative than sulfur , and if you think about this pie bond in here , there 's ineffective overlap of those p orbitals , and that 's because sulfur and oxygen are in different periods on the periodic table , so sulfur is in the third period , so it has a larger p orbital than oxygen . oxygen 's in the second period . you get ineffective overlap of these orbitals here , and so that 's another reason why this is going to contribute to the overall structure . plus , you have these chlorines here , withdrawing some electron density from the sulfur , so chlorine is more electronegative than the sulfur , so that the end result of all this is going to make this sulfur very electrophilic right here , and so therefore , our carboxylic acid is able to act as a nucleophile , and so if these electrons move into here , these electrons can attack our sulfur so the nucleophile attacks our electrophile , and then these electrons kick off onto the oxygen , so let 's go ahead and show that . we would have our r group bonded to our carbon , and then that 's bonded to an oxygen . the oxygen has two loan pairs of electrons on it . the oxygen formed a bond with the sulfur , and now the sulfur 's bonded to this oxygen , which gets a negative one formal charge . this sulfur is bonded to two chlorines , so we draw in our chlorines here with all the loan pairs of electrons , and there 's still a loan pair of electrons on our sulfur . we now have a double bond between the carbon and this oxygen , and one loan pair of electrons on this oxygen gives it a plus one formal charge . following some of our electrons , these electrons in magenta move in here , they 'd form our double bond , and then we can think about these electrons in blue , forming the bonds between the oxygen and the sulfur . finally , we could think about these electrons in here in green moving off onto our oxygen like that . so for the next step , we could think about these electrons moving in here to form our double bond between oxygen and sulfur . that would kick off the chloride anion as a leaving group , and we know the chloride anion is an excellent leaving group because it 's stable on it 's own . when we draw the results of that , we would have our group . we would have our carbon , we would have that carbon double-bonded to our oxygen here with a loan pair of electrons plus one formal charge , and then we would have our oxygen with two loan pairs of electrons , and our sulfur is now double-bonded to this oxygen . all right , there 's still a loan pair of electrons on that sulfur . only one chlorine bonded to this sulfur now , so we lost the chloride anion . let 's go ahead and show some of those electrons . let 's say these electrons in red here move in to reform the dull bond between oxygen and sulfur , and then we had some electrons kick off onto the chlorine , so we have the chloride anion that forms . let 's go ahead and show that as well . these electrons in here come off onto chlorine , then we have the chloride anions . let 's draw in the chloride anion . let 's get some more space down here . we also have the chloride anion that forms , so we draw that in here like that . a negative , more informal charge . at this stage , we need to consider whether the chloride anion is going to function as a base or a nucleophile , and it could do both , so let 's first think about the chloride anion functioning as a base . if it functions as a base , it could take this proton , leaving these electrons behind on the oxygen , so let 's go ahead and draw the lewd form . lewd form are carbonile with two loan pairs of electrons on the oxygen , and then we would have this oxygen here with two loan pairs , and then this sulfur , right when it would still would be double-bonded , one loan pair , and then the chlorine , like that . we 're saying that in this acid-based reaction , these electrons in here move back onto here to form the carbonile . you can consider this to be the intermediate for this mechanism . you 'll see some versions of this mechanism take this intermediate and continue on to form your product . i 'm going to show the chloride anion functioning as a nucleophile over here for this . it 's an acid-based reaction , so it 's possible to , once again , protonate your carbonile . that 's going to activate it . this carbon right here becomes more electrophilic , and that means that the chloride anion can function as a nucleophile and attack our electrophile since the chloride anion is n't a great nucleophile on it 's own , so if the chloride anion attacks here , that would push these electrons off onto the oxygen , and we could go ahead and draw what we would form , so we would have our r group , we would have our carbon , we would have this oxygen with two loan pairs of electrons bonded to hydrogen here , so let 's show those electrons in blue . if these electrons in blue kick off onto the oxygen , we could say that those are these electrons . then we could also say that these electrons here in green , on the chloride anion are going to form a bond between the chlorine and the carbon , so we can go ahead and draw in the bond between the chlorine and the carbon like that . we still have our oxygen right here bonded to our sulfur , double-bonded to this oxygen , and then we have our loan pair of electrons here . then we have our chlorine . what all this does is make a much better leaving group than what we started off with so you could think about this as our leaving group . this is a leaving group , and then the next step of the mechanism , and if we go back up here , that 's a much better leaving group than the oh that we started out with . if that 's going to be our leaving group , we could show these electrons in here moving into form our double bond which would kick these electrons off onto our oxygen . let 's go ahead and show what we would make from that . we would now form our r group , our carbon would be double-bonded to an oxygen like this , loan pair of electrons plus one formal charge , and then we 'd have our chlorine over here with all of its electrons too . let 's go ahead and show where those electrons came from so if these electrons move in here , that would reform our carbonile , so i can go ahead and draw that in . then we still have this bond in green , here , just to clarify . then we would have some electrons , and now let 's make a magenta , so these electrons in here are going to kick off onto the oxygen , so we could go ahead and draw that . we would have our oxygen . so we 'd have three loan pairs of electrons . one of them would be the magenta ones , so we could go ahead and do that . that gives our oxygen a negative one formal charge because it 's bonded to the sulfur here , the sulfur 's still double-bonded to this oxygen , and once again , still a loan pair of electrons and a chlorine over here like that . the reason why this is a good leaving group is because we can push these electrons into here , push these electrons off onto here , and we can form sulfur dioxide as a gas . if we go ahead and draw the result of that , we would now have this oxygen double-bonded to this sulfur , double-bonded to this oxygen , so draw in loan pairs of electrons here , and we still have a loan pair of electrons on our sulfur . if these electrons in red here move in to form this bond , we 've now formed sulfur dioxide , and we also formed a chloride anion , so it once again follows some electrons . if these electrons move off onto chlorine , we have the chloride anion , so now we have our chloride anion which could function as a base and in the last step of our mechanism , deprotonate , so take this proton , leave these electrons behind and finally , we formed our acyl chloride . we have our carbon double-bonded to an oxygen with two loan pairs of electrons , and we have our chlorine over here , like that . if we follow those final electrons , these electrons in blue , move off onto your carbonile . you would also form hcl by this , so you form hcl , which is also a gas , so the formation of these gases , the formation of sulfur dioxide and hcl drives the reaction to completion here . that was a long mechanism . once again , you 'll see some slight variations on this , so if i go back up to here , you could show this step , the nucleophile attacking the electrophile , and directly form here if you wanted to , so as your nucleophile attacks your sulfur , these electrons could kick off onto chlorine to form the chloride anion at this step , and at this step , and then you would get this at this point . once again , i talked about the possibility of using this as your intermediate . depending on which textbook you look in , you might see some slight variations on this mechanism . it is a little bit of a long one here . all right , let 's look at two other ways to make an acyl chloride , so starting with a carboxylic acid , you could add phosphorus , a pentachloride , or a phosphorus trichloride , and both of those will give you an acyl chloride as well . the mechanism is pretty similar and also we could think about , instead of phosphorus trichloride , we could add a pbr3 , so a phosphorus tribromide instead of putting a chlorine here . that would put a bromine , and so we could form an acyl bromide this way , and we 'll see a use for this reaction in a later video .
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that 's going to activate it . this carbon right here becomes more electrophilic , and that means that the chloride anion can function as a nucleophile and attack our electrophile since the chloride anion is n't a great nucleophile on it 's own , so if the chloride anion attacks here , that would push these electrons off onto the oxygen , and we could go ahead and draw what we would form , so we would have our r group , we would have our carbon , we would have this oxygen with two loan pairs of electrons bonded to hydrogen here , so let 's show those electrons in blue . if these electrons in blue kick off onto the oxygen , we could say that those are these electrons .
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why did n't the lone pair of carbon attached to the hydrogen , directly attack the positively charged sulphur ?
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: here 's a general structure for an acyl chloride , also called an acid chloride , and it 's a carboxylic acid derivative , so we can form them from carboxylic acid , so if we start with a carboxylic acid and add thionyl chloride , we can form our acyl chlorides , and we would also form a sulfur dioxide and hcl in this reaction . let 's look at the structure of thionyl chloride . here we have the dot structure right here and we could draw a resonance structure for this , so we could show these electrons in here moving off onto the oxygen , so let 's go ahead and draw what we would form from that . our oxygen would now have three loan pairs of electrons on it , giving it a negative formal charge . our sulfur would still be barred to these chlorines here . it would solve a loan pair of electrons and it would get a plus one formal charge like that . this is a major contributor to the overall structure . oxygen is more electronegative than sulfur , and if you think about this pie bond in here , there 's ineffective overlap of those p orbitals , and that 's because sulfur and oxygen are in different periods on the periodic table , so sulfur is in the third period , so it has a larger p orbital than oxygen . oxygen 's in the second period . you get ineffective overlap of these orbitals here , and so that 's another reason why this is going to contribute to the overall structure . plus , you have these chlorines here , withdrawing some electron density from the sulfur , so chlorine is more electronegative than the sulfur , so that the end result of all this is going to make this sulfur very electrophilic right here , and so therefore , our carboxylic acid is able to act as a nucleophile , and so if these electrons move into here , these electrons can attack our sulfur so the nucleophile attacks our electrophile , and then these electrons kick off onto the oxygen , so let 's go ahead and show that . we would have our r group bonded to our carbon , and then that 's bonded to an oxygen . the oxygen has two loan pairs of electrons on it . the oxygen formed a bond with the sulfur , and now the sulfur 's bonded to this oxygen , which gets a negative one formal charge . this sulfur is bonded to two chlorines , so we draw in our chlorines here with all the loan pairs of electrons , and there 's still a loan pair of electrons on our sulfur . we now have a double bond between the carbon and this oxygen , and one loan pair of electrons on this oxygen gives it a plus one formal charge . following some of our electrons , these electrons in magenta move in here , they 'd form our double bond , and then we can think about these electrons in blue , forming the bonds between the oxygen and the sulfur . finally , we could think about these electrons in here in green moving off onto our oxygen like that . so for the next step , we could think about these electrons moving in here to form our double bond between oxygen and sulfur . that would kick off the chloride anion as a leaving group , and we know the chloride anion is an excellent leaving group because it 's stable on it 's own . when we draw the results of that , we would have our group . we would have our carbon , we would have that carbon double-bonded to our oxygen here with a loan pair of electrons plus one formal charge , and then we would have our oxygen with two loan pairs of electrons , and our sulfur is now double-bonded to this oxygen . all right , there 's still a loan pair of electrons on that sulfur . only one chlorine bonded to this sulfur now , so we lost the chloride anion . let 's go ahead and show some of those electrons . let 's say these electrons in red here move in to reform the dull bond between oxygen and sulfur , and then we had some electrons kick off onto the chlorine , so we have the chloride anion that forms . let 's go ahead and show that as well . these electrons in here come off onto chlorine , then we have the chloride anions . let 's draw in the chloride anion . let 's get some more space down here . we also have the chloride anion that forms , so we draw that in here like that . a negative , more informal charge . at this stage , we need to consider whether the chloride anion is going to function as a base or a nucleophile , and it could do both , so let 's first think about the chloride anion functioning as a base . if it functions as a base , it could take this proton , leaving these electrons behind on the oxygen , so let 's go ahead and draw the lewd form . lewd form are carbonile with two loan pairs of electrons on the oxygen , and then we would have this oxygen here with two loan pairs , and then this sulfur , right when it would still would be double-bonded , one loan pair , and then the chlorine , like that . we 're saying that in this acid-based reaction , these electrons in here move back onto here to form the carbonile . you can consider this to be the intermediate for this mechanism . you 'll see some versions of this mechanism take this intermediate and continue on to form your product . i 'm going to show the chloride anion functioning as a nucleophile over here for this . it 's an acid-based reaction , so it 's possible to , once again , protonate your carbonile . that 's going to activate it . this carbon right here becomes more electrophilic , and that means that the chloride anion can function as a nucleophile and attack our electrophile since the chloride anion is n't a great nucleophile on it 's own , so if the chloride anion attacks here , that would push these electrons off onto the oxygen , and we could go ahead and draw what we would form , so we would have our r group , we would have our carbon , we would have this oxygen with two loan pairs of electrons bonded to hydrogen here , so let 's show those electrons in blue . if these electrons in blue kick off onto the oxygen , we could say that those are these electrons . then we could also say that these electrons here in green , on the chloride anion are going to form a bond between the chlorine and the carbon , so we can go ahead and draw in the bond between the chlorine and the carbon like that . we still have our oxygen right here bonded to our sulfur , double-bonded to this oxygen , and then we have our loan pair of electrons here . then we have our chlorine . what all this does is make a much better leaving group than what we started off with so you could think about this as our leaving group . this is a leaving group , and then the next step of the mechanism , and if we go back up here , that 's a much better leaving group than the oh that we started out with . if that 's going to be our leaving group , we could show these electrons in here moving into form our double bond which would kick these electrons off onto our oxygen . let 's go ahead and show what we would make from that . we would now form our r group , our carbon would be double-bonded to an oxygen like this , loan pair of electrons plus one formal charge , and then we 'd have our chlorine over here with all of its electrons too . let 's go ahead and show where those electrons came from so if these electrons move in here , that would reform our carbonile , so i can go ahead and draw that in . then we still have this bond in green , here , just to clarify . then we would have some electrons , and now let 's make a magenta , so these electrons in here are going to kick off onto the oxygen , so we could go ahead and draw that . we would have our oxygen . so we 'd have three loan pairs of electrons . one of them would be the magenta ones , so we could go ahead and do that . that gives our oxygen a negative one formal charge because it 's bonded to the sulfur here , the sulfur 's still double-bonded to this oxygen , and once again , still a loan pair of electrons and a chlorine over here like that . the reason why this is a good leaving group is because we can push these electrons into here , push these electrons off onto here , and we can form sulfur dioxide as a gas . if we go ahead and draw the result of that , we would now have this oxygen double-bonded to this sulfur , double-bonded to this oxygen , so draw in loan pairs of electrons here , and we still have a loan pair of electrons on our sulfur . if these electrons in red here move in to form this bond , we 've now formed sulfur dioxide , and we also formed a chloride anion , so it once again follows some electrons . if these electrons move off onto chlorine , we have the chloride anion , so now we have our chloride anion which could function as a base and in the last step of our mechanism , deprotonate , so take this proton , leave these electrons behind and finally , we formed our acyl chloride . we have our carbon double-bonded to an oxygen with two loan pairs of electrons , and we have our chlorine over here , like that . if we follow those final electrons , these electrons in blue , move off onto your carbonile . you would also form hcl by this , so you form hcl , which is also a gas , so the formation of these gases , the formation of sulfur dioxide and hcl drives the reaction to completion here . that was a long mechanism . once again , you 'll see some slight variations on this , so if i go back up to here , you could show this step , the nucleophile attacking the electrophile , and directly form here if you wanted to , so as your nucleophile attacks your sulfur , these electrons could kick off onto chlorine to form the chloride anion at this step , and at this step , and then you would get this at this point . once again , i talked about the possibility of using this as your intermediate . depending on which textbook you look in , you might see some slight variations on this mechanism . it is a little bit of a long one here . all right , let 's look at two other ways to make an acyl chloride , so starting with a carboxylic acid , you could add phosphorus , a pentachloride , or a phosphorus trichloride , and both of those will give you an acyl chloride as well . the mechanism is pretty similar and also we could think about , instead of phosphorus trichloride , we could add a pbr3 , so a phosphorus tribromide instead of putting a chlorine here . that would put a bromine , and so we could form an acyl bromide this way , and we 'll see a use for this reaction in a later video .
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: here 's a general structure for an acyl chloride , also called an acid chloride , and it 's a carboxylic acid derivative , so we can form them from carboxylic acid , so if we start with a carboxylic acid and add thionyl chloride , we can form our acyl chlorides , and we would also form a sulfur dioxide and hcl in this reaction . let 's look at the structure of thionyl chloride .
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does the mechanism change if you deprotonate the carboxylic acid using pyridine ?
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( light jazz piano music ) we 're at the musee d'orsay and we 're looking at four of over 30 canvases that monet made of rouen cathedral , which is a little more than an hour 's drive north of paris . over two late winters and early springs 1892 and 1893 , he went to the space across from the cathedral and he did the cathedral in different effects of light . so what he did was he had several canvases going at once , each for a different moment of the day and a different effect of light . well , that makes sense . if monet is trying to define this ephemeral quality of light , then as the sun moves , he would need to change canvases . right . he ca n't paint that fast . no . and then he would come back to it day after day , but in also different weather effects , and having his temporary studio across the street allowed him to paint in the rain , early in the morning , etc . there 's a lot of paint on these canvases , and so this is not something that was done quickly . monet was always interested in capturing the fleeting effects of something that he saw , but here it 's become the exact subject of the painting . the irony is that as he 's capturing something that 's fleeting , he takes longer and longer to paint it , and to finish it , not outside , but to finish it in the studio . there 's another irony here which is that if the subject is really about light and the way light constructs form , and i think that really is the subject , he 's picked a pretty potent thing to render that . yes . that is to say , a medieval cathedral which with all of its religious connotations , its historical connotations , and is solid in the extreme , and yet in the rendering by monet these are not such solid forms . no , they really appear very light , almost filigree forms . they lack a sense of heavy three-dimensionality . the subject of a gothic cathedral is divine light itself . so why would he be interested , in a just formal sense , in a gothic cathedral ? and i always thought it had to do with the enormous complexity of the surface . there 's no doubt it 's the complexity of light and shadow on the facade of a cathedral like rouen cathedral that was appealing to him . but i do n't think it 's simply because the gothic church has a fabulous facade , i mean , he 's choosing something very identified with france , the gothic style . there feels to me like there 's something nationalistic here , there feels to me like there 's something poignant here . this is in a sense taking that grand history , taking all of the power that these function as symbolically , and in a sense understanding them through the lens of the late nineteenth century . they are meant to be seen together , and he exhibited them together . they 're very beautiful and one really does get the sense of optical effects of different times of day , the morning mist , the sun coming out , the heat of the afternoon sun . what happens to my eyes as i move across the canvas , is different parts of the cathedral protrude and recede in different ways and different light , and in a sense the physical stone itself becomes really this mutable experience in that the building is shaped and reshaped by the way the light hits it . right . and that the very architecture is transformed , and in a sense it is a triumph of the optical over the physical . which is something very different than the gothic architects would have thought about the church , because what could be seen was really a symbol for what could n't be seen , and in a way , what monet seems to be telling us here in the end of the nineteenth century is what we see is what there is . that there is truth to our experiential . ( jazz piano riffs )
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( light jazz piano music ) we 're at the musee d'orsay and we 're looking at four of over 30 canvases that monet made of rouen cathedral , which is a little more than an hour 's drive north of paris . over two late winters and early springs 1892 and 1893 , he went to the space across from the cathedral and he did the cathedral in different effects of light .
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why do these cathedrals look so blurry ?
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( light jazz piano music ) we 're at the musee d'orsay and we 're looking at four of over 30 canvases that monet made of rouen cathedral , which is a little more than an hour 's drive north of paris . over two late winters and early springs 1892 and 1893 , he went to the space across from the cathedral and he did the cathedral in different effects of light . so what he did was he had several canvases going at once , each for a different moment of the day and a different effect of light . well , that makes sense . if monet is trying to define this ephemeral quality of light , then as the sun moves , he would need to change canvases . right . he ca n't paint that fast . no . and then he would come back to it day after day , but in also different weather effects , and having his temporary studio across the street allowed him to paint in the rain , early in the morning , etc . there 's a lot of paint on these canvases , and so this is not something that was done quickly . monet was always interested in capturing the fleeting effects of something that he saw , but here it 's become the exact subject of the painting . the irony is that as he 's capturing something that 's fleeting , he takes longer and longer to paint it , and to finish it , not outside , but to finish it in the studio . there 's another irony here which is that if the subject is really about light and the way light constructs form , and i think that really is the subject , he 's picked a pretty potent thing to render that . yes . that is to say , a medieval cathedral which with all of its religious connotations , its historical connotations , and is solid in the extreme , and yet in the rendering by monet these are not such solid forms . no , they really appear very light , almost filigree forms . they lack a sense of heavy three-dimensionality . the subject of a gothic cathedral is divine light itself . so why would he be interested , in a just formal sense , in a gothic cathedral ? and i always thought it had to do with the enormous complexity of the surface . there 's no doubt it 's the complexity of light and shadow on the facade of a cathedral like rouen cathedral that was appealing to him . but i do n't think it 's simply because the gothic church has a fabulous facade , i mean , he 's choosing something very identified with france , the gothic style . there feels to me like there 's something nationalistic here , there feels to me like there 's something poignant here . this is in a sense taking that grand history , taking all of the power that these function as symbolically , and in a sense understanding them through the lens of the late nineteenth century . they are meant to be seen together , and he exhibited them together . they 're very beautiful and one really does get the sense of optical effects of different times of day , the morning mist , the sun coming out , the heat of the afternoon sun . what happens to my eyes as i move across the canvas , is different parts of the cathedral protrude and recede in different ways and different light , and in a sense the physical stone itself becomes really this mutable experience in that the building is shaped and reshaped by the way the light hits it . right . and that the very architecture is transformed , and in a sense it is a triumph of the optical over the physical . which is something very different than the gothic architects would have thought about the church , because what could be seen was really a symbol for what could n't be seen , and in a way , what monet seems to be telling us here in the end of the nineteenth century is what we see is what there is . that there is truth to our experiential . ( jazz piano riffs )
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there 's a lot of paint on these canvases , and so this is not something that was done quickly . monet was always interested in capturing the fleeting effects of something that he saw , but here it 's become the exact subject of the painting . the irony is that as he 's capturing something that 's fleeting , he takes longer and longer to paint it , and to finish it , not outside , but to finish it in the studio .
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what is monet 's style of brushstrokes in this painting ?
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the standard question you often get in your algebra class is they will give you this equation and it 'll say identify the conic section and graph it if you can . and the equation they give you wo n't be in the standard form , because if it was you could just kind of pattern match with what i showed in some of the previous videos and you 'd be able to get it . so let 's do a question like and let 's see if we can figure it out . so what i have here is 9x squared plus 4y squared plus 54x minus 8y plus 49 is equal to 0 . and once again , i mean who knows what this is it 's just not in the standard form . and actually one quick clue to tell you what this is you look at the x squared and the y squared terms if there are . if there 's only an x squared term and then there 's just a y and not a y squared term , then you 're probably dealing with a parabola , and we 'll go into that more later . or if it 's the other way around , if it 's just an x term and a y squared term , it 's probably a parabola . but assuming that we 're dealing with a circle , an ellipse , or a hyperbola , there will be an x squared term and a y squared term . if they both kind of have the same number in front of them , that 's a pretty good clue that we 're going to be dealing with a circle . if they both have different numbers , but they 're both positive in front of them , that 's a pretty good clue we 're probably going to be dealing with an ellipse . if one of them has a negative number in front of them and the other one has a positive number , that tells you that we 're probably going to be dealing with a hyperbola . but with that said , i mean that might help you identify things very quickly at this level , but it does n't help you graph it or get into the standard form . so let 's get it in the standard form . and the key to getting it in the standard form is really just completing the square . and i encourage you to re-watch the completing the square video , because that 's all we 're going to do right here to get it into the standard form . so the first thing i like to do to complete the square , and you 're going to have to do it for the x variables and for the y terms , is group the x and y terms . let 's see . the x terms are 9x squared plus 54x . and let 's do the y terms in magenta . so then you have plus 4y squared minus 8y and then you have -- let me do this in a different color -- plus 49 is equal to 0 . and so the easy thing to do when you complete the square , the thing i like to do is , it 's very clear we can factor out a 9 out of both of these numbers , and we can factor out a 4 out of both of those . let 's do that , because that will help us complete the square . so this is the same thing is 9 times x squared plus 9 times 6 is 54 , 6x . i 'm going to add something else here , but i 'll leave it blank for now . plus 4 times y squared minus 2y i 'm probably going to add something here too , so i 'll leave it blank for now . plus 49 is equal to 0 . so what are we going to add here ? we 're going to complete the square . we want to add some number here so that this whole three term expression becomes a perfect square . likewise , we 're going to add some number here , so this three term number expression becomes a perfect square . and of course whatever we add on the side , we 're going to have to multiply it by 9 , because we 're really adding nine times that . and add it on to that side . whatever we add here , we 're going to have to multiply it times 4 and add it on that side . if i put a 1 here , it 's really like as if i had a 4 here , because 1 times 4 is 4 and if i had a 1 here it 's 1 times 9 . so 9 there . let 's do that . when we complete the square , we just take half of this coefficient . this coefficient is 6 , we take half of it is 3 , we square it , we get a 9 . remember it 's an equation , so what you do to one side , you have to do to the other . so if we added a 9 here , we 're actually adding 9 times 9 to the left-hand side of the equation , so we have to add 81 to the right-hand side to make the equation still hold . and you could kind of view it if we go back up here . this is the same thing , just to make that clear as if i added plus 81 right here . of course i would have had to add plus 81 up here . now let 's go to the y terms . you take half of this coefficient is minus 2 , half of that is minus 1 . you square it , you get plus 1 . 1 times 4 , so we 're really adding 4 to the left-hand side of the equation . and just so you understand what i did here . this is equivalent as if i just added a 4 here , and then i later just factored out this 4 . and so what does this become ? this expression is 9 times what ? this is the square of -- you could factor this , but we did it on purpose -- it 's x plus 3 squared and then we have plus 4 times -- what is this right here ? that 's y minus 1 squared . you might want to review factoring of polynomial or completing the square if you found that step a little daunting . and then we have plus 49 is equal to 0 plus 81 plus 84 is equal to 85 . all right , so now we have 9 times plus 3 squared plus 4 times y minus 1 squared . and let 's subtract 49 from both sides . that is equal to -- let 's see if i subtract 50 from 85 i get 35 , so if i subtract 49 , i get 36 . and now we are getting close to the standard form of something , but remember all the standard forms we did except for the circle -- we had a y -- and we know this is n't a circle , because we have these weird coefficients , well not weird but different coefficients in front of these terms . so to get the 1 on the right-hand side let 's divide everything by 36 . if you divide everything by 36 , this term becomes x plus 3 squared over see 9 over 36 is the same thing as 1 over 4 , and then you have plus y minus 1 squared 4 over 36 is the same thing as 1 over 9 and all of that is equal to 1 . and there you go . we have it in the standard form , and you can see our intuition at the beginning the problem was correct . this is indeed an ellipse , and now we can actually graph it . so first of all , actually good place to start , where 's the center of the this ellipse going to be ? it 's going to be x is equal to negative 3 . what x value makes this whole terms 0 ? so it 's going to be x is equal to minus 3 , and y is going to be equal to 1 . what y value makes this term 0 ? y is equal to 1 . that 's our center . so let 's graph that , and then we can draw the ellipse . it 's going to be in the negative quadrant . this is our x-axis and this is our y-axis . and then the center of our ellipse is at minus 3 and positive 1 , so that 's the center . and then , what is the radius in the x direction ? we just take the square root of this , so it 's 2 . so in the x direction we go two to the right . we go two to the left . and in the y direction , what do we do ? well we go up three and down three . the square root of this . let me do that . remember you have to take the square root of both of those . the vertical axis is actually the major radius or the semi-major axis is 3 , because that 's the longer one . and then the 2 is the minor radius , because that 's the shorter one . and now we 're ready to draw this ellipse . i 'll draw it in brown . let me see if i can do this properly . i have a shaky hand . all right , it looks something like that . and there you go . we took this kind of crazy looking thing , and all we did is algebraically manipulate it . we just completed the squares with the x 's and the y terms . and then we divided both sides by this number right here and we got it into the standard form . we said oh this is an ellipse . we have both of these terms , they 're both positive , we 're adding we 're not subtracting , they have different coefficients underneath here . so we 're ready to go over the ellipse , and we realized that the center was at minus 3,1 , and then we just drew the major radius , or the major axis and the minor axis . see you in the next video .
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so to get the 1 on the right-hand side let 's divide everything by 36 . if you divide everything by 36 , this term becomes x plus 3 squared over see 9 over 36 is the same thing as 1 over 4 , and then you have plus y minus 1 squared 4 over 36 is the same thing as 1 over 9 and all of that is equal to 1 . and there you go .
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so ( x+3 ) ^2/4 equals 0 , ( y-1 ) ^2/9=0 , therefore ( x+3 ) ^2/4 + ( y-1 ) ^2/9 = 0 ?
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the standard question you often get in your algebra class is they will give you this equation and it 'll say identify the conic section and graph it if you can . and the equation they give you wo n't be in the standard form , because if it was you could just kind of pattern match with what i showed in some of the previous videos and you 'd be able to get it . so let 's do a question like and let 's see if we can figure it out . so what i have here is 9x squared plus 4y squared plus 54x minus 8y plus 49 is equal to 0 . and once again , i mean who knows what this is it 's just not in the standard form . and actually one quick clue to tell you what this is you look at the x squared and the y squared terms if there are . if there 's only an x squared term and then there 's just a y and not a y squared term , then you 're probably dealing with a parabola , and we 'll go into that more later . or if it 's the other way around , if it 's just an x term and a y squared term , it 's probably a parabola . but assuming that we 're dealing with a circle , an ellipse , or a hyperbola , there will be an x squared term and a y squared term . if they both kind of have the same number in front of them , that 's a pretty good clue that we 're going to be dealing with a circle . if they both have different numbers , but they 're both positive in front of them , that 's a pretty good clue we 're probably going to be dealing with an ellipse . if one of them has a negative number in front of them and the other one has a positive number , that tells you that we 're probably going to be dealing with a hyperbola . but with that said , i mean that might help you identify things very quickly at this level , but it does n't help you graph it or get into the standard form . so let 's get it in the standard form . and the key to getting it in the standard form is really just completing the square . and i encourage you to re-watch the completing the square video , because that 's all we 're going to do right here to get it into the standard form . so the first thing i like to do to complete the square , and you 're going to have to do it for the x variables and for the y terms , is group the x and y terms . let 's see . the x terms are 9x squared plus 54x . and let 's do the y terms in magenta . so then you have plus 4y squared minus 8y and then you have -- let me do this in a different color -- plus 49 is equal to 0 . and so the easy thing to do when you complete the square , the thing i like to do is , it 's very clear we can factor out a 9 out of both of these numbers , and we can factor out a 4 out of both of those . let 's do that , because that will help us complete the square . so this is the same thing is 9 times x squared plus 9 times 6 is 54 , 6x . i 'm going to add something else here , but i 'll leave it blank for now . plus 4 times y squared minus 2y i 'm probably going to add something here too , so i 'll leave it blank for now . plus 49 is equal to 0 . so what are we going to add here ? we 're going to complete the square . we want to add some number here so that this whole three term expression becomes a perfect square . likewise , we 're going to add some number here , so this three term number expression becomes a perfect square . and of course whatever we add on the side , we 're going to have to multiply it by 9 , because we 're really adding nine times that . and add it on to that side . whatever we add here , we 're going to have to multiply it times 4 and add it on that side . if i put a 1 here , it 's really like as if i had a 4 here , because 1 times 4 is 4 and if i had a 1 here it 's 1 times 9 . so 9 there . let 's do that . when we complete the square , we just take half of this coefficient . this coefficient is 6 , we take half of it is 3 , we square it , we get a 9 . remember it 's an equation , so what you do to one side , you have to do to the other . so if we added a 9 here , we 're actually adding 9 times 9 to the left-hand side of the equation , so we have to add 81 to the right-hand side to make the equation still hold . and you could kind of view it if we go back up here . this is the same thing , just to make that clear as if i added plus 81 right here . of course i would have had to add plus 81 up here . now let 's go to the y terms . you take half of this coefficient is minus 2 , half of that is minus 1 . you square it , you get plus 1 . 1 times 4 , so we 're really adding 4 to the left-hand side of the equation . and just so you understand what i did here . this is equivalent as if i just added a 4 here , and then i later just factored out this 4 . and so what does this become ? this expression is 9 times what ? this is the square of -- you could factor this , but we did it on purpose -- it 's x plus 3 squared and then we have plus 4 times -- what is this right here ? that 's y minus 1 squared . you might want to review factoring of polynomial or completing the square if you found that step a little daunting . and then we have plus 49 is equal to 0 plus 81 plus 84 is equal to 85 . all right , so now we have 9 times plus 3 squared plus 4 times y minus 1 squared . and let 's subtract 49 from both sides . that is equal to -- let 's see if i subtract 50 from 85 i get 35 , so if i subtract 49 , i get 36 . and now we are getting close to the standard form of something , but remember all the standard forms we did except for the circle -- we had a y -- and we know this is n't a circle , because we have these weird coefficients , well not weird but different coefficients in front of these terms . so to get the 1 on the right-hand side let 's divide everything by 36 . if you divide everything by 36 , this term becomes x plus 3 squared over see 9 over 36 is the same thing as 1 over 4 , and then you have plus y minus 1 squared 4 over 36 is the same thing as 1 over 9 and all of that is equal to 1 . and there you go . we have it in the standard form , and you can see our intuition at the beginning the problem was correct . this is indeed an ellipse , and now we can actually graph it . so first of all , actually good place to start , where 's the center of the this ellipse going to be ? it 's going to be x is equal to negative 3 . what x value makes this whole terms 0 ? so it 's going to be x is equal to minus 3 , and y is going to be equal to 1 . what y value makes this term 0 ? y is equal to 1 . that 's our center . so let 's graph that , and then we can draw the ellipse . it 's going to be in the negative quadrant . this is our x-axis and this is our y-axis . and then the center of our ellipse is at minus 3 and positive 1 , so that 's the center . and then , what is the radius in the x direction ? we just take the square root of this , so it 's 2 . so in the x direction we go two to the right . we go two to the left . and in the y direction , what do we do ? well we go up three and down three . the square root of this . let me do that . remember you have to take the square root of both of those . the vertical axis is actually the major radius or the semi-major axis is 3 , because that 's the longer one . and then the 2 is the minor radius , because that 's the shorter one . and now we 're ready to draw this ellipse . i 'll draw it in brown . let me see if i can do this properly . i have a shaky hand . all right , it looks something like that . and there you go . we took this kind of crazy looking thing , and all we did is algebraically manipulate it . we just completed the squares with the x 's and the y terms . and then we divided both sides by this number right here and we got it into the standard form . we said oh this is an ellipse . we have both of these terms , they 're both positive , we 're adding we 're not subtracting , they have different coefficients underneath here . so we 're ready to go over the ellipse , and we realized that the center was at minus 3,1 , and then we just drew the major radius , or the major axis and the minor axis . see you in the next video .
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if one of them has a negative number in front of them and the other one has a positive number , that tells you that we 're probably going to be dealing with a hyperbola . but with that said , i mean that might help you identify things very quickly at this level , but it does n't help you graph it or get into the standard form . so let 's get it in the standard form .
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can someone explain to me how you identify the a from the b ?
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the standard question you often get in your algebra class is they will give you this equation and it 'll say identify the conic section and graph it if you can . and the equation they give you wo n't be in the standard form , because if it was you could just kind of pattern match with what i showed in some of the previous videos and you 'd be able to get it . so let 's do a question like and let 's see if we can figure it out . so what i have here is 9x squared plus 4y squared plus 54x minus 8y plus 49 is equal to 0 . and once again , i mean who knows what this is it 's just not in the standard form . and actually one quick clue to tell you what this is you look at the x squared and the y squared terms if there are . if there 's only an x squared term and then there 's just a y and not a y squared term , then you 're probably dealing with a parabola , and we 'll go into that more later . or if it 's the other way around , if it 's just an x term and a y squared term , it 's probably a parabola . but assuming that we 're dealing with a circle , an ellipse , or a hyperbola , there will be an x squared term and a y squared term . if they both kind of have the same number in front of them , that 's a pretty good clue that we 're going to be dealing with a circle . if they both have different numbers , but they 're both positive in front of them , that 's a pretty good clue we 're probably going to be dealing with an ellipse . if one of them has a negative number in front of them and the other one has a positive number , that tells you that we 're probably going to be dealing with a hyperbola . but with that said , i mean that might help you identify things very quickly at this level , but it does n't help you graph it or get into the standard form . so let 's get it in the standard form . and the key to getting it in the standard form is really just completing the square . and i encourage you to re-watch the completing the square video , because that 's all we 're going to do right here to get it into the standard form . so the first thing i like to do to complete the square , and you 're going to have to do it for the x variables and for the y terms , is group the x and y terms . let 's see . the x terms are 9x squared plus 54x . and let 's do the y terms in magenta . so then you have plus 4y squared minus 8y and then you have -- let me do this in a different color -- plus 49 is equal to 0 . and so the easy thing to do when you complete the square , the thing i like to do is , it 's very clear we can factor out a 9 out of both of these numbers , and we can factor out a 4 out of both of those . let 's do that , because that will help us complete the square . so this is the same thing is 9 times x squared plus 9 times 6 is 54 , 6x . i 'm going to add something else here , but i 'll leave it blank for now . plus 4 times y squared minus 2y i 'm probably going to add something here too , so i 'll leave it blank for now . plus 49 is equal to 0 . so what are we going to add here ? we 're going to complete the square . we want to add some number here so that this whole three term expression becomes a perfect square . likewise , we 're going to add some number here , so this three term number expression becomes a perfect square . and of course whatever we add on the side , we 're going to have to multiply it by 9 , because we 're really adding nine times that . and add it on to that side . whatever we add here , we 're going to have to multiply it times 4 and add it on that side . if i put a 1 here , it 's really like as if i had a 4 here , because 1 times 4 is 4 and if i had a 1 here it 's 1 times 9 . so 9 there . let 's do that . when we complete the square , we just take half of this coefficient . this coefficient is 6 , we take half of it is 3 , we square it , we get a 9 . remember it 's an equation , so what you do to one side , you have to do to the other . so if we added a 9 here , we 're actually adding 9 times 9 to the left-hand side of the equation , so we have to add 81 to the right-hand side to make the equation still hold . and you could kind of view it if we go back up here . this is the same thing , just to make that clear as if i added plus 81 right here . of course i would have had to add plus 81 up here . now let 's go to the y terms . you take half of this coefficient is minus 2 , half of that is minus 1 . you square it , you get plus 1 . 1 times 4 , so we 're really adding 4 to the left-hand side of the equation . and just so you understand what i did here . this is equivalent as if i just added a 4 here , and then i later just factored out this 4 . and so what does this become ? this expression is 9 times what ? this is the square of -- you could factor this , but we did it on purpose -- it 's x plus 3 squared and then we have plus 4 times -- what is this right here ? that 's y minus 1 squared . you might want to review factoring of polynomial or completing the square if you found that step a little daunting . and then we have plus 49 is equal to 0 plus 81 plus 84 is equal to 85 . all right , so now we have 9 times plus 3 squared plus 4 times y minus 1 squared . and let 's subtract 49 from both sides . that is equal to -- let 's see if i subtract 50 from 85 i get 35 , so if i subtract 49 , i get 36 . and now we are getting close to the standard form of something , but remember all the standard forms we did except for the circle -- we had a y -- and we know this is n't a circle , because we have these weird coefficients , well not weird but different coefficients in front of these terms . so to get the 1 on the right-hand side let 's divide everything by 36 . if you divide everything by 36 , this term becomes x plus 3 squared over see 9 over 36 is the same thing as 1 over 4 , and then you have plus y minus 1 squared 4 over 36 is the same thing as 1 over 9 and all of that is equal to 1 . and there you go . we have it in the standard form , and you can see our intuition at the beginning the problem was correct . this is indeed an ellipse , and now we can actually graph it . so first of all , actually good place to start , where 's the center of the this ellipse going to be ? it 's going to be x is equal to negative 3 . what x value makes this whole terms 0 ? so it 's going to be x is equal to minus 3 , and y is going to be equal to 1 . what y value makes this term 0 ? y is equal to 1 . that 's our center . so let 's graph that , and then we can draw the ellipse . it 's going to be in the negative quadrant . this is our x-axis and this is our y-axis . and then the center of our ellipse is at minus 3 and positive 1 , so that 's the center . and then , what is the radius in the x direction ? we just take the square root of this , so it 's 2 . so in the x direction we go two to the right . we go two to the left . and in the y direction , what do we do ? well we go up three and down three . the square root of this . let me do that . remember you have to take the square root of both of those . the vertical axis is actually the major radius or the semi-major axis is 3 , because that 's the longer one . and then the 2 is the minor radius , because that 's the shorter one . and now we 're ready to draw this ellipse . i 'll draw it in brown . let me see if i can do this properly . i have a shaky hand . all right , it looks something like that . and there you go . we took this kind of crazy looking thing , and all we did is algebraically manipulate it . we just completed the squares with the x 's and the y terms . and then we divided both sides by this number right here and we got it into the standard form . we said oh this is an ellipse . we have both of these terms , they 're both positive , we 're adding we 're not subtracting , they have different coefficients underneath here . so we 're ready to go over the ellipse , and we realized that the center was at minus 3,1 , and then we just drew the major radius , or the major axis and the minor axis . see you in the next video .
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and then we divided both sides by this number right here and we got it into the standard form . we said oh this is an ellipse . we have both of these terms , they 're both positive , we 're adding we 're not subtracting , they have different coefficients underneath here .
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how exactly , though , does sal know that it is an ellipse ?
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the standard question you often get in your algebra class is they will give you this equation and it 'll say identify the conic section and graph it if you can . and the equation they give you wo n't be in the standard form , because if it was you could just kind of pattern match with what i showed in some of the previous videos and you 'd be able to get it . so let 's do a question like and let 's see if we can figure it out . so what i have here is 9x squared plus 4y squared plus 54x minus 8y plus 49 is equal to 0 . and once again , i mean who knows what this is it 's just not in the standard form . and actually one quick clue to tell you what this is you look at the x squared and the y squared terms if there are . if there 's only an x squared term and then there 's just a y and not a y squared term , then you 're probably dealing with a parabola , and we 'll go into that more later . or if it 's the other way around , if it 's just an x term and a y squared term , it 's probably a parabola . but assuming that we 're dealing with a circle , an ellipse , or a hyperbola , there will be an x squared term and a y squared term . if they both kind of have the same number in front of them , that 's a pretty good clue that we 're going to be dealing with a circle . if they both have different numbers , but they 're both positive in front of them , that 's a pretty good clue we 're probably going to be dealing with an ellipse . if one of them has a negative number in front of them and the other one has a positive number , that tells you that we 're probably going to be dealing with a hyperbola . but with that said , i mean that might help you identify things very quickly at this level , but it does n't help you graph it or get into the standard form . so let 's get it in the standard form . and the key to getting it in the standard form is really just completing the square . and i encourage you to re-watch the completing the square video , because that 's all we 're going to do right here to get it into the standard form . so the first thing i like to do to complete the square , and you 're going to have to do it for the x variables and for the y terms , is group the x and y terms . let 's see . the x terms are 9x squared plus 54x . and let 's do the y terms in magenta . so then you have plus 4y squared minus 8y and then you have -- let me do this in a different color -- plus 49 is equal to 0 . and so the easy thing to do when you complete the square , the thing i like to do is , it 's very clear we can factor out a 9 out of both of these numbers , and we can factor out a 4 out of both of those . let 's do that , because that will help us complete the square . so this is the same thing is 9 times x squared plus 9 times 6 is 54 , 6x . i 'm going to add something else here , but i 'll leave it blank for now . plus 4 times y squared minus 2y i 'm probably going to add something here too , so i 'll leave it blank for now . plus 49 is equal to 0 . so what are we going to add here ? we 're going to complete the square . we want to add some number here so that this whole three term expression becomes a perfect square . likewise , we 're going to add some number here , so this three term number expression becomes a perfect square . and of course whatever we add on the side , we 're going to have to multiply it by 9 , because we 're really adding nine times that . and add it on to that side . whatever we add here , we 're going to have to multiply it times 4 and add it on that side . if i put a 1 here , it 's really like as if i had a 4 here , because 1 times 4 is 4 and if i had a 1 here it 's 1 times 9 . so 9 there . let 's do that . when we complete the square , we just take half of this coefficient . this coefficient is 6 , we take half of it is 3 , we square it , we get a 9 . remember it 's an equation , so what you do to one side , you have to do to the other . so if we added a 9 here , we 're actually adding 9 times 9 to the left-hand side of the equation , so we have to add 81 to the right-hand side to make the equation still hold . and you could kind of view it if we go back up here . this is the same thing , just to make that clear as if i added plus 81 right here . of course i would have had to add plus 81 up here . now let 's go to the y terms . you take half of this coefficient is minus 2 , half of that is minus 1 . you square it , you get plus 1 . 1 times 4 , so we 're really adding 4 to the left-hand side of the equation . and just so you understand what i did here . this is equivalent as if i just added a 4 here , and then i later just factored out this 4 . and so what does this become ? this expression is 9 times what ? this is the square of -- you could factor this , but we did it on purpose -- it 's x plus 3 squared and then we have plus 4 times -- what is this right here ? that 's y minus 1 squared . you might want to review factoring of polynomial or completing the square if you found that step a little daunting . and then we have plus 49 is equal to 0 plus 81 plus 84 is equal to 85 . all right , so now we have 9 times plus 3 squared plus 4 times y minus 1 squared . and let 's subtract 49 from both sides . that is equal to -- let 's see if i subtract 50 from 85 i get 35 , so if i subtract 49 , i get 36 . and now we are getting close to the standard form of something , but remember all the standard forms we did except for the circle -- we had a y -- and we know this is n't a circle , because we have these weird coefficients , well not weird but different coefficients in front of these terms . so to get the 1 on the right-hand side let 's divide everything by 36 . if you divide everything by 36 , this term becomes x plus 3 squared over see 9 over 36 is the same thing as 1 over 4 , and then you have plus y minus 1 squared 4 over 36 is the same thing as 1 over 9 and all of that is equal to 1 . and there you go . we have it in the standard form , and you can see our intuition at the beginning the problem was correct . this is indeed an ellipse , and now we can actually graph it . so first of all , actually good place to start , where 's the center of the this ellipse going to be ? it 's going to be x is equal to negative 3 . what x value makes this whole terms 0 ? so it 's going to be x is equal to minus 3 , and y is going to be equal to 1 . what y value makes this term 0 ? y is equal to 1 . that 's our center . so let 's graph that , and then we can draw the ellipse . it 's going to be in the negative quadrant . this is our x-axis and this is our y-axis . and then the center of our ellipse is at minus 3 and positive 1 , so that 's the center . and then , what is the radius in the x direction ? we just take the square root of this , so it 's 2 . so in the x direction we go two to the right . we go two to the left . and in the y direction , what do we do ? well we go up three and down three . the square root of this . let me do that . remember you have to take the square root of both of those . the vertical axis is actually the major radius or the semi-major axis is 3 , because that 's the longer one . and then the 2 is the minor radius , because that 's the shorter one . and now we 're ready to draw this ellipse . i 'll draw it in brown . let me see if i can do this properly . i have a shaky hand . all right , it looks something like that . and there you go . we took this kind of crazy looking thing , and all we did is algebraically manipulate it . we just completed the squares with the x 's and the y terms . and then we divided both sides by this number right here and we got it into the standard form . we said oh this is an ellipse . we have both of these terms , they 're both positive , we 're adding we 're not subtracting , they have different coefficients underneath here . so we 're ready to go over the ellipse , and we realized that the center was at minus 3,1 , and then we just drew the major radius , or the major axis and the minor axis . see you in the next video .
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and then we divided both sides by this number right here and we got it into the standard form . we said oh this is an ellipse . we have both of these terms , they 're both positive , we 're adding we 're not subtracting , they have different coefficients underneath here .
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how do i know if an equation is representing an ellipse but not a circle ?
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the standard question you often get in your algebra class is they will give you this equation and it 'll say identify the conic section and graph it if you can . and the equation they give you wo n't be in the standard form , because if it was you could just kind of pattern match with what i showed in some of the previous videos and you 'd be able to get it . so let 's do a question like and let 's see if we can figure it out . so what i have here is 9x squared plus 4y squared plus 54x minus 8y plus 49 is equal to 0 . and once again , i mean who knows what this is it 's just not in the standard form . and actually one quick clue to tell you what this is you look at the x squared and the y squared terms if there are . if there 's only an x squared term and then there 's just a y and not a y squared term , then you 're probably dealing with a parabola , and we 'll go into that more later . or if it 's the other way around , if it 's just an x term and a y squared term , it 's probably a parabola . but assuming that we 're dealing with a circle , an ellipse , or a hyperbola , there will be an x squared term and a y squared term . if they both kind of have the same number in front of them , that 's a pretty good clue that we 're going to be dealing with a circle . if they both have different numbers , but they 're both positive in front of them , that 's a pretty good clue we 're probably going to be dealing with an ellipse . if one of them has a negative number in front of them and the other one has a positive number , that tells you that we 're probably going to be dealing with a hyperbola . but with that said , i mean that might help you identify things very quickly at this level , but it does n't help you graph it or get into the standard form . so let 's get it in the standard form . and the key to getting it in the standard form is really just completing the square . and i encourage you to re-watch the completing the square video , because that 's all we 're going to do right here to get it into the standard form . so the first thing i like to do to complete the square , and you 're going to have to do it for the x variables and for the y terms , is group the x and y terms . let 's see . the x terms are 9x squared plus 54x . and let 's do the y terms in magenta . so then you have plus 4y squared minus 8y and then you have -- let me do this in a different color -- plus 49 is equal to 0 . and so the easy thing to do when you complete the square , the thing i like to do is , it 's very clear we can factor out a 9 out of both of these numbers , and we can factor out a 4 out of both of those . let 's do that , because that will help us complete the square . so this is the same thing is 9 times x squared plus 9 times 6 is 54 , 6x . i 'm going to add something else here , but i 'll leave it blank for now . plus 4 times y squared minus 2y i 'm probably going to add something here too , so i 'll leave it blank for now . plus 49 is equal to 0 . so what are we going to add here ? we 're going to complete the square . we want to add some number here so that this whole three term expression becomes a perfect square . likewise , we 're going to add some number here , so this three term number expression becomes a perfect square . and of course whatever we add on the side , we 're going to have to multiply it by 9 , because we 're really adding nine times that . and add it on to that side . whatever we add here , we 're going to have to multiply it times 4 and add it on that side . if i put a 1 here , it 's really like as if i had a 4 here , because 1 times 4 is 4 and if i had a 1 here it 's 1 times 9 . so 9 there . let 's do that . when we complete the square , we just take half of this coefficient . this coefficient is 6 , we take half of it is 3 , we square it , we get a 9 . remember it 's an equation , so what you do to one side , you have to do to the other . so if we added a 9 here , we 're actually adding 9 times 9 to the left-hand side of the equation , so we have to add 81 to the right-hand side to make the equation still hold . and you could kind of view it if we go back up here . this is the same thing , just to make that clear as if i added plus 81 right here . of course i would have had to add plus 81 up here . now let 's go to the y terms . you take half of this coefficient is minus 2 , half of that is minus 1 . you square it , you get plus 1 . 1 times 4 , so we 're really adding 4 to the left-hand side of the equation . and just so you understand what i did here . this is equivalent as if i just added a 4 here , and then i later just factored out this 4 . and so what does this become ? this expression is 9 times what ? this is the square of -- you could factor this , but we did it on purpose -- it 's x plus 3 squared and then we have plus 4 times -- what is this right here ? that 's y minus 1 squared . you might want to review factoring of polynomial or completing the square if you found that step a little daunting . and then we have plus 49 is equal to 0 plus 81 plus 84 is equal to 85 . all right , so now we have 9 times plus 3 squared plus 4 times y minus 1 squared . and let 's subtract 49 from both sides . that is equal to -- let 's see if i subtract 50 from 85 i get 35 , so if i subtract 49 , i get 36 . and now we are getting close to the standard form of something , but remember all the standard forms we did except for the circle -- we had a y -- and we know this is n't a circle , because we have these weird coefficients , well not weird but different coefficients in front of these terms . so to get the 1 on the right-hand side let 's divide everything by 36 . if you divide everything by 36 , this term becomes x plus 3 squared over see 9 over 36 is the same thing as 1 over 4 , and then you have plus y minus 1 squared 4 over 36 is the same thing as 1 over 9 and all of that is equal to 1 . and there you go . we have it in the standard form , and you can see our intuition at the beginning the problem was correct . this is indeed an ellipse , and now we can actually graph it . so first of all , actually good place to start , where 's the center of the this ellipse going to be ? it 's going to be x is equal to negative 3 . what x value makes this whole terms 0 ? so it 's going to be x is equal to minus 3 , and y is going to be equal to 1 . what y value makes this term 0 ? y is equal to 1 . that 's our center . so let 's graph that , and then we can draw the ellipse . it 's going to be in the negative quadrant . this is our x-axis and this is our y-axis . and then the center of our ellipse is at minus 3 and positive 1 , so that 's the center . and then , what is the radius in the x direction ? we just take the square root of this , so it 's 2 . so in the x direction we go two to the right . we go two to the left . and in the y direction , what do we do ? well we go up three and down three . the square root of this . let me do that . remember you have to take the square root of both of those . the vertical axis is actually the major radius or the semi-major axis is 3 , because that 's the longer one . and then the 2 is the minor radius , because that 's the shorter one . and now we 're ready to draw this ellipse . i 'll draw it in brown . let me see if i can do this properly . i have a shaky hand . all right , it looks something like that . and there you go . we took this kind of crazy looking thing , and all we did is algebraically manipulate it . we just completed the squares with the x 's and the y terms . and then we divided both sides by this number right here and we got it into the standard form . we said oh this is an ellipse . we have both of these terms , they 're both positive , we 're adding we 're not subtracting , they have different coefficients underneath here . so we 're ready to go over the ellipse , and we realized that the center was at minus 3,1 , and then we just drew the major radius , or the major axis and the minor axis . see you in the next video .
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well we go up three and down three . the square root of this . let me do that .
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should you take only the principal square root of the numbers at the end ?
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the standard question you often get in your algebra class is they will give you this equation and it 'll say identify the conic section and graph it if you can . and the equation they give you wo n't be in the standard form , because if it was you could just kind of pattern match with what i showed in some of the previous videos and you 'd be able to get it . so let 's do a question like and let 's see if we can figure it out . so what i have here is 9x squared plus 4y squared plus 54x minus 8y plus 49 is equal to 0 . and once again , i mean who knows what this is it 's just not in the standard form . and actually one quick clue to tell you what this is you look at the x squared and the y squared terms if there are . if there 's only an x squared term and then there 's just a y and not a y squared term , then you 're probably dealing with a parabola , and we 'll go into that more later . or if it 's the other way around , if it 's just an x term and a y squared term , it 's probably a parabola . but assuming that we 're dealing with a circle , an ellipse , or a hyperbola , there will be an x squared term and a y squared term . if they both kind of have the same number in front of them , that 's a pretty good clue that we 're going to be dealing with a circle . if they both have different numbers , but they 're both positive in front of them , that 's a pretty good clue we 're probably going to be dealing with an ellipse . if one of them has a negative number in front of them and the other one has a positive number , that tells you that we 're probably going to be dealing with a hyperbola . but with that said , i mean that might help you identify things very quickly at this level , but it does n't help you graph it or get into the standard form . so let 's get it in the standard form . and the key to getting it in the standard form is really just completing the square . and i encourage you to re-watch the completing the square video , because that 's all we 're going to do right here to get it into the standard form . so the first thing i like to do to complete the square , and you 're going to have to do it for the x variables and for the y terms , is group the x and y terms . let 's see . the x terms are 9x squared plus 54x . and let 's do the y terms in magenta . so then you have plus 4y squared minus 8y and then you have -- let me do this in a different color -- plus 49 is equal to 0 . and so the easy thing to do when you complete the square , the thing i like to do is , it 's very clear we can factor out a 9 out of both of these numbers , and we can factor out a 4 out of both of those . let 's do that , because that will help us complete the square . so this is the same thing is 9 times x squared plus 9 times 6 is 54 , 6x . i 'm going to add something else here , but i 'll leave it blank for now . plus 4 times y squared minus 2y i 'm probably going to add something here too , so i 'll leave it blank for now . plus 49 is equal to 0 . so what are we going to add here ? we 're going to complete the square . we want to add some number here so that this whole three term expression becomes a perfect square . likewise , we 're going to add some number here , so this three term number expression becomes a perfect square . and of course whatever we add on the side , we 're going to have to multiply it by 9 , because we 're really adding nine times that . and add it on to that side . whatever we add here , we 're going to have to multiply it times 4 and add it on that side . if i put a 1 here , it 's really like as if i had a 4 here , because 1 times 4 is 4 and if i had a 1 here it 's 1 times 9 . so 9 there . let 's do that . when we complete the square , we just take half of this coefficient . this coefficient is 6 , we take half of it is 3 , we square it , we get a 9 . remember it 's an equation , so what you do to one side , you have to do to the other . so if we added a 9 here , we 're actually adding 9 times 9 to the left-hand side of the equation , so we have to add 81 to the right-hand side to make the equation still hold . and you could kind of view it if we go back up here . this is the same thing , just to make that clear as if i added plus 81 right here . of course i would have had to add plus 81 up here . now let 's go to the y terms . you take half of this coefficient is minus 2 , half of that is minus 1 . you square it , you get plus 1 . 1 times 4 , so we 're really adding 4 to the left-hand side of the equation . and just so you understand what i did here . this is equivalent as if i just added a 4 here , and then i later just factored out this 4 . and so what does this become ? this expression is 9 times what ? this is the square of -- you could factor this , but we did it on purpose -- it 's x plus 3 squared and then we have plus 4 times -- what is this right here ? that 's y minus 1 squared . you might want to review factoring of polynomial or completing the square if you found that step a little daunting . and then we have plus 49 is equal to 0 plus 81 plus 84 is equal to 85 . all right , so now we have 9 times plus 3 squared plus 4 times y minus 1 squared . and let 's subtract 49 from both sides . that is equal to -- let 's see if i subtract 50 from 85 i get 35 , so if i subtract 49 , i get 36 . and now we are getting close to the standard form of something , but remember all the standard forms we did except for the circle -- we had a y -- and we know this is n't a circle , because we have these weird coefficients , well not weird but different coefficients in front of these terms . so to get the 1 on the right-hand side let 's divide everything by 36 . if you divide everything by 36 , this term becomes x plus 3 squared over see 9 over 36 is the same thing as 1 over 4 , and then you have plus y minus 1 squared 4 over 36 is the same thing as 1 over 9 and all of that is equal to 1 . and there you go . we have it in the standard form , and you can see our intuition at the beginning the problem was correct . this is indeed an ellipse , and now we can actually graph it . so first of all , actually good place to start , where 's the center of the this ellipse going to be ? it 's going to be x is equal to negative 3 . what x value makes this whole terms 0 ? so it 's going to be x is equal to minus 3 , and y is going to be equal to 1 . what y value makes this term 0 ? y is equal to 1 . that 's our center . so let 's graph that , and then we can draw the ellipse . it 's going to be in the negative quadrant . this is our x-axis and this is our y-axis . and then the center of our ellipse is at minus 3 and positive 1 , so that 's the center . and then , what is the radius in the x direction ? we just take the square root of this , so it 's 2 . so in the x direction we go two to the right . we go two to the left . and in the y direction , what do we do ? well we go up three and down three . the square root of this . let me do that . remember you have to take the square root of both of those . the vertical axis is actually the major radius or the semi-major axis is 3 , because that 's the longer one . and then the 2 is the minor radius , because that 's the shorter one . and now we 're ready to draw this ellipse . i 'll draw it in brown . let me see if i can do this properly . i have a shaky hand . all right , it looks something like that . and there you go . we took this kind of crazy looking thing , and all we did is algebraically manipulate it . we just completed the squares with the x 's and the y terms . and then we divided both sides by this number right here and we got it into the standard form . we said oh this is an ellipse . we have both of these terms , they 're both positive , we 're adding we 're not subtracting , they have different coefficients underneath here . so we 're ready to go over the ellipse , and we realized that the center was at minus 3,1 , and then we just drew the major radius , or the major axis and the minor axis . see you in the next video .
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well we go up three and down three . the square root of this . let me do that .
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or should you also take the negative square root ?
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the standard question you often get in your algebra class is they will give you this equation and it 'll say identify the conic section and graph it if you can . and the equation they give you wo n't be in the standard form , because if it was you could just kind of pattern match with what i showed in some of the previous videos and you 'd be able to get it . so let 's do a question like and let 's see if we can figure it out . so what i have here is 9x squared plus 4y squared plus 54x minus 8y plus 49 is equal to 0 . and once again , i mean who knows what this is it 's just not in the standard form . and actually one quick clue to tell you what this is you look at the x squared and the y squared terms if there are . if there 's only an x squared term and then there 's just a y and not a y squared term , then you 're probably dealing with a parabola , and we 'll go into that more later . or if it 's the other way around , if it 's just an x term and a y squared term , it 's probably a parabola . but assuming that we 're dealing with a circle , an ellipse , or a hyperbola , there will be an x squared term and a y squared term . if they both kind of have the same number in front of them , that 's a pretty good clue that we 're going to be dealing with a circle . if they both have different numbers , but they 're both positive in front of them , that 's a pretty good clue we 're probably going to be dealing with an ellipse . if one of them has a negative number in front of them and the other one has a positive number , that tells you that we 're probably going to be dealing with a hyperbola . but with that said , i mean that might help you identify things very quickly at this level , but it does n't help you graph it or get into the standard form . so let 's get it in the standard form . and the key to getting it in the standard form is really just completing the square . and i encourage you to re-watch the completing the square video , because that 's all we 're going to do right here to get it into the standard form . so the first thing i like to do to complete the square , and you 're going to have to do it for the x variables and for the y terms , is group the x and y terms . let 's see . the x terms are 9x squared plus 54x . and let 's do the y terms in magenta . so then you have plus 4y squared minus 8y and then you have -- let me do this in a different color -- plus 49 is equal to 0 . and so the easy thing to do when you complete the square , the thing i like to do is , it 's very clear we can factor out a 9 out of both of these numbers , and we can factor out a 4 out of both of those . let 's do that , because that will help us complete the square . so this is the same thing is 9 times x squared plus 9 times 6 is 54 , 6x . i 'm going to add something else here , but i 'll leave it blank for now . plus 4 times y squared minus 2y i 'm probably going to add something here too , so i 'll leave it blank for now . plus 49 is equal to 0 . so what are we going to add here ? we 're going to complete the square . we want to add some number here so that this whole three term expression becomes a perfect square . likewise , we 're going to add some number here , so this three term number expression becomes a perfect square . and of course whatever we add on the side , we 're going to have to multiply it by 9 , because we 're really adding nine times that . and add it on to that side . whatever we add here , we 're going to have to multiply it times 4 and add it on that side . if i put a 1 here , it 's really like as if i had a 4 here , because 1 times 4 is 4 and if i had a 1 here it 's 1 times 9 . so 9 there . let 's do that . when we complete the square , we just take half of this coefficient . this coefficient is 6 , we take half of it is 3 , we square it , we get a 9 . remember it 's an equation , so what you do to one side , you have to do to the other . so if we added a 9 here , we 're actually adding 9 times 9 to the left-hand side of the equation , so we have to add 81 to the right-hand side to make the equation still hold . and you could kind of view it if we go back up here . this is the same thing , just to make that clear as if i added plus 81 right here . of course i would have had to add plus 81 up here . now let 's go to the y terms . you take half of this coefficient is minus 2 , half of that is minus 1 . you square it , you get plus 1 . 1 times 4 , so we 're really adding 4 to the left-hand side of the equation . and just so you understand what i did here . this is equivalent as if i just added a 4 here , and then i later just factored out this 4 . and so what does this become ? this expression is 9 times what ? this is the square of -- you could factor this , but we did it on purpose -- it 's x plus 3 squared and then we have plus 4 times -- what is this right here ? that 's y minus 1 squared . you might want to review factoring of polynomial or completing the square if you found that step a little daunting . and then we have plus 49 is equal to 0 plus 81 plus 84 is equal to 85 . all right , so now we have 9 times plus 3 squared plus 4 times y minus 1 squared . and let 's subtract 49 from both sides . that is equal to -- let 's see if i subtract 50 from 85 i get 35 , so if i subtract 49 , i get 36 . and now we are getting close to the standard form of something , but remember all the standard forms we did except for the circle -- we had a y -- and we know this is n't a circle , because we have these weird coefficients , well not weird but different coefficients in front of these terms . so to get the 1 on the right-hand side let 's divide everything by 36 . if you divide everything by 36 , this term becomes x plus 3 squared over see 9 over 36 is the same thing as 1 over 4 , and then you have plus y minus 1 squared 4 over 36 is the same thing as 1 over 9 and all of that is equal to 1 . and there you go . we have it in the standard form , and you can see our intuition at the beginning the problem was correct . this is indeed an ellipse , and now we can actually graph it . so first of all , actually good place to start , where 's the center of the this ellipse going to be ? it 's going to be x is equal to negative 3 . what x value makes this whole terms 0 ? so it 's going to be x is equal to minus 3 , and y is going to be equal to 1 . what y value makes this term 0 ? y is equal to 1 . that 's our center . so let 's graph that , and then we can draw the ellipse . it 's going to be in the negative quadrant . this is our x-axis and this is our y-axis . and then the center of our ellipse is at minus 3 and positive 1 , so that 's the center . and then , what is the radius in the x direction ? we just take the square root of this , so it 's 2 . so in the x direction we go two to the right . we go two to the left . and in the y direction , what do we do ? well we go up three and down three . the square root of this . let me do that . remember you have to take the square root of both of those . the vertical axis is actually the major radius or the semi-major axis is 3 , because that 's the longer one . and then the 2 is the minor radius , because that 's the shorter one . and now we 're ready to draw this ellipse . i 'll draw it in brown . let me see if i can do this properly . i have a shaky hand . all right , it looks something like that . and there you go . we took this kind of crazy looking thing , and all we did is algebraically manipulate it . we just completed the squares with the x 's and the y terms . and then we divided both sides by this number right here and we got it into the standard form . we said oh this is an ellipse . we have both of these terms , they 're both positive , we 're adding we 're not subtracting , they have different coefficients underneath here . so we 're ready to go over the ellipse , and we realized that the center was at minus 3,1 , and then we just drew the major radius , or the major axis and the minor axis . see you in the next video .
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this is the same thing , just to make that clear as if i added plus 81 right here . of course i would have had to add plus 81 up here . now let 's go to the y terms .
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how would you find the domain and range of the graph ?
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the standard question you often get in your algebra class is they will give you this equation and it 'll say identify the conic section and graph it if you can . and the equation they give you wo n't be in the standard form , because if it was you could just kind of pattern match with what i showed in some of the previous videos and you 'd be able to get it . so let 's do a question like and let 's see if we can figure it out . so what i have here is 9x squared plus 4y squared plus 54x minus 8y plus 49 is equal to 0 . and once again , i mean who knows what this is it 's just not in the standard form . and actually one quick clue to tell you what this is you look at the x squared and the y squared terms if there are . if there 's only an x squared term and then there 's just a y and not a y squared term , then you 're probably dealing with a parabola , and we 'll go into that more later . or if it 's the other way around , if it 's just an x term and a y squared term , it 's probably a parabola . but assuming that we 're dealing with a circle , an ellipse , or a hyperbola , there will be an x squared term and a y squared term . if they both kind of have the same number in front of them , that 's a pretty good clue that we 're going to be dealing with a circle . if they both have different numbers , but they 're both positive in front of them , that 's a pretty good clue we 're probably going to be dealing with an ellipse . if one of them has a negative number in front of them and the other one has a positive number , that tells you that we 're probably going to be dealing with a hyperbola . but with that said , i mean that might help you identify things very quickly at this level , but it does n't help you graph it or get into the standard form . so let 's get it in the standard form . and the key to getting it in the standard form is really just completing the square . and i encourage you to re-watch the completing the square video , because that 's all we 're going to do right here to get it into the standard form . so the first thing i like to do to complete the square , and you 're going to have to do it for the x variables and for the y terms , is group the x and y terms . let 's see . the x terms are 9x squared plus 54x . and let 's do the y terms in magenta . so then you have plus 4y squared minus 8y and then you have -- let me do this in a different color -- plus 49 is equal to 0 . and so the easy thing to do when you complete the square , the thing i like to do is , it 's very clear we can factor out a 9 out of both of these numbers , and we can factor out a 4 out of both of those . let 's do that , because that will help us complete the square . so this is the same thing is 9 times x squared plus 9 times 6 is 54 , 6x . i 'm going to add something else here , but i 'll leave it blank for now . plus 4 times y squared minus 2y i 'm probably going to add something here too , so i 'll leave it blank for now . plus 49 is equal to 0 . so what are we going to add here ? we 're going to complete the square . we want to add some number here so that this whole three term expression becomes a perfect square . likewise , we 're going to add some number here , so this three term number expression becomes a perfect square . and of course whatever we add on the side , we 're going to have to multiply it by 9 , because we 're really adding nine times that . and add it on to that side . whatever we add here , we 're going to have to multiply it times 4 and add it on that side . if i put a 1 here , it 's really like as if i had a 4 here , because 1 times 4 is 4 and if i had a 1 here it 's 1 times 9 . so 9 there . let 's do that . when we complete the square , we just take half of this coefficient . this coefficient is 6 , we take half of it is 3 , we square it , we get a 9 . remember it 's an equation , so what you do to one side , you have to do to the other . so if we added a 9 here , we 're actually adding 9 times 9 to the left-hand side of the equation , so we have to add 81 to the right-hand side to make the equation still hold . and you could kind of view it if we go back up here . this is the same thing , just to make that clear as if i added plus 81 right here . of course i would have had to add plus 81 up here . now let 's go to the y terms . you take half of this coefficient is minus 2 , half of that is minus 1 . you square it , you get plus 1 . 1 times 4 , so we 're really adding 4 to the left-hand side of the equation . and just so you understand what i did here . this is equivalent as if i just added a 4 here , and then i later just factored out this 4 . and so what does this become ? this expression is 9 times what ? this is the square of -- you could factor this , but we did it on purpose -- it 's x plus 3 squared and then we have plus 4 times -- what is this right here ? that 's y minus 1 squared . you might want to review factoring of polynomial or completing the square if you found that step a little daunting . and then we have plus 49 is equal to 0 plus 81 plus 84 is equal to 85 . all right , so now we have 9 times plus 3 squared plus 4 times y minus 1 squared . and let 's subtract 49 from both sides . that is equal to -- let 's see if i subtract 50 from 85 i get 35 , so if i subtract 49 , i get 36 . and now we are getting close to the standard form of something , but remember all the standard forms we did except for the circle -- we had a y -- and we know this is n't a circle , because we have these weird coefficients , well not weird but different coefficients in front of these terms . so to get the 1 on the right-hand side let 's divide everything by 36 . if you divide everything by 36 , this term becomes x plus 3 squared over see 9 over 36 is the same thing as 1 over 4 , and then you have plus y minus 1 squared 4 over 36 is the same thing as 1 over 9 and all of that is equal to 1 . and there you go . we have it in the standard form , and you can see our intuition at the beginning the problem was correct . this is indeed an ellipse , and now we can actually graph it . so first of all , actually good place to start , where 's the center of the this ellipse going to be ? it 's going to be x is equal to negative 3 . what x value makes this whole terms 0 ? so it 's going to be x is equal to minus 3 , and y is going to be equal to 1 . what y value makes this term 0 ? y is equal to 1 . that 's our center . so let 's graph that , and then we can draw the ellipse . it 's going to be in the negative quadrant . this is our x-axis and this is our y-axis . and then the center of our ellipse is at minus 3 and positive 1 , so that 's the center . and then , what is the radius in the x direction ? we just take the square root of this , so it 's 2 . so in the x direction we go two to the right . we go two to the left . and in the y direction , what do we do ? well we go up three and down three . the square root of this . let me do that . remember you have to take the square root of both of those . the vertical axis is actually the major radius or the semi-major axis is 3 , because that 's the longer one . and then the 2 is the minor radius , because that 's the shorter one . and now we 're ready to draw this ellipse . i 'll draw it in brown . let me see if i can do this properly . i have a shaky hand . all right , it looks something like that . and there you go . we took this kind of crazy looking thing , and all we did is algebraically manipulate it . we just completed the squares with the x 's and the y terms . and then we divided both sides by this number right here and we got it into the standard form . we said oh this is an ellipse . we have both of these terms , they 're both positive , we 're adding we 're not subtracting , they have different coefficients underneath here . so we 're ready to go over the ellipse , and we realized that the center was at minus 3,1 , and then we just drew the major radius , or the major axis and the minor axis . see you in the next video .
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so let 's do a question like and let 's see if we can figure it out . so what i have here is 9x squared plus 4y squared plus 54x minus 8y plus 49 is equal to 0 . and once again , i mean who knows what this is it 's just not in the standard form .
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why does n't sal factor out 9x and 4y ?
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the standard question you often get in your algebra class is they will give you this equation and it 'll say identify the conic section and graph it if you can . and the equation they give you wo n't be in the standard form , because if it was you could just kind of pattern match with what i showed in some of the previous videos and you 'd be able to get it . so let 's do a question like and let 's see if we can figure it out . so what i have here is 9x squared plus 4y squared plus 54x minus 8y plus 49 is equal to 0 . and once again , i mean who knows what this is it 's just not in the standard form . and actually one quick clue to tell you what this is you look at the x squared and the y squared terms if there are . if there 's only an x squared term and then there 's just a y and not a y squared term , then you 're probably dealing with a parabola , and we 'll go into that more later . or if it 's the other way around , if it 's just an x term and a y squared term , it 's probably a parabola . but assuming that we 're dealing with a circle , an ellipse , or a hyperbola , there will be an x squared term and a y squared term . if they both kind of have the same number in front of them , that 's a pretty good clue that we 're going to be dealing with a circle . if they both have different numbers , but they 're both positive in front of them , that 's a pretty good clue we 're probably going to be dealing with an ellipse . if one of them has a negative number in front of them and the other one has a positive number , that tells you that we 're probably going to be dealing with a hyperbola . but with that said , i mean that might help you identify things very quickly at this level , but it does n't help you graph it or get into the standard form . so let 's get it in the standard form . and the key to getting it in the standard form is really just completing the square . and i encourage you to re-watch the completing the square video , because that 's all we 're going to do right here to get it into the standard form . so the first thing i like to do to complete the square , and you 're going to have to do it for the x variables and for the y terms , is group the x and y terms . let 's see . the x terms are 9x squared plus 54x . and let 's do the y terms in magenta . so then you have plus 4y squared minus 8y and then you have -- let me do this in a different color -- plus 49 is equal to 0 . and so the easy thing to do when you complete the square , the thing i like to do is , it 's very clear we can factor out a 9 out of both of these numbers , and we can factor out a 4 out of both of those . let 's do that , because that will help us complete the square . so this is the same thing is 9 times x squared plus 9 times 6 is 54 , 6x . i 'm going to add something else here , but i 'll leave it blank for now . plus 4 times y squared minus 2y i 'm probably going to add something here too , so i 'll leave it blank for now . plus 49 is equal to 0 . so what are we going to add here ? we 're going to complete the square . we want to add some number here so that this whole three term expression becomes a perfect square . likewise , we 're going to add some number here , so this three term number expression becomes a perfect square . and of course whatever we add on the side , we 're going to have to multiply it by 9 , because we 're really adding nine times that . and add it on to that side . whatever we add here , we 're going to have to multiply it times 4 and add it on that side . if i put a 1 here , it 's really like as if i had a 4 here , because 1 times 4 is 4 and if i had a 1 here it 's 1 times 9 . so 9 there . let 's do that . when we complete the square , we just take half of this coefficient . this coefficient is 6 , we take half of it is 3 , we square it , we get a 9 . remember it 's an equation , so what you do to one side , you have to do to the other . so if we added a 9 here , we 're actually adding 9 times 9 to the left-hand side of the equation , so we have to add 81 to the right-hand side to make the equation still hold . and you could kind of view it if we go back up here . this is the same thing , just to make that clear as if i added plus 81 right here . of course i would have had to add plus 81 up here . now let 's go to the y terms . you take half of this coefficient is minus 2 , half of that is minus 1 . you square it , you get plus 1 . 1 times 4 , so we 're really adding 4 to the left-hand side of the equation . and just so you understand what i did here . this is equivalent as if i just added a 4 here , and then i later just factored out this 4 . and so what does this become ? this expression is 9 times what ? this is the square of -- you could factor this , but we did it on purpose -- it 's x plus 3 squared and then we have plus 4 times -- what is this right here ? that 's y minus 1 squared . you might want to review factoring of polynomial or completing the square if you found that step a little daunting . and then we have plus 49 is equal to 0 plus 81 plus 84 is equal to 85 . all right , so now we have 9 times plus 3 squared plus 4 times y minus 1 squared . and let 's subtract 49 from both sides . that is equal to -- let 's see if i subtract 50 from 85 i get 35 , so if i subtract 49 , i get 36 . and now we are getting close to the standard form of something , but remember all the standard forms we did except for the circle -- we had a y -- and we know this is n't a circle , because we have these weird coefficients , well not weird but different coefficients in front of these terms . so to get the 1 on the right-hand side let 's divide everything by 36 . if you divide everything by 36 , this term becomes x plus 3 squared over see 9 over 36 is the same thing as 1 over 4 , and then you have plus y minus 1 squared 4 over 36 is the same thing as 1 over 9 and all of that is equal to 1 . and there you go . we have it in the standard form , and you can see our intuition at the beginning the problem was correct . this is indeed an ellipse , and now we can actually graph it . so first of all , actually good place to start , where 's the center of the this ellipse going to be ? it 's going to be x is equal to negative 3 . what x value makes this whole terms 0 ? so it 's going to be x is equal to minus 3 , and y is going to be equal to 1 . what y value makes this term 0 ? y is equal to 1 . that 's our center . so let 's graph that , and then we can draw the ellipse . it 's going to be in the negative quadrant . this is our x-axis and this is our y-axis . and then the center of our ellipse is at minus 3 and positive 1 , so that 's the center . and then , what is the radius in the x direction ? we just take the square root of this , so it 's 2 . so in the x direction we go two to the right . we go two to the left . and in the y direction , what do we do ? well we go up three and down three . the square root of this . let me do that . remember you have to take the square root of both of those . the vertical axis is actually the major radius or the semi-major axis is 3 , because that 's the longer one . and then the 2 is the minor radius , because that 's the shorter one . and now we 're ready to draw this ellipse . i 'll draw it in brown . let me see if i can do this properly . i have a shaky hand . all right , it looks something like that . and there you go . we took this kind of crazy looking thing , and all we did is algebraically manipulate it . we just completed the squares with the x 's and the y terms . and then we divided both sides by this number right here and we got it into the standard form . we said oh this is an ellipse . we have both of these terms , they 're both positive , we 're adding we 're not subtracting , they have different coefficients underneath here . so we 're ready to go over the ellipse , and we realized that the center was at minus 3,1 , and then we just drew the major radius , or the major axis and the minor axis . see you in the next video .
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the standard question you often get in your algebra class is they will give you this equation and it 'll say identify the conic section and graph it if you can . and the equation they give you wo n't be in the standard form , because if it was you could just kind of pattern match with what i showed in some of the previous videos and you 'd be able to get it .
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at what time does sal explain the way to identify the conic section based on the equation ?
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the standard question you often get in your algebra class is they will give you this equation and it 'll say identify the conic section and graph it if you can . and the equation they give you wo n't be in the standard form , because if it was you could just kind of pattern match with what i showed in some of the previous videos and you 'd be able to get it . so let 's do a question like and let 's see if we can figure it out . so what i have here is 9x squared plus 4y squared plus 54x minus 8y plus 49 is equal to 0 . and once again , i mean who knows what this is it 's just not in the standard form . and actually one quick clue to tell you what this is you look at the x squared and the y squared terms if there are . if there 's only an x squared term and then there 's just a y and not a y squared term , then you 're probably dealing with a parabola , and we 'll go into that more later . or if it 's the other way around , if it 's just an x term and a y squared term , it 's probably a parabola . but assuming that we 're dealing with a circle , an ellipse , or a hyperbola , there will be an x squared term and a y squared term . if they both kind of have the same number in front of them , that 's a pretty good clue that we 're going to be dealing with a circle . if they both have different numbers , but they 're both positive in front of them , that 's a pretty good clue we 're probably going to be dealing with an ellipse . if one of them has a negative number in front of them and the other one has a positive number , that tells you that we 're probably going to be dealing with a hyperbola . but with that said , i mean that might help you identify things very quickly at this level , but it does n't help you graph it or get into the standard form . so let 's get it in the standard form . and the key to getting it in the standard form is really just completing the square . and i encourage you to re-watch the completing the square video , because that 's all we 're going to do right here to get it into the standard form . so the first thing i like to do to complete the square , and you 're going to have to do it for the x variables and for the y terms , is group the x and y terms . let 's see . the x terms are 9x squared plus 54x . and let 's do the y terms in magenta . so then you have plus 4y squared minus 8y and then you have -- let me do this in a different color -- plus 49 is equal to 0 . and so the easy thing to do when you complete the square , the thing i like to do is , it 's very clear we can factor out a 9 out of both of these numbers , and we can factor out a 4 out of both of those . let 's do that , because that will help us complete the square . so this is the same thing is 9 times x squared plus 9 times 6 is 54 , 6x . i 'm going to add something else here , but i 'll leave it blank for now . plus 4 times y squared minus 2y i 'm probably going to add something here too , so i 'll leave it blank for now . plus 49 is equal to 0 . so what are we going to add here ? we 're going to complete the square . we want to add some number here so that this whole three term expression becomes a perfect square . likewise , we 're going to add some number here , so this three term number expression becomes a perfect square . and of course whatever we add on the side , we 're going to have to multiply it by 9 , because we 're really adding nine times that . and add it on to that side . whatever we add here , we 're going to have to multiply it times 4 and add it on that side . if i put a 1 here , it 's really like as if i had a 4 here , because 1 times 4 is 4 and if i had a 1 here it 's 1 times 9 . so 9 there . let 's do that . when we complete the square , we just take half of this coefficient . this coefficient is 6 , we take half of it is 3 , we square it , we get a 9 . remember it 's an equation , so what you do to one side , you have to do to the other . so if we added a 9 here , we 're actually adding 9 times 9 to the left-hand side of the equation , so we have to add 81 to the right-hand side to make the equation still hold . and you could kind of view it if we go back up here . this is the same thing , just to make that clear as if i added plus 81 right here . of course i would have had to add plus 81 up here . now let 's go to the y terms . you take half of this coefficient is minus 2 , half of that is minus 1 . you square it , you get plus 1 . 1 times 4 , so we 're really adding 4 to the left-hand side of the equation . and just so you understand what i did here . this is equivalent as if i just added a 4 here , and then i later just factored out this 4 . and so what does this become ? this expression is 9 times what ? this is the square of -- you could factor this , but we did it on purpose -- it 's x plus 3 squared and then we have plus 4 times -- what is this right here ? that 's y minus 1 squared . you might want to review factoring of polynomial or completing the square if you found that step a little daunting . and then we have plus 49 is equal to 0 plus 81 plus 84 is equal to 85 . all right , so now we have 9 times plus 3 squared plus 4 times y minus 1 squared . and let 's subtract 49 from both sides . that is equal to -- let 's see if i subtract 50 from 85 i get 35 , so if i subtract 49 , i get 36 . and now we are getting close to the standard form of something , but remember all the standard forms we did except for the circle -- we had a y -- and we know this is n't a circle , because we have these weird coefficients , well not weird but different coefficients in front of these terms . so to get the 1 on the right-hand side let 's divide everything by 36 . if you divide everything by 36 , this term becomes x plus 3 squared over see 9 over 36 is the same thing as 1 over 4 , and then you have plus y minus 1 squared 4 over 36 is the same thing as 1 over 9 and all of that is equal to 1 . and there you go . we have it in the standard form , and you can see our intuition at the beginning the problem was correct . this is indeed an ellipse , and now we can actually graph it . so first of all , actually good place to start , where 's the center of the this ellipse going to be ? it 's going to be x is equal to negative 3 . what x value makes this whole terms 0 ? so it 's going to be x is equal to minus 3 , and y is going to be equal to 1 . what y value makes this term 0 ? y is equal to 1 . that 's our center . so let 's graph that , and then we can draw the ellipse . it 's going to be in the negative quadrant . this is our x-axis and this is our y-axis . and then the center of our ellipse is at minus 3 and positive 1 , so that 's the center . and then , what is the radius in the x direction ? we just take the square root of this , so it 's 2 . so in the x direction we go two to the right . we go two to the left . and in the y direction , what do we do ? well we go up three and down three . the square root of this . let me do that . remember you have to take the square root of both of those . the vertical axis is actually the major radius or the semi-major axis is 3 , because that 's the longer one . and then the 2 is the minor radius , because that 's the shorter one . and now we 're ready to draw this ellipse . i 'll draw it in brown . let me see if i can do this properly . i have a shaky hand . all right , it looks something like that . and there you go . we took this kind of crazy looking thing , and all we did is algebraically manipulate it . we just completed the squares with the x 's and the y terms . and then we divided both sides by this number right here and we got it into the standard form . we said oh this is an ellipse . we have both of these terms , they 're both positive , we 're adding we 're not subtracting , they have different coefficients underneath here . so we 're ready to go over the ellipse , and we realized that the center was at minus 3,1 , and then we just drew the major radius , or the major axis and the minor axis . see you in the next video .
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the standard question you often get in your algebra class is they will give you this equation and it 'll say identify the conic section and graph it if you can . and the equation they give you wo n't be in the standard form , because if it was you could just kind of pattern match with what i showed in some of the previous videos and you 'd be able to get it .
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what is the general procedure to determine the conic ?
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the standard question you often get in your algebra class is they will give you this equation and it 'll say identify the conic section and graph it if you can . and the equation they give you wo n't be in the standard form , because if it was you could just kind of pattern match with what i showed in some of the previous videos and you 'd be able to get it . so let 's do a question like and let 's see if we can figure it out . so what i have here is 9x squared plus 4y squared plus 54x minus 8y plus 49 is equal to 0 . and once again , i mean who knows what this is it 's just not in the standard form . and actually one quick clue to tell you what this is you look at the x squared and the y squared terms if there are . if there 's only an x squared term and then there 's just a y and not a y squared term , then you 're probably dealing with a parabola , and we 'll go into that more later . or if it 's the other way around , if it 's just an x term and a y squared term , it 's probably a parabola . but assuming that we 're dealing with a circle , an ellipse , or a hyperbola , there will be an x squared term and a y squared term . if they both kind of have the same number in front of them , that 's a pretty good clue that we 're going to be dealing with a circle . if they both have different numbers , but they 're both positive in front of them , that 's a pretty good clue we 're probably going to be dealing with an ellipse . if one of them has a negative number in front of them and the other one has a positive number , that tells you that we 're probably going to be dealing with a hyperbola . but with that said , i mean that might help you identify things very quickly at this level , but it does n't help you graph it or get into the standard form . so let 's get it in the standard form . and the key to getting it in the standard form is really just completing the square . and i encourage you to re-watch the completing the square video , because that 's all we 're going to do right here to get it into the standard form . so the first thing i like to do to complete the square , and you 're going to have to do it for the x variables and for the y terms , is group the x and y terms . let 's see . the x terms are 9x squared plus 54x . and let 's do the y terms in magenta . so then you have plus 4y squared minus 8y and then you have -- let me do this in a different color -- plus 49 is equal to 0 . and so the easy thing to do when you complete the square , the thing i like to do is , it 's very clear we can factor out a 9 out of both of these numbers , and we can factor out a 4 out of both of those . let 's do that , because that will help us complete the square . so this is the same thing is 9 times x squared plus 9 times 6 is 54 , 6x . i 'm going to add something else here , but i 'll leave it blank for now . plus 4 times y squared minus 2y i 'm probably going to add something here too , so i 'll leave it blank for now . plus 49 is equal to 0 . so what are we going to add here ? we 're going to complete the square . we want to add some number here so that this whole three term expression becomes a perfect square . likewise , we 're going to add some number here , so this three term number expression becomes a perfect square . and of course whatever we add on the side , we 're going to have to multiply it by 9 , because we 're really adding nine times that . and add it on to that side . whatever we add here , we 're going to have to multiply it times 4 and add it on that side . if i put a 1 here , it 's really like as if i had a 4 here , because 1 times 4 is 4 and if i had a 1 here it 's 1 times 9 . so 9 there . let 's do that . when we complete the square , we just take half of this coefficient . this coefficient is 6 , we take half of it is 3 , we square it , we get a 9 . remember it 's an equation , so what you do to one side , you have to do to the other . so if we added a 9 here , we 're actually adding 9 times 9 to the left-hand side of the equation , so we have to add 81 to the right-hand side to make the equation still hold . and you could kind of view it if we go back up here . this is the same thing , just to make that clear as if i added plus 81 right here . of course i would have had to add plus 81 up here . now let 's go to the y terms . you take half of this coefficient is minus 2 , half of that is minus 1 . you square it , you get plus 1 . 1 times 4 , so we 're really adding 4 to the left-hand side of the equation . and just so you understand what i did here . this is equivalent as if i just added a 4 here , and then i later just factored out this 4 . and so what does this become ? this expression is 9 times what ? this is the square of -- you could factor this , but we did it on purpose -- it 's x plus 3 squared and then we have plus 4 times -- what is this right here ? that 's y minus 1 squared . you might want to review factoring of polynomial or completing the square if you found that step a little daunting . and then we have plus 49 is equal to 0 plus 81 plus 84 is equal to 85 . all right , so now we have 9 times plus 3 squared plus 4 times y minus 1 squared . and let 's subtract 49 from both sides . that is equal to -- let 's see if i subtract 50 from 85 i get 35 , so if i subtract 49 , i get 36 . and now we are getting close to the standard form of something , but remember all the standard forms we did except for the circle -- we had a y -- and we know this is n't a circle , because we have these weird coefficients , well not weird but different coefficients in front of these terms . so to get the 1 on the right-hand side let 's divide everything by 36 . if you divide everything by 36 , this term becomes x plus 3 squared over see 9 over 36 is the same thing as 1 over 4 , and then you have plus y minus 1 squared 4 over 36 is the same thing as 1 over 9 and all of that is equal to 1 . and there you go . we have it in the standard form , and you can see our intuition at the beginning the problem was correct . this is indeed an ellipse , and now we can actually graph it . so first of all , actually good place to start , where 's the center of the this ellipse going to be ? it 's going to be x is equal to negative 3 . what x value makes this whole terms 0 ? so it 's going to be x is equal to minus 3 , and y is going to be equal to 1 . what y value makes this term 0 ? y is equal to 1 . that 's our center . so let 's graph that , and then we can draw the ellipse . it 's going to be in the negative quadrant . this is our x-axis and this is our y-axis . and then the center of our ellipse is at minus 3 and positive 1 , so that 's the center . and then , what is the radius in the x direction ? we just take the square root of this , so it 's 2 . so in the x direction we go two to the right . we go two to the left . and in the y direction , what do we do ? well we go up three and down three . the square root of this . let me do that . remember you have to take the square root of both of those . the vertical axis is actually the major radius or the semi-major axis is 3 , because that 's the longer one . and then the 2 is the minor radius , because that 's the shorter one . and now we 're ready to draw this ellipse . i 'll draw it in brown . let me see if i can do this properly . i have a shaky hand . all right , it looks something like that . and there you go . we took this kind of crazy looking thing , and all we did is algebraically manipulate it . we just completed the squares with the x 's and the y terms . and then we divided both sides by this number right here and we got it into the standard form . we said oh this is an ellipse . we have both of these terms , they 're both positive , we 're adding we 're not subtracting , they have different coefficients underneath here . so we 're ready to go over the ellipse , and we realized that the center was at minus 3,1 , and then we just drew the major radius , or the major axis and the minor axis . see you in the next video .
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the standard question you often get in your algebra class is they will give you this equation and it 'll say identify the conic section and graph it if you can . and the equation they give you wo n't be in the standard form , because if it was you could just kind of pattern match with what i showed in some of the previous videos and you 'd be able to get it . so let 's do a question like and let 's see if we can figure it out .
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is there a way to derive the end equation ( ) equation into y= form , so that you could put it into a graphing calculator ?
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the standard question you often get in your algebra class is they will give you this equation and it 'll say identify the conic section and graph it if you can . and the equation they give you wo n't be in the standard form , because if it was you could just kind of pattern match with what i showed in some of the previous videos and you 'd be able to get it . so let 's do a question like and let 's see if we can figure it out . so what i have here is 9x squared plus 4y squared plus 54x minus 8y plus 49 is equal to 0 . and once again , i mean who knows what this is it 's just not in the standard form . and actually one quick clue to tell you what this is you look at the x squared and the y squared terms if there are . if there 's only an x squared term and then there 's just a y and not a y squared term , then you 're probably dealing with a parabola , and we 'll go into that more later . or if it 's the other way around , if it 's just an x term and a y squared term , it 's probably a parabola . but assuming that we 're dealing with a circle , an ellipse , or a hyperbola , there will be an x squared term and a y squared term . if they both kind of have the same number in front of them , that 's a pretty good clue that we 're going to be dealing with a circle . if they both have different numbers , but they 're both positive in front of them , that 's a pretty good clue we 're probably going to be dealing with an ellipse . if one of them has a negative number in front of them and the other one has a positive number , that tells you that we 're probably going to be dealing with a hyperbola . but with that said , i mean that might help you identify things very quickly at this level , but it does n't help you graph it or get into the standard form . so let 's get it in the standard form . and the key to getting it in the standard form is really just completing the square . and i encourage you to re-watch the completing the square video , because that 's all we 're going to do right here to get it into the standard form . so the first thing i like to do to complete the square , and you 're going to have to do it for the x variables and for the y terms , is group the x and y terms . let 's see . the x terms are 9x squared plus 54x . and let 's do the y terms in magenta . so then you have plus 4y squared minus 8y and then you have -- let me do this in a different color -- plus 49 is equal to 0 . and so the easy thing to do when you complete the square , the thing i like to do is , it 's very clear we can factor out a 9 out of both of these numbers , and we can factor out a 4 out of both of those . let 's do that , because that will help us complete the square . so this is the same thing is 9 times x squared plus 9 times 6 is 54 , 6x . i 'm going to add something else here , but i 'll leave it blank for now . plus 4 times y squared minus 2y i 'm probably going to add something here too , so i 'll leave it blank for now . plus 49 is equal to 0 . so what are we going to add here ? we 're going to complete the square . we want to add some number here so that this whole three term expression becomes a perfect square . likewise , we 're going to add some number here , so this three term number expression becomes a perfect square . and of course whatever we add on the side , we 're going to have to multiply it by 9 , because we 're really adding nine times that . and add it on to that side . whatever we add here , we 're going to have to multiply it times 4 and add it on that side . if i put a 1 here , it 's really like as if i had a 4 here , because 1 times 4 is 4 and if i had a 1 here it 's 1 times 9 . so 9 there . let 's do that . when we complete the square , we just take half of this coefficient . this coefficient is 6 , we take half of it is 3 , we square it , we get a 9 . remember it 's an equation , so what you do to one side , you have to do to the other . so if we added a 9 here , we 're actually adding 9 times 9 to the left-hand side of the equation , so we have to add 81 to the right-hand side to make the equation still hold . and you could kind of view it if we go back up here . this is the same thing , just to make that clear as if i added plus 81 right here . of course i would have had to add plus 81 up here . now let 's go to the y terms . you take half of this coefficient is minus 2 , half of that is minus 1 . you square it , you get plus 1 . 1 times 4 , so we 're really adding 4 to the left-hand side of the equation . and just so you understand what i did here . this is equivalent as if i just added a 4 here , and then i later just factored out this 4 . and so what does this become ? this expression is 9 times what ? this is the square of -- you could factor this , but we did it on purpose -- it 's x plus 3 squared and then we have plus 4 times -- what is this right here ? that 's y minus 1 squared . you might want to review factoring of polynomial or completing the square if you found that step a little daunting . and then we have plus 49 is equal to 0 plus 81 plus 84 is equal to 85 . all right , so now we have 9 times plus 3 squared plus 4 times y minus 1 squared . and let 's subtract 49 from both sides . that is equal to -- let 's see if i subtract 50 from 85 i get 35 , so if i subtract 49 , i get 36 . and now we are getting close to the standard form of something , but remember all the standard forms we did except for the circle -- we had a y -- and we know this is n't a circle , because we have these weird coefficients , well not weird but different coefficients in front of these terms . so to get the 1 on the right-hand side let 's divide everything by 36 . if you divide everything by 36 , this term becomes x plus 3 squared over see 9 over 36 is the same thing as 1 over 4 , and then you have plus y minus 1 squared 4 over 36 is the same thing as 1 over 9 and all of that is equal to 1 . and there you go . we have it in the standard form , and you can see our intuition at the beginning the problem was correct . this is indeed an ellipse , and now we can actually graph it . so first of all , actually good place to start , where 's the center of the this ellipse going to be ? it 's going to be x is equal to negative 3 . what x value makes this whole terms 0 ? so it 's going to be x is equal to minus 3 , and y is going to be equal to 1 . what y value makes this term 0 ? y is equal to 1 . that 's our center . so let 's graph that , and then we can draw the ellipse . it 's going to be in the negative quadrant . this is our x-axis and this is our y-axis . and then the center of our ellipse is at minus 3 and positive 1 , so that 's the center . and then , what is the radius in the x direction ? we just take the square root of this , so it 's 2 . so in the x direction we go two to the right . we go two to the left . and in the y direction , what do we do ? well we go up three and down three . the square root of this . let me do that . remember you have to take the square root of both of those . the vertical axis is actually the major radius or the semi-major axis is 3 , because that 's the longer one . and then the 2 is the minor radius , because that 's the shorter one . and now we 're ready to draw this ellipse . i 'll draw it in brown . let me see if i can do this properly . i have a shaky hand . all right , it looks something like that . and there you go . we took this kind of crazy looking thing , and all we did is algebraically manipulate it . we just completed the squares with the x 's and the y terms . and then we divided both sides by this number right here and we got it into the standard form . we said oh this is an ellipse . we have both of these terms , they 're both positive , we 're adding we 're not subtracting , they have different coefficients underneath here . so we 're ready to go over the ellipse , and we realized that the center was at minus 3,1 , and then we just drew the major radius , or the major axis and the minor axis . see you in the next video .
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the standard question you often get in your algebra class is they will give you this equation and it 'll say identify the conic section and graph it if you can . and the equation they give you wo n't be in the standard form , because if it was you could just kind of pattern match with what i showed in some of the previous videos and you 'd be able to get it .
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also , is there a way to identify the intercepts from the standard equation , or must you use a graphing calculator for those too ?
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the standard question you often get in your algebra class is they will give you this equation and it 'll say identify the conic section and graph it if you can . and the equation they give you wo n't be in the standard form , because if it was you could just kind of pattern match with what i showed in some of the previous videos and you 'd be able to get it . so let 's do a question like and let 's see if we can figure it out . so what i have here is 9x squared plus 4y squared plus 54x minus 8y plus 49 is equal to 0 . and once again , i mean who knows what this is it 's just not in the standard form . and actually one quick clue to tell you what this is you look at the x squared and the y squared terms if there are . if there 's only an x squared term and then there 's just a y and not a y squared term , then you 're probably dealing with a parabola , and we 'll go into that more later . or if it 's the other way around , if it 's just an x term and a y squared term , it 's probably a parabola . but assuming that we 're dealing with a circle , an ellipse , or a hyperbola , there will be an x squared term and a y squared term . if they both kind of have the same number in front of them , that 's a pretty good clue that we 're going to be dealing with a circle . if they both have different numbers , but they 're both positive in front of them , that 's a pretty good clue we 're probably going to be dealing with an ellipse . if one of them has a negative number in front of them and the other one has a positive number , that tells you that we 're probably going to be dealing with a hyperbola . but with that said , i mean that might help you identify things very quickly at this level , but it does n't help you graph it or get into the standard form . so let 's get it in the standard form . and the key to getting it in the standard form is really just completing the square . and i encourage you to re-watch the completing the square video , because that 's all we 're going to do right here to get it into the standard form . so the first thing i like to do to complete the square , and you 're going to have to do it for the x variables and for the y terms , is group the x and y terms . let 's see . the x terms are 9x squared plus 54x . and let 's do the y terms in magenta . so then you have plus 4y squared minus 8y and then you have -- let me do this in a different color -- plus 49 is equal to 0 . and so the easy thing to do when you complete the square , the thing i like to do is , it 's very clear we can factor out a 9 out of both of these numbers , and we can factor out a 4 out of both of those . let 's do that , because that will help us complete the square . so this is the same thing is 9 times x squared plus 9 times 6 is 54 , 6x . i 'm going to add something else here , but i 'll leave it blank for now . plus 4 times y squared minus 2y i 'm probably going to add something here too , so i 'll leave it blank for now . plus 49 is equal to 0 . so what are we going to add here ? we 're going to complete the square . we want to add some number here so that this whole three term expression becomes a perfect square . likewise , we 're going to add some number here , so this three term number expression becomes a perfect square . and of course whatever we add on the side , we 're going to have to multiply it by 9 , because we 're really adding nine times that . and add it on to that side . whatever we add here , we 're going to have to multiply it times 4 and add it on that side . if i put a 1 here , it 's really like as if i had a 4 here , because 1 times 4 is 4 and if i had a 1 here it 's 1 times 9 . so 9 there . let 's do that . when we complete the square , we just take half of this coefficient . this coefficient is 6 , we take half of it is 3 , we square it , we get a 9 . remember it 's an equation , so what you do to one side , you have to do to the other . so if we added a 9 here , we 're actually adding 9 times 9 to the left-hand side of the equation , so we have to add 81 to the right-hand side to make the equation still hold . and you could kind of view it if we go back up here . this is the same thing , just to make that clear as if i added plus 81 right here . of course i would have had to add plus 81 up here . now let 's go to the y terms . you take half of this coefficient is minus 2 , half of that is minus 1 . you square it , you get plus 1 . 1 times 4 , so we 're really adding 4 to the left-hand side of the equation . and just so you understand what i did here . this is equivalent as if i just added a 4 here , and then i later just factored out this 4 . and so what does this become ? this expression is 9 times what ? this is the square of -- you could factor this , but we did it on purpose -- it 's x plus 3 squared and then we have plus 4 times -- what is this right here ? that 's y minus 1 squared . you might want to review factoring of polynomial or completing the square if you found that step a little daunting . and then we have plus 49 is equal to 0 plus 81 plus 84 is equal to 85 . all right , so now we have 9 times plus 3 squared plus 4 times y minus 1 squared . and let 's subtract 49 from both sides . that is equal to -- let 's see if i subtract 50 from 85 i get 35 , so if i subtract 49 , i get 36 . and now we are getting close to the standard form of something , but remember all the standard forms we did except for the circle -- we had a y -- and we know this is n't a circle , because we have these weird coefficients , well not weird but different coefficients in front of these terms . so to get the 1 on the right-hand side let 's divide everything by 36 . if you divide everything by 36 , this term becomes x plus 3 squared over see 9 over 36 is the same thing as 1 over 4 , and then you have plus y minus 1 squared 4 over 36 is the same thing as 1 over 9 and all of that is equal to 1 . and there you go . we have it in the standard form , and you can see our intuition at the beginning the problem was correct . this is indeed an ellipse , and now we can actually graph it . so first of all , actually good place to start , where 's the center of the this ellipse going to be ? it 's going to be x is equal to negative 3 . what x value makes this whole terms 0 ? so it 's going to be x is equal to minus 3 , and y is going to be equal to 1 . what y value makes this term 0 ? y is equal to 1 . that 's our center . so let 's graph that , and then we can draw the ellipse . it 's going to be in the negative quadrant . this is our x-axis and this is our y-axis . and then the center of our ellipse is at minus 3 and positive 1 , so that 's the center . and then , what is the radius in the x direction ? we just take the square root of this , so it 's 2 . so in the x direction we go two to the right . we go two to the left . and in the y direction , what do we do ? well we go up three and down three . the square root of this . let me do that . remember you have to take the square root of both of those . the vertical axis is actually the major radius or the semi-major axis is 3 , because that 's the longer one . and then the 2 is the minor radius , because that 's the shorter one . and now we 're ready to draw this ellipse . i 'll draw it in brown . let me see if i can do this properly . i have a shaky hand . all right , it looks something like that . and there you go . we took this kind of crazy looking thing , and all we did is algebraically manipulate it . we just completed the squares with the x 's and the y terms . and then we divided both sides by this number right here and we got it into the standard form . we said oh this is an ellipse . we have both of these terms , they 're both positive , we 're adding we 're not subtracting , they have different coefficients underneath here . so we 're ready to go over the ellipse , and we realized that the center was at minus 3,1 , and then we just drew the major radius , or the major axis and the minor axis . see you in the next video .
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let me do that . remember you have to take the square root of both of those . the vertical axis is actually the major radius or the semi-major axis is 3 , because that 's the longer one .
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what is an easy way to remember how to solve conic equations ?
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the standard question you often get in your algebra class is they will give you this equation and it 'll say identify the conic section and graph it if you can . and the equation they give you wo n't be in the standard form , because if it was you could just kind of pattern match with what i showed in some of the previous videos and you 'd be able to get it . so let 's do a question like and let 's see if we can figure it out . so what i have here is 9x squared plus 4y squared plus 54x minus 8y plus 49 is equal to 0 . and once again , i mean who knows what this is it 's just not in the standard form . and actually one quick clue to tell you what this is you look at the x squared and the y squared terms if there are . if there 's only an x squared term and then there 's just a y and not a y squared term , then you 're probably dealing with a parabola , and we 'll go into that more later . or if it 's the other way around , if it 's just an x term and a y squared term , it 's probably a parabola . but assuming that we 're dealing with a circle , an ellipse , or a hyperbola , there will be an x squared term and a y squared term . if they both kind of have the same number in front of them , that 's a pretty good clue that we 're going to be dealing with a circle . if they both have different numbers , but they 're both positive in front of them , that 's a pretty good clue we 're probably going to be dealing with an ellipse . if one of them has a negative number in front of them and the other one has a positive number , that tells you that we 're probably going to be dealing with a hyperbola . but with that said , i mean that might help you identify things very quickly at this level , but it does n't help you graph it or get into the standard form . so let 's get it in the standard form . and the key to getting it in the standard form is really just completing the square . and i encourage you to re-watch the completing the square video , because that 's all we 're going to do right here to get it into the standard form . so the first thing i like to do to complete the square , and you 're going to have to do it for the x variables and for the y terms , is group the x and y terms . let 's see . the x terms are 9x squared plus 54x . and let 's do the y terms in magenta . so then you have plus 4y squared minus 8y and then you have -- let me do this in a different color -- plus 49 is equal to 0 . and so the easy thing to do when you complete the square , the thing i like to do is , it 's very clear we can factor out a 9 out of both of these numbers , and we can factor out a 4 out of both of those . let 's do that , because that will help us complete the square . so this is the same thing is 9 times x squared plus 9 times 6 is 54 , 6x . i 'm going to add something else here , but i 'll leave it blank for now . plus 4 times y squared minus 2y i 'm probably going to add something here too , so i 'll leave it blank for now . plus 49 is equal to 0 . so what are we going to add here ? we 're going to complete the square . we want to add some number here so that this whole three term expression becomes a perfect square . likewise , we 're going to add some number here , so this three term number expression becomes a perfect square . and of course whatever we add on the side , we 're going to have to multiply it by 9 , because we 're really adding nine times that . and add it on to that side . whatever we add here , we 're going to have to multiply it times 4 and add it on that side . if i put a 1 here , it 's really like as if i had a 4 here , because 1 times 4 is 4 and if i had a 1 here it 's 1 times 9 . so 9 there . let 's do that . when we complete the square , we just take half of this coefficient . this coefficient is 6 , we take half of it is 3 , we square it , we get a 9 . remember it 's an equation , so what you do to one side , you have to do to the other . so if we added a 9 here , we 're actually adding 9 times 9 to the left-hand side of the equation , so we have to add 81 to the right-hand side to make the equation still hold . and you could kind of view it if we go back up here . this is the same thing , just to make that clear as if i added plus 81 right here . of course i would have had to add plus 81 up here . now let 's go to the y terms . you take half of this coefficient is minus 2 , half of that is minus 1 . you square it , you get plus 1 . 1 times 4 , so we 're really adding 4 to the left-hand side of the equation . and just so you understand what i did here . this is equivalent as if i just added a 4 here , and then i later just factored out this 4 . and so what does this become ? this expression is 9 times what ? this is the square of -- you could factor this , but we did it on purpose -- it 's x plus 3 squared and then we have plus 4 times -- what is this right here ? that 's y minus 1 squared . you might want to review factoring of polynomial or completing the square if you found that step a little daunting . and then we have plus 49 is equal to 0 plus 81 plus 84 is equal to 85 . all right , so now we have 9 times plus 3 squared plus 4 times y minus 1 squared . and let 's subtract 49 from both sides . that is equal to -- let 's see if i subtract 50 from 85 i get 35 , so if i subtract 49 , i get 36 . and now we are getting close to the standard form of something , but remember all the standard forms we did except for the circle -- we had a y -- and we know this is n't a circle , because we have these weird coefficients , well not weird but different coefficients in front of these terms . so to get the 1 on the right-hand side let 's divide everything by 36 . if you divide everything by 36 , this term becomes x plus 3 squared over see 9 over 36 is the same thing as 1 over 4 , and then you have plus y minus 1 squared 4 over 36 is the same thing as 1 over 9 and all of that is equal to 1 . and there you go . we have it in the standard form , and you can see our intuition at the beginning the problem was correct . this is indeed an ellipse , and now we can actually graph it . so first of all , actually good place to start , where 's the center of the this ellipse going to be ? it 's going to be x is equal to negative 3 . what x value makes this whole terms 0 ? so it 's going to be x is equal to minus 3 , and y is going to be equal to 1 . what y value makes this term 0 ? y is equal to 1 . that 's our center . so let 's graph that , and then we can draw the ellipse . it 's going to be in the negative quadrant . this is our x-axis and this is our y-axis . and then the center of our ellipse is at minus 3 and positive 1 , so that 's the center . and then , what is the radius in the x direction ? we just take the square root of this , so it 's 2 . so in the x direction we go two to the right . we go two to the left . and in the y direction , what do we do ? well we go up three and down three . the square root of this . let me do that . remember you have to take the square root of both of those . the vertical axis is actually the major radius or the semi-major axis is 3 , because that 's the longer one . and then the 2 is the minor radius , because that 's the shorter one . and now we 're ready to draw this ellipse . i 'll draw it in brown . let me see if i can do this properly . i have a shaky hand . all right , it looks something like that . and there you go . we took this kind of crazy looking thing , and all we did is algebraically manipulate it . we just completed the squares with the x 's and the y terms . and then we divided both sides by this number right here and we got it into the standard form . we said oh this is an ellipse . we have both of these terms , they 're both positive , we 're adding we 're not subtracting , they have different coefficients underneath here . so we 're ready to go over the ellipse , and we realized that the center was at minus 3,1 , and then we just drew the major radius , or the major axis and the minor axis . see you in the next video .
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and then we divided both sides by this number right here and we got it into the standard form . we said oh this is an ellipse . we have both of these terms , they 're both positive , we 're adding we 're not subtracting , they have different coefficients underneath here .
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how to proof that ellipse is a conic section ?
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the standard question you often get in your algebra class is they will give you this equation and it 'll say identify the conic section and graph it if you can . and the equation they give you wo n't be in the standard form , because if it was you could just kind of pattern match with what i showed in some of the previous videos and you 'd be able to get it . so let 's do a question like and let 's see if we can figure it out . so what i have here is 9x squared plus 4y squared plus 54x minus 8y plus 49 is equal to 0 . and once again , i mean who knows what this is it 's just not in the standard form . and actually one quick clue to tell you what this is you look at the x squared and the y squared terms if there are . if there 's only an x squared term and then there 's just a y and not a y squared term , then you 're probably dealing with a parabola , and we 'll go into that more later . or if it 's the other way around , if it 's just an x term and a y squared term , it 's probably a parabola . but assuming that we 're dealing with a circle , an ellipse , or a hyperbola , there will be an x squared term and a y squared term . if they both kind of have the same number in front of them , that 's a pretty good clue that we 're going to be dealing with a circle . if they both have different numbers , but they 're both positive in front of them , that 's a pretty good clue we 're probably going to be dealing with an ellipse . if one of them has a negative number in front of them and the other one has a positive number , that tells you that we 're probably going to be dealing with a hyperbola . but with that said , i mean that might help you identify things very quickly at this level , but it does n't help you graph it or get into the standard form . so let 's get it in the standard form . and the key to getting it in the standard form is really just completing the square . and i encourage you to re-watch the completing the square video , because that 's all we 're going to do right here to get it into the standard form . so the first thing i like to do to complete the square , and you 're going to have to do it for the x variables and for the y terms , is group the x and y terms . let 's see . the x terms are 9x squared plus 54x . and let 's do the y terms in magenta . so then you have plus 4y squared minus 8y and then you have -- let me do this in a different color -- plus 49 is equal to 0 . and so the easy thing to do when you complete the square , the thing i like to do is , it 's very clear we can factor out a 9 out of both of these numbers , and we can factor out a 4 out of both of those . let 's do that , because that will help us complete the square . so this is the same thing is 9 times x squared plus 9 times 6 is 54 , 6x . i 'm going to add something else here , but i 'll leave it blank for now . plus 4 times y squared minus 2y i 'm probably going to add something here too , so i 'll leave it blank for now . plus 49 is equal to 0 . so what are we going to add here ? we 're going to complete the square . we want to add some number here so that this whole three term expression becomes a perfect square . likewise , we 're going to add some number here , so this three term number expression becomes a perfect square . and of course whatever we add on the side , we 're going to have to multiply it by 9 , because we 're really adding nine times that . and add it on to that side . whatever we add here , we 're going to have to multiply it times 4 and add it on that side . if i put a 1 here , it 's really like as if i had a 4 here , because 1 times 4 is 4 and if i had a 1 here it 's 1 times 9 . so 9 there . let 's do that . when we complete the square , we just take half of this coefficient . this coefficient is 6 , we take half of it is 3 , we square it , we get a 9 . remember it 's an equation , so what you do to one side , you have to do to the other . so if we added a 9 here , we 're actually adding 9 times 9 to the left-hand side of the equation , so we have to add 81 to the right-hand side to make the equation still hold . and you could kind of view it if we go back up here . this is the same thing , just to make that clear as if i added plus 81 right here . of course i would have had to add plus 81 up here . now let 's go to the y terms . you take half of this coefficient is minus 2 , half of that is minus 1 . you square it , you get plus 1 . 1 times 4 , so we 're really adding 4 to the left-hand side of the equation . and just so you understand what i did here . this is equivalent as if i just added a 4 here , and then i later just factored out this 4 . and so what does this become ? this expression is 9 times what ? this is the square of -- you could factor this , but we did it on purpose -- it 's x plus 3 squared and then we have plus 4 times -- what is this right here ? that 's y minus 1 squared . you might want to review factoring of polynomial or completing the square if you found that step a little daunting . and then we have plus 49 is equal to 0 plus 81 plus 84 is equal to 85 . all right , so now we have 9 times plus 3 squared plus 4 times y minus 1 squared . and let 's subtract 49 from both sides . that is equal to -- let 's see if i subtract 50 from 85 i get 35 , so if i subtract 49 , i get 36 . and now we are getting close to the standard form of something , but remember all the standard forms we did except for the circle -- we had a y -- and we know this is n't a circle , because we have these weird coefficients , well not weird but different coefficients in front of these terms . so to get the 1 on the right-hand side let 's divide everything by 36 . if you divide everything by 36 , this term becomes x plus 3 squared over see 9 over 36 is the same thing as 1 over 4 , and then you have plus y minus 1 squared 4 over 36 is the same thing as 1 over 9 and all of that is equal to 1 . and there you go . we have it in the standard form , and you can see our intuition at the beginning the problem was correct . this is indeed an ellipse , and now we can actually graph it . so first of all , actually good place to start , where 's the center of the this ellipse going to be ? it 's going to be x is equal to negative 3 . what x value makes this whole terms 0 ? so it 's going to be x is equal to minus 3 , and y is going to be equal to 1 . what y value makes this term 0 ? y is equal to 1 . that 's our center . so let 's graph that , and then we can draw the ellipse . it 's going to be in the negative quadrant . this is our x-axis and this is our y-axis . and then the center of our ellipse is at minus 3 and positive 1 , so that 's the center . and then , what is the radius in the x direction ? we just take the square root of this , so it 's 2 . so in the x direction we go two to the right . we go two to the left . and in the y direction , what do we do ? well we go up three and down three . the square root of this . let me do that . remember you have to take the square root of both of those . the vertical axis is actually the major radius or the semi-major axis is 3 , because that 's the longer one . and then the 2 is the minor radius , because that 's the shorter one . and now we 're ready to draw this ellipse . i 'll draw it in brown . let me see if i can do this properly . i have a shaky hand . all right , it looks something like that . and there you go . we took this kind of crazy looking thing , and all we did is algebraically manipulate it . we just completed the squares with the x 's and the y terms . and then we divided both sides by this number right here and we got it into the standard form . we said oh this is an ellipse . we have both of these terms , they 're both positive , we 're adding we 're not subtracting , they have different coefficients underneath here . so we 're ready to go over the ellipse , and we realized that the center was at minus 3,1 , and then we just drew the major radius , or the major axis and the minor axis . see you in the next video .
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and then we divided both sides by this number right here and we got it into the standard form . we said oh this is an ellipse . we have both of these terms , they 're both positive , we 're adding we 're not subtracting , they have different coefficients underneath here .
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why was the equation an ellipse and not a hyperbola ?
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the standard question you often get in your algebra class is they will give you this equation and it 'll say identify the conic section and graph it if you can . and the equation they give you wo n't be in the standard form , because if it was you could just kind of pattern match with what i showed in some of the previous videos and you 'd be able to get it . so let 's do a question like and let 's see if we can figure it out . so what i have here is 9x squared plus 4y squared plus 54x minus 8y plus 49 is equal to 0 . and once again , i mean who knows what this is it 's just not in the standard form . and actually one quick clue to tell you what this is you look at the x squared and the y squared terms if there are . if there 's only an x squared term and then there 's just a y and not a y squared term , then you 're probably dealing with a parabola , and we 'll go into that more later . or if it 's the other way around , if it 's just an x term and a y squared term , it 's probably a parabola . but assuming that we 're dealing with a circle , an ellipse , or a hyperbola , there will be an x squared term and a y squared term . if they both kind of have the same number in front of them , that 's a pretty good clue that we 're going to be dealing with a circle . if they both have different numbers , but they 're both positive in front of them , that 's a pretty good clue we 're probably going to be dealing with an ellipse . if one of them has a negative number in front of them and the other one has a positive number , that tells you that we 're probably going to be dealing with a hyperbola . but with that said , i mean that might help you identify things very quickly at this level , but it does n't help you graph it or get into the standard form . so let 's get it in the standard form . and the key to getting it in the standard form is really just completing the square . and i encourage you to re-watch the completing the square video , because that 's all we 're going to do right here to get it into the standard form . so the first thing i like to do to complete the square , and you 're going to have to do it for the x variables and for the y terms , is group the x and y terms . let 's see . the x terms are 9x squared plus 54x . and let 's do the y terms in magenta . so then you have plus 4y squared minus 8y and then you have -- let me do this in a different color -- plus 49 is equal to 0 . and so the easy thing to do when you complete the square , the thing i like to do is , it 's very clear we can factor out a 9 out of both of these numbers , and we can factor out a 4 out of both of those . let 's do that , because that will help us complete the square . so this is the same thing is 9 times x squared plus 9 times 6 is 54 , 6x . i 'm going to add something else here , but i 'll leave it blank for now . plus 4 times y squared minus 2y i 'm probably going to add something here too , so i 'll leave it blank for now . plus 49 is equal to 0 . so what are we going to add here ? we 're going to complete the square . we want to add some number here so that this whole three term expression becomes a perfect square . likewise , we 're going to add some number here , so this three term number expression becomes a perfect square . and of course whatever we add on the side , we 're going to have to multiply it by 9 , because we 're really adding nine times that . and add it on to that side . whatever we add here , we 're going to have to multiply it times 4 and add it on that side . if i put a 1 here , it 's really like as if i had a 4 here , because 1 times 4 is 4 and if i had a 1 here it 's 1 times 9 . so 9 there . let 's do that . when we complete the square , we just take half of this coefficient . this coefficient is 6 , we take half of it is 3 , we square it , we get a 9 . remember it 's an equation , so what you do to one side , you have to do to the other . so if we added a 9 here , we 're actually adding 9 times 9 to the left-hand side of the equation , so we have to add 81 to the right-hand side to make the equation still hold . and you could kind of view it if we go back up here . this is the same thing , just to make that clear as if i added plus 81 right here . of course i would have had to add plus 81 up here . now let 's go to the y terms . you take half of this coefficient is minus 2 , half of that is minus 1 . you square it , you get plus 1 . 1 times 4 , so we 're really adding 4 to the left-hand side of the equation . and just so you understand what i did here . this is equivalent as if i just added a 4 here , and then i later just factored out this 4 . and so what does this become ? this expression is 9 times what ? this is the square of -- you could factor this , but we did it on purpose -- it 's x plus 3 squared and then we have plus 4 times -- what is this right here ? that 's y minus 1 squared . you might want to review factoring of polynomial or completing the square if you found that step a little daunting . and then we have plus 49 is equal to 0 plus 81 plus 84 is equal to 85 . all right , so now we have 9 times plus 3 squared plus 4 times y minus 1 squared . and let 's subtract 49 from both sides . that is equal to -- let 's see if i subtract 50 from 85 i get 35 , so if i subtract 49 , i get 36 . and now we are getting close to the standard form of something , but remember all the standard forms we did except for the circle -- we had a y -- and we know this is n't a circle , because we have these weird coefficients , well not weird but different coefficients in front of these terms . so to get the 1 on the right-hand side let 's divide everything by 36 . if you divide everything by 36 , this term becomes x plus 3 squared over see 9 over 36 is the same thing as 1 over 4 , and then you have plus y minus 1 squared 4 over 36 is the same thing as 1 over 9 and all of that is equal to 1 . and there you go . we have it in the standard form , and you can see our intuition at the beginning the problem was correct . this is indeed an ellipse , and now we can actually graph it . so first of all , actually good place to start , where 's the center of the this ellipse going to be ? it 's going to be x is equal to negative 3 . what x value makes this whole terms 0 ? so it 's going to be x is equal to minus 3 , and y is going to be equal to 1 . what y value makes this term 0 ? y is equal to 1 . that 's our center . so let 's graph that , and then we can draw the ellipse . it 's going to be in the negative quadrant . this is our x-axis and this is our y-axis . and then the center of our ellipse is at minus 3 and positive 1 , so that 's the center . and then , what is the radius in the x direction ? we just take the square root of this , so it 's 2 . so in the x direction we go two to the right . we go two to the left . and in the y direction , what do we do ? well we go up three and down three . the square root of this . let me do that . remember you have to take the square root of both of those . the vertical axis is actually the major radius or the semi-major axis is 3 , because that 's the longer one . and then the 2 is the minor radius , because that 's the shorter one . and now we 're ready to draw this ellipse . i 'll draw it in brown . let me see if i can do this properly . i have a shaky hand . all right , it looks something like that . and there you go . we took this kind of crazy looking thing , and all we did is algebraically manipulate it . we just completed the squares with the x 's and the y terms . and then we divided both sides by this number right here and we got it into the standard form . we said oh this is an ellipse . we have both of these terms , they 're both positive , we 're adding we 're not subtracting , they have different coefficients underneath here . so we 're ready to go over the ellipse , and we realized that the center was at minus 3,1 , and then we just drew the major radius , or the major axis and the minor axis . see you in the next video .
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but with that said , i mean that might help you identify things very quickly at this level , but it does n't help you graph it or get into the standard form . so let 's get it in the standard form . and the key to getting it in the standard form is really just completing the square .
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when sal mentioned the standard form , is n't the standard form where a and b are not fractions ?
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the standard question you often get in your algebra class is they will give you this equation and it 'll say identify the conic section and graph it if you can . and the equation they give you wo n't be in the standard form , because if it was you could just kind of pattern match with what i showed in some of the previous videos and you 'd be able to get it . so let 's do a question like and let 's see if we can figure it out . so what i have here is 9x squared plus 4y squared plus 54x minus 8y plus 49 is equal to 0 . and once again , i mean who knows what this is it 's just not in the standard form . and actually one quick clue to tell you what this is you look at the x squared and the y squared terms if there are . if there 's only an x squared term and then there 's just a y and not a y squared term , then you 're probably dealing with a parabola , and we 'll go into that more later . or if it 's the other way around , if it 's just an x term and a y squared term , it 's probably a parabola . but assuming that we 're dealing with a circle , an ellipse , or a hyperbola , there will be an x squared term and a y squared term . if they both kind of have the same number in front of them , that 's a pretty good clue that we 're going to be dealing with a circle . if they both have different numbers , but they 're both positive in front of them , that 's a pretty good clue we 're probably going to be dealing with an ellipse . if one of them has a negative number in front of them and the other one has a positive number , that tells you that we 're probably going to be dealing with a hyperbola . but with that said , i mean that might help you identify things very quickly at this level , but it does n't help you graph it or get into the standard form . so let 's get it in the standard form . and the key to getting it in the standard form is really just completing the square . and i encourage you to re-watch the completing the square video , because that 's all we 're going to do right here to get it into the standard form . so the first thing i like to do to complete the square , and you 're going to have to do it for the x variables and for the y terms , is group the x and y terms . let 's see . the x terms are 9x squared plus 54x . and let 's do the y terms in magenta . so then you have plus 4y squared minus 8y and then you have -- let me do this in a different color -- plus 49 is equal to 0 . and so the easy thing to do when you complete the square , the thing i like to do is , it 's very clear we can factor out a 9 out of both of these numbers , and we can factor out a 4 out of both of those . let 's do that , because that will help us complete the square . so this is the same thing is 9 times x squared plus 9 times 6 is 54 , 6x . i 'm going to add something else here , but i 'll leave it blank for now . plus 4 times y squared minus 2y i 'm probably going to add something here too , so i 'll leave it blank for now . plus 49 is equal to 0 . so what are we going to add here ? we 're going to complete the square . we want to add some number here so that this whole three term expression becomes a perfect square . likewise , we 're going to add some number here , so this three term number expression becomes a perfect square . and of course whatever we add on the side , we 're going to have to multiply it by 9 , because we 're really adding nine times that . and add it on to that side . whatever we add here , we 're going to have to multiply it times 4 and add it on that side . if i put a 1 here , it 's really like as if i had a 4 here , because 1 times 4 is 4 and if i had a 1 here it 's 1 times 9 . so 9 there . let 's do that . when we complete the square , we just take half of this coefficient . this coefficient is 6 , we take half of it is 3 , we square it , we get a 9 . remember it 's an equation , so what you do to one side , you have to do to the other . so if we added a 9 here , we 're actually adding 9 times 9 to the left-hand side of the equation , so we have to add 81 to the right-hand side to make the equation still hold . and you could kind of view it if we go back up here . this is the same thing , just to make that clear as if i added plus 81 right here . of course i would have had to add plus 81 up here . now let 's go to the y terms . you take half of this coefficient is minus 2 , half of that is minus 1 . you square it , you get plus 1 . 1 times 4 , so we 're really adding 4 to the left-hand side of the equation . and just so you understand what i did here . this is equivalent as if i just added a 4 here , and then i later just factored out this 4 . and so what does this become ? this expression is 9 times what ? this is the square of -- you could factor this , but we did it on purpose -- it 's x plus 3 squared and then we have plus 4 times -- what is this right here ? that 's y minus 1 squared . you might want to review factoring of polynomial or completing the square if you found that step a little daunting . and then we have plus 49 is equal to 0 plus 81 plus 84 is equal to 85 . all right , so now we have 9 times plus 3 squared plus 4 times y minus 1 squared . and let 's subtract 49 from both sides . that is equal to -- let 's see if i subtract 50 from 85 i get 35 , so if i subtract 49 , i get 36 . and now we are getting close to the standard form of something , but remember all the standard forms we did except for the circle -- we had a y -- and we know this is n't a circle , because we have these weird coefficients , well not weird but different coefficients in front of these terms . so to get the 1 on the right-hand side let 's divide everything by 36 . if you divide everything by 36 , this term becomes x plus 3 squared over see 9 over 36 is the same thing as 1 over 4 , and then you have plus y minus 1 squared 4 over 36 is the same thing as 1 over 9 and all of that is equal to 1 . and there you go . we have it in the standard form , and you can see our intuition at the beginning the problem was correct . this is indeed an ellipse , and now we can actually graph it . so first of all , actually good place to start , where 's the center of the this ellipse going to be ? it 's going to be x is equal to negative 3 . what x value makes this whole terms 0 ? so it 's going to be x is equal to minus 3 , and y is going to be equal to 1 . what y value makes this term 0 ? y is equal to 1 . that 's our center . so let 's graph that , and then we can draw the ellipse . it 's going to be in the negative quadrant . this is our x-axis and this is our y-axis . and then the center of our ellipse is at minus 3 and positive 1 , so that 's the center . and then , what is the radius in the x direction ? we just take the square root of this , so it 's 2 . so in the x direction we go two to the right . we go two to the left . and in the y direction , what do we do ? well we go up three and down three . the square root of this . let me do that . remember you have to take the square root of both of those . the vertical axis is actually the major radius or the semi-major axis is 3 , because that 's the longer one . and then the 2 is the minor radius , because that 's the shorter one . and now we 're ready to draw this ellipse . i 'll draw it in brown . let me see if i can do this properly . i have a shaky hand . all right , it looks something like that . and there you go . we took this kind of crazy looking thing , and all we did is algebraically manipulate it . we just completed the squares with the x 's and the y terms . and then we divided both sides by this number right here and we got it into the standard form . we said oh this is an ellipse . we have both of these terms , they 're both positive , we 're adding we 're not subtracting , they have different coefficients underneath here . so we 're ready to go over the ellipse , and we realized that the center was at minus 3,1 , and then we just drew the major radius , or the major axis and the minor axis . see you in the next video .
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but with that said , i mean that might help you identify things very quickly at this level , but it does n't help you graph it or get into the standard form . so let 's get it in the standard form . and the key to getting it in the standard form is really just completing the square .
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if yes , is n't sal 's standard form not a standard form ?
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the standard question you often get in your algebra class is they will give you this equation and it 'll say identify the conic section and graph it if you can . and the equation they give you wo n't be in the standard form , because if it was you could just kind of pattern match with what i showed in some of the previous videos and you 'd be able to get it . so let 's do a question like and let 's see if we can figure it out . so what i have here is 9x squared plus 4y squared plus 54x minus 8y plus 49 is equal to 0 . and once again , i mean who knows what this is it 's just not in the standard form . and actually one quick clue to tell you what this is you look at the x squared and the y squared terms if there are . if there 's only an x squared term and then there 's just a y and not a y squared term , then you 're probably dealing with a parabola , and we 'll go into that more later . or if it 's the other way around , if it 's just an x term and a y squared term , it 's probably a parabola . but assuming that we 're dealing with a circle , an ellipse , or a hyperbola , there will be an x squared term and a y squared term . if they both kind of have the same number in front of them , that 's a pretty good clue that we 're going to be dealing with a circle . if they both have different numbers , but they 're both positive in front of them , that 's a pretty good clue we 're probably going to be dealing with an ellipse . if one of them has a negative number in front of them and the other one has a positive number , that tells you that we 're probably going to be dealing with a hyperbola . but with that said , i mean that might help you identify things very quickly at this level , but it does n't help you graph it or get into the standard form . so let 's get it in the standard form . and the key to getting it in the standard form is really just completing the square . and i encourage you to re-watch the completing the square video , because that 's all we 're going to do right here to get it into the standard form . so the first thing i like to do to complete the square , and you 're going to have to do it for the x variables and for the y terms , is group the x and y terms . let 's see . the x terms are 9x squared plus 54x . and let 's do the y terms in magenta . so then you have plus 4y squared minus 8y and then you have -- let me do this in a different color -- plus 49 is equal to 0 . and so the easy thing to do when you complete the square , the thing i like to do is , it 's very clear we can factor out a 9 out of both of these numbers , and we can factor out a 4 out of both of those . let 's do that , because that will help us complete the square . so this is the same thing is 9 times x squared plus 9 times 6 is 54 , 6x . i 'm going to add something else here , but i 'll leave it blank for now . plus 4 times y squared minus 2y i 'm probably going to add something here too , so i 'll leave it blank for now . plus 49 is equal to 0 . so what are we going to add here ? we 're going to complete the square . we want to add some number here so that this whole three term expression becomes a perfect square . likewise , we 're going to add some number here , so this three term number expression becomes a perfect square . and of course whatever we add on the side , we 're going to have to multiply it by 9 , because we 're really adding nine times that . and add it on to that side . whatever we add here , we 're going to have to multiply it times 4 and add it on that side . if i put a 1 here , it 's really like as if i had a 4 here , because 1 times 4 is 4 and if i had a 1 here it 's 1 times 9 . so 9 there . let 's do that . when we complete the square , we just take half of this coefficient . this coefficient is 6 , we take half of it is 3 , we square it , we get a 9 . remember it 's an equation , so what you do to one side , you have to do to the other . so if we added a 9 here , we 're actually adding 9 times 9 to the left-hand side of the equation , so we have to add 81 to the right-hand side to make the equation still hold . and you could kind of view it if we go back up here . this is the same thing , just to make that clear as if i added plus 81 right here . of course i would have had to add plus 81 up here . now let 's go to the y terms . you take half of this coefficient is minus 2 , half of that is minus 1 . you square it , you get plus 1 . 1 times 4 , so we 're really adding 4 to the left-hand side of the equation . and just so you understand what i did here . this is equivalent as if i just added a 4 here , and then i later just factored out this 4 . and so what does this become ? this expression is 9 times what ? this is the square of -- you could factor this , but we did it on purpose -- it 's x plus 3 squared and then we have plus 4 times -- what is this right here ? that 's y minus 1 squared . you might want to review factoring of polynomial or completing the square if you found that step a little daunting . and then we have plus 49 is equal to 0 plus 81 plus 84 is equal to 85 . all right , so now we have 9 times plus 3 squared plus 4 times y minus 1 squared . and let 's subtract 49 from both sides . that is equal to -- let 's see if i subtract 50 from 85 i get 35 , so if i subtract 49 , i get 36 . and now we are getting close to the standard form of something , but remember all the standard forms we did except for the circle -- we had a y -- and we know this is n't a circle , because we have these weird coefficients , well not weird but different coefficients in front of these terms . so to get the 1 on the right-hand side let 's divide everything by 36 . if you divide everything by 36 , this term becomes x plus 3 squared over see 9 over 36 is the same thing as 1 over 4 , and then you have plus y minus 1 squared 4 over 36 is the same thing as 1 over 9 and all of that is equal to 1 . and there you go . we have it in the standard form , and you can see our intuition at the beginning the problem was correct . this is indeed an ellipse , and now we can actually graph it . so first of all , actually good place to start , where 's the center of the this ellipse going to be ? it 's going to be x is equal to negative 3 . what x value makes this whole terms 0 ? so it 's going to be x is equal to minus 3 , and y is going to be equal to 1 . what y value makes this term 0 ? y is equal to 1 . that 's our center . so let 's graph that , and then we can draw the ellipse . it 's going to be in the negative quadrant . this is our x-axis and this is our y-axis . and then the center of our ellipse is at minus 3 and positive 1 , so that 's the center . and then , what is the radius in the x direction ? we just take the square root of this , so it 's 2 . so in the x direction we go two to the right . we go two to the left . and in the y direction , what do we do ? well we go up three and down three . the square root of this . let me do that . remember you have to take the square root of both of those . the vertical axis is actually the major radius or the semi-major axis is 3 , because that 's the longer one . and then the 2 is the minor radius , because that 's the shorter one . and now we 're ready to draw this ellipse . i 'll draw it in brown . let me see if i can do this properly . i have a shaky hand . all right , it looks something like that . and there you go . we took this kind of crazy looking thing , and all we did is algebraically manipulate it . we just completed the squares with the x 's and the y terms . and then we divided both sides by this number right here and we got it into the standard form . we said oh this is an ellipse . we have both of these terms , they 're both positive , we 're adding we 're not subtracting , they have different coefficients underneath here . so we 're ready to go over the ellipse , and we realized that the center was at minus 3,1 , and then we just drew the major radius , or the major axis and the minor axis . see you in the next video .
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what y value makes this term 0 ? y is equal to 1 . that 's our center .
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are you allowed to plot the ellipse when the two expressions are equal to a number other than 1 ?
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the standard question you often get in your algebra class is they will give you this equation and it 'll say identify the conic section and graph it if you can . and the equation they give you wo n't be in the standard form , because if it was you could just kind of pattern match with what i showed in some of the previous videos and you 'd be able to get it . so let 's do a question like and let 's see if we can figure it out . so what i have here is 9x squared plus 4y squared plus 54x minus 8y plus 49 is equal to 0 . and once again , i mean who knows what this is it 's just not in the standard form . and actually one quick clue to tell you what this is you look at the x squared and the y squared terms if there are . if there 's only an x squared term and then there 's just a y and not a y squared term , then you 're probably dealing with a parabola , and we 'll go into that more later . or if it 's the other way around , if it 's just an x term and a y squared term , it 's probably a parabola . but assuming that we 're dealing with a circle , an ellipse , or a hyperbola , there will be an x squared term and a y squared term . if they both kind of have the same number in front of them , that 's a pretty good clue that we 're going to be dealing with a circle . if they both have different numbers , but they 're both positive in front of them , that 's a pretty good clue we 're probably going to be dealing with an ellipse . if one of them has a negative number in front of them and the other one has a positive number , that tells you that we 're probably going to be dealing with a hyperbola . but with that said , i mean that might help you identify things very quickly at this level , but it does n't help you graph it or get into the standard form . so let 's get it in the standard form . and the key to getting it in the standard form is really just completing the square . and i encourage you to re-watch the completing the square video , because that 's all we 're going to do right here to get it into the standard form . so the first thing i like to do to complete the square , and you 're going to have to do it for the x variables and for the y terms , is group the x and y terms . let 's see . the x terms are 9x squared plus 54x . and let 's do the y terms in magenta . so then you have plus 4y squared minus 8y and then you have -- let me do this in a different color -- plus 49 is equal to 0 . and so the easy thing to do when you complete the square , the thing i like to do is , it 's very clear we can factor out a 9 out of both of these numbers , and we can factor out a 4 out of both of those . let 's do that , because that will help us complete the square . so this is the same thing is 9 times x squared plus 9 times 6 is 54 , 6x . i 'm going to add something else here , but i 'll leave it blank for now . plus 4 times y squared minus 2y i 'm probably going to add something here too , so i 'll leave it blank for now . plus 49 is equal to 0 . so what are we going to add here ? we 're going to complete the square . we want to add some number here so that this whole three term expression becomes a perfect square . likewise , we 're going to add some number here , so this three term number expression becomes a perfect square . and of course whatever we add on the side , we 're going to have to multiply it by 9 , because we 're really adding nine times that . and add it on to that side . whatever we add here , we 're going to have to multiply it times 4 and add it on that side . if i put a 1 here , it 's really like as if i had a 4 here , because 1 times 4 is 4 and if i had a 1 here it 's 1 times 9 . so 9 there . let 's do that . when we complete the square , we just take half of this coefficient . this coefficient is 6 , we take half of it is 3 , we square it , we get a 9 . remember it 's an equation , so what you do to one side , you have to do to the other . so if we added a 9 here , we 're actually adding 9 times 9 to the left-hand side of the equation , so we have to add 81 to the right-hand side to make the equation still hold . and you could kind of view it if we go back up here . this is the same thing , just to make that clear as if i added plus 81 right here . of course i would have had to add plus 81 up here . now let 's go to the y terms . you take half of this coefficient is minus 2 , half of that is minus 1 . you square it , you get plus 1 . 1 times 4 , so we 're really adding 4 to the left-hand side of the equation . and just so you understand what i did here . this is equivalent as if i just added a 4 here , and then i later just factored out this 4 . and so what does this become ? this expression is 9 times what ? this is the square of -- you could factor this , but we did it on purpose -- it 's x plus 3 squared and then we have plus 4 times -- what is this right here ? that 's y minus 1 squared . you might want to review factoring of polynomial or completing the square if you found that step a little daunting . and then we have plus 49 is equal to 0 plus 81 plus 84 is equal to 85 . all right , so now we have 9 times plus 3 squared plus 4 times y minus 1 squared . and let 's subtract 49 from both sides . that is equal to -- let 's see if i subtract 50 from 85 i get 35 , so if i subtract 49 , i get 36 . and now we are getting close to the standard form of something , but remember all the standard forms we did except for the circle -- we had a y -- and we know this is n't a circle , because we have these weird coefficients , well not weird but different coefficients in front of these terms . so to get the 1 on the right-hand side let 's divide everything by 36 . if you divide everything by 36 , this term becomes x plus 3 squared over see 9 over 36 is the same thing as 1 over 4 , and then you have plus y minus 1 squared 4 over 36 is the same thing as 1 over 9 and all of that is equal to 1 . and there you go . we have it in the standard form , and you can see our intuition at the beginning the problem was correct . this is indeed an ellipse , and now we can actually graph it . so first of all , actually good place to start , where 's the center of the this ellipse going to be ? it 's going to be x is equal to negative 3 . what x value makes this whole terms 0 ? so it 's going to be x is equal to minus 3 , and y is going to be equal to 1 . what y value makes this term 0 ? y is equal to 1 . that 's our center . so let 's graph that , and then we can draw the ellipse . it 's going to be in the negative quadrant . this is our x-axis and this is our y-axis . and then the center of our ellipse is at minus 3 and positive 1 , so that 's the center . and then , what is the radius in the x direction ? we just take the square root of this , so it 's 2 . so in the x direction we go two to the right . we go two to the left . and in the y direction , what do we do ? well we go up three and down three . the square root of this . let me do that . remember you have to take the square root of both of those . the vertical axis is actually the major radius or the semi-major axis is 3 , because that 's the longer one . and then the 2 is the minor radius , because that 's the shorter one . and now we 're ready to draw this ellipse . i 'll draw it in brown . let me see if i can do this properly . i have a shaky hand . all right , it looks something like that . and there you go . we took this kind of crazy looking thing , and all we did is algebraically manipulate it . we just completed the squares with the x 's and the y terms . and then we divided both sides by this number right here and we got it into the standard form . we said oh this is an ellipse . we have both of these terms , they 're both positive , we 're adding we 're not subtracting , they have different coefficients underneath here . so we 're ready to go over the ellipse , and we realized that the center was at minus 3,1 , and then we just drew the major radius , or the major axis and the minor axis . see you in the next video .
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and now we are getting close to the standard form of something , but remember all the standard forms we did except for the circle -- we had a y -- and we know this is n't a circle , because we have these weird coefficients , well not weird but different coefficients in front of these terms . so to get the 1 on the right-hand side let 's divide everything by 36 . if you divide everything by 36 , this term becomes x plus 3 squared over see 9 over 36 is the same thing as 1 over 4 , and then you have plus y minus 1 squared 4 over 36 is the same thing as 1 over 9 and all of that is equal to 1 . and there you go .
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for example , can you plot ( x+3 ) 2 + ( x-1 ) 2 = 36 or must it be in standard form ?
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the standard question you often get in your algebra class is they will give you this equation and it 'll say identify the conic section and graph it if you can . and the equation they give you wo n't be in the standard form , because if it was you could just kind of pattern match with what i showed in some of the previous videos and you 'd be able to get it . so let 's do a question like and let 's see if we can figure it out . so what i have here is 9x squared plus 4y squared plus 54x minus 8y plus 49 is equal to 0 . and once again , i mean who knows what this is it 's just not in the standard form . and actually one quick clue to tell you what this is you look at the x squared and the y squared terms if there are . if there 's only an x squared term and then there 's just a y and not a y squared term , then you 're probably dealing with a parabola , and we 'll go into that more later . or if it 's the other way around , if it 's just an x term and a y squared term , it 's probably a parabola . but assuming that we 're dealing with a circle , an ellipse , or a hyperbola , there will be an x squared term and a y squared term . if they both kind of have the same number in front of them , that 's a pretty good clue that we 're going to be dealing with a circle . if they both have different numbers , but they 're both positive in front of them , that 's a pretty good clue we 're probably going to be dealing with an ellipse . if one of them has a negative number in front of them and the other one has a positive number , that tells you that we 're probably going to be dealing with a hyperbola . but with that said , i mean that might help you identify things very quickly at this level , but it does n't help you graph it or get into the standard form . so let 's get it in the standard form . and the key to getting it in the standard form is really just completing the square . and i encourage you to re-watch the completing the square video , because that 's all we 're going to do right here to get it into the standard form . so the first thing i like to do to complete the square , and you 're going to have to do it for the x variables and for the y terms , is group the x and y terms . let 's see . the x terms are 9x squared plus 54x . and let 's do the y terms in magenta . so then you have plus 4y squared minus 8y and then you have -- let me do this in a different color -- plus 49 is equal to 0 . and so the easy thing to do when you complete the square , the thing i like to do is , it 's very clear we can factor out a 9 out of both of these numbers , and we can factor out a 4 out of both of those . let 's do that , because that will help us complete the square . so this is the same thing is 9 times x squared plus 9 times 6 is 54 , 6x . i 'm going to add something else here , but i 'll leave it blank for now . plus 4 times y squared minus 2y i 'm probably going to add something here too , so i 'll leave it blank for now . plus 49 is equal to 0 . so what are we going to add here ? we 're going to complete the square . we want to add some number here so that this whole three term expression becomes a perfect square . likewise , we 're going to add some number here , so this three term number expression becomes a perfect square . and of course whatever we add on the side , we 're going to have to multiply it by 9 , because we 're really adding nine times that . and add it on to that side . whatever we add here , we 're going to have to multiply it times 4 and add it on that side . if i put a 1 here , it 's really like as if i had a 4 here , because 1 times 4 is 4 and if i had a 1 here it 's 1 times 9 . so 9 there . let 's do that . when we complete the square , we just take half of this coefficient . this coefficient is 6 , we take half of it is 3 , we square it , we get a 9 . remember it 's an equation , so what you do to one side , you have to do to the other . so if we added a 9 here , we 're actually adding 9 times 9 to the left-hand side of the equation , so we have to add 81 to the right-hand side to make the equation still hold . and you could kind of view it if we go back up here . this is the same thing , just to make that clear as if i added plus 81 right here . of course i would have had to add plus 81 up here . now let 's go to the y terms . you take half of this coefficient is minus 2 , half of that is minus 1 . you square it , you get plus 1 . 1 times 4 , so we 're really adding 4 to the left-hand side of the equation . and just so you understand what i did here . this is equivalent as if i just added a 4 here , and then i later just factored out this 4 . and so what does this become ? this expression is 9 times what ? this is the square of -- you could factor this , but we did it on purpose -- it 's x plus 3 squared and then we have plus 4 times -- what is this right here ? that 's y minus 1 squared . you might want to review factoring of polynomial or completing the square if you found that step a little daunting . and then we have plus 49 is equal to 0 plus 81 plus 84 is equal to 85 . all right , so now we have 9 times plus 3 squared plus 4 times y minus 1 squared . and let 's subtract 49 from both sides . that is equal to -- let 's see if i subtract 50 from 85 i get 35 , so if i subtract 49 , i get 36 . and now we are getting close to the standard form of something , but remember all the standard forms we did except for the circle -- we had a y -- and we know this is n't a circle , because we have these weird coefficients , well not weird but different coefficients in front of these terms . so to get the 1 on the right-hand side let 's divide everything by 36 . if you divide everything by 36 , this term becomes x plus 3 squared over see 9 over 36 is the same thing as 1 over 4 , and then you have plus y minus 1 squared 4 over 36 is the same thing as 1 over 9 and all of that is equal to 1 . and there you go . we have it in the standard form , and you can see our intuition at the beginning the problem was correct . this is indeed an ellipse , and now we can actually graph it . so first of all , actually good place to start , where 's the center of the this ellipse going to be ? it 's going to be x is equal to negative 3 . what x value makes this whole terms 0 ? so it 's going to be x is equal to minus 3 , and y is going to be equal to 1 . what y value makes this term 0 ? y is equal to 1 . that 's our center . so let 's graph that , and then we can draw the ellipse . it 's going to be in the negative quadrant . this is our x-axis and this is our y-axis . and then the center of our ellipse is at minus 3 and positive 1 , so that 's the center . and then , what is the radius in the x direction ? we just take the square root of this , so it 's 2 . so in the x direction we go two to the right . we go two to the left . and in the y direction , what do we do ? well we go up three and down three . the square root of this . let me do that . remember you have to take the square root of both of those . the vertical axis is actually the major radius or the semi-major axis is 3 , because that 's the longer one . and then the 2 is the minor radius , because that 's the shorter one . and now we 're ready to draw this ellipse . i 'll draw it in brown . let me see if i can do this properly . i have a shaky hand . all right , it looks something like that . and there you go . we took this kind of crazy looking thing , and all we did is algebraically manipulate it . we just completed the squares with the x 's and the y terms . and then we divided both sides by this number right here and we got it into the standard form . we said oh this is an ellipse . we have both of these terms , they 're both positive , we 're adding we 're not subtracting , they have different coefficients underneath here . so we 're ready to go over the ellipse , and we realized that the center was at minus 3,1 , and then we just drew the major radius , or the major axis and the minor axis . see you in the next video .
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so it 's going to be x is equal to minus 3 , and y is going to be equal to 1 . what y value makes this term 0 ? y is equal to 1 .
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what if you have an `` xy '' term in the equation ?
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the standard question you often get in your algebra class is they will give you this equation and it 'll say identify the conic section and graph it if you can . and the equation they give you wo n't be in the standard form , because if it was you could just kind of pattern match with what i showed in some of the previous videos and you 'd be able to get it . so let 's do a question like and let 's see if we can figure it out . so what i have here is 9x squared plus 4y squared plus 54x minus 8y plus 49 is equal to 0 . and once again , i mean who knows what this is it 's just not in the standard form . and actually one quick clue to tell you what this is you look at the x squared and the y squared terms if there are . if there 's only an x squared term and then there 's just a y and not a y squared term , then you 're probably dealing with a parabola , and we 'll go into that more later . or if it 's the other way around , if it 's just an x term and a y squared term , it 's probably a parabola . but assuming that we 're dealing with a circle , an ellipse , or a hyperbola , there will be an x squared term and a y squared term . if they both kind of have the same number in front of them , that 's a pretty good clue that we 're going to be dealing with a circle . if they both have different numbers , but they 're both positive in front of them , that 's a pretty good clue we 're probably going to be dealing with an ellipse . if one of them has a negative number in front of them and the other one has a positive number , that tells you that we 're probably going to be dealing with a hyperbola . but with that said , i mean that might help you identify things very quickly at this level , but it does n't help you graph it or get into the standard form . so let 's get it in the standard form . and the key to getting it in the standard form is really just completing the square . and i encourage you to re-watch the completing the square video , because that 's all we 're going to do right here to get it into the standard form . so the first thing i like to do to complete the square , and you 're going to have to do it for the x variables and for the y terms , is group the x and y terms . let 's see . the x terms are 9x squared plus 54x . and let 's do the y terms in magenta . so then you have plus 4y squared minus 8y and then you have -- let me do this in a different color -- plus 49 is equal to 0 . and so the easy thing to do when you complete the square , the thing i like to do is , it 's very clear we can factor out a 9 out of both of these numbers , and we can factor out a 4 out of both of those . let 's do that , because that will help us complete the square . so this is the same thing is 9 times x squared plus 9 times 6 is 54 , 6x . i 'm going to add something else here , but i 'll leave it blank for now . plus 4 times y squared minus 2y i 'm probably going to add something here too , so i 'll leave it blank for now . plus 49 is equal to 0 . so what are we going to add here ? we 're going to complete the square . we want to add some number here so that this whole three term expression becomes a perfect square . likewise , we 're going to add some number here , so this three term number expression becomes a perfect square . and of course whatever we add on the side , we 're going to have to multiply it by 9 , because we 're really adding nine times that . and add it on to that side . whatever we add here , we 're going to have to multiply it times 4 and add it on that side . if i put a 1 here , it 's really like as if i had a 4 here , because 1 times 4 is 4 and if i had a 1 here it 's 1 times 9 . so 9 there . let 's do that . when we complete the square , we just take half of this coefficient . this coefficient is 6 , we take half of it is 3 , we square it , we get a 9 . remember it 's an equation , so what you do to one side , you have to do to the other . so if we added a 9 here , we 're actually adding 9 times 9 to the left-hand side of the equation , so we have to add 81 to the right-hand side to make the equation still hold . and you could kind of view it if we go back up here . this is the same thing , just to make that clear as if i added plus 81 right here . of course i would have had to add plus 81 up here . now let 's go to the y terms . you take half of this coefficient is minus 2 , half of that is minus 1 . you square it , you get plus 1 . 1 times 4 , so we 're really adding 4 to the left-hand side of the equation . and just so you understand what i did here . this is equivalent as if i just added a 4 here , and then i later just factored out this 4 . and so what does this become ? this expression is 9 times what ? this is the square of -- you could factor this , but we did it on purpose -- it 's x plus 3 squared and then we have plus 4 times -- what is this right here ? that 's y minus 1 squared . you might want to review factoring of polynomial or completing the square if you found that step a little daunting . and then we have plus 49 is equal to 0 plus 81 plus 84 is equal to 85 . all right , so now we have 9 times plus 3 squared plus 4 times y minus 1 squared . and let 's subtract 49 from both sides . that is equal to -- let 's see if i subtract 50 from 85 i get 35 , so if i subtract 49 , i get 36 . and now we are getting close to the standard form of something , but remember all the standard forms we did except for the circle -- we had a y -- and we know this is n't a circle , because we have these weird coefficients , well not weird but different coefficients in front of these terms . so to get the 1 on the right-hand side let 's divide everything by 36 . if you divide everything by 36 , this term becomes x plus 3 squared over see 9 over 36 is the same thing as 1 over 4 , and then you have plus y minus 1 squared 4 over 36 is the same thing as 1 over 9 and all of that is equal to 1 . and there you go . we have it in the standard form , and you can see our intuition at the beginning the problem was correct . this is indeed an ellipse , and now we can actually graph it . so first of all , actually good place to start , where 's the center of the this ellipse going to be ? it 's going to be x is equal to negative 3 . what x value makes this whole terms 0 ? so it 's going to be x is equal to minus 3 , and y is going to be equal to 1 . what y value makes this term 0 ? y is equal to 1 . that 's our center . so let 's graph that , and then we can draw the ellipse . it 's going to be in the negative quadrant . this is our x-axis and this is our y-axis . and then the center of our ellipse is at minus 3 and positive 1 , so that 's the center . and then , what is the radius in the x direction ? we just take the square root of this , so it 's 2 . so in the x direction we go two to the right . we go two to the left . and in the y direction , what do we do ? well we go up three and down three . the square root of this . let me do that . remember you have to take the square root of both of those . the vertical axis is actually the major radius or the semi-major axis is 3 , because that 's the longer one . and then the 2 is the minor radius , because that 's the shorter one . and now we 're ready to draw this ellipse . i 'll draw it in brown . let me see if i can do this properly . i have a shaky hand . all right , it looks something like that . and there you go . we took this kind of crazy looking thing , and all we did is algebraically manipulate it . we just completed the squares with the x 's and the y terms . and then we divided both sides by this number right here and we got it into the standard form . we said oh this is an ellipse . we have both of these terms , they 're both positive , we 're adding we 're not subtracting , they have different coefficients underneath here . so we 're ready to go over the ellipse , and we realized that the center was at minus 3,1 , and then we just drew the major radius , or the major axis and the minor axis . see you in the next video .
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this is the same thing , just to make that clear as if i added plus 81 right here . of course i would have had to add plus 81 up here . now let 's go to the y terms .
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what would the foci of this equation be ?
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the standard question you often get in your algebra class is they will give you this equation and it 'll say identify the conic section and graph it if you can . and the equation they give you wo n't be in the standard form , because if it was you could just kind of pattern match with what i showed in some of the previous videos and you 'd be able to get it . so let 's do a question like and let 's see if we can figure it out . so what i have here is 9x squared plus 4y squared plus 54x minus 8y plus 49 is equal to 0 . and once again , i mean who knows what this is it 's just not in the standard form . and actually one quick clue to tell you what this is you look at the x squared and the y squared terms if there are . if there 's only an x squared term and then there 's just a y and not a y squared term , then you 're probably dealing with a parabola , and we 'll go into that more later . or if it 's the other way around , if it 's just an x term and a y squared term , it 's probably a parabola . but assuming that we 're dealing with a circle , an ellipse , or a hyperbola , there will be an x squared term and a y squared term . if they both kind of have the same number in front of them , that 's a pretty good clue that we 're going to be dealing with a circle . if they both have different numbers , but they 're both positive in front of them , that 's a pretty good clue we 're probably going to be dealing with an ellipse . if one of them has a negative number in front of them and the other one has a positive number , that tells you that we 're probably going to be dealing with a hyperbola . but with that said , i mean that might help you identify things very quickly at this level , but it does n't help you graph it or get into the standard form . so let 's get it in the standard form . and the key to getting it in the standard form is really just completing the square . and i encourage you to re-watch the completing the square video , because that 's all we 're going to do right here to get it into the standard form . so the first thing i like to do to complete the square , and you 're going to have to do it for the x variables and for the y terms , is group the x and y terms . let 's see . the x terms are 9x squared plus 54x . and let 's do the y terms in magenta . so then you have plus 4y squared minus 8y and then you have -- let me do this in a different color -- plus 49 is equal to 0 . and so the easy thing to do when you complete the square , the thing i like to do is , it 's very clear we can factor out a 9 out of both of these numbers , and we can factor out a 4 out of both of those . let 's do that , because that will help us complete the square . so this is the same thing is 9 times x squared plus 9 times 6 is 54 , 6x . i 'm going to add something else here , but i 'll leave it blank for now . plus 4 times y squared minus 2y i 'm probably going to add something here too , so i 'll leave it blank for now . plus 49 is equal to 0 . so what are we going to add here ? we 're going to complete the square . we want to add some number here so that this whole three term expression becomes a perfect square . likewise , we 're going to add some number here , so this three term number expression becomes a perfect square . and of course whatever we add on the side , we 're going to have to multiply it by 9 , because we 're really adding nine times that . and add it on to that side . whatever we add here , we 're going to have to multiply it times 4 and add it on that side . if i put a 1 here , it 's really like as if i had a 4 here , because 1 times 4 is 4 and if i had a 1 here it 's 1 times 9 . so 9 there . let 's do that . when we complete the square , we just take half of this coefficient . this coefficient is 6 , we take half of it is 3 , we square it , we get a 9 . remember it 's an equation , so what you do to one side , you have to do to the other . so if we added a 9 here , we 're actually adding 9 times 9 to the left-hand side of the equation , so we have to add 81 to the right-hand side to make the equation still hold . and you could kind of view it if we go back up here . this is the same thing , just to make that clear as if i added plus 81 right here . of course i would have had to add plus 81 up here . now let 's go to the y terms . you take half of this coefficient is minus 2 , half of that is minus 1 . you square it , you get plus 1 . 1 times 4 , so we 're really adding 4 to the left-hand side of the equation . and just so you understand what i did here . this is equivalent as if i just added a 4 here , and then i later just factored out this 4 . and so what does this become ? this expression is 9 times what ? this is the square of -- you could factor this , but we did it on purpose -- it 's x plus 3 squared and then we have plus 4 times -- what is this right here ? that 's y minus 1 squared . you might want to review factoring of polynomial or completing the square if you found that step a little daunting . and then we have plus 49 is equal to 0 plus 81 plus 84 is equal to 85 . all right , so now we have 9 times plus 3 squared plus 4 times y minus 1 squared . and let 's subtract 49 from both sides . that is equal to -- let 's see if i subtract 50 from 85 i get 35 , so if i subtract 49 , i get 36 . and now we are getting close to the standard form of something , but remember all the standard forms we did except for the circle -- we had a y -- and we know this is n't a circle , because we have these weird coefficients , well not weird but different coefficients in front of these terms . so to get the 1 on the right-hand side let 's divide everything by 36 . if you divide everything by 36 , this term becomes x plus 3 squared over see 9 over 36 is the same thing as 1 over 4 , and then you have plus y minus 1 squared 4 over 36 is the same thing as 1 over 9 and all of that is equal to 1 . and there you go . we have it in the standard form , and you can see our intuition at the beginning the problem was correct . this is indeed an ellipse , and now we can actually graph it . so first of all , actually good place to start , where 's the center of the this ellipse going to be ? it 's going to be x is equal to negative 3 . what x value makes this whole terms 0 ? so it 's going to be x is equal to minus 3 , and y is going to be equal to 1 . what y value makes this term 0 ? y is equal to 1 . that 's our center . so let 's graph that , and then we can draw the ellipse . it 's going to be in the negative quadrant . this is our x-axis and this is our y-axis . and then the center of our ellipse is at minus 3 and positive 1 , so that 's the center . and then , what is the radius in the x direction ? we just take the square root of this , so it 's 2 . so in the x direction we go two to the right . we go two to the left . and in the y direction , what do we do ? well we go up three and down three . the square root of this . let me do that . remember you have to take the square root of both of those . the vertical axis is actually the major radius or the semi-major axis is 3 , because that 's the longer one . and then the 2 is the minor radius , because that 's the shorter one . and now we 're ready to draw this ellipse . i 'll draw it in brown . let me see if i can do this properly . i have a shaky hand . all right , it looks something like that . and there you go . we took this kind of crazy looking thing , and all we did is algebraically manipulate it . we just completed the squares with the x 's and the y terms . and then we divided both sides by this number right here and we got it into the standard form . we said oh this is an ellipse . we have both of these terms , they 're both positive , we 're adding we 're not subtracting , they have different coefficients underneath here . so we 're ready to go over the ellipse , and we realized that the center was at minus 3,1 , and then we just drew the major radius , or the major axis and the minor axis . see you in the next video .
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so in the x direction we go two to the right . we go two to the left . and in the y direction , what do we do ?
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if you have an equationin two variables , how can you quickly identify it as an equation for an ellipse , hyperbola or parabola ?
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the standard question you often get in your algebra class is they will give you this equation and it 'll say identify the conic section and graph it if you can . and the equation they give you wo n't be in the standard form , because if it was you could just kind of pattern match with what i showed in some of the previous videos and you 'd be able to get it . so let 's do a question like and let 's see if we can figure it out . so what i have here is 9x squared plus 4y squared plus 54x minus 8y plus 49 is equal to 0 . and once again , i mean who knows what this is it 's just not in the standard form . and actually one quick clue to tell you what this is you look at the x squared and the y squared terms if there are . if there 's only an x squared term and then there 's just a y and not a y squared term , then you 're probably dealing with a parabola , and we 'll go into that more later . or if it 's the other way around , if it 's just an x term and a y squared term , it 's probably a parabola . but assuming that we 're dealing with a circle , an ellipse , or a hyperbola , there will be an x squared term and a y squared term . if they both kind of have the same number in front of them , that 's a pretty good clue that we 're going to be dealing with a circle . if they both have different numbers , but they 're both positive in front of them , that 's a pretty good clue we 're probably going to be dealing with an ellipse . if one of them has a negative number in front of them and the other one has a positive number , that tells you that we 're probably going to be dealing with a hyperbola . but with that said , i mean that might help you identify things very quickly at this level , but it does n't help you graph it or get into the standard form . so let 's get it in the standard form . and the key to getting it in the standard form is really just completing the square . and i encourage you to re-watch the completing the square video , because that 's all we 're going to do right here to get it into the standard form . so the first thing i like to do to complete the square , and you 're going to have to do it for the x variables and for the y terms , is group the x and y terms . let 's see . the x terms are 9x squared plus 54x . and let 's do the y terms in magenta . so then you have plus 4y squared minus 8y and then you have -- let me do this in a different color -- plus 49 is equal to 0 . and so the easy thing to do when you complete the square , the thing i like to do is , it 's very clear we can factor out a 9 out of both of these numbers , and we can factor out a 4 out of both of those . let 's do that , because that will help us complete the square . so this is the same thing is 9 times x squared plus 9 times 6 is 54 , 6x . i 'm going to add something else here , but i 'll leave it blank for now . plus 4 times y squared minus 2y i 'm probably going to add something here too , so i 'll leave it blank for now . plus 49 is equal to 0 . so what are we going to add here ? we 're going to complete the square . we want to add some number here so that this whole three term expression becomes a perfect square . likewise , we 're going to add some number here , so this three term number expression becomes a perfect square . and of course whatever we add on the side , we 're going to have to multiply it by 9 , because we 're really adding nine times that . and add it on to that side . whatever we add here , we 're going to have to multiply it times 4 and add it on that side . if i put a 1 here , it 's really like as if i had a 4 here , because 1 times 4 is 4 and if i had a 1 here it 's 1 times 9 . so 9 there . let 's do that . when we complete the square , we just take half of this coefficient . this coefficient is 6 , we take half of it is 3 , we square it , we get a 9 . remember it 's an equation , so what you do to one side , you have to do to the other . so if we added a 9 here , we 're actually adding 9 times 9 to the left-hand side of the equation , so we have to add 81 to the right-hand side to make the equation still hold . and you could kind of view it if we go back up here . this is the same thing , just to make that clear as if i added plus 81 right here . of course i would have had to add plus 81 up here . now let 's go to the y terms . you take half of this coefficient is minus 2 , half of that is minus 1 . you square it , you get plus 1 . 1 times 4 , so we 're really adding 4 to the left-hand side of the equation . and just so you understand what i did here . this is equivalent as if i just added a 4 here , and then i later just factored out this 4 . and so what does this become ? this expression is 9 times what ? this is the square of -- you could factor this , but we did it on purpose -- it 's x plus 3 squared and then we have plus 4 times -- what is this right here ? that 's y minus 1 squared . you might want to review factoring of polynomial or completing the square if you found that step a little daunting . and then we have plus 49 is equal to 0 plus 81 plus 84 is equal to 85 . all right , so now we have 9 times plus 3 squared plus 4 times y minus 1 squared . and let 's subtract 49 from both sides . that is equal to -- let 's see if i subtract 50 from 85 i get 35 , so if i subtract 49 , i get 36 . and now we are getting close to the standard form of something , but remember all the standard forms we did except for the circle -- we had a y -- and we know this is n't a circle , because we have these weird coefficients , well not weird but different coefficients in front of these terms . so to get the 1 on the right-hand side let 's divide everything by 36 . if you divide everything by 36 , this term becomes x plus 3 squared over see 9 over 36 is the same thing as 1 over 4 , and then you have plus y minus 1 squared 4 over 36 is the same thing as 1 over 9 and all of that is equal to 1 . and there you go . we have it in the standard form , and you can see our intuition at the beginning the problem was correct . this is indeed an ellipse , and now we can actually graph it . so first of all , actually good place to start , where 's the center of the this ellipse going to be ? it 's going to be x is equal to negative 3 . what x value makes this whole terms 0 ? so it 's going to be x is equal to minus 3 , and y is going to be equal to 1 . what y value makes this term 0 ? y is equal to 1 . that 's our center . so let 's graph that , and then we can draw the ellipse . it 's going to be in the negative quadrant . this is our x-axis and this is our y-axis . and then the center of our ellipse is at minus 3 and positive 1 , so that 's the center . and then , what is the radius in the x direction ? we just take the square root of this , so it 's 2 . so in the x direction we go two to the right . we go two to the left . and in the y direction , what do we do ? well we go up three and down three . the square root of this . let me do that . remember you have to take the square root of both of those . the vertical axis is actually the major radius or the semi-major axis is 3 , because that 's the longer one . and then the 2 is the minor radius , because that 's the shorter one . and now we 're ready to draw this ellipse . i 'll draw it in brown . let me see if i can do this properly . i have a shaky hand . all right , it looks something like that . and there you go . we took this kind of crazy looking thing , and all we did is algebraically manipulate it . we just completed the squares with the x 's and the y terms . and then we divided both sides by this number right here and we got it into the standard form . we said oh this is an ellipse . we have both of these terms , they 're both positive , we 're adding we 're not subtracting , they have different coefficients underneath here . so we 're ready to go over the ellipse , and we realized that the center was at minus 3,1 , and then we just drew the major radius , or the major axis and the minor axis . see you in the next video .
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but with that said , i mean that might help you identify things very quickly at this level , but it does n't help you graph it or get into the standard form . so let 's get it in the standard form . and the key to getting it in the standard form is really just completing the square .
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if you 're graphing an equation from standard form , what do the variables stand for ?
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the standard question you often get in your algebra class is they will give you this equation and it 'll say identify the conic section and graph it if you can . and the equation they give you wo n't be in the standard form , because if it was you could just kind of pattern match with what i showed in some of the previous videos and you 'd be able to get it . so let 's do a question like and let 's see if we can figure it out . so what i have here is 9x squared plus 4y squared plus 54x minus 8y plus 49 is equal to 0 . and once again , i mean who knows what this is it 's just not in the standard form . and actually one quick clue to tell you what this is you look at the x squared and the y squared terms if there are . if there 's only an x squared term and then there 's just a y and not a y squared term , then you 're probably dealing with a parabola , and we 'll go into that more later . or if it 's the other way around , if it 's just an x term and a y squared term , it 's probably a parabola . but assuming that we 're dealing with a circle , an ellipse , or a hyperbola , there will be an x squared term and a y squared term . if they both kind of have the same number in front of them , that 's a pretty good clue that we 're going to be dealing with a circle . if they both have different numbers , but they 're both positive in front of them , that 's a pretty good clue we 're probably going to be dealing with an ellipse . if one of them has a negative number in front of them and the other one has a positive number , that tells you that we 're probably going to be dealing with a hyperbola . but with that said , i mean that might help you identify things very quickly at this level , but it does n't help you graph it or get into the standard form . so let 's get it in the standard form . and the key to getting it in the standard form is really just completing the square . and i encourage you to re-watch the completing the square video , because that 's all we 're going to do right here to get it into the standard form . so the first thing i like to do to complete the square , and you 're going to have to do it for the x variables and for the y terms , is group the x and y terms . let 's see . the x terms are 9x squared plus 54x . and let 's do the y terms in magenta . so then you have plus 4y squared minus 8y and then you have -- let me do this in a different color -- plus 49 is equal to 0 . and so the easy thing to do when you complete the square , the thing i like to do is , it 's very clear we can factor out a 9 out of both of these numbers , and we can factor out a 4 out of both of those . let 's do that , because that will help us complete the square . so this is the same thing is 9 times x squared plus 9 times 6 is 54 , 6x . i 'm going to add something else here , but i 'll leave it blank for now . plus 4 times y squared minus 2y i 'm probably going to add something here too , so i 'll leave it blank for now . plus 49 is equal to 0 . so what are we going to add here ? we 're going to complete the square . we want to add some number here so that this whole three term expression becomes a perfect square . likewise , we 're going to add some number here , so this three term number expression becomes a perfect square . and of course whatever we add on the side , we 're going to have to multiply it by 9 , because we 're really adding nine times that . and add it on to that side . whatever we add here , we 're going to have to multiply it times 4 and add it on that side . if i put a 1 here , it 's really like as if i had a 4 here , because 1 times 4 is 4 and if i had a 1 here it 's 1 times 9 . so 9 there . let 's do that . when we complete the square , we just take half of this coefficient . this coefficient is 6 , we take half of it is 3 , we square it , we get a 9 . remember it 's an equation , so what you do to one side , you have to do to the other . so if we added a 9 here , we 're actually adding 9 times 9 to the left-hand side of the equation , so we have to add 81 to the right-hand side to make the equation still hold . and you could kind of view it if we go back up here . this is the same thing , just to make that clear as if i added plus 81 right here . of course i would have had to add plus 81 up here . now let 's go to the y terms . you take half of this coefficient is minus 2 , half of that is minus 1 . you square it , you get plus 1 . 1 times 4 , so we 're really adding 4 to the left-hand side of the equation . and just so you understand what i did here . this is equivalent as if i just added a 4 here , and then i later just factored out this 4 . and so what does this become ? this expression is 9 times what ? this is the square of -- you could factor this , but we did it on purpose -- it 's x plus 3 squared and then we have plus 4 times -- what is this right here ? that 's y minus 1 squared . you might want to review factoring of polynomial or completing the square if you found that step a little daunting . and then we have plus 49 is equal to 0 plus 81 plus 84 is equal to 85 . all right , so now we have 9 times plus 3 squared plus 4 times y minus 1 squared . and let 's subtract 49 from both sides . that is equal to -- let 's see if i subtract 50 from 85 i get 35 , so if i subtract 49 , i get 36 . and now we are getting close to the standard form of something , but remember all the standard forms we did except for the circle -- we had a y -- and we know this is n't a circle , because we have these weird coefficients , well not weird but different coefficients in front of these terms . so to get the 1 on the right-hand side let 's divide everything by 36 . if you divide everything by 36 , this term becomes x plus 3 squared over see 9 over 36 is the same thing as 1 over 4 , and then you have plus y minus 1 squared 4 over 36 is the same thing as 1 over 9 and all of that is equal to 1 . and there you go . we have it in the standard form , and you can see our intuition at the beginning the problem was correct . this is indeed an ellipse , and now we can actually graph it . so first of all , actually good place to start , where 's the center of the this ellipse going to be ? it 's going to be x is equal to negative 3 . what x value makes this whole terms 0 ? so it 's going to be x is equal to minus 3 , and y is going to be equal to 1 . what y value makes this term 0 ? y is equal to 1 . that 's our center . so let 's graph that , and then we can draw the ellipse . it 's going to be in the negative quadrant . this is our x-axis and this is our y-axis . and then the center of our ellipse is at minus 3 and positive 1 , so that 's the center . and then , what is the radius in the x direction ? we just take the square root of this , so it 's 2 . so in the x direction we go two to the right . we go two to the left . and in the y direction , what do we do ? well we go up three and down three . the square root of this . let me do that . remember you have to take the square root of both of those . the vertical axis is actually the major radius or the semi-major axis is 3 , because that 's the longer one . and then the 2 is the minor radius , because that 's the shorter one . and now we 're ready to draw this ellipse . i 'll draw it in brown . let me see if i can do this properly . i have a shaky hand . all right , it looks something like that . and there you go . we took this kind of crazy looking thing , and all we did is algebraically manipulate it . we just completed the squares with the x 's and the y terms . and then we divided both sides by this number right here and we got it into the standard form . we said oh this is an ellipse . we have both of these terms , they 're both positive , we 're adding we 're not subtracting , they have different coefficients underneath here . so we 're ready to go over the ellipse , and we realized that the center was at minus 3,1 , and then we just drew the major radius , or the major axis and the minor axis . see you in the next video .
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that is equal to -- let 's see if i subtract 50 from 85 i get 35 , so if i subtract 49 , i get 36 . and now we are getting close to the standard form of something , but remember all the standard forms we did except for the circle -- we had a y -- and we know this is n't a circle , because we have these weird coefficients , well not weird but different coefficients in front of these terms . so to get the 1 on the right-hand side let 's divide everything by 36 .
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do i have to memorize all the formulas for mathematical forms ?
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the standard question you often get in your algebra class is they will give you this equation and it 'll say identify the conic section and graph it if you can . and the equation they give you wo n't be in the standard form , because if it was you could just kind of pattern match with what i showed in some of the previous videos and you 'd be able to get it . so let 's do a question like and let 's see if we can figure it out . so what i have here is 9x squared plus 4y squared plus 54x minus 8y plus 49 is equal to 0 . and once again , i mean who knows what this is it 's just not in the standard form . and actually one quick clue to tell you what this is you look at the x squared and the y squared terms if there are . if there 's only an x squared term and then there 's just a y and not a y squared term , then you 're probably dealing with a parabola , and we 'll go into that more later . or if it 's the other way around , if it 's just an x term and a y squared term , it 's probably a parabola . but assuming that we 're dealing with a circle , an ellipse , or a hyperbola , there will be an x squared term and a y squared term . if they both kind of have the same number in front of them , that 's a pretty good clue that we 're going to be dealing with a circle . if they both have different numbers , but they 're both positive in front of them , that 's a pretty good clue we 're probably going to be dealing with an ellipse . if one of them has a negative number in front of them and the other one has a positive number , that tells you that we 're probably going to be dealing with a hyperbola . but with that said , i mean that might help you identify things very quickly at this level , but it does n't help you graph it or get into the standard form . so let 's get it in the standard form . and the key to getting it in the standard form is really just completing the square . and i encourage you to re-watch the completing the square video , because that 's all we 're going to do right here to get it into the standard form . so the first thing i like to do to complete the square , and you 're going to have to do it for the x variables and for the y terms , is group the x and y terms . let 's see . the x terms are 9x squared plus 54x . and let 's do the y terms in magenta . so then you have plus 4y squared minus 8y and then you have -- let me do this in a different color -- plus 49 is equal to 0 . and so the easy thing to do when you complete the square , the thing i like to do is , it 's very clear we can factor out a 9 out of both of these numbers , and we can factor out a 4 out of both of those . let 's do that , because that will help us complete the square . so this is the same thing is 9 times x squared plus 9 times 6 is 54 , 6x . i 'm going to add something else here , but i 'll leave it blank for now . plus 4 times y squared minus 2y i 'm probably going to add something here too , so i 'll leave it blank for now . plus 49 is equal to 0 . so what are we going to add here ? we 're going to complete the square . we want to add some number here so that this whole three term expression becomes a perfect square . likewise , we 're going to add some number here , so this three term number expression becomes a perfect square . and of course whatever we add on the side , we 're going to have to multiply it by 9 , because we 're really adding nine times that . and add it on to that side . whatever we add here , we 're going to have to multiply it times 4 and add it on that side . if i put a 1 here , it 's really like as if i had a 4 here , because 1 times 4 is 4 and if i had a 1 here it 's 1 times 9 . so 9 there . let 's do that . when we complete the square , we just take half of this coefficient . this coefficient is 6 , we take half of it is 3 , we square it , we get a 9 . remember it 's an equation , so what you do to one side , you have to do to the other . so if we added a 9 here , we 're actually adding 9 times 9 to the left-hand side of the equation , so we have to add 81 to the right-hand side to make the equation still hold . and you could kind of view it if we go back up here . this is the same thing , just to make that clear as if i added plus 81 right here . of course i would have had to add plus 81 up here . now let 's go to the y terms . you take half of this coefficient is minus 2 , half of that is minus 1 . you square it , you get plus 1 . 1 times 4 , so we 're really adding 4 to the left-hand side of the equation . and just so you understand what i did here . this is equivalent as if i just added a 4 here , and then i later just factored out this 4 . and so what does this become ? this expression is 9 times what ? this is the square of -- you could factor this , but we did it on purpose -- it 's x plus 3 squared and then we have plus 4 times -- what is this right here ? that 's y minus 1 squared . you might want to review factoring of polynomial or completing the square if you found that step a little daunting . and then we have plus 49 is equal to 0 plus 81 plus 84 is equal to 85 . all right , so now we have 9 times plus 3 squared plus 4 times y minus 1 squared . and let 's subtract 49 from both sides . that is equal to -- let 's see if i subtract 50 from 85 i get 35 , so if i subtract 49 , i get 36 . and now we are getting close to the standard form of something , but remember all the standard forms we did except for the circle -- we had a y -- and we know this is n't a circle , because we have these weird coefficients , well not weird but different coefficients in front of these terms . so to get the 1 on the right-hand side let 's divide everything by 36 . if you divide everything by 36 , this term becomes x plus 3 squared over see 9 over 36 is the same thing as 1 over 4 , and then you have plus y minus 1 squared 4 over 36 is the same thing as 1 over 9 and all of that is equal to 1 . and there you go . we have it in the standard form , and you can see our intuition at the beginning the problem was correct . this is indeed an ellipse , and now we can actually graph it . so first of all , actually good place to start , where 's the center of the this ellipse going to be ? it 's going to be x is equal to negative 3 . what x value makes this whole terms 0 ? so it 's going to be x is equal to minus 3 , and y is going to be equal to 1 . what y value makes this term 0 ? y is equal to 1 . that 's our center . so let 's graph that , and then we can draw the ellipse . it 's going to be in the negative quadrant . this is our x-axis and this is our y-axis . and then the center of our ellipse is at minus 3 and positive 1 , so that 's the center . and then , what is the radius in the x direction ? we just take the square root of this , so it 's 2 . so in the x direction we go two to the right . we go two to the left . and in the y direction , what do we do ? well we go up three and down three . the square root of this . let me do that . remember you have to take the square root of both of those . the vertical axis is actually the major radius or the semi-major axis is 3 , because that 's the longer one . and then the 2 is the minor radius , because that 's the shorter one . and now we 're ready to draw this ellipse . i 'll draw it in brown . let me see if i can do this properly . i have a shaky hand . all right , it looks something like that . and there you go . we took this kind of crazy looking thing , and all we did is algebraically manipulate it . we just completed the squares with the x 's and the y terms . and then we divided both sides by this number right here and we got it into the standard form . we said oh this is an ellipse . we have both of these terms , they 're both positive , we 're adding we 're not subtracting , they have different coefficients underneath here . so we 're ready to go over the ellipse , and we realized that the center was at minus 3,1 , and then we just drew the major radius , or the major axis and the minor axis . see you in the next video .
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so it 's going to be x is equal to minus 3 , and y is going to be equal to 1 . what y value makes this term 0 ? y is equal to 1 .
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what happens if there is a xy term ?
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the standard question you often get in your algebra class is they will give you this equation and it 'll say identify the conic section and graph it if you can . and the equation they give you wo n't be in the standard form , because if it was you could just kind of pattern match with what i showed in some of the previous videos and you 'd be able to get it . so let 's do a question like and let 's see if we can figure it out . so what i have here is 9x squared plus 4y squared plus 54x minus 8y plus 49 is equal to 0 . and once again , i mean who knows what this is it 's just not in the standard form . and actually one quick clue to tell you what this is you look at the x squared and the y squared terms if there are . if there 's only an x squared term and then there 's just a y and not a y squared term , then you 're probably dealing with a parabola , and we 'll go into that more later . or if it 's the other way around , if it 's just an x term and a y squared term , it 's probably a parabola . but assuming that we 're dealing with a circle , an ellipse , or a hyperbola , there will be an x squared term and a y squared term . if they both kind of have the same number in front of them , that 's a pretty good clue that we 're going to be dealing with a circle . if they both have different numbers , but they 're both positive in front of them , that 's a pretty good clue we 're probably going to be dealing with an ellipse . if one of them has a negative number in front of them and the other one has a positive number , that tells you that we 're probably going to be dealing with a hyperbola . but with that said , i mean that might help you identify things very quickly at this level , but it does n't help you graph it or get into the standard form . so let 's get it in the standard form . and the key to getting it in the standard form is really just completing the square . and i encourage you to re-watch the completing the square video , because that 's all we 're going to do right here to get it into the standard form . so the first thing i like to do to complete the square , and you 're going to have to do it for the x variables and for the y terms , is group the x and y terms . let 's see . the x terms are 9x squared plus 54x . and let 's do the y terms in magenta . so then you have plus 4y squared minus 8y and then you have -- let me do this in a different color -- plus 49 is equal to 0 . and so the easy thing to do when you complete the square , the thing i like to do is , it 's very clear we can factor out a 9 out of both of these numbers , and we can factor out a 4 out of both of those . let 's do that , because that will help us complete the square . so this is the same thing is 9 times x squared plus 9 times 6 is 54 , 6x . i 'm going to add something else here , but i 'll leave it blank for now . plus 4 times y squared minus 2y i 'm probably going to add something here too , so i 'll leave it blank for now . plus 49 is equal to 0 . so what are we going to add here ? we 're going to complete the square . we want to add some number here so that this whole three term expression becomes a perfect square . likewise , we 're going to add some number here , so this three term number expression becomes a perfect square . and of course whatever we add on the side , we 're going to have to multiply it by 9 , because we 're really adding nine times that . and add it on to that side . whatever we add here , we 're going to have to multiply it times 4 and add it on that side . if i put a 1 here , it 's really like as if i had a 4 here , because 1 times 4 is 4 and if i had a 1 here it 's 1 times 9 . so 9 there . let 's do that . when we complete the square , we just take half of this coefficient . this coefficient is 6 , we take half of it is 3 , we square it , we get a 9 . remember it 's an equation , so what you do to one side , you have to do to the other . so if we added a 9 here , we 're actually adding 9 times 9 to the left-hand side of the equation , so we have to add 81 to the right-hand side to make the equation still hold . and you could kind of view it if we go back up here . this is the same thing , just to make that clear as if i added plus 81 right here . of course i would have had to add plus 81 up here . now let 's go to the y terms . you take half of this coefficient is minus 2 , half of that is minus 1 . you square it , you get plus 1 . 1 times 4 , so we 're really adding 4 to the left-hand side of the equation . and just so you understand what i did here . this is equivalent as if i just added a 4 here , and then i later just factored out this 4 . and so what does this become ? this expression is 9 times what ? this is the square of -- you could factor this , but we did it on purpose -- it 's x plus 3 squared and then we have plus 4 times -- what is this right here ? that 's y minus 1 squared . you might want to review factoring of polynomial or completing the square if you found that step a little daunting . and then we have plus 49 is equal to 0 plus 81 plus 84 is equal to 85 . all right , so now we have 9 times plus 3 squared plus 4 times y minus 1 squared . and let 's subtract 49 from both sides . that is equal to -- let 's see if i subtract 50 from 85 i get 35 , so if i subtract 49 , i get 36 . and now we are getting close to the standard form of something , but remember all the standard forms we did except for the circle -- we had a y -- and we know this is n't a circle , because we have these weird coefficients , well not weird but different coefficients in front of these terms . so to get the 1 on the right-hand side let 's divide everything by 36 . if you divide everything by 36 , this term becomes x plus 3 squared over see 9 over 36 is the same thing as 1 over 4 , and then you have plus y minus 1 squared 4 over 36 is the same thing as 1 over 9 and all of that is equal to 1 . and there you go . we have it in the standard form , and you can see our intuition at the beginning the problem was correct . this is indeed an ellipse , and now we can actually graph it . so first of all , actually good place to start , where 's the center of the this ellipse going to be ? it 's going to be x is equal to negative 3 . what x value makes this whole terms 0 ? so it 's going to be x is equal to minus 3 , and y is going to be equal to 1 . what y value makes this term 0 ? y is equal to 1 . that 's our center . so let 's graph that , and then we can draw the ellipse . it 's going to be in the negative quadrant . this is our x-axis and this is our y-axis . and then the center of our ellipse is at minus 3 and positive 1 , so that 's the center . and then , what is the radius in the x direction ? we just take the square root of this , so it 's 2 . so in the x direction we go two to the right . we go two to the left . and in the y direction , what do we do ? well we go up three and down three . the square root of this . let me do that . remember you have to take the square root of both of those . the vertical axis is actually the major radius or the semi-major axis is 3 , because that 's the longer one . and then the 2 is the minor radius , because that 's the shorter one . and now we 're ready to draw this ellipse . i 'll draw it in brown . let me see if i can do this properly . i have a shaky hand . all right , it looks something like that . and there you go . we took this kind of crazy looking thing , and all we did is algebraically manipulate it . we just completed the squares with the x 's and the y terms . and then we divided both sides by this number right here and we got it into the standard form . we said oh this is an ellipse . we have both of these terms , they 're both positive , we 're adding we 're not subtracting , they have different coefficients underneath here . so we 're ready to go over the ellipse , and we realized that the center was at minus 3,1 , and then we just drew the major radius , or the major axis and the minor axis . see you in the next video .
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the standard question you often get in your algebra class is they will give you this equation and it 'll say identify the conic section and graph it if you can . and the equation they give you wo n't be in the standard form , because if it was you could just kind of pattern match with what i showed in some of the previous videos and you 'd be able to get it .
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what do you do when you have an equation that you need to identify the conic section such as 4x^2 - 9y^2=36 ?
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the standard question you often get in your algebra class is they will give you this equation and it 'll say identify the conic section and graph it if you can . and the equation they give you wo n't be in the standard form , because if it was you could just kind of pattern match with what i showed in some of the previous videos and you 'd be able to get it . so let 's do a question like and let 's see if we can figure it out . so what i have here is 9x squared plus 4y squared plus 54x minus 8y plus 49 is equal to 0 . and once again , i mean who knows what this is it 's just not in the standard form . and actually one quick clue to tell you what this is you look at the x squared and the y squared terms if there are . if there 's only an x squared term and then there 's just a y and not a y squared term , then you 're probably dealing with a parabola , and we 'll go into that more later . or if it 's the other way around , if it 's just an x term and a y squared term , it 's probably a parabola . but assuming that we 're dealing with a circle , an ellipse , or a hyperbola , there will be an x squared term and a y squared term . if they both kind of have the same number in front of them , that 's a pretty good clue that we 're going to be dealing with a circle . if they both have different numbers , but they 're both positive in front of them , that 's a pretty good clue we 're probably going to be dealing with an ellipse . if one of them has a negative number in front of them and the other one has a positive number , that tells you that we 're probably going to be dealing with a hyperbola . but with that said , i mean that might help you identify things very quickly at this level , but it does n't help you graph it or get into the standard form . so let 's get it in the standard form . and the key to getting it in the standard form is really just completing the square . and i encourage you to re-watch the completing the square video , because that 's all we 're going to do right here to get it into the standard form . so the first thing i like to do to complete the square , and you 're going to have to do it for the x variables and for the y terms , is group the x and y terms . let 's see . the x terms are 9x squared plus 54x . and let 's do the y terms in magenta . so then you have plus 4y squared minus 8y and then you have -- let me do this in a different color -- plus 49 is equal to 0 . and so the easy thing to do when you complete the square , the thing i like to do is , it 's very clear we can factor out a 9 out of both of these numbers , and we can factor out a 4 out of both of those . let 's do that , because that will help us complete the square . so this is the same thing is 9 times x squared plus 9 times 6 is 54 , 6x . i 'm going to add something else here , but i 'll leave it blank for now . plus 4 times y squared minus 2y i 'm probably going to add something here too , so i 'll leave it blank for now . plus 49 is equal to 0 . so what are we going to add here ? we 're going to complete the square . we want to add some number here so that this whole three term expression becomes a perfect square . likewise , we 're going to add some number here , so this three term number expression becomes a perfect square . and of course whatever we add on the side , we 're going to have to multiply it by 9 , because we 're really adding nine times that . and add it on to that side . whatever we add here , we 're going to have to multiply it times 4 and add it on that side . if i put a 1 here , it 's really like as if i had a 4 here , because 1 times 4 is 4 and if i had a 1 here it 's 1 times 9 . so 9 there . let 's do that . when we complete the square , we just take half of this coefficient . this coefficient is 6 , we take half of it is 3 , we square it , we get a 9 . remember it 's an equation , so what you do to one side , you have to do to the other . so if we added a 9 here , we 're actually adding 9 times 9 to the left-hand side of the equation , so we have to add 81 to the right-hand side to make the equation still hold . and you could kind of view it if we go back up here . this is the same thing , just to make that clear as if i added plus 81 right here . of course i would have had to add plus 81 up here . now let 's go to the y terms . you take half of this coefficient is minus 2 , half of that is minus 1 . you square it , you get plus 1 . 1 times 4 , so we 're really adding 4 to the left-hand side of the equation . and just so you understand what i did here . this is equivalent as if i just added a 4 here , and then i later just factored out this 4 . and so what does this become ? this expression is 9 times what ? this is the square of -- you could factor this , but we did it on purpose -- it 's x plus 3 squared and then we have plus 4 times -- what is this right here ? that 's y minus 1 squared . you might want to review factoring of polynomial or completing the square if you found that step a little daunting . and then we have plus 49 is equal to 0 plus 81 plus 84 is equal to 85 . all right , so now we have 9 times plus 3 squared plus 4 times y minus 1 squared . and let 's subtract 49 from both sides . that is equal to -- let 's see if i subtract 50 from 85 i get 35 , so if i subtract 49 , i get 36 . and now we are getting close to the standard form of something , but remember all the standard forms we did except for the circle -- we had a y -- and we know this is n't a circle , because we have these weird coefficients , well not weird but different coefficients in front of these terms . so to get the 1 on the right-hand side let 's divide everything by 36 . if you divide everything by 36 , this term becomes x plus 3 squared over see 9 over 36 is the same thing as 1 over 4 , and then you have plus y minus 1 squared 4 over 36 is the same thing as 1 over 9 and all of that is equal to 1 . and there you go . we have it in the standard form , and you can see our intuition at the beginning the problem was correct . this is indeed an ellipse , and now we can actually graph it . so first of all , actually good place to start , where 's the center of the this ellipse going to be ? it 's going to be x is equal to negative 3 . what x value makes this whole terms 0 ? so it 's going to be x is equal to minus 3 , and y is going to be equal to 1 . what y value makes this term 0 ? y is equal to 1 . that 's our center . so let 's graph that , and then we can draw the ellipse . it 's going to be in the negative quadrant . this is our x-axis and this is our y-axis . and then the center of our ellipse is at minus 3 and positive 1 , so that 's the center . and then , what is the radius in the x direction ? we just take the square root of this , so it 's 2 . so in the x direction we go two to the right . we go two to the left . and in the y direction , what do we do ? well we go up three and down three . the square root of this . let me do that . remember you have to take the square root of both of those . the vertical axis is actually the major radius or the semi-major axis is 3 , because that 's the longer one . and then the 2 is the minor radius , because that 's the shorter one . and now we 're ready to draw this ellipse . i 'll draw it in brown . let me see if i can do this properly . i have a shaky hand . all right , it looks something like that . and there you go . we took this kind of crazy looking thing , and all we did is algebraically manipulate it . we just completed the squares with the x 's and the y terms . and then we divided both sides by this number right here and we got it into the standard form . we said oh this is an ellipse . we have both of these terms , they 're both positive , we 're adding we 're not subtracting , they have different coefficients underneath here . so we 're ready to go over the ellipse , and we realized that the center was at minus 3,1 , and then we just drew the major radius , or the major axis and the minor axis . see you in the next video .
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but with that said , i mean that might help you identify things very quickly at this level , but it does n't help you graph it or get into the standard form . so let 's get it in the standard form . and the key to getting it in the standard form is really just completing the square .
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how do we convert the following equation into standard form of an ellipse ?
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the standard question you often get in your algebra class is they will give you this equation and it 'll say identify the conic section and graph it if you can . and the equation they give you wo n't be in the standard form , because if it was you could just kind of pattern match with what i showed in some of the previous videos and you 'd be able to get it . so let 's do a question like and let 's see if we can figure it out . so what i have here is 9x squared plus 4y squared plus 54x minus 8y plus 49 is equal to 0 . and once again , i mean who knows what this is it 's just not in the standard form . and actually one quick clue to tell you what this is you look at the x squared and the y squared terms if there are . if there 's only an x squared term and then there 's just a y and not a y squared term , then you 're probably dealing with a parabola , and we 'll go into that more later . or if it 's the other way around , if it 's just an x term and a y squared term , it 's probably a parabola . but assuming that we 're dealing with a circle , an ellipse , or a hyperbola , there will be an x squared term and a y squared term . if they both kind of have the same number in front of them , that 's a pretty good clue that we 're going to be dealing with a circle . if they both have different numbers , but they 're both positive in front of them , that 's a pretty good clue we 're probably going to be dealing with an ellipse . if one of them has a negative number in front of them and the other one has a positive number , that tells you that we 're probably going to be dealing with a hyperbola . but with that said , i mean that might help you identify things very quickly at this level , but it does n't help you graph it or get into the standard form . so let 's get it in the standard form . and the key to getting it in the standard form is really just completing the square . and i encourage you to re-watch the completing the square video , because that 's all we 're going to do right here to get it into the standard form . so the first thing i like to do to complete the square , and you 're going to have to do it for the x variables and for the y terms , is group the x and y terms . let 's see . the x terms are 9x squared plus 54x . and let 's do the y terms in magenta . so then you have plus 4y squared minus 8y and then you have -- let me do this in a different color -- plus 49 is equal to 0 . and so the easy thing to do when you complete the square , the thing i like to do is , it 's very clear we can factor out a 9 out of both of these numbers , and we can factor out a 4 out of both of those . let 's do that , because that will help us complete the square . so this is the same thing is 9 times x squared plus 9 times 6 is 54 , 6x . i 'm going to add something else here , but i 'll leave it blank for now . plus 4 times y squared minus 2y i 'm probably going to add something here too , so i 'll leave it blank for now . plus 49 is equal to 0 . so what are we going to add here ? we 're going to complete the square . we want to add some number here so that this whole three term expression becomes a perfect square . likewise , we 're going to add some number here , so this three term number expression becomes a perfect square . and of course whatever we add on the side , we 're going to have to multiply it by 9 , because we 're really adding nine times that . and add it on to that side . whatever we add here , we 're going to have to multiply it times 4 and add it on that side . if i put a 1 here , it 's really like as if i had a 4 here , because 1 times 4 is 4 and if i had a 1 here it 's 1 times 9 . so 9 there . let 's do that . when we complete the square , we just take half of this coefficient . this coefficient is 6 , we take half of it is 3 , we square it , we get a 9 . remember it 's an equation , so what you do to one side , you have to do to the other . so if we added a 9 here , we 're actually adding 9 times 9 to the left-hand side of the equation , so we have to add 81 to the right-hand side to make the equation still hold . and you could kind of view it if we go back up here . this is the same thing , just to make that clear as if i added plus 81 right here . of course i would have had to add plus 81 up here . now let 's go to the y terms . you take half of this coefficient is minus 2 , half of that is minus 1 . you square it , you get plus 1 . 1 times 4 , so we 're really adding 4 to the left-hand side of the equation . and just so you understand what i did here . this is equivalent as if i just added a 4 here , and then i later just factored out this 4 . and so what does this become ? this expression is 9 times what ? this is the square of -- you could factor this , but we did it on purpose -- it 's x plus 3 squared and then we have plus 4 times -- what is this right here ? that 's y minus 1 squared . you might want to review factoring of polynomial or completing the square if you found that step a little daunting . and then we have plus 49 is equal to 0 plus 81 plus 84 is equal to 85 . all right , so now we have 9 times plus 3 squared plus 4 times y minus 1 squared . and let 's subtract 49 from both sides . that is equal to -- let 's see if i subtract 50 from 85 i get 35 , so if i subtract 49 , i get 36 . and now we are getting close to the standard form of something , but remember all the standard forms we did except for the circle -- we had a y -- and we know this is n't a circle , because we have these weird coefficients , well not weird but different coefficients in front of these terms . so to get the 1 on the right-hand side let 's divide everything by 36 . if you divide everything by 36 , this term becomes x plus 3 squared over see 9 over 36 is the same thing as 1 over 4 , and then you have plus y minus 1 squared 4 over 36 is the same thing as 1 over 9 and all of that is equal to 1 . and there you go . we have it in the standard form , and you can see our intuition at the beginning the problem was correct . this is indeed an ellipse , and now we can actually graph it . so first of all , actually good place to start , where 's the center of the this ellipse going to be ? it 's going to be x is equal to negative 3 . what x value makes this whole terms 0 ? so it 's going to be x is equal to minus 3 , and y is going to be equal to 1 . what y value makes this term 0 ? y is equal to 1 . that 's our center . so let 's graph that , and then we can draw the ellipse . it 's going to be in the negative quadrant . this is our x-axis and this is our y-axis . and then the center of our ellipse is at minus 3 and positive 1 , so that 's the center . and then , what is the radius in the x direction ? we just take the square root of this , so it 's 2 . so in the x direction we go two to the right . we go two to the left . and in the y direction , what do we do ? well we go up three and down three . the square root of this . let me do that . remember you have to take the square root of both of those . the vertical axis is actually the major radius or the semi-major axis is 3 , because that 's the longer one . and then the 2 is the minor radius , because that 's the shorter one . and now we 're ready to draw this ellipse . i 'll draw it in brown . let me see if i can do this properly . i have a shaky hand . all right , it looks something like that . and there you go . we took this kind of crazy looking thing , and all we did is algebraically manipulate it . we just completed the squares with the x 's and the y terms . and then we divided both sides by this number right here and we got it into the standard form . we said oh this is an ellipse . we have both of these terms , they 're both positive , we 're adding we 're not subtracting , they have different coefficients underneath here . so we 're ready to go over the ellipse , and we realized that the center was at minus 3,1 , and then we just drew the major radius , or the major axis and the minor axis . see you in the next video .
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so let 's get it in the standard form . and the key to getting it in the standard form is really just completing the square . and i encourage you to re-watch the completing the square video , because that 's all we 're going to do right here to get it into the standard form .
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how do you determine what the key numbers are ?
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the standard question you often get in your algebra class is they will give you this equation and it 'll say identify the conic section and graph it if you can . and the equation they give you wo n't be in the standard form , because if it was you could just kind of pattern match with what i showed in some of the previous videos and you 'd be able to get it . so let 's do a question like and let 's see if we can figure it out . so what i have here is 9x squared plus 4y squared plus 54x minus 8y plus 49 is equal to 0 . and once again , i mean who knows what this is it 's just not in the standard form . and actually one quick clue to tell you what this is you look at the x squared and the y squared terms if there are . if there 's only an x squared term and then there 's just a y and not a y squared term , then you 're probably dealing with a parabola , and we 'll go into that more later . or if it 's the other way around , if it 's just an x term and a y squared term , it 's probably a parabola . but assuming that we 're dealing with a circle , an ellipse , or a hyperbola , there will be an x squared term and a y squared term . if they both kind of have the same number in front of them , that 's a pretty good clue that we 're going to be dealing with a circle . if they both have different numbers , but they 're both positive in front of them , that 's a pretty good clue we 're probably going to be dealing with an ellipse . if one of them has a negative number in front of them and the other one has a positive number , that tells you that we 're probably going to be dealing with a hyperbola . but with that said , i mean that might help you identify things very quickly at this level , but it does n't help you graph it or get into the standard form . so let 's get it in the standard form . and the key to getting it in the standard form is really just completing the square . and i encourage you to re-watch the completing the square video , because that 's all we 're going to do right here to get it into the standard form . so the first thing i like to do to complete the square , and you 're going to have to do it for the x variables and for the y terms , is group the x and y terms . let 's see . the x terms are 9x squared plus 54x . and let 's do the y terms in magenta . so then you have plus 4y squared minus 8y and then you have -- let me do this in a different color -- plus 49 is equal to 0 . and so the easy thing to do when you complete the square , the thing i like to do is , it 's very clear we can factor out a 9 out of both of these numbers , and we can factor out a 4 out of both of those . let 's do that , because that will help us complete the square . so this is the same thing is 9 times x squared plus 9 times 6 is 54 , 6x . i 'm going to add something else here , but i 'll leave it blank for now . plus 4 times y squared minus 2y i 'm probably going to add something here too , so i 'll leave it blank for now . plus 49 is equal to 0 . so what are we going to add here ? we 're going to complete the square . we want to add some number here so that this whole three term expression becomes a perfect square . likewise , we 're going to add some number here , so this three term number expression becomes a perfect square . and of course whatever we add on the side , we 're going to have to multiply it by 9 , because we 're really adding nine times that . and add it on to that side . whatever we add here , we 're going to have to multiply it times 4 and add it on that side . if i put a 1 here , it 's really like as if i had a 4 here , because 1 times 4 is 4 and if i had a 1 here it 's 1 times 9 . so 9 there . let 's do that . when we complete the square , we just take half of this coefficient . this coefficient is 6 , we take half of it is 3 , we square it , we get a 9 . remember it 's an equation , so what you do to one side , you have to do to the other . so if we added a 9 here , we 're actually adding 9 times 9 to the left-hand side of the equation , so we have to add 81 to the right-hand side to make the equation still hold . and you could kind of view it if we go back up here . this is the same thing , just to make that clear as if i added plus 81 right here . of course i would have had to add plus 81 up here . now let 's go to the y terms . you take half of this coefficient is minus 2 , half of that is minus 1 . you square it , you get plus 1 . 1 times 4 , so we 're really adding 4 to the left-hand side of the equation . and just so you understand what i did here . this is equivalent as if i just added a 4 here , and then i later just factored out this 4 . and so what does this become ? this expression is 9 times what ? this is the square of -- you could factor this , but we did it on purpose -- it 's x plus 3 squared and then we have plus 4 times -- what is this right here ? that 's y minus 1 squared . you might want to review factoring of polynomial or completing the square if you found that step a little daunting . and then we have plus 49 is equal to 0 plus 81 plus 84 is equal to 85 . all right , so now we have 9 times plus 3 squared plus 4 times y minus 1 squared . and let 's subtract 49 from both sides . that is equal to -- let 's see if i subtract 50 from 85 i get 35 , so if i subtract 49 , i get 36 . and now we are getting close to the standard form of something , but remember all the standard forms we did except for the circle -- we had a y -- and we know this is n't a circle , because we have these weird coefficients , well not weird but different coefficients in front of these terms . so to get the 1 on the right-hand side let 's divide everything by 36 . if you divide everything by 36 , this term becomes x plus 3 squared over see 9 over 36 is the same thing as 1 over 4 , and then you have plus y minus 1 squared 4 over 36 is the same thing as 1 over 9 and all of that is equal to 1 . and there you go . we have it in the standard form , and you can see our intuition at the beginning the problem was correct . this is indeed an ellipse , and now we can actually graph it . so first of all , actually good place to start , where 's the center of the this ellipse going to be ? it 's going to be x is equal to negative 3 . what x value makes this whole terms 0 ? so it 's going to be x is equal to minus 3 , and y is going to be equal to 1 . what y value makes this term 0 ? y is equal to 1 . that 's our center . so let 's graph that , and then we can draw the ellipse . it 's going to be in the negative quadrant . this is our x-axis and this is our y-axis . and then the center of our ellipse is at minus 3 and positive 1 , so that 's the center . and then , what is the radius in the x direction ? we just take the square root of this , so it 's 2 . so in the x direction we go two to the right . we go two to the left . and in the y direction , what do we do ? well we go up three and down three . the square root of this . let me do that . remember you have to take the square root of both of those . the vertical axis is actually the major radius or the semi-major axis is 3 , because that 's the longer one . and then the 2 is the minor radius , because that 's the shorter one . and now we 're ready to draw this ellipse . i 'll draw it in brown . let me see if i can do this properly . i have a shaky hand . all right , it looks something like that . and there you go . we took this kind of crazy looking thing , and all we did is algebraically manipulate it . we just completed the squares with the x 's and the y terms . and then we divided both sides by this number right here and we got it into the standard form . we said oh this is an ellipse . we have both of these terms , they 're both positive , we 're adding we 're not subtracting , they have different coefficients underneath here . so we 're ready to go over the ellipse , and we realized that the center was at minus 3,1 , and then we just drew the major radius , or the major axis and the minor axis . see you in the next video .
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plus 4 times y squared minus 2y i 'm probably going to add something here too , so i 'll leave it blank for now . plus 49 is equal to 0 . so what are we going to add here ?
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... how did sal know it was an ellipse ?
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the standard question you often get in your algebra class is they will give you this equation and it 'll say identify the conic section and graph it if you can . and the equation they give you wo n't be in the standard form , because if it was you could just kind of pattern match with what i showed in some of the previous videos and you 'd be able to get it . so let 's do a question like and let 's see if we can figure it out . so what i have here is 9x squared plus 4y squared plus 54x minus 8y plus 49 is equal to 0 . and once again , i mean who knows what this is it 's just not in the standard form . and actually one quick clue to tell you what this is you look at the x squared and the y squared terms if there are . if there 's only an x squared term and then there 's just a y and not a y squared term , then you 're probably dealing with a parabola , and we 'll go into that more later . or if it 's the other way around , if it 's just an x term and a y squared term , it 's probably a parabola . but assuming that we 're dealing with a circle , an ellipse , or a hyperbola , there will be an x squared term and a y squared term . if they both kind of have the same number in front of them , that 's a pretty good clue that we 're going to be dealing with a circle . if they both have different numbers , but they 're both positive in front of them , that 's a pretty good clue we 're probably going to be dealing with an ellipse . if one of them has a negative number in front of them and the other one has a positive number , that tells you that we 're probably going to be dealing with a hyperbola . but with that said , i mean that might help you identify things very quickly at this level , but it does n't help you graph it or get into the standard form . so let 's get it in the standard form . and the key to getting it in the standard form is really just completing the square . and i encourage you to re-watch the completing the square video , because that 's all we 're going to do right here to get it into the standard form . so the first thing i like to do to complete the square , and you 're going to have to do it for the x variables and for the y terms , is group the x and y terms . let 's see . the x terms are 9x squared plus 54x . and let 's do the y terms in magenta . so then you have plus 4y squared minus 8y and then you have -- let me do this in a different color -- plus 49 is equal to 0 . and so the easy thing to do when you complete the square , the thing i like to do is , it 's very clear we can factor out a 9 out of both of these numbers , and we can factor out a 4 out of both of those . let 's do that , because that will help us complete the square . so this is the same thing is 9 times x squared plus 9 times 6 is 54 , 6x . i 'm going to add something else here , but i 'll leave it blank for now . plus 4 times y squared minus 2y i 'm probably going to add something here too , so i 'll leave it blank for now . plus 49 is equal to 0 . so what are we going to add here ? we 're going to complete the square . we want to add some number here so that this whole three term expression becomes a perfect square . likewise , we 're going to add some number here , so this three term number expression becomes a perfect square . and of course whatever we add on the side , we 're going to have to multiply it by 9 , because we 're really adding nine times that . and add it on to that side . whatever we add here , we 're going to have to multiply it times 4 and add it on that side . if i put a 1 here , it 's really like as if i had a 4 here , because 1 times 4 is 4 and if i had a 1 here it 's 1 times 9 . so 9 there . let 's do that . when we complete the square , we just take half of this coefficient . this coefficient is 6 , we take half of it is 3 , we square it , we get a 9 . remember it 's an equation , so what you do to one side , you have to do to the other . so if we added a 9 here , we 're actually adding 9 times 9 to the left-hand side of the equation , so we have to add 81 to the right-hand side to make the equation still hold . and you could kind of view it if we go back up here . this is the same thing , just to make that clear as if i added plus 81 right here . of course i would have had to add plus 81 up here . now let 's go to the y terms . you take half of this coefficient is minus 2 , half of that is minus 1 . you square it , you get plus 1 . 1 times 4 , so we 're really adding 4 to the left-hand side of the equation . and just so you understand what i did here . this is equivalent as if i just added a 4 here , and then i later just factored out this 4 . and so what does this become ? this expression is 9 times what ? this is the square of -- you could factor this , but we did it on purpose -- it 's x plus 3 squared and then we have plus 4 times -- what is this right here ? that 's y minus 1 squared . you might want to review factoring of polynomial or completing the square if you found that step a little daunting . and then we have plus 49 is equal to 0 plus 81 plus 84 is equal to 85 . all right , so now we have 9 times plus 3 squared plus 4 times y minus 1 squared . and let 's subtract 49 from both sides . that is equal to -- let 's see if i subtract 50 from 85 i get 35 , so if i subtract 49 , i get 36 . and now we are getting close to the standard form of something , but remember all the standard forms we did except for the circle -- we had a y -- and we know this is n't a circle , because we have these weird coefficients , well not weird but different coefficients in front of these terms . so to get the 1 on the right-hand side let 's divide everything by 36 . if you divide everything by 36 , this term becomes x plus 3 squared over see 9 over 36 is the same thing as 1 over 4 , and then you have plus y minus 1 squared 4 over 36 is the same thing as 1 over 9 and all of that is equal to 1 . and there you go . we have it in the standard form , and you can see our intuition at the beginning the problem was correct . this is indeed an ellipse , and now we can actually graph it . so first of all , actually good place to start , where 's the center of the this ellipse going to be ? it 's going to be x is equal to negative 3 . what x value makes this whole terms 0 ? so it 's going to be x is equal to minus 3 , and y is going to be equal to 1 . what y value makes this term 0 ? y is equal to 1 . that 's our center . so let 's graph that , and then we can draw the ellipse . it 's going to be in the negative quadrant . this is our x-axis and this is our y-axis . and then the center of our ellipse is at minus 3 and positive 1 , so that 's the center . and then , what is the radius in the x direction ? we just take the square root of this , so it 's 2 . so in the x direction we go two to the right . we go two to the left . and in the y direction , what do we do ? well we go up three and down three . the square root of this . let me do that . remember you have to take the square root of both of those . the vertical axis is actually the major radius or the semi-major axis is 3 , because that 's the longer one . and then the 2 is the minor radius , because that 's the shorter one . and now we 're ready to draw this ellipse . i 'll draw it in brown . let me see if i can do this properly . i have a shaky hand . all right , it looks something like that . and there you go . we took this kind of crazy looking thing , and all we did is algebraically manipulate it . we just completed the squares with the x 's and the y terms . and then we divided both sides by this number right here and we got it into the standard form . we said oh this is an ellipse . we have both of these terms , they 're both positive , we 're adding we 're not subtracting , they have different coefficients underneath here . so we 're ready to go over the ellipse , and we realized that the center was at minus 3,1 , and then we just drew the major radius , or the major axis and the minor axis . see you in the next video .
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so to get the 1 on the right-hand side let 's divide everything by 36 . if you divide everything by 36 , this term becomes x plus 3 squared over see 9 over 36 is the same thing as 1 over 4 , and then you have plus y minus 1 squared 4 over 36 is the same thing as 1 over 9 and all of that is equal to 1 . and there you go .
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now , during the step when sal changes 9 ( x^2+6x+9 ) +4 ( y^2-2y+1 ) +49=0 to 9 ( 3+x ) ^2+4 ( y-1 ) 2+49=85 , how did he do that ?
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the standard question you often get in your algebra class is they will give you this equation and it 'll say identify the conic section and graph it if you can . and the equation they give you wo n't be in the standard form , because if it was you could just kind of pattern match with what i showed in some of the previous videos and you 'd be able to get it . so let 's do a question like and let 's see if we can figure it out . so what i have here is 9x squared plus 4y squared plus 54x minus 8y plus 49 is equal to 0 . and once again , i mean who knows what this is it 's just not in the standard form . and actually one quick clue to tell you what this is you look at the x squared and the y squared terms if there are . if there 's only an x squared term and then there 's just a y and not a y squared term , then you 're probably dealing with a parabola , and we 'll go into that more later . or if it 's the other way around , if it 's just an x term and a y squared term , it 's probably a parabola . but assuming that we 're dealing with a circle , an ellipse , or a hyperbola , there will be an x squared term and a y squared term . if they both kind of have the same number in front of them , that 's a pretty good clue that we 're going to be dealing with a circle . if they both have different numbers , but they 're both positive in front of them , that 's a pretty good clue we 're probably going to be dealing with an ellipse . if one of them has a negative number in front of them and the other one has a positive number , that tells you that we 're probably going to be dealing with a hyperbola . but with that said , i mean that might help you identify things very quickly at this level , but it does n't help you graph it or get into the standard form . so let 's get it in the standard form . and the key to getting it in the standard form is really just completing the square . and i encourage you to re-watch the completing the square video , because that 's all we 're going to do right here to get it into the standard form . so the first thing i like to do to complete the square , and you 're going to have to do it for the x variables and for the y terms , is group the x and y terms . let 's see . the x terms are 9x squared plus 54x . and let 's do the y terms in magenta . so then you have plus 4y squared minus 8y and then you have -- let me do this in a different color -- plus 49 is equal to 0 . and so the easy thing to do when you complete the square , the thing i like to do is , it 's very clear we can factor out a 9 out of both of these numbers , and we can factor out a 4 out of both of those . let 's do that , because that will help us complete the square . so this is the same thing is 9 times x squared plus 9 times 6 is 54 , 6x . i 'm going to add something else here , but i 'll leave it blank for now . plus 4 times y squared minus 2y i 'm probably going to add something here too , so i 'll leave it blank for now . plus 49 is equal to 0 . so what are we going to add here ? we 're going to complete the square . we want to add some number here so that this whole three term expression becomes a perfect square . likewise , we 're going to add some number here , so this three term number expression becomes a perfect square . and of course whatever we add on the side , we 're going to have to multiply it by 9 , because we 're really adding nine times that . and add it on to that side . whatever we add here , we 're going to have to multiply it times 4 and add it on that side . if i put a 1 here , it 's really like as if i had a 4 here , because 1 times 4 is 4 and if i had a 1 here it 's 1 times 9 . so 9 there . let 's do that . when we complete the square , we just take half of this coefficient . this coefficient is 6 , we take half of it is 3 , we square it , we get a 9 . remember it 's an equation , so what you do to one side , you have to do to the other . so if we added a 9 here , we 're actually adding 9 times 9 to the left-hand side of the equation , so we have to add 81 to the right-hand side to make the equation still hold . and you could kind of view it if we go back up here . this is the same thing , just to make that clear as if i added plus 81 right here . of course i would have had to add plus 81 up here . now let 's go to the y terms . you take half of this coefficient is minus 2 , half of that is minus 1 . you square it , you get plus 1 . 1 times 4 , so we 're really adding 4 to the left-hand side of the equation . and just so you understand what i did here . this is equivalent as if i just added a 4 here , and then i later just factored out this 4 . and so what does this become ? this expression is 9 times what ? this is the square of -- you could factor this , but we did it on purpose -- it 's x plus 3 squared and then we have plus 4 times -- what is this right here ? that 's y minus 1 squared . you might want to review factoring of polynomial or completing the square if you found that step a little daunting . and then we have plus 49 is equal to 0 plus 81 plus 84 is equal to 85 . all right , so now we have 9 times plus 3 squared plus 4 times y minus 1 squared . and let 's subtract 49 from both sides . that is equal to -- let 's see if i subtract 50 from 85 i get 35 , so if i subtract 49 , i get 36 . and now we are getting close to the standard form of something , but remember all the standard forms we did except for the circle -- we had a y -- and we know this is n't a circle , because we have these weird coefficients , well not weird but different coefficients in front of these terms . so to get the 1 on the right-hand side let 's divide everything by 36 . if you divide everything by 36 , this term becomes x plus 3 squared over see 9 over 36 is the same thing as 1 over 4 , and then you have plus y minus 1 squared 4 over 36 is the same thing as 1 over 9 and all of that is equal to 1 . and there you go . we have it in the standard form , and you can see our intuition at the beginning the problem was correct . this is indeed an ellipse , and now we can actually graph it . so first of all , actually good place to start , where 's the center of the this ellipse going to be ? it 's going to be x is equal to negative 3 . what x value makes this whole terms 0 ? so it 's going to be x is equal to minus 3 , and y is going to be equal to 1 . what y value makes this term 0 ? y is equal to 1 . that 's our center . so let 's graph that , and then we can draw the ellipse . it 's going to be in the negative quadrant . this is our x-axis and this is our y-axis . and then the center of our ellipse is at minus 3 and positive 1 , so that 's the center . and then , what is the radius in the x direction ? we just take the square root of this , so it 's 2 . so in the x direction we go two to the right . we go two to the left . and in the y direction , what do we do ? well we go up three and down three . the square root of this . let me do that . remember you have to take the square root of both of those . the vertical axis is actually the major radius or the semi-major axis is 3 , because that 's the longer one . and then the 2 is the minor radius , because that 's the shorter one . and now we 're ready to draw this ellipse . i 'll draw it in brown . let me see if i can do this properly . i have a shaky hand . all right , it looks something like that . and there you go . we took this kind of crazy looking thing , and all we did is algebraically manipulate it . we just completed the squares with the x 's and the y terms . and then we divided both sides by this number right here and we got it into the standard form . we said oh this is an ellipse . we have both of these terms , they 're both positive , we 're adding we 're not subtracting , they have different coefficients underneath here . so we 're ready to go over the ellipse , and we realized that the center was at minus 3,1 , and then we just drew the major radius , or the major axis and the minor axis . see you in the next video .
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the standard question you often get in your algebra class is they will give you this equation and it 'll say identify the conic section and graph it if you can . and the equation they give you wo n't be in the standard form , because if it was you could just kind of pattern match with what i showed in some of the previous videos and you 'd be able to get it .
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is there another word for conics ?
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the standard question you often get in your algebra class is they will give you this equation and it 'll say identify the conic section and graph it if you can . and the equation they give you wo n't be in the standard form , because if it was you could just kind of pattern match with what i showed in some of the previous videos and you 'd be able to get it . so let 's do a question like and let 's see if we can figure it out . so what i have here is 9x squared plus 4y squared plus 54x minus 8y plus 49 is equal to 0 . and once again , i mean who knows what this is it 's just not in the standard form . and actually one quick clue to tell you what this is you look at the x squared and the y squared terms if there are . if there 's only an x squared term and then there 's just a y and not a y squared term , then you 're probably dealing with a parabola , and we 'll go into that more later . or if it 's the other way around , if it 's just an x term and a y squared term , it 's probably a parabola . but assuming that we 're dealing with a circle , an ellipse , or a hyperbola , there will be an x squared term and a y squared term . if they both kind of have the same number in front of them , that 's a pretty good clue that we 're going to be dealing with a circle . if they both have different numbers , but they 're both positive in front of them , that 's a pretty good clue we 're probably going to be dealing with an ellipse . if one of them has a negative number in front of them and the other one has a positive number , that tells you that we 're probably going to be dealing with a hyperbola . but with that said , i mean that might help you identify things very quickly at this level , but it does n't help you graph it or get into the standard form . so let 's get it in the standard form . and the key to getting it in the standard form is really just completing the square . and i encourage you to re-watch the completing the square video , because that 's all we 're going to do right here to get it into the standard form . so the first thing i like to do to complete the square , and you 're going to have to do it for the x variables and for the y terms , is group the x and y terms . let 's see . the x terms are 9x squared plus 54x . and let 's do the y terms in magenta . so then you have plus 4y squared minus 8y and then you have -- let me do this in a different color -- plus 49 is equal to 0 . and so the easy thing to do when you complete the square , the thing i like to do is , it 's very clear we can factor out a 9 out of both of these numbers , and we can factor out a 4 out of both of those . let 's do that , because that will help us complete the square . so this is the same thing is 9 times x squared plus 9 times 6 is 54 , 6x . i 'm going to add something else here , but i 'll leave it blank for now . plus 4 times y squared minus 2y i 'm probably going to add something here too , so i 'll leave it blank for now . plus 49 is equal to 0 . so what are we going to add here ? we 're going to complete the square . we want to add some number here so that this whole three term expression becomes a perfect square . likewise , we 're going to add some number here , so this three term number expression becomes a perfect square . and of course whatever we add on the side , we 're going to have to multiply it by 9 , because we 're really adding nine times that . and add it on to that side . whatever we add here , we 're going to have to multiply it times 4 and add it on that side . if i put a 1 here , it 's really like as if i had a 4 here , because 1 times 4 is 4 and if i had a 1 here it 's 1 times 9 . so 9 there . let 's do that . when we complete the square , we just take half of this coefficient . this coefficient is 6 , we take half of it is 3 , we square it , we get a 9 . remember it 's an equation , so what you do to one side , you have to do to the other . so if we added a 9 here , we 're actually adding 9 times 9 to the left-hand side of the equation , so we have to add 81 to the right-hand side to make the equation still hold . and you could kind of view it if we go back up here . this is the same thing , just to make that clear as if i added plus 81 right here . of course i would have had to add plus 81 up here . now let 's go to the y terms . you take half of this coefficient is minus 2 , half of that is minus 1 . you square it , you get plus 1 . 1 times 4 , so we 're really adding 4 to the left-hand side of the equation . and just so you understand what i did here . this is equivalent as if i just added a 4 here , and then i later just factored out this 4 . and so what does this become ? this expression is 9 times what ? this is the square of -- you could factor this , but we did it on purpose -- it 's x plus 3 squared and then we have plus 4 times -- what is this right here ? that 's y minus 1 squared . you might want to review factoring of polynomial or completing the square if you found that step a little daunting . and then we have plus 49 is equal to 0 plus 81 plus 84 is equal to 85 . all right , so now we have 9 times plus 3 squared plus 4 times y minus 1 squared . and let 's subtract 49 from both sides . that is equal to -- let 's see if i subtract 50 from 85 i get 35 , so if i subtract 49 , i get 36 . and now we are getting close to the standard form of something , but remember all the standard forms we did except for the circle -- we had a y -- and we know this is n't a circle , because we have these weird coefficients , well not weird but different coefficients in front of these terms . so to get the 1 on the right-hand side let 's divide everything by 36 . if you divide everything by 36 , this term becomes x plus 3 squared over see 9 over 36 is the same thing as 1 over 4 , and then you have plus y minus 1 squared 4 over 36 is the same thing as 1 over 9 and all of that is equal to 1 . and there you go . we have it in the standard form , and you can see our intuition at the beginning the problem was correct . this is indeed an ellipse , and now we can actually graph it . so first of all , actually good place to start , where 's the center of the this ellipse going to be ? it 's going to be x is equal to negative 3 . what x value makes this whole terms 0 ? so it 's going to be x is equal to minus 3 , and y is going to be equal to 1 . what y value makes this term 0 ? y is equal to 1 . that 's our center . so let 's graph that , and then we can draw the ellipse . it 's going to be in the negative quadrant . this is our x-axis and this is our y-axis . and then the center of our ellipse is at minus 3 and positive 1 , so that 's the center . and then , what is the radius in the x direction ? we just take the square root of this , so it 's 2 . so in the x direction we go two to the right . we go two to the left . and in the y direction , what do we do ? well we go up three and down three . the square root of this . let me do that . remember you have to take the square root of both of those . the vertical axis is actually the major radius or the semi-major axis is 3 , because that 's the longer one . and then the 2 is the minor radius , because that 's the shorter one . and now we 're ready to draw this ellipse . i 'll draw it in brown . let me see if i can do this properly . i have a shaky hand . all right , it looks something like that . and there you go . we took this kind of crazy looking thing , and all we did is algebraically manipulate it . we just completed the squares with the x 's and the y terms . and then we divided both sides by this number right here and we got it into the standard form . we said oh this is an ellipse . we have both of these terms , they 're both positive , we 're adding we 're not subtracting , they have different coefficients underneath here . so we 're ready to go over the ellipse , and we realized that the center was at minus 3,1 , and then we just drew the major radius , or the major axis and the minor axis . see you in the next video .
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this is the same thing , just to make that clear as if i added plus 81 right here . of course i would have had to add plus 81 up here . now let 's go to the y terms .
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would ellipses always be perfect squares ?
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the standard question you often get in your algebra class is they will give you this equation and it 'll say identify the conic section and graph it if you can . and the equation they give you wo n't be in the standard form , because if it was you could just kind of pattern match with what i showed in some of the previous videos and you 'd be able to get it . so let 's do a question like and let 's see if we can figure it out . so what i have here is 9x squared plus 4y squared plus 54x minus 8y plus 49 is equal to 0 . and once again , i mean who knows what this is it 's just not in the standard form . and actually one quick clue to tell you what this is you look at the x squared and the y squared terms if there are . if there 's only an x squared term and then there 's just a y and not a y squared term , then you 're probably dealing with a parabola , and we 'll go into that more later . or if it 's the other way around , if it 's just an x term and a y squared term , it 's probably a parabola . but assuming that we 're dealing with a circle , an ellipse , or a hyperbola , there will be an x squared term and a y squared term . if they both kind of have the same number in front of them , that 's a pretty good clue that we 're going to be dealing with a circle . if they both have different numbers , but they 're both positive in front of them , that 's a pretty good clue we 're probably going to be dealing with an ellipse . if one of them has a negative number in front of them and the other one has a positive number , that tells you that we 're probably going to be dealing with a hyperbola . but with that said , i mean that might help you identify things very quickly at this level , but it does n't help you graph it or get into the standard form . so let 's get it in the standard form . and the key to getting it in the standard form is really just completing the square . and i encourage you to re-watch the completing the square video , because that 's all we 're going to do right here to get it into the standard form . so the first thing i like to do to complete the square , and you 're going to have to do it for the x variables and for the y terms , is group the x and y terms . let 's see . the x terms are 9x squared plus 54x . and let 's do the y terms in magenta . so then you have plus 4y squared minus 8y and then you have -- let me do this in a different color -- plus 49 is equal to 0 . and so the easy thing to do when you complete the square , the thing i like to do is , it 's very clear we can factor out a 9 out of both of these numbers , and we can factor out a 4 out of both of those . let 's do that , because that will help us complete the square . so this is the same thing is 9 times x squared plus 9 times 6 is 54 , 6x . i 'm going to add something else here , but i 'll leave it blank for now . plus 4 times y squared minus 2y i 'm probably going to add something here too , so i 'll leave it blank for now . plus 49 is equal to 0 . so what are we going to add here ? we 're going to complete the square . we want to add some number here so that this whole three term expression becomes a perfect square . likewise , we 're going to add some number here , so this three term number expression becomes a perfect square . and of course whatever we add on the side , we 're going to have to multiply it by 9 , because we 're really adding nine times that . and add it on to that side . whatever we add here , we 're going to have to multiply it times 4 and add it on that side . if i put a 1 here , it 's really like as if i had a 4 here , because 1 times 4 is 4 and if i had a 1 here it 's 1 times 9 . so 9 there . let 's do that . when we complete the square , we just take half of this coefficient . this coefficient is 6 , we take half of it is 3 , we square it , we get a 9 . remember it 's an equation , so what you do to one side , you have to do to the other . so if we added a 9 here , we 're actually adding 9 times 9 to the left-hand side of the equation , so we have to add 81 to the right-hand side to make the equation still hold . and you could kind of view it if we go back up here . this is the same thing , just to make that clear as if i added plus 81 right here . of course i would have had to add plus 81 up here . now let 's go to the y terms . you take half of this coefficient is minus 2 , half of that is minus 1 . you square it , you get plus 1 . 1 times 4 , so we 're really adding 4 to the left-hand side of the equation . and just so you understand what i did here . this is equivalent as if i just added a 4 here , and then i later just factored out this 4 . and so what does this become ? this expression is 9 times what ? this is the square of -- you could factor this , but we did it on purpose -- it 's x plus 3 squared and then we have plus 4 times -- what is this right here ? that 's y minus 1 squared . you might want to review factoring of polynomial or completing the square if you found that step a little daunting . and then we have plus 49 is equal to 0 plus 81 plus 84 is equal to 85 . all right , so now we have 9 times plus 3 squared plus 4 times y minus 1 squared . and let 's subtract 49 from both sides . that is equal to -- let 's see if i subtract 50 from 85 i get 35 , so if i subtract 49 , i get 36 . and now we are getting close to the standard form of something , but remember all the standard forms we did except for the circle -- we had a y -- and we know this is n't a circle , because we have these weird coefficients , well not weird but different coefficients in front of these terms . so to get the 1 on the right-hand side let 's divide everything by 36 . if you divide everything by 36 , this term becomes x plus 3 squared over see 9 over 36 is the same thing as 1 over 4 , and then you have plus y minus 1 squared 4 over 36 is the same thing as 1 over 9 and all of that is equal to 1 . and there you go . we have it in the standard form , and you can see our intuition at the beginning the problem was correct . this is indeed an ellipse , and now we can actually graph it . so first of all , actually good place to start , where 's the center of the this ellipse going to be ? it 's going to be x is equal to negative 3 . what x value makes this whole terms 0 ? so it 's going to be x is equal to minus 3 , and y is going to be equal to 1 . what y value makes this term 0 ? y is equal to 1 . that 's our center . so let 's graph that , and then we can draw the ellipse . it 's going to be in the negative quadrant . this is our x-axis and this is our y-axis . and then the center of our ellipse is at minus 3 and positive 1 , so that 's the center . and then , what is the radius in the x direction ? we just take the square root of this , so it 's 2 . so in the x direction we go two to the right . we go two to the left . and in the y direction , what do we do ? well we go up three and down three . the square root of this . let me do that . remember you have to take the square root of both of those . the vertical axis is actually the major radius or the semi-major axis is 3 , because that 's the longer one . and then the 2 is the minor radius , because that 's the shorter one . and now we 're ready to draw this ellipse . i 'll draw it in brown . let me see if i can do this properly . i have a shaky hand . all right , it looks something like that . and there you go . we took this kind of crazy looking thing , and all we did is algebraically manipulate it . we just completed the squares with the x 's and the y terms . and then we divided both sides by this number right here and we got it into the standard form . we said oh this is an ellipse . we have both of these terms , they 're both positive , we 're adding we 're not subtracting , they have different coefficients underneath here . so we 're ready to go over the ellipse , and we realized that the center was at minus 3,1 , and then we just drew the major radius , or the major axis and the minor axis . see you in the next video .
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the standard question you often get in your algebra class is they will give you this equation and it 'll say identify the conic section and graph it if you can . and the equation they give you wo n't be in the standard form , because if it was you could just kind of pattern match with what i showed in some of the previous videos and you 'd be able to get it .
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what are the exponents for 14 ?
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the standard question you often get in your algebra class is they will give you this equation and it 'll say identify the conic section and graph it if you can . and the equation they give you wo n't be in the standard form , because if it was you could just kind of pattern match with what i showed in some of the previous videos and you 'd be able to get it . so let 's do a question like and let 's see if we can figure it out . so what i have here is 9x squared plus 4y squared plus 54x minus 8y plus 49 is equal to 0 . and once again , i mean who knows what this is it 's just not in the standard form . and actually one quick clue to tell you what this is you look at the x squared and the y squared terms if there are . if there 's only an x squared term and then there 's just a y and not a y squared term , then you 're probably dealing with a parabola , and we 'll go into that more later . or if it 's the other way around , if it 's just an x term and a y squared term , it 's probably a parabola . but assuming that we 're dealing with a circle , an ellipse , or a hyperbola , there will be an x squared term and a y squared term . if they both kind of have the same number in front of them , that 's a pretty good clue that we 're going to be dealing with a circle . if they both have different numbers , but they 're both positive in front of them , that 's a pretty good clue we 're probably going to be dealing with an ellipse . if one of them has a negative number in front of them and the other one has a positive number , that tells you that we 're probably going to be dealing with a hyperbola . but with that said , i mean that might help you identify things very quickly at this level , but it does n't help you graph it or get into the standard form . so let 's get it in the standard form . and the key to getting it in the standard form is really just completing the square . and i encourage you to re-watch the completing the square video , because that 's all we 're going to do right here to get it into the standard form . so the first thing i like to do to complete the square , and you 're going to have to do it for the x variables and for the y terms , is group the x and y terms . let 's see . the x terms are 9x squared plus 54x . and let 's do the y terms in magenta . so then you have plus 4y squared minus 8y and then you have -- let me do this in a different color -- plus 49 is equal to 0 . and so the easy thing to do when you complete the square , the thing i like to do is , it 's very clear we can factor out a 9 out of both of these numbers , and we can factor out a 4 out of both of those . let 's do that , because that will help us complete the square . so this is the same thing is 9 times x squared plus 9 times 6 is 54 , 6x . i 'm going to add something else here , but i 'll leave it blank for now . plus 4 times y squared minus 2y i 'm probably going to add something here too , so i 'll leave it blank for now . plus 49 is equal to 0 . so what are we going to add here ? we 're going to complete the square . we want to add some number here so that this whole three term expression becomes a perfect square . likewise , we 're going to add some number here , so this three term number expression becomes a perfect square . and of course whatever we add on the side , we 're going to have to multiply it by 9 , because we 're really adding nine times that . and add it on to that side . whatever we add here , we 're going to have to multiply it times 4 and add it on that side . if i put a 1 here , it 's really like as if i had a 4 here , because 1 times 4 is 4 and if i had a 1 here it 's 1 times 9 . so 9 there . let 's do that . when we complete the square , we just take half of this coefficient . this coefficient is 6 , we take half of it is 3 , we square it , we get a 9 . remember it 's an equation , so what you do to one side , you have to do to the other . so if we added a 9 here , we 're actually adding 9 times 9 to the left-hand side of the equation , so we have to add 81 to the right-hand side to make the equation still hold . and you could kind of view it if we go back up here . this is the same thing , just to make that clear as if i added plus 81 right here . of course i would have had to add plus 81 up here . now let 's go to the y terms . you take half of this coefficient is minus 2 , half of that is minus 1 . you square it , you get plus 1 . 1 times 4 , so we 're really adding 4 to the left-hand side of the equation . and just so you understand what i did here . this is equivalent as if i just added a 4 here , and then i later just factored out this 4 . and so what does this become ? this expression is 9 times what ? this is the square of -- you could factor this , but we did it on purpose -- it 's x plus 3 squared and then we have plus 4 times -- what is this right here ? that 's y minus 1 squared . you might want to review factoring of polynomial or completing the square if you found that step a little daunting . and then we have plus 49 is equal to 0 plus 81 plus 84 is equal to 85 . all right , so now we have 9 times plus 3 squared plus 4 times y minus 1 squared . and let 's subtract 49 from both sides . that is equal to -- let 's see if i subtract 50 from 85 i get 35 , so if i subtract 49 , i get 36 . and now we are getting close to the standard form of something , but remember all the standard forms we did except for the circle -- we had a y -- and we know this is n't a circle , because we have these weird coefficients , well not weird but different coefficients in front of these terms . so to get the 1 on the right-hand side let 's divide everything by 36 . if you divide everything by 36 , this term becomes x plus 3 squared over see 9 over 36 is the same thing as 1 over 4 , and then you have plus y minus 1 squared 4 over 36 is the same thing as 1 over 9 and all of that is equal to 1 . and there you go . we have it in the standard form , and you can see our intuition at the beginning the problem was correct . this is indeed an ellipse , and now we can actually graph it . so first of all , actually good place to start , where 's the center of the this ellipse going to be ? it 's going to be x is equal to negative 3 . what x value makes this whole terms 0 ? so it 's going to be x is equal to minus 3 , and y is going to be equal to 1 . what y value makes this term 0 ? y is equal to 1 . that 's our center . so let 's graph that , and then we can draw the ellipse . it 's going to be in the negative quadrant . this is our x-axis and this is our y-axis . and then the center of our ellipse is at minus 3 and positive 1 , so that 's the center . and then , what is the radius in the x direction ? we just take the square root of this , so it 's 2 . so in the x direction we go two to the right . we go two to the left . and in the y direction , what do we do ? well we go up three and down three . the square root of this . let me do that . remember you have to take the square root of both of those . the vertical axis is actually the major radius or the semi-major axis is 3 , because that 's the longer one . and then the 2 is the minor radius , because that 's the shorter one . and now we 're ready to draw this ellipse . i 'll draw it in brown . let me see if i can do this properly . i have a shaky hand . all right , it looks something like that . and there you go . we took this kind of crazy looking thing , and all we did is algebraically manipulate it . we just completed the squares with the x 's and the y terms . and then we divided both sides by this number right here and we got it into the standard form . we said oh this is an ellipse . we have both of these terms , they 're both positive , we 're adding we 're not subtracting , they have different coefficients underneath here . so we 're ready to go over the ellipse , and we realized that the center was at minus 3,1 , and then we just drew the major radius , or the major axis and the minor axis . see you in the next video .
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whatever we add here , we 're going to have to multiply it times 4 and add it on that side . if i put a 1 here , it 's really like as if i had a 4 here , because 1 times 4 is 4 and if i had a 1 here it 's 1 times 9 . so 9 there .
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when we put x= -3 and y = 1 , , , , , does n't it make 0=1 ?
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: when you step back and look at the sociology content here , you might be wondering how in the world any of this applies to medicine . you might say , it has nothing to do with the physical health of a person , so why do you need to know all of this ? why do these social theories and social structures matter to someone in the field of medicine ? let 's go through the different theories and figure this out . first off , we have functionalism . remember , functionalism is the theory that different institutions in a society adjust to minor changes to keep the society stable and functioning . if we look at the function of medicine in society from a functionalist point of view . we 're asking what is the purpose of medicine . well when people become ill medicine ensures that they return to a functional state , so they can contribute to society . being sick is detrimental to the well being of the society as a whole and when you 're sick people can usually tell . they tell you to go home and get better . the assumption is that you 're not supposed to participate in society when you 're sick . this upsets the stability of the society on a small scale at least . the doctor is there to get you better again , so you can get back to participating in society . on a bigger scale , the institution of medicine helps us stabilize the social system in emergency situations like hurricanes or earthquakes , where hospitals and medical professionals take over large spaces like school gyms to provide the medical assistance needed by the many people who are injured . in day-to-day life medicine helps to improve the quality of life for the aging population , allowing them to contribute to society for longer . okay , that seems to make sense . medicine keeps people healthy and participating in society . what about conflict theory ? how do conflicting groups in a society affect the health of an individual ? as we know , conflict theory is all about the inequality between different groups . in the case of medicine , this could have quite a significant impact on who has access to medical care , meaning both access to hospitals and the ability to be covered by insurance . wealthier citizens can pay for the best medical care , but people that are scraping by can not afford hospital bills without insurance . sometimes people can afford health insurance when it is n't provided by their employer or they ca n't afford the deductibles , so they skip the hospital visit and try to heal on their own . meaning they are sick for longer or perhaps they never get better . the unequal access to valuable resources in society , like education , housing or well-paying jobs , leads to health disparities and limited access to medical care . even the power struggle between different interest groups can affect the health of an individual . take a look at air pollution regulations . factories want more lax regulations to reduce costs , while the people living near those factories want stricter regulations for their own health and well being . asthma rates rise in areas with higher levels of particulate matter in the air . stricter air pollution regulations keep the residents healthier , in terms of asthma rates at least , but put a dent in the income of factories . all right , two down . let 's take a look at the theory of social constructionism now . in case you forgot , social constructionism is the idea that society gives value to everything . a diamond was just a rock until society agreed that it should have value . in regards to medicine , it means that as a society , we have attached different meanings to different behaviors , and we have different preconceptions of different people . in short it means stereotypes , we have assumptions about people based on their appearance or actions and we treat people differently because of those assumptions . we have preconceptions about different races , ages , genders , even subcultures like medal heads . in the past , if we saw someone talking to themselves on the street we would assume they were mentally unstable so we would give them a wide berth . today , we know people might be talking on a bluetooth device and so we assume they are n't crazy . assumptions can be very dangerous to a medical professional . they can affect how you treat your patient or your diagnosis . but interaction between the patient and the doctor is influenced by stereotype assumptions on both sides . perhaps the patient feels some symptom is not important enough to mention to the doctor or perhaps the doctor makes a false assumption based on how the patient appears . assumptions also affect how the health system views the patient . there are people who argue that someone who ca n't afford health care does n't deserve it because they do n't work hard enough . you ca n't declare a characteristic of a person based on their circumstance . there are people who do n't work who can still afford health care , while some people work hard at minimum wage who ca n't spare the money . you also have to be aware of medicalization , where patients or doctors will construct an illness out out ordinary behavior . a child who ca n't sit still in class does n't necessarily have add . they might just need to get out on the playground and run . now that we know to be aware of social based assumptions , let 's check out how symbolic interactionism applies to medicine . remember the symbolic interactionism states that individuals give the world meaning by interacting with it . one person could consider a bridge a way to cross over a body of water , while another person considers it a good shelter from the rain . there are many ways we can see how this applies to medicine . let 's take a look at two . for one , we have the doctor-patient relationship . the meaning given to simple objects , like a lab coat or a stethoscope , can affect the interaction . it is important for the doctor to realize the meaning the patient 's given to the tools of medicine . the patient may see the lab coat as a sign of authority , giving the doctor the power to diagnose and treat them . is the stethoscope a way for the doctor to connect with the patient , or is it just a tool that decorates a doctor 's neck ? second , we have the changes in society . recently , there has been a medicalization of society , where everything from beauty to just being fidgety now has a medical fix . standards of beauty have encouraged many people to undergo unnecessary plastic surgery . people can choose to have c-sections when giving birth . which can effect both the mother and child later in life . normal behaviors are being shown as illnesses . one of the most prevalent examples of illness manufacturing is in the case of depression . while depression is a serious condition it 's importance and severity have been marginalized . it seems like every other person today is depressed . when you 're sad , society views that as there being something wrong with you , but in reality sadness is a natural biological function . we 're suppose to be sad sometimes . all right , let 's take a look at something a bit more specific . feminist theory is an offshoot of conflict theory that focuses on the inequalities between men and women in society . these inequalities are pretty apparent when looking at the field of medicine . though the admittance of women into med school is on the rise , it is still am male-dominated field . the heads of hospitals and doctors in general still tend to be men . there 's a disparity in the jobs and salary between male and female doctors . men more often occupy higher paid positions . women are more often found in family medicine rather than specialized fields . this disparity in health care positions translates into a disparity in power . if you take a look at medicine from the perspective of rational choice and exchange theories , you can observe big worldview issues of power . rational choice and exchange theory assume that people behave rationally according to their best interests . and that you can break down any social institution into the self-interest of interactions between individuals . so , let 's see how this applies . when you look at the medical system as a whole , you can ask , what is the purpose of the medical system ? does it really exist to keep people healthy or is there some other reason ? perhaps it ’ s a capitalist competition to earn the most money . perhaps the structure of our medical system benefits private companies more than it does the sick people that it ’ s supposed to be helping . people run every aspect of the medical system and those people will make decisions that benefit themselves more than a random sick stranger . and perhaps , that effects why people go to the doctor or not when they 're sick . will going to the doctor benefit them the most in the long run ? or will it cost them an arm and a leg ? some people avoid doctor visits for minor things , because they can not afford the expense . but that could allow something that could be easily treated to become a much larger problem . the self interested behavior people in charge of different aspects of the medical system will trickle down to eventually effect the well being of a patient . well look at that bridging the gap between sociology and medicine seemed a near impossible task when we started but now we have quite a few examples . to be honest , there are probably so many more ways that sociology 's involved in medicine . let 's take a look outside the specific theories too . where you live can affect you health . there are urban areas called food deserts , where there are no grocery stores within a reasonable distance . the only places to eat are fast food restaurants or , perhaps , grabbing a snack at a gas station . it is nearly impossible to get the nutrition a body needs from only these sources , and malnutrition can lead to a host of other problems . some neighborhoods have no gyms or playgrounds , nowhere for residents to exercise . you can use these examples to come up with your own examples of other places where sociology applies to medicine .
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if you take a look at medicine from the perspective of rational choice and exchange theories , you can observe big worldview issues of power . rational choice and exchange theory assume that people behave rationally according to their best interests . and that you can break down any social institution into the self-interest of interactions between individuals .
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is happiness not accounted for in rational exchange theory ?
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sal : let 's see if we can calculate the definite integral from zero to one of x squared times two to the x to the third power d x . like always i encourage you to pause this video and see if you can figure this out on your own . i 'm assuming you 've had a go at it . there 's a couple of interesting things here . the first thing , at least that my brain does , it says , `` i 'm used to taking derivatives and anti-derivatives of e to the x , not some other base to the x . '' we know that the derivative with respect to x of e to the x is e to the x , or we could say that the anti-derivative of e to the x is equal to e to the x plus c. since i 'm dealing with something raised to , this particular situation , something raised to a function of x , it seems like i might want to put , i might want to change the base here , but how do i do that ? the way i would do that is re-express two in terms of e. what would be two in terms of e ? two is equal to e , is equal to e raised to the power that you need to raise e to to get to two . what 's the power that you have to raise two to to get to two ? well that 's the natural log of two . once again the natural log of two is the exponent that you have to raise e to to get to two . if you actually raise e to it you 're going to get two . this is what two is . now what is two to the x to the third ? well if we raise both sides of this to the x to the third power , we raise both sides to the x to the third power , two to the x to the third is equal to , if i raise something to an exponent and then raise that to an exponent , it 's going to be equal to e to the x to the third , x to the third , times the natural log of two , times the natural log of two . that already seems pretty interesting . let 's rewrite this , and actually what i 'm going to do , let 's just focus on the indefinite integral first , see if we can figure that out . then we can apply , then we can take , we can evaluate the definite ones . let 's just think about this , let 's think about the indefinite integral of x squared times two to the x to the third power d x. i really want to find the anti-derivative of this . well this is going to be the exact same thing as the integral of , i 'll write my x squared still , but instead of two to the x to the third i 'm going to write all of this business . let me just copy and paste that . we already established that this is the same thing as two to the x to the third power . copy and paste , just like that . then let me close it with a d x. i was able to get it in terms of e as a base . that makes me a little bit more comfortable but it still seems pretty complicated . you might be saying , `` okay , look . `` maybe u substitution could be at play here . '' because i have this crazy expression , x to the third times the natural log of two , but what 's the derivative of that ? well that 's going to be three x squared times the natural log of two , or three times the natural log of two times x squared . that 's just a constant times x squared . we already have a x squared here so maybe we can engineer this a little bit to have the constant there as well . let 's think about that . if we made this , if we defined this as u , if we said u is equal to x to the third times the natural log of two , what is du going to be ? du is going to be , it 's going to be , well natural log of two is just a constant so it 's going to be three x squared times the natural log of two . we could actually just change the order we 're multiplying a little bit . we could say that this is the same thing as x squared times three natural log of two , which is the same thing just using logarithm properties , as x squared times the natural log of two to the third power . three natural log of two is the same thing as the natural log of two to the third power . this is equal to x squared times the natural log of eight . let 's see , if this is u , where is du ? oh , and of course we ca n't forget the dx . this is a dx right over here , dx , dx , dx . where is the du ? well we have a dx . let me circle things . you have a dx here , you have a dx there . you have an x squared here , you have an x squared here . so really all we need is , all we need here is the natural log of eight . ideally we would have the natural log of eight right over here , and we could put it there as long as we also , we could multiply by the natural log of eight as long as we also divide by a natural log of eight . we can do it like right over here , we could divide by natural log of eight . but we know that the anti-derivative of some constant times a function is the same thing as a constant times the anti-derivative of that function . we could just take that on the outside . it 's one over the natural log of eight . let 's write this in terms of u and du . this simplifies to one over the natural log of eight times the anti-derivative of e to the u , e to the u , that 's the u , du . this times this times that is du , du . and this is straightforward , we know what this is going to be . this is going to be equal to , let me just write the one over natural log of eight out here , one over natural log of eight times e to the u , and of course if we 're thinking in terms of just anti-derivative there would be some constant out there . then we would just reverse the substitution . we already know what u is . this is going to be equal to , the anti-derivative of this expression is one over the natural log of eight times e to the , instead of u , we know that u is x to the third times the natural log of two . and of course we could put a plus c there . now , going back to the original problem . we just need to evaluate the anti-derivative of this at each of these points . let 's rewrite that . given what we just figured out , let me copy and paste that . this is just going to be equal to , it 's going to be equal to the anti-derivative evaluated at one minus the anti-derivative evaluated at zero . we do n't have to worry about the constants because those will cancel out . so we are going to get , we are going to get one -- let me evaluate it first at one . you 're going to get one over the natural log of eight times e to the one to the third power , which is just one , times the natural log of two , natural log of two , that 's evaluated at one . then we 're going to have minus it evaluated it at zero . it 's going to be one over the natural log of eight times e to the , well when x is zero this whole thing is going to be zero . well e to the zero is just one , and e to the natural log of two , well that 's just going to be two , we already established that early on , this is just going to be equal to two . we are left with two over the natural log of eight minus one over the natural log of eight , which is just going to be equal to one over the natural log of eight . and we are , and we are done .
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it 's one over the natural log of eight . let 's write this in terms of u and du . this simplifies to one over the natural log of eight times the anti-derivative of e to the u , e to the u , that 's the u , du .
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did sal forget to change the boundaries in terms of u ?
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sal : let 's see if we can calculate the definite integral from zero to one of x squared times two to the x to the third power d x . like always i encourage you to pause this video and see if you can figure this out on your own . i 'm assuming you 've had a go at it . there 's a couple of interesting things here . the first thing , at least that my brain does , it says , `` i 'm used to taking derivatives and anti-derivatives of e to the x , not some other base to the x . '' we know that the derivative with respect to x of e to the x is e to the x , or we could say that the anti-derivative of e to the x is equal to e to the x plus c. since i 'm dealing with something raised to , this particular situation , something raised to a function of x , it seems like i might want to put , i might want to change the base here , but how do i do that ? the way i would do that is re-express two in terms of e. what would be two in terms of e ? two is equal to e , is equal to e raised to the power that you need to raise e to to get to two . what 's the power that you have to raise two to to get to two ? well that 's the natural log of two . once again the natural log of two is the exponent that you have to raise e to to get to two . if you actually raise e to it you 're going to get two . this is what two is . now what is two to the x to the third ? well if we raise both sides of this to the x to the third power , we raise both sides to the x to the third power , two to the x to the third is equal to , if i raise something to an exponent and then raise that to an exponent , it 's going to be equal to e to the x to the third , x to the third , times the natural log of two , times the natural log of two . that already seems pretty interesting . let 's rewrite this , and actually what i 'm going to do , let 's just focus on the indefinite integral first , see if we can figure that out . then we can apply , then we can take , we can evaluate the definite ones . let 's just think about this , let 's think about the indefinite integral of x squared times two to the x to the third power d x. i really want to find the anti-derivative of this . well this is going to be the exact same thing as the integral of , i 'll write my x squared still , but instead of two to the x to the third i 'm going to write all of this business . let me just copy and paste that . we already established that this is the same thing as two to the x to the third power . copy and paste , just like that . then let me close it with a d x. i was able to get it in terms of e as a base . that makes me a little bit more comfortable but it still seems pretty complicated . you might be saying , `` okay , look . `` maybe u substitution could be at play here . '' because i have this crazy expression , x to the third times the natural log of two , but what 's the derivative of that ? well that 's going to be three x squared times the natural log of two , or three times the natural log of two times x squared . that 's just a constant times x squared . we already have a x squared here so maybe we can engineer this a little bit to have the constant there as well . let 's think about that . if we made this , if we defined this as u , if we said u is equal to x to the third times the natural log of two , what is du going to be ? du is going to be , it 's going to be , well natural log of two is just a constant so it 's going to be three x squared times the natural log of two . we could actually just change the order we 're multiplying a little bit . we could say that this is the same thing as x squared times three natural log of two , which is the same thing just using logarithm properties , as x squared times the natural log of two to the third power . three natural log of two is the same thing as the natural log of two to the third power . this is equal to x squared times the natural log of eight . let 's see , if this is u , where is du ? oh , and of course we ca n't forget the dx . this is a dx right over here , dx , dx , dx . where is the du ? well we have a dx . let me circle things . you have a dx here , you have a dx there . you have an x squared here , you have an x squared here . so really all we need is , all we need here is the natural log of eight . ideally we would have the natural log of eight right over here , and we could put it there as long as we also , we could multiply by the natural log of eight as long as we also divide by a natural log of eight . we can do it like right over here , we could divide by natural log of eight . but we know that the anti-derivative of some constant times a function is the same thing as a constant times the anti-derivative of that function . we could just take that on the outside . it 's one over the natural log of eight . let 's write this in terms of u and du . this simplifies to one over the natural log of eight times the anti-derivative of e to the u , e to the u , that 's the u , du . this times this times that is du , du . and this is straightforward , we know what this is going to be . this is going to be equal to , let me just write the one over natural log of eight out here , one over natural log of eight times e to the u , and of course if we 're thinking in terms of just anti-derivative there would be some constant out there . then we would just reverse the substitution . we already know what u is . this is going to be equal to , the anti-derivative of this expression is one over the natural log of eight times e to the , instead of u , we know that u is x to the third times the natural log of two . and of course we could put a plus c there . now , going back to the original problem . we just need to evaluate the anti-derivative of this at each of these points . let 's rewrite that . given what we just figured out , let me copy and paste that . this is just going to be equal to , it 's going to be equal to the anti-derivative evaluated at one minus the anti-derivative evaluated at zero . we do n't have to worry about the constants because those will cancel out . so we are going to get , we are going to get one -- let me evaluate it first at one . you 're going to get one over the natural log of eight times e to the one to the third power , which is just one , times the natural log of two , natural log of two , that 's evaluated at one . then we 're going to have minus it evaluated it at zero . it 's going to be one over the natural log of eight times e to the , well when x is zero this whole thing is going to be zero . well e to the zero is just one , and e to the natural log of two , well that 's just going to be two , we already established that early on , this is just going to be equal to two . we are left with two over the natural log of eight minus one over the natural log of eight , which is just going to be equal to one over the natural log of eight . and we are , and we are done .
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then we can apply , then we can take , we can evaluate the definite ones . let 's just think about this , let 's think about the indefinite integral of x squared times two to the x to the third power d x. i really want to find the anti-derivative of this . well this is going to be the exact same thing as the integral of , i 'll write my x squared still , but instead of two to the x to the third i 'm going to write all of this business .
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0 , why is n't the chain rule applied when taking the derivative of x^3ln ( 2 ) ?
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sal : let 's see if we can calculate the definite integral from zero to one of x squared times two to the x to the third power d x . like always i encourage you to pause this video and see if you can figure this out on your own . i 'm assuming you 've had a go at it . there 's a couple of interesting things here . the first thing , at least that my brain does , it says , `` i 'm used to taking derivatives and anti-derivatives of e to the x , not some other base to the x . '' we know that the derivative with respect to x of e to the x is e to the x , or we could say that the anti-derivative of e to the x is equal to e to the x plus c. since i 'm dealing with something raised to , this particular situation , something raised to a function of x , it seems like i might want to put , i might want to change the base here , but how do i do that ? the way i would do that is re-express two in terms of e. what would be two in terms of e ? two is equal to e , is equal to e raised to the power that you need to raise e to to get to two . what 's the power that you have to raise two to to get to two ? well that 's the natural log of two . once again the natural log of two is the exponent that you have to raise e to to get to two . if you actually raise e to it you 're going to get two . this is what two is . now what is two to the x to the third ? well if we raise both sides of this to the x to the third power , we raise both sides to the x to the third power , two to the x to the third is equal to , if i raise something to an exponent and then raise that to an exponent , it 's going to be equal to e to the x to the third , x to the third , times the natural log of two , times the natural log of two . that already seems pretty interesting . let 's rewrite this , and actually what i 'm going to do , let 's just focus on the indefinite integral first , see if we can figure that out . then we can apply , then we can take , we can evaluate the definite ones . let 's just think about this , let 's think about the indefinite integral of x squared times two to the x to the third power d x. i really want to find the anti-derivative of this . well this is going to be the exact same thing as the integral of , i 'll write my x squared still , but instead of two to the x to the third i 'm going to write all of this business . let me just copy and paste that . we already established that this is the same thing as two to the x to the third power . copy and paste , just like that . then let me close it with a d x. i was able to get it in terms of e as a base . that makes me a little bit more comfortable but it still seems pretty complicated . you might be saying , `` okay , look . `` maybe u substitution could be at play here . '' because i have this crazy expression , x to the third times the natural log of two , but what 's the derivative of that ? well that 's going to be three x squared times the natural log of two , or three times the natural log of two times x squared . that 's just a constant times x squared . we already have a x squared here so maybe we can engineer this a little bit to have the constant there as well . let 's think about that . if we made this , if we defined this as u , if we said u is equal to x to the third times the natural log of two , what is du going to be ? du is going to be , it 's going to be , well natural log of two is just a constant so it 's going to be three x squared times the natural log of two . we could actually just change the order we 're multiplying a little bit . we could say that this is the same thing as x squared times three natural log of two , which is the same thing just using logarithm properties , as x squared times the natural log of two to the third power . three natural log of two is the same thing as the natural log of two to the third power . this is equal to x squared times the natural log of eight . let 's see , if this is u , where is du ? oh , and of course we ca n't forget the dx . this is a dx right over here , dx , dx , dx . where is the du ? well we have a dx . let me circle things . you have a dx here , you have a dx there . you have an x squared here , you have an x squared here . so really all we need is , all we need here is the natural log of eight . ideally we would have the natural log of eight right over here , and we could put it there as long as we also , we could multiply by the natural log of eight as long as we also divide by a natural log of eight . we can do it like right over here , we could divide by natural log of eight . but we know that the anti-derivative of some constant times a function is the same thing as a constant times the anti-derivative of that function . we could just take that on the outside . it 's one over the natural log of eight . let 's write this in terms of u and du . this simplifies to one over the natural log of eight times the anti-derivative of e to the u , e to the u , that 's the u , du . this times this times that is du , du . and this is straightforward , we know what this is going to be . this is going to be equal to , let me just write the one over natural log of eight out here , one over natural log of eight times e to the u , and of course if we 're thinking in terms of just anti-derivative there would be some constant out there . then we would just reverse the substitution . we already know what u is . this is going to be equal to , the anti-derivative of this expression is one over the natural log of eight times e to the , instead of u , we know that u is x to the third times the natural log of two . and of course we could put a plus c there . now , going back to the original problem . we just need to evaluate the anti-derivative of this at each of these points . let 's rewrite that . given what we just figured out , let me copy and paste that . this is just going to be equal to , it 's going to be equal to the anti-derivative evaluated at one minus the anti-derivative evaluated at zero . we do n't have to worry about the constants because those will cancel out . so we are going to get , we are going to get one -- let me evaluate it first at one . you 're going to get one over the natural log of eight times e to the one to the third power , which is just one , times the natural log of two , natural log of two , that 's evaluated at one . then we 're going to have minus it evaluated it at zero . it 's going to be one over the natural log of eight times e to the , well when x is zero this whole thing is going to be zero . well e to the zero is just one , and e to the natural log of two , well that 's just going to be two , we already established that early on , this is just going to be equal to two . we are left with two over the natural log of eight minus one over the natural log of eight , which is just going to be equal to one over the natural log of eight . and we are , and we are done .
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let 's see , if this is u , where is du ? oh , and of course we ca n't forget the dx . this is a dx right over here , dx , dx , dx .
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does n't the product rule of derivatives apply here ?
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sal : let 's see if we can calculate the definite integral from zero to one of x squared times two to the x to the third power d x . like always i encourage you to pause this video and see if you can figure this out on your own . i 'm assuming you 've had a go at it . there 's a couple of interesting things here . the first thing , at least that my brain does , it says , `` i 'm used to taking derivatives and anti-derivatives of e to the x , not some other base to the x . '' we know that the derivative with respect to x of e to the x is e to the x , or we could say that the anti-derivative of e to the x is equal to e to the x plus c. since i 'm dealing with something raised to , this particular situation , something raised to a function of x , it seems like i might want to put , i might want to change the base here , but how do i do that ? the way i would do that is re-express two in terms of e. what would be two in terms of e ? two is equal to e , is equal to e raised to the power that you need to raise e to to get to two . what 's the power that you have to raise two to to get to two ? well that 's the natural log of two . once again the natural log of two is the exponent that you have to raise e to to get to two . if you actually raise e to it you 're going to get two . this is what two is . now what is two to the x to the third ? well if we raise both sides of this to the x to the third power , we raise both sides to the x to the third power , two to the x to the third is equal to , if i raise something to an exponent and then raise that to an exponent , it 's going to be equal to e to the x to the third , x to the third , times the natural log of two , times the natural log of two . that already seems pretty interesting . let 's rewrite this , and actually what i 'm going to do , let 's just focus on the indefinite integral first , see if we can figure that out . then we can apply , then we can take , we can evaluate the definite ones . let 's just think about this , let 's think about the indefinite integral of x squared times two to the x to the third power d x. i really want to find the anti-derivative of this . well this is going to be the exact same thing as the integral of , i 'll write my x squared still , but instead of two to the x to the third i 'm going to write all of this business . let me just copy and paste that . we already established that this is the same thing as two to the x to the third power . copy and paste , just like that . then let me close it with a d x. i was able to get it in terms of e as a base . that makes me a little bit more comfortable but it still seems pretty complicated . you might be saying , `` okay , look . `` maybe u substitution could be at play here . '' because i have this crazy expression , x to the third times the natural log of two , but what 's the derivative of that ? well that 's going to be three x squared times the natural log of two , or three times the natural log of two times x squared . that 's just a constant times x squared . we already have a x squared here so maybe we can engineer this a little bit to have the constant there as well . let 's think about that . if we made this , if we defined this as u , if we said u is equal to x to the third times the natural log of two , what is du going to be ? du is going to be , it 's going to be , well natural log of two is just a constant so it 's going to be three x squared times the natural log of two . we could actually just change the order we 're multiplying a little bit . we could say that this is the same thing as x squared times three natural log of two , which is the same thing just using logarithm properties , as x squared times the natural log of two to the third power . three natural log of two is the same thing as the natural log of two to the third power . this is equal to x squared times the natural log of eight . let 's see , if this is u , where is du ? oh , and of course we ca n't forget the dx . this is a dx right over here , dx , dx , dx . where is the du ? well we have a dx . let me circle things . you have a dx here , you have a dx there . you have an x squared here , you have an x squared here . so really all we need is , all we need here is the natural log of eight . ideally we would have the natural log of eight right over here , and we could put it there as long as we also , we could multiply by the natural log of eight as long as we also divide by a natural log of eight . we can do it like right over here , we could divide by natural log of eight . but we know that the anti-derivative of some constant times a function is the same thing as a constant times the anti-derivative of that function . we could just take that on the outside . it 's one over the natural log of eight . let 's write this in terms of u and du . this simplifies to one over the natural log of eight times the anti-derivative of e to the u , e to the u , that 's the u , du . this times this times that is du , du . and this is straightforward , we know what this is going to be . this is going to be equal to , let me just write the one over natural log of eight out here , one over natural log of eight times e to the u , and of course if we 're thinking in terms of just anti-derivative there would be some constant out there . then we would just reverse the substitution . we already know what u is . this is going to be equal to , the anti-derivative of this expression is one over the natural log of eight times e to the , instead of u , we know that u is x to the third times the natural log of two . and of course we could put a plus c there . now , going back to the original problem . we just need to evaluate the anti-derivative of this at each of these points . let 's rewrite that . given what we just figured out , let me copy and paste that . this is just going to be equal to , it 's going to be equal to the anti-derivative evaluated at one minus the anti-derivative evaluated at zero . we do n't have to worry about the constants because those will cancel out . so we are going to get , we are going to get one -- let me evaluate it first at one . you 're going to get one over the natural log of eight times e to the one to the third power , which is just one , times the natural log of two , natural log of two , that 's evaluated at one . then we 're going to have minus it evaluated it at zero . it 's going to be one over the natural log of eight times e to the , well when x is zero this whole thing is going to be zero . well e to the zero is just one , and e to the natural log of two , well that 's just going to be two , we already established that early on , this is just going to be equal to two . we are left with two over the natural log of eight minus one over the natural log of eight , which is just going to be equal to one over the natural log of eight . and we are , and we are done .
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we know that the derivative with respect to x of e to the x is e to the x , or we could say that the anti-derivative of e to the x is equal to e to the x plus c. since i 'm dealing with something raised to , this particular situation , something raised to a function of x , it seems like i might want to put , i might want to change the base here , but how do i do that ? the way i would do that is re-express two in terms of e. what would be two in terms of e ? two is equal to e , is equal to e raised to the power that you need to raise e to to get to two .
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how is e^ln2 =2 ?
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sal : let 's see if we can calculate the definite integral from zero to one of x squared times two to the x to the third power d x . like always i encourage you to pause this video and see if you can figure this out on your own . i 'm assuming you 've had a go at it . there 's a couple of interesting things here . the first thing , at least that my brain does , it says , `` i 'm used to taking derivatives and anti-derivatives of e to the x , not some other base to the x . '' we know that the derivative with respect to x of e to the x is e to the x , or we could say that the anti-derivative of e to the x is equal to e to the x plus c. since i 'm dealing with something raised to , this particular situation , something raised to a function of x , it seems like i might want to put , i might want to change the base here , but how do i do that ? the way i would do that is re-express two in terms of e. what would be two in terms of e ? two is equal to e , is equal to e raised to the power that you need to raise e to to get to two . what 's the power that you have to raise two to to get to two ? well that 's the natural log of two . once again the natural log of two is the exponent that you have to raise e to to get to two . if you actually raise e to it you 're going to get two . this is what two is . now what is two to the x to the third ? well if we raise both sides of this to the x to the third power , we raise both sides to the x to the third power , two to the x to the third is equal to , if i raise something to an exponent and then raise that to an exponent , it 's going to be equal to e to the x to the third , x to the third , times the natural log of two , times the natural log of two . that already seems pretty interesting . let 's rewrite this , and actually what i 'm going to do , let 's just focus on the indefinite integral first , see if we can figure that out . then we can apply , then we can take , we can evaluate the definite ones . let 's just think about this , let 's think about the indefinite integral of x squared times two to the x to the third power d x. i really want to find the anti-derivative of this . well this is going to be the exact same thing as the integral of , i 'll write my x squared still , but instead of two to the x to the third i 'm going to write all of this business . let me just copy and paste that . we already established that this is the same thing as two to the x to the third power . copy and paste , just like that . then let me close it with a d x. i was able to get it in terms of e as a base . that makes me a little bit more comfortable but it still seems pretty complicated . you might be saying , `` okay , look . `` maybe u substitution could be at play here . '' because i have this crazy expression , x to the third times the natural log of two , but what 's the derivative of that ? well that 's going to be three x squared times the natural log of two , or three times the natural log of two times x squared . that 's just a constant times x squared . we already have a x squared here so maybe we can engineer this a little bit to have the constant there as well . let 's think about that . if we made this , if we defined this as u , if we said u is equal to x to the third times the natural log of two , what is du going to be ? du is going to be , it 's going to be , well natural log of two is just a constant so it 's going to be three x squared times the natural log of two . we could actually just change the order we 're multiplying a little bit . we could say that this is the same thing as x squared times three natural log of two , which is the same thing just using logarithm properties , as x squared times the natural log of two to the third power . three natural log of two is the same thing as the natural log of two to the third power . this is equal to x squared times the natural log of eight . let 's see , if this is u , where is du ? oh , and of course we ca n't forget the dx . this is a dx right over here , dx , dx , dx . where is the du ? well we have a dx . let me circle things . you have a dx here , you have a dx there . you have an x squared here , you have an x squared here . so really all we need is , all we need here is the natural log of eight . ideally we would have the natural log of eight right over here , and we could put it there as long as we also , we could multiply by the natural log of eight as long as we also divide by a natural log of eight . we can do it like right over here , we could divide by natural log of eight . but we know that the anti-derivative of some constant times a function is the same thing as a constant times the anti-derivative of that function . we could just take that on the outside . it 's one over the natural log of eight . let 's write this in terms of u and du . this simplifies to one over the natural log of eight times the anti-derivative of e to the u , e to the u , that 's the u , du . this times this times that is du , du . and this is straightforward , we know what this is going to be . this is going to be equal to , let me just write the one over natural log of eight out here , one over natural log of eight times e to the u , and of course if we 're thinking in terms of just anti-derivative there would be some constant out there . then we would just reverse the substitution . we already know what u is . this is going to be equal to , the anti-derivative of this expression is one over the natural log of eight times e to the , instead of u , we know that u is x to the third times the natural log of two . and of course we could put a plus c there . now , going back to the original problem . we just need to evaluate the anti-derivative of this at each of these points . let 's rewrite that . given what we just figured out , let me copy and paste that . this is just going to be equal to , it 's going to be equal to the anti-derivative evaluated at one minus the anti-derivative evaluated at zero . we do n't have to worry about the constants because those will cancel out . so we are going to get , we are going to get one -- let me evaluate it first at one . you 're going to get one over the natural log of eight times e to the one to the third power , which is just one , times the natural log of two , natural log of two , that 's evaluated at one . then we 're going to have minus it evaluated it at zero . it 's going to be one over the natural log of eight times e to the , well when x is zero this whole thing is going to be zero . well e to the zero is just one , and e to the natural log of two , well that 's just going to be two , we already established that early on , this is just going to be equal to two . we are left with two over the natural log of eight minus one over the natural log of eight , which is just going to be equal to one over the natural log of eight . and we are , and we are done .
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the first thing , at least that my brain does , it says , `` i 'm used to taking derivatives and anti-derivatives of e to the x , not some other base to the x . '' we know that the derivative with respect to x of e to the x is e to the x , or we could say that the anti-derivative of e to the x is equal to e to the x plus c. since i 'm dealing with something raised to , this particular situation , something raised to a function of x , it seems like i might want to put , i might want to change the base here , but how do i do that ? the way i would do that is re-express two in terms of e. what would be two in terms of e ? two is equal to e , is equal to e raised to the power that you need to raise e to to get to two .
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could the exponential function be simplified back into the one given in the question , i.e ( e^ ( ln2 ( x^3 ) = ( 2^x^3 ) ?
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sal : let 's see if we can calculate the definite integral from zero to one of x squared times two to the x to the third power d x . like always i encourage you to pause this video and see if you can figure this out on your own . i 'm assuming you 've had a go at it . there 's a couple of interesting things here . the first thing , at least that my brain does , it says , `` i 'm used to taking derivatives and anti-derivatives of e to the x , not some other base to the x . '' we know that the derivative with respect to x of e to the x is e to the x , or we could say that the anti-derivative of e to the x is equal to e to the x plus c. since i 'm dealing with something raised to , this particular situation , something raised to a function of x , it seems like i might want to put , i might want to change the base here , but how do i do that ? the way i would do that is re-express two in terms of e. what would be two in terms of e ? two is equal to e , is equal to e raised to the power that you need to raise e to to get to two . what 's the power that you have to raise two to to get to two ? well that 's the natural log of two . once again the natural log of two is the exponent that you have to raise e to to get to two . if you actually raise e to it you 're going to get two . this is what two is . now what is two to the x to the third ? well if we raise both sides of this to the x to the third power , we raise both sides to the x to the third power , two to the x to the third is equal to , if i raise something to an exponent and then raise that to an exponent , it 's going to be equal to e to the x to the third , x to the third , times the natural log of two , times the natural log of two . that already seems pretty interesting . let 's rewrite this , and actually what i 'm going to do , let 's just focus on the indefinite integral first , see if we can figure that out . then we can apply , then we can take , we can evaluate the definite ones . let 's just think about this , let 's think about the indefinite integral of x squared times two to the x to the third power d x. i really want to find the anti-derivative of this . well this is going to be the exact same thing as the integral of , i 'll write my x squared still , but instead of two to the x to the third i 'm going to write all of this business . let me just copy and paste that . we already established that this is the same thing as two to the x to the third power . copy and paste , just like that . then let me close it with a d x. i was able to get it in terms of e as a base . that makes me a little bit more comfortable but it still seems pretty complicated . you might be saying , `` okay , look . `` maybe u substitution could be at play here . '' because i have this crazy expression , x to the third times the natural log of two , but what 's the derivative of that ? well that 's going to be three x squared times the natural log of two , or three times the natural log of two times x squared . that 's just a constant times x squared . we already have a x squared here so maybe we can engineer this a little bit to have the constant there as well . let 's think about that . if we made this , if we defined this as u , if we said u is equal to x to the third times the natural log of two , what is du going to be ? du is going to be , it 's going to be , well natural log of two is just a constant so it 's going to be three x squared times the natural log of two . we could actually just change the order we 're multiplying a little bit . we could say that this is the same thing as x squared times three natural log of two , which is the same thing just using logarithm properties , as x squared times the natural log of two to the third power . three natural log of two is the same thing as the natural log of two to the third power . this is equal to x squared times the natural log of eight . let 's see , if this is u , where is du ? oh , and of course we ca n't forget the dx . this is a dx right over here , dx , dx , dx . where is the du ? well we have a dx . let me circle things . you have a dx here , you have a dx there . you have an x squared here , you have an x squared here . so really all we need is , all we need here is the natural log of eight . ideally we would have the natural log of eight right over here , and we could put it there as long as we also , we could multiply by the natural log of eight as long as we also divide by a natural log of eight . we can do it like right over here , we could divide by natural log of eight . but we know that the anti-derivative of some constant times a function is the same thing as a constant times the anti-derivative of that function . we could just take that on the outside . it 's one over the natural log of eight . let 's write this in terms of u and du . this simplifies to one over the natural log of eight times the anti-derivative of e to the u , e to the u , that 's the u , du . this times this times that is du , du . and this is straightforward , we know what this is going to be . this is going to be equal to , let me just write the one over natural log of eight out here , one over natural log of eight times e to the u , and of course if we 're thinking in terms of just anti-derivative there would be some constant out there . then we would just reverse the substitution . we already know what u is . this is going to be equal to , the anti-derivative of this expression is one over the natural log of eight times e to the , instead of u , we know that u is x to the third times the natural log of two . and of course we could put a plus c there . now , going back to the original problem . we just need to evaluate the anti-derivative of this at each of these points . let 's rewrite that . given what we just figured out , let me copy and paste that . this is just going to be equal to , it 's going to be equal to the anti-derivative evaluated at one minus the anti-derivative evaluated at zero . we do n't have to worry about the constants because those will cancel out . so we are going to get , we are going to get one -- let me evaluate it first at one . you 're going to get one over the natural log of eight times e to the one to the third power , which is just one , times the natural log of two , natural log of two , that 's evaluated at one . then we 're going to have minus it evaluated it at zero . it 's going to be one over the natural log of eight times e to the , well when x is zero this whole thing is going to be zero . well e to the zero is just one , and e to the natural log of two , well that 's just going to be two , we already established that early on , this is just going to be equal to two . we are left with two over the natural log of eight minus one over the natural log of eight , which is just going to be equal to one over the natural log of eight . and we are , and we are done .
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let 's see , if this is u , where is du ? oh , and of course we ca n't forget the dx . this is a dx right over here , dx , dx , dx . where is the du ?
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when taking the derivative of u , why do n't you isolate dx and plug what you get into the dx in the integral instead of trying to imagine a combination that will create du ?
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sal : let 's see if we can calculate the definite integral from zero to one of x squared times two to the x to the third power d x . like always i encourage you to pause this video and see if you can figure this out on your own . i 'm assuming you 've had a go at it . there 's a couple of interesting things here . the first thing , at least that my brain does , it says , `` i 'm used to taking derivatives and anti-derivatives of e to the x , not some other base to the x . '' we know that the derivative with respect to x of e to the x is e to the x , or we could say that the anti-derivative of e to the x is equal to e to the x plus c. since i 'm dealing with something raised to , this particular situation , something raised to a function of x , it seems like i might want to put , i might want to change the base here , but how do i do that ? the way i would do that is re-express two in terms of e. what would be two in terms of e ? two is equal to e , is equal to e raised to the power that you need to raise e to to get to two . what 's the power that you have to raise two to to get to two ? well that 's the natural log of two . once again the natural log of two is the exponent that you have to raise e to to get to two . if you actually raise e to it you 're going to get two . this is what two is . now what is two to the x to the third ? well if we raise both sides of this to the x to the third power , we raise both sides to the x to the third power , two to the x to the third is equal to , if i raise something to an exponent and then raise that to an exponent , it 's going to be equal to e to the x to the third , x to the third , times the natural log of two , times the natural log of two . that already seems pretty interesting . let 's rewrite this , and actually what i 'm going to do , let 's just focus on the indefinite integral first , see if we can figure that out . then we can apply , then we can take , we can evaluate the definite ones . let 's just think about this , let 's think about the indefinite integral of x squared times two to the x to the third power d x. i really want to find the anti-derivative of this . well this is going to be the exact same thing as the integral of , i 'll write my x squared still , but instead of two to the x to the third i 'm going to write all of this business . let me just copy and paste that . we already established that this is the same thing as two to the x to the third power . copy and paste , just like that . then let me close it with a d x. i was able to get it in terms of e as a base . that makes me a little bit more comfortable but it still seems pretty complicated . you might be saying , `` okay , look . `` maybe u substitution could be at play here . '' because i have this crazy expression , x to the third times the natural log of two , but what 's the derivative of that ? well that 's going to be three x squared times the natural log of two , or three times the natural log of two times x squared . that 's just a constant times x squared . we already have a x squared here so maybe we can engineer this a little bit to have the constant there as well . let 's think about that . if we made this , if we defined this as u , if we said u is equal to x to the third times the natural log of two , what is du going to be ? du is going to be , it 's going to be , well natural log of two is just a constant so it 's going to be three x squared times the natural log of two . we could actually just change the order we 're multiplying a little bit . we could say that this is the same thing as x squared times three natural log of two , which is the same thing just using logarithm properties , as x squared times the natural log of two to the third power . three natural log of two is the same thing as the natural log of two to the third power . this is equal to x squared times the natural log of eight . let 's see , if this is u , where is du ? oh , and of course we ca n't forget the dx . this is a dx right over here , dx , dx , dx . where is the du ? well we have a dx . let me circle things . you have a dx here , you have a dx there . you have an x squared here , you have an x squared here . so really all we need is , all we need here is the natural log of eight . ideally we would have the natural log of eight right over here , and we could put it there as long as we also , we could multiply by the natural log of eight as long as we also divide by a natural log of eight . we can do it like right over here , we could divide by natural log of eight . but we know that the anti-derivative of some constant times a function is the same thing as a constant times the anti-derivative of that function . we could just take that on the outside . it 's one over the natural log of eight . let 's write this in terms of u and du . this simplifies to one over the natural log of eight times the anti-derivative of e to the u , e to the u , that 's the u , du . this times this times that is du , du . and this is straightforward , we know what this is going to be . this is going to be equal to , let me just write the one over natural log of eight out here , one over natural log of eight times e to the u , and of course if we 're thinking in terms of just anti-derivative there would be some constant out there . then we would just reverse the substitution . we already know what u is . this is going to be equal to , the anti-derivative of this expression is one over the natural log of eight times e to the , instead of u , we know that u is x to the third times the natural log of two . and of course we could put a plus c there . now , going back to the original problem . we just need to evaluate the anti-derivative of this at each of these points . let 's rewrite that . given what we just figured out , let me copy and paste that . this is just going to be equal to , it 's going to be equal to the anti-derivative evaluated at one minus the anti-derivative evaluated at zero . we do n't have to worry about the constants because those will cancel out . so we are going to get , we are going to get one -- let me evaluate it first at one . you 're going to get one over the natural log of eight times e to the one to the third power , which is just one , times the natural log of two , natural log of two , that 's evaluated at one . then we 're going to have minus it evaluated it at zero . it 's going to be one over the natural log of eight times e to the , well when x is zero this whole thing is going to be zero . well e to the zero is just one , and e to the natural log of two , well that 's just going to be two , we already established that early on , this is just going to be equal to two . we are left with two over the natural log of eight minus one over the natural log of eight , which is just going to be equal to one over the natural log of eight . and we are , and we are done .
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we can do it like right over here , we could divide by natural log of eight . but we know that the anti-derivative of some constant times a function is the same thing as a constant times the anti-derivative of that function . we could just take that on the outside .
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isnt the derivative of a constant 0 ?
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sal : let 's see if we can calculate the definite integral from zero to one of x squared times two to the x to the third power d x . like always i encourage you to pause this video and see if you can figure this out on your own . i 'm assuming you 've had a go at it . there 's a couple of interesting things here . the first thing , at least that my brain does , it says , `` i 'm used to taking derivatives and anti-derivatives of e to the x , not some other base to the x . '' we know that the derivative with respect to x of e to the x is e to the x , or we could say that the anti-derivative of e to the x is equal to e to the x plus c. since i 'm dealing with something raised to , this particular situation , something raised to a function of x , it seems like i might want to put , i might want to change the base here , but how do i do that ? the way i would do that is re-express two in terms of e. what would be two in terms of e ? two is equal to e , is equal to e raised to the power that you need to raise e to to get to two . what 's the power that you have to raise two to to get to two ? well that 's the natural log of two . once again the natural log of two is the exponent that you have to raise e to to get to two . if you actually raise e to it you 're going to get two . this is what two is . now what is two to the x to the third ? well if we raise both sides of this to the x to the third power , we raise both sides to the x to the third power , two to the x to the third is equal to , if i raise something to an exponent and then raise that to an exponent , it 's going to be equal to e to the x to the third , x to the third , times the natural log of two , times the natural log of two . that already seems pretty interesting . let 's rewrite this , and actually what i 'm going to do , let 's just focus on the indefinite integral first , see if we can figure that out . then we can apply , then we can take , we can evaluate the definite ones . let 's just think about this , let 's think about the indefinite integral of x squared times two to the x to the third power d x. i really want to find the anti-derivative of this . well this is going to be the exact same thing as the integral of , i 'll write my x squared still , but instead of two to the x to the third i 'm going to write all of this business . let me just copy and paste that . we already established that this is the same thing as two to the x to the third power . copy and paste , just like that . then let me close it with a d x. i was able to get it in terms of e as a base . that makes me a little bit more comfortable but it still seems pretty complicated . you might be saying , `` okay , look . `` maybe u substitution could be at play here . '' because i have this crazy expression , x to the third times the natural log of two , but what 's the derivative of that ? well that 's going to be three x squared times the natural log of two , or three times the natural log of two times x squared . that 's just a constant times x squared . we already have a x squared here so maybe we can engineer this a little bit to have the constant there as well . let 's think about that . if we made this , if we defined this as u , if we said u is equal to x to the third times the natural log of two , what is du going to be ? du is going to be , it 's going to be , well natural log of two is just a constant so it 's going to be three x squared times the natural log of two . we could actually just change the order we 're multiplying a little bit . we could say that this is the same thing as x squared times three natural log of two , which is the same thing just using logarithm properties , as x squared times the natural log of two to the third power . three natural log of two is the same thing as the natural log of two to the third power . this is equal to x squared times the natural log of eight . let 's see , if this is u , where is du ? oh , and of course we ca n't forget the dx . this is a dx right over here , dx , dx , dx . where is the du ? well we have a dx . let me circle things . you have a dx here , you have a dx there . you have an x squared here , you have an x squared here . so really all we need is , all we need here is the natural log of eight . ideally we would have the natural log of eight right over here , and we could put it there as long as we also , we could multiply by the natural log of eight as long as we also divide by a natural log of eight . we can do it like right over here , we could divide by natural log of eight . but we know that the anti-derivative of some constant times a function is the same thing as a constant times the anti-derivative of that function . we could just take that on the outside . it 's one over the natural log of eight . let 's write this in terms of u and du . this simplifies to one over the natural log of eight times the anti-derivative of e to the u , e to the u , that 's the u , du . this times this times that is du , du . and this is straightforward , we know what this is going to be . this is going to be equal to , let me just write the one over natural log of eight out here , one over natural log of eight times e to the u , and of course if we 're thinking in terms of just anti-derivative there would be some constant out there . then we would just reverse the substitution . we already know what u is . this is going to be equal to , the anti-derivative of this expression is one over the natural log of eight times e to the , instead of u , we know that u is x to the third times the natural log of two . and of course we could put a plus c there . now , going back to the original problem . we just need to evaluate the anti-derivative of this at each of these points . let 's rewrite that . given what we just figured out , let me copy and paste that . this is just going to be equal to , it 's going to be equal to the anti-derivative evaluated at one minus the anti-derivative evaluated at zero . we do n't have to worry about the constants because those will cancel out . so we are going to get , we are going to get one -- let me evaluate it first at one . you 're going to get one over the natural log of eight times e to the one to the third power , which is just one , times the natural log of two , natural log of two , that 's evaluated at one . then we 're going to have minus it evaluated it at zero . it 's going to be one over the natural log of eight times e to the , well when x is zero this whole thing is going to be zero . well e to the zero is just one , and e to the natural log of two , well that 's just going to be two , we already established that early on , this is just going to be equal to two . we are left with two over the natural log of eight minus one over the natural log of eight , which is just going to be equal to one over the natural log of eight . and we are , and we are done .
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well that 's going to be three x squared times the natural log of two , or three times the natural log of two times x squared . that 's just a constant times x squared . we already have a x squared here so maybe we can engineer this a little bit to have the constant there as well .
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i thought ln ( 2 ) is a constant that goes away when differentiating for x ?
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sal : let 's see if we can calculate the definite integral from zero to one of x squared times two to the x to the third power d x . like always i encourage you to pause this video and see if you can figure this out on your own . i 'm assuming you 've had a go at it . there 's a couple of interesting things here . the first thing , at least that my brain does , it says , `` i 'm used to taking derivatives and anti-derivatives of e to the x , not some other base to the x . '' we know that the derivative with respect to x of e to the x is e to the x , or we could say that the anti-derivative of e to the x is equal to e to the x plus c. since i 'm dealing with something raised to , this particular situation , something raised to a function of x , it seems like i might want to put , i might want to change the base here , but how do i do that ? the way i would do that is re-express two in terms of e. what would be two in terms of e ? two is equal to e , is equal to e raised to the power that you need to raise e to to get to two . what 's the power that you have to raise two to to get to two ? well that 's the natural log of two . once again the natural log of two is the exponent that you have to raise e to to get to two . if you actually raise e to it you 're going to get two . this is what two is . now what is two to the x to the third ? well if we raise both sides of this to the x to the third power , we raise both sides to the x to the third power , two to the x to the third is equal to , if i raise something to an exponent and then raise that to an exponent , it 's going to be equal to e to the x to the third , x to the third , times the natural log of two , times the natural log of two . that already seems pretty interesting . let 's rewrite this , and actually what i 'm going to do , let 's just focus on the indefinite integral first , see if we can figure that out . then we can apply , then we can take , we can evaluate the definite ones . let 's just think about this , let 's think about the indefinite integral of x squared times two to the x to the third power d x. i really want to find the anti-derivative of this . well this is going to be the exact same thing as the integral of , i 'll write my x squared still , but instead of two to the x to the third i 'm going to write all of this business . let me just copy and paste that . we already established that this is the same thing as two to the x to the third power . copy and paste , just like that . then let me close it with a d x. i was able to get it in terms of e as a base . that makes me a little bit more comfortable but it still seems pretty complicated . you might be saying , `` okay , look . `` maybe u substitution could be at play here . '' because i have this crazy expression , x to the third times the natural log of two , but what 's the derivative of that ? well that 's going to be three x squared times the natural log of two , or three times the natural log of two times x squared . that 's just a constant times x squared . we already have a x squared here so maybe we can engineer this a little bit to have the constant there as well . let 's think about that . if we made this , if we defined this as u , if we said u is equal to x to the third times the natural log of two , what is du going to be ? du is going to be , it 's going to be , well natural log of two is just a constant so it 's going to be three x squared times the natural log of two . we could actually just change the order we 're multiplying a little bit . we could say that this is the same thing as x squared times three natural log of two , which is the same thing just using logarithm properties , as x squared times the natural log of two to the third power . three natural log of two is the same thing as the natural log of two to the third power . this is equal to x squared times the natural log of eight . let 's see , if this is u , where is du ? oh , and of course we ca n't forget the dx . this is a dx right over here , dx , dx , dx . where is the du ? well we have a dx . let me circle things . you have a dx here , you have a dx there . you have an x squared here , you have an x squared here . so really all we need is , all we need here is the natural log of eight . ideally we would have the natural log of eight right over here , and we could put it there as long as we also , we could multiply by the natural log of eight as long as we also divide by a natural log of eight . we can do it like right over here , we could divide by natural log of eight . but we know that the anti-derivative of some constant times a function is the same thing as a constant times the anti-derivative of that function . we could just take that on the outside . it 's one over the natural log of eight . let 's write this in terms of u and du . this simplifies to one over the natural log of eight times the anti-derivative of e to the u , e to the u , that 's the u , du . this times this times that is du , du . and this is straightforward , we know what this is going to be . this is going to be equal to , let me just write the one over natural log of eight out here , one over natural log of eight times e to the u , and of course if we 're thinking in terms of just anti-derivative there would be some constant out there . then we would just reverse the substitution . we already know what u is . this is going to be equal to , the anti-derivative of this expression is one over the natural log of eight times e to the , instead of u , we know that u is x to the third times the natural log of two . and of course we could put a plus c there . now , going back to the original problem . we just need to evaluate the anti-derivative of this at each of these points . let 's rewrite that . given what we just figured out , let me copy and paste that . this is just going to be equal to , it 's going to be equal to the anti-derivative evaluated at one minus the anti-derivative evaluated at zero . we do n't have to worry about the constants because those will cancel out . so we are going to get , we are going to get one -- let me evaluate it first at one . you 're going to get one over the natural log of eight times e to the one to the third power , which is just one , times the natural log of two , natural log of two , that 's evaluated at one . then we 're going to have minus it evaluated it at zero . it 's going to be one over the natural log of eight times e to the , well when x is zero this whole thing is going to be zero . well e to the zero is just one , and e to the natural log of two , well that 's just going to be two , we already established that early on , this is just going to be equal to two . we are left with two over the natural log of eight minus one over the natural log of eight , which is just going to be equal to one over the natural log of eight . and we are , and we are done .
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but we know that the anti-derivative of some constant times a function is the same thing as a constant times the anti-derivative of that function . we could just take that on the outside . it 's one over the natural log of eight .
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can we take the constant out of definite integral ( multiply outside of the integral , not inside ) ?
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sal : let 's see if we can calculate the definite integral from zero to one of x squared times two to the x to the third power d x . like always i encourage you to pause this video and see if you can figure this out on your own . i 'm assuming you 've had a go at it . there 's a couple of interesting things here . the first thing , at least that my brain does , it says , `` i 'm used to taking derivatives and anti-derivatives of e to the x , not some other base to the x . '' we know that the derivative with respect to x of e to the x is e to the x , or we could say that the anti-derivative of e to the x is equal to e to the x plus c. since i 'm dealing with something raised to , this particular situation , something raised to a function of x , it seems like i might want to put , i might want to change the base here , but how do i do that ? the way i would do that is re-express two in terms of e. what would be two in terms of e ? two is equal to e , is equal to e raised to the power that you need to raise e to to get to two . what 's the power that you have to raise two to to get to two ? well that 's the natural log of two . once again the natural log of two is the exponent that you have to raise e to to get to two . if you actually raise e to it you 're going to get two . this is what two is . now what is two to the x to the third ? well if we raise both sides of this to the x to the third power , we raise both sides to the x to the third power , two to the x to the third is equal to , if i raise something to an exponent and then raise that to an exponent , it 's going to be equal to e to the x to the third , x to the third , times the natural log of two , times the natural log of two . that already seems pretty interesting . let 's rewrite this , and actually what i 'm going to do , let 's just focus on the indefinite integral first , see if we can figure that out . then we can apply , then we can take , we can evaluate the definite ones . let 's just think about this , let 's think about the indefinite integral of x squared times two to the x to the third power d x. i really want to find the anti-derivative of this . well this is going to be the exact same thing as the integral of , i 'll write my x squared still , but instead of two to the x to the third i 'm going to write all of this business . let me just copy and paste that . we already established that this is the same thing as two to the x to the third power . copy and paste , just like that . then let me close it with a d x. i was able to get it in terms of e as a base . that makes me a little bit more comfortable but it still seems pretty complicated . you might be saying , `` okay , look . `` maybe u substitution could be at play here . '' because i have this crazy expression , x to the third times the natural log of two , but what 's the derivative of that ? well that 's going to be three x squared times the natural log of two , or three times the natural log of two times x squared . that 's just a constant times x squared . we already have a x squared here so maybe we can engineer this a little bit to have the constant there as well . let 's think about that . if we made this , if we defined this as u , if we said u is equal to x to the third times the natural log of two , what is du going to be ? du is going to be , it 's going to be , well natural log of two is just a constant so it 's going to be three x squared times the natural log of two . we could actually just change the order we 're multiplying a little bit . we could say that this is the same thing as x squared times three natural log of two , which is the same thing just using logarithm properties , as x squared times the natural log of two to the third power . three natural log of two is the same thing as the natural log of two to the third power . this is equal to x squared times the natural log of eight . let 's see , if this is u , where is du ? oh , and of course we ca n't forget the dx . this is a dx right over here , dx , dx , dx . where is the du ? well we have a dx . let me circle things . you have a dx here , you have a dx there . you have an x squared here , you have an x squared here . so really all we need is , all we need here is the natural log of eight . ideally we would have the natural log of eight right over here , and we could put it there as long as we also , we could multiply by the natural log of eight as long as we also divide by a natural log of eight . we can do it like right over here , we could divide by natural log of eight . but we know that the anti-derivative of some constant times a function is the same thing as a constant times the anti-derivative of that function . we could just take that on the outside . it 's one over the natural log of eight . let 's write this in terms of u and du . this simplifies to one over the natural log of eight times the anti-derivative of e to the u , e to the u , that 's the u , du . this times this times that is du , du . and this is straightforward , we know what this is going to be . this is going to be equal to , let me just write the one over natural log of eight out here , one over natural log of eight times e to the u , and of course if we 're thinking in terms of just anti-derivative there would be some constant out there . then we would just reverse the substitution . we already know what u is . this is going to be equal to , the anti-derivative of this expression is one over the natural log of eight times e to the , instead of u , we know that u is x to the third times the natural log of two . and of course we could put a plus c there . now , going back to the original problem . we just need to evaluate the anti-derivative of this at each of these points . let 's rewrite that . given what we just figured out , let me copy and paste that . this is just going to be equal to , it 's going to be equal to the anti-derivative evaluated at one minus the anti-derivative evaluated at zero . we do n't have to worry about the constants because those will cancel out . so we are going to get , we are going to get one -- let me evaluate it first at one . you 're going to get one over the natural log of eight times e to the one to the third power , which is just one , times the natural log of two , natural log of two , that 's evaluated at one . then we 're going to have minus it evaluated it at zero . it 's going to be one over the natural log of eight times e to the , well when x is zero this whole thing is going to be zero . well e to the zero is just one , and e to the natural log of two , well that 's just going to be two , we already established that early on , this is just going to be equal to two . we are left with two over the natural log of eight minus one over the natural log of eight , which is just going to be equal to one over the natural log of eight . and we are , and we are done .
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oh , and of course we ca n't forget the dx . this is a dx right over here , dx , dx , dx . where is the du ?
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i 'll explain : so y=2^x ln ( y ) =ln ( 2^x ) ln ( y ) =xln ( 2 ) then use implicit differentiation : 1/y * dy/dx=ln ( 2 ) dy/dx=y*ln ( 2 ) dy/dx=2^x*ln ( 2 ) that 's how i would find the derivative of 2^x , so would it be possible ( or i guess not too hard ) to use this method to find the antiderivative of this function ?
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sal : let 's see if we can calculate the definite integral from zero to one of x squared times two to the x to the third power d x . like always i encourage you to pause this video and see if you can figure this out on your own . i 'm assuming you 've had a go at it . there 's a couple of interesting things here . the first thing , at least that my brain does , it says , `` i 'm used to taking derivatives and anti-derivatives of e to the x , not some other base to the x . '' we know that the derivative with respect to x of e to the x is e to the x , or we could say that the anti-derivative of e to the x is equal to e to the x plus c. since i 'm dealing with something raised to , this particular situation , something raised to a function of x , it seems like i might want to put , i might want to change the base here , but how do i do that ? the way i would do that is re-express two in terms of e. what would be two in terms of e ? two is equal to e , is equal to e raised to the power that you need to raise e to to get to two . what 's the power that you have to raise two to to get to two ? well that 's the natural log of two . once again the natural log of two is the exponent that you have to raise e to to get to two . if you actually raise e to it you 're going to get two . this is what two is . now what is two to the x to the third ? well if we raise both sides of this to the x to the third power , we raise both sides to the x to the third power , two to the x to the third is equal to , if i raise something to an exponent and then raise that to an exponent , it 's going to be equal to e to the x to the third , x to the third , times the natural log of two , times the natural log of two . that already seems pretty interesting . let 's rewrite this , and actually what i 'm going to do , let 's just focus on the indefinite integral first , see if we can figure that out . then we can apply , then we can take , we can evaluate the definite ones . let 's just think about this , let 's think about the indefinite integral of x squared times two to the x to the third power d x. i really want to find the anti-derivative of this . well this is going to be the exact same thing as the integral of , i 'll write my x squared still , but instead of two to the x to the third i 'm going to write all of this business . let me just copy and paste that . we already established that this is the same thing as two to the x to the third power . copy and paste , just like that . then let me close it with a d x. i was able to get it in terms of e as a base . that makes me a little bit more comfortable but it still seems pretty complicated . you might be saying , `` okay , look . `` maybe u substitution could be at play here . '' because i have this crazy expression , x to the third times the natural log of two , but what 's the derivative of that ? well that 's going to be three x squared times the natural log of two , or three times the natural log of two times x squared . that 's just a constant times x squared . we already have a x squared here so maybe we can engineer this a little bit to have the constant there as well . let 's think about that . if we made this , if we defined this as u , if we said u is equal to x to the third times the natural log of two , what is du going to be ? du is going to be , it 's going to be , well natural log of two is just a constant so it 's going to be three x squared times the natural log of two . we could actually just change the order we 're multiplying a little bit . we could say that this is the same thing as x squared times three natural log of two , which is the same thing just using logarithm properties , as x squared times the natural log of two to the third power . three natural log of two is the same thing as the natural log of two to the third power . this is equal to x squared times the natural log of eight . let 's see , if this is u , where is du ? oh , and of course we ca n't forget the dx . this is a dx right over here , dx , dx , dx . where is the du ? well we have a dx . let me circle things . you have a dx here , you have a dx there . you have an x squared here , you have an x squared here . so really all we need is , all we need here is the natural log of eight . ideally we would have the natural log of eight right over here , and we could put it there as long as we also , we could multiply by the natural log of eight as long as we also divide by a natural log of eight . we can do it like right over here , we could divide by natural log of eight . but we know that the anti-derivative of some constant times a function is the same thing as a constant times the anti-derivative of that function . we could just take that on the outside . it 's one over the natural log of eight . let 's write this in terms of u and du . this simplifies to one over the natural log of eight times the anti-derivative of e to the u , e to the u , that 's the u , du . this times this times that is du , du . and this is straightforward , we know what this is going to be . this is going to be equal to , let me just write the one over natural log of eight out here , one over natural log of eight times e to the u , and of course if we 're thinking in terms of just anti-derivative there would be some constant out there . then we would just reverse the substitution . we already know what u is . this is going to be equal to , the anti-derivative of this expression is one over the natural log of eight times e to the , instead of u , we know that u is x to the third times the natural log of two . and of course we could put a plus c there . now , going back to the original problem . we just need to evaluate the anti-derivative of this at each of these points . let 's rewrite that . given what we just figured out , let me copy and paste that . this is just going to be equal to , it 's going to be equal to the anti-derivative evaluated at one minus the anti-derivative evaluated at zero . we do n't have to worry about the constants because those will cancel out . so we are going to get , we are going to get one -- let me evaluate it first at one . you 're going to get one over the natural log of eight times e to the one to the third power , which is just one , times the natural log of two , natural log of two , that 's evaluated at one . then we 're going to have minus it evaluated it at zero . it 's going to be one over the natural log of eight times e to the , well when x is zero this whole thing is going to be zero . well e to the zero is just one , and e to the natural log of two , well that 's just going to be two , we already established that early on , this is just going to be equal to two . we are left with two over the natural log of eight minus one over the natural log of eight , which is just going to be equal to one over the natural log of eight . and we are , and we are done .
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where is the du ? well we have a dx . let me circle things .
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does the final answer `` 1/ln8 '' require the `` + c `` as well ?
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sal : let 's see if we can calculate the definite integral from zero to one of x squared times two to the x to the third power d x . like always i encourage you to pause this video and see if you can figure this out on your own . i 'm assuming you 've had a go at it . there 's a couple of interesting things here . the first thing , at least that my brain does , it says , `` i 'm used to taking derivatives and anti-derivatives of e to the x , not some other base to the x . '' we know that the derivative with respect to x of e to the x is e to the x , or we could say that the anti-derivative of e to the x is equal to e to the x plus c. since i 'm dealing with something raised to , this particular situation , something raised to a function of x , it seems like i might want to put , i might want to change the base here , but how do i do that ? the way i would do that is re-express two in terms of e. what would be two in terms of e ? two is equal to e , is equal to e raised to the power that you need to raise e to to get to two . what 's the power that you have to raise two to to get to two ? well that 's the natural log of two . once again the natural log of two is the exponent that you have to raise e to to get to two . if you actually raise e to it you 're going to get two . this is what two is . now what is two to the x to the third ? well if we raise both sides of this to the x to the third power , we raise both sides to the x to the third power , two to the x to the third is equal to , if i raise something to an exponent and then raise that to an exponent , it 's going to be equal to e to the x to the third , x to the third , times the natural log of two , times the natural log of two . that already seems pretty interesting . let 's rewrite this , and actually what i 'm going to do , let 's just focus on the indefinite integral first , see if we can figure that out . then we can apply , then we can take , we can evaluate the definite ones . let 's just think about this , let 's think about the indefinite integral of x squared times two to the x to the third power d x. i really want to find the anti-derivative of this . well this is going to be the exact same thing as the integral of , i 'll write my x squared still , but instead of two to the x to the third i 'm going to write all of this business . let me just copy and paste that . we already established that this is the same thing as two to the x to the third power . copy and paste , just like that . then let me close it with a d x. i was able to get it in terms of e as a base . that makes me a little bit more comfortable but it still seems pretty complicated . you might be saying , `` okay , look . `` maybe u substitution could be at play here . '' because i have this crazy expression , x to the third times the natural log of two , but what 's the derivative of that ? well that 's going to be three x squared times the natural log of two , or three times the natural log of two times x squared . that 's just a constant times x squared . we already have a x squared here so maybe we can engineer this a little bit to have the constant there as well . let 's think about that . if we made this , if we defined this as u , if we said u is equal to x to the third times the natural log of two , what is du going to be ? du is going to be , it 's going to be , well natural log of two is just a constant so it 's going to be three x squared times the natural log of two . we could actually just change the order we 're multiplying a little bit . we could say that this is the same thing as x squared times three natural log of two , which is the same thing just using logarithm properties , as x squared times the natural log of two to the third power . three natural log of two is the same thing as the natural log of two to the third power . this is equal to x squared times the natural log of eight . let 's see , if this is u , where is du ? oh , and of course we ca n't forget the dx . this is a dx right over here , dx , dx , dx . where is the du ? well we have a dx . let me circle things . you have a dx here , you have a dx there . you have an x squared here , you have an x squared here . so really all we need is , all we need here is the natural log of eight . ideally we would have the natural log of eight right over here , and we could put it there as long as we also , we could multiply by the natural log of eight as long as we also divide by a natural log of eight . we can do it like right over here , we could divide by natural log of eight . but we know that the anti-derivative of some constant times a function is the same thing as a constant times the anti-derivative of that function . we could just take that on the outside . it 's one over the natural log of eight . let 's write this in terms of u and du . this simplifies to one over the natural log of eight times the anti-derivative of e to the u , e to the u , that 's the u , du . this times this times that is du , du . and this is straightforward , we know what this is going to be . this is going to be equal to , let me just write the one over natural log of eight out here , one over natural log of eight times e to the u , and of course if we 're thinking in terms of just anti-derivative there would be some constant out there . then we would just reverse the substitution . we already know what u is . this is going to be equal to , the anti-derivative of this expression is one over the natural log of eight times e to the , instead of u , we know that u is x to the third times the natural log of two . and of course we could put a plus c there . now , going back to the original problem . we just need to evaluate the anti-derivative of this at each of these points . let 's rewrite that . given what we just figured out , let me copy and paste that . this is just going to be equal to , it 's going to be equal to the anti-derivative evaluated at one minus the anti-derivative evaluated at zero . we do n't have to worry about the constants because those will cancel out . so we are going to get , we are going to get one -- let me evaluate it first at one . you 're going to get one over the natural log of eight times e to the one to the third power , which is just one , times the natural log of two , natural log of two , that 's evaluated at one . then we 're going to have minus it evaluated it at zero . it 's going to be one over the natural log of eight times e to the , well when x is zero this whole thing is going to be zero . well e to the zero is just one , and e to the natural log of two , well that 's just going to be two , we already established that early on , this is just going to be equal to two . we are left with two over the natural log of eight minus one over the natural log of eight , which is just going to be equal to one over the natural log of eight . and we are , and we are done .
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sal : let 's see if we can calculate the definite integral from zero to one of x squared times two to the x to the third power d x . like always i encourage you to pause this video and see if you can figure this out on your own .
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please answer my question .. what is i have the integration of x^2 from -2 to 2 ... and i wanted to solve it using substitution t=x^2 ..i will end up with these limits from 4 to 4 ..which will lead up to 0 ..how is this possible ?
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sal : let 's see if we can calculate the definite integral from zero to one of x squared times two to the x to the third power d x . like always i encourage you to pause this video and see if you can figure this out on your own . i 'm assuming you 've had a go at it . there 's a couple of interesting things here . the first thing , at least that my brain does , it says , `` i 'm used to taking derivatives and anti-derivatives of e to the x , not some other base to the x . '' we know that the derivative with respect to x of e to the x is e to the x , or we could say that the anti-derivative of e to the x is equal to e to the x plus c. since i 'm dealing with something raised to , this particular situation , something raised to a function of x , it seems like i might want to put , i might want to change the base here , but how do i do that ? the way i would do that is re-express two in terms of e. what would be two in terms of e ? two is equal to e , is equal to e raised to the power that you need to raise e to to get to two . what 's the power that you have to raise two to to get to two ? well that 's the natural log of two . once again the natural log of two is the exponent that you have to raise e to to get to two . if you actually raise e to it you 're going to get two . this is what two is . now what is two to the x to the third ? well if we raise both sides of this to the x to the third power , we raise both sides to the x to the third power , two to the x to the third is equal to , if i raise something to an exponent and then raise that to an exponent , it 's going to be equal to e to the x to the third , x to the third , times the natural log of two , times the natural log of two . that already seems pretty interesting . let 's rewrite this , and actually what i 'm going to do , let 's just focus on the indefinite integral first , see if we can figure that out . then we can apply , then we can take , we can evaluate the definite ones . let 's just think about this , let 's think about the indefinite integral of x squared times two to the x to the third power d x. i really want to find the anti-derivative of this . well this is going to be the exact same thing as the integral of , i 'll write my x squared still , but instead of two to the x to the third i 'm going to write all of this business . let me just copy and paste that . we already established that this is the same thing as two to the x to the third power . copy and paste , just like that . then let me close it with a d x. i was able to get it in terms of e as a base . that makes me a little bit more comfortable but it still seems pretty complicated . you might be saying , `` okay , look . `` maybe u substitution could be at play here . '' because i have this crazy expression , x to the third times the natural log of two , but what 's the derivative of that ? well that 's going to be three x squared times the natural log of two , or three times the natural log of two times x squared . that 's just a constant times x squared . we already have a x squared here so maybe we can engineer this a little bit to have the constant there as well . let 's think about that . if we made this , if we defined this as u , if we said u is equal to x to the third times the natural log of two , what is du going to be ? du is going to be , it 's going to be , well natural log of two is just a constant so it 's going to be three x squared times the natural log of two . we could actually just change the order we 're multiplying a little bit . we could say that this is the same thing as x squared times three natural log of two , which is the same thing just using logarithm properties , as x squared times the natural log of two to the third power . three natural log of two is the same thing as the natural log of two to the third power . this is equal to x squared times the natural log of eight . let 's see , if this is u , where is du ? oh , and of course we ca n't forget the dx . this is a dx right over here , dx , dx , dx . where is the du ? well we have a dx . let me circle things . you have a dx here , you have a dx there . you have an x squared here , you have an x squared here . so really all we need is , all we need here is the natural log of eight . ideally we would have the natural log of eight right over here , and we could put it there as long as we also , we could multiply by the natural log of eight as long as we also divide by a natural log of eight . we can do it like right over here , we could divide by natural log of eight . but we know that the anti-derivative of some constant times a function is the same thing as a constant times the anti-derivative of that function . we could just take that on the outside . it 's one over the natural log of eight . let 's write this in terms of u and du . this simplifies to one over the natural log of eight times the anti-derivative of e to the u , e to the u , that 's the u , du . this times this times that is du , du . and this is straightforward , we know what this is going to be . this is going to be equal to , let me just write the one over natural log of eight out here , one over natural log of eight times e to the u , and of course if we 're thinking in terms of just anti-derivative there would be some constant out there . then we would just reverse the substitution . we already know what u is . this is going to be equal to , the anti-derivative of this expression is one over the natural log of eight times e to the , instead of u , we know that u is x to the third times the natural log of two . and of course we could put a plus c there . now , going back to the original problem . we just need to evaluate the anti-derivative of this at each of these points . let 's rewrite that . given what we just figured out , let me copy and paste that . this is just going to be equal to , it 's going to be equal to the anti-derivative evaluated at one minus the anti-derivative evaluated at zero . we do n't have to worry about the constants because those will cancel out . so we are going to get , we are going to get one -- let me evaluate it first at one . you 're going to get one over the natural log of eight times e to the one to the third power , which is just one , times the natural log of two , natural log of two , that 's evaluated at one . then we 're going to have minus it evaluated it at zero . it 's going to be one over the natural log of eight times e to the , well when x is zero this whole thing is going to be zero . well e to the zero is just one , and e to the natural log of two , well that 's just going to be two , we already established that early on , this is just going to be equal to two . we are left with two over the natural log of eight minus one over the natural log of eight , which is just going to be equal to one over the natural log of eight . and we are , and we are done .
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we can do it like right over here , we could divide by natural log of eight . but we know that the anti-derivative of some constant times a function is the same thing as a constant times the anti-derivative of that function . we could just take that on the outside .
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why does sal keep the ln 2 as part of the derivative if ln 2 is a constant ?
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sal : let 's see if we can calculate the definite integral from zero to one of x squared times two to the x to the third power d x . like always i encourage you to pause this video and see if you can figure this out on your own . i 'm assuming you 've had a go at it . there 's a couple of interesting things here . the first thing , at least that my brain does , it says , `` i 'm used to taking derivatives and anti-derivatives of e to the x , not some other base to the x . '' we know that the derivative with respect to x of e to the x is e to the x , or we could say that the anti-derivative of e to the x is equal to e to the x plus c. since i 'm dealing with something raised to , this particular situation , something raised to a function of x , it seems like i might want to put , i might want to change the base here , but how do i do that ? the way i would do that is re-express two in terms of e. what would be two in terms of e ? two is equal to e , is equal to e raised to the power that you need to raise e to to get to two . what 's the power that you have to raise two to to get to two ? well that 's the natural log of two . once again the natural log of two is the exponent that you have to raise e to to get to two . if you actually raise e to it you 're going to get two . this is what two is . now what is two to the x to the third ? well if we raise both sides of this to the x to the third power , we raise both sides to the x to the third power , two to the x to the third is equal to , if i raise something to an exponent and then raise that to an exponent , it 's going to be equal to e to the x to the third , x to the third , times the natural log of two , times the natural log of two . that already seems pretty interesting . let 's rewrite this , and actually what i 'm going to do , let 's just focus on the indefinite integral first , see if we can figure that out . then we can apply , then we can take , we can evaluate the definite ones . let 's just think about this , let 's think about the indefinite integral of x squared times two to the x to the third power d x. i really want to find the anti-derivative of this . well this is going to be the exact same thing as the integral of , i 'll write my x squared still , but instead of two to the x to the third i 'm going to write all of this business . let me just copy and paste that . we already established that this is the same thing as two to the x to the third power . copy and paste , just like that . then let me close it with a d x. i was able to get it in terms of e as a base . that makes me a little bit more comfortable but it still seems pretty complicated . you might be saying , `` okay , look . `` maybe u substitution could be at play here . '' because i have this crazy expression , x to the third times the natural log of two , but what 's the derivative of that ? well that 's going to be three x squared times the natural log of two , or three times the natural log of two times x squared . that 's just a constant times x squared . we already have a x squared here so maybe we can engineer this a little bit to have the constant there as well . let 's think about that . if we made this , if we defined this as u , if we said u is equal to x to the third times the natural log of two , what is du going to be ? du is going to be , it 's going to be , well natural log of two is just a constant so it 's going to be three x squared times the natural log of two . we could actually just change the order we 're multiplying a little bit . we could say that this is the same thing as x squared times three natural log of two , which is the same thing just using logarithm properties , as x squared times the natural log of two to the third power . three natural log of two is the same thing as the natural log of two to the third power . this is equal to x squared times the natural log of eight . let 's see , if this is u , where is du ? oh , and of course we ca n't forget the dx . this is a dx right over here , dx , dx , dx . where is the du ? well we have a dx . let me circle things . you have a dx here , you have a dx there . you have an x squared here , you have an x squared here . so really all we need is , all we need here is the natural log of eight . ideally we would have the natural log of eight right over here , and we could put it there as long as we also , we could multiply by the natural log of eight as long as we also divide by a natural log of eight . we can do it like right over here , we could divide by natural log of eight . but we know that the anti-derivative of some constant times a function is the same thing as a constant times the anti-derivative of that function . we could just take that on the outside . it 's one over the natural log of eight . let 's write this in terms of u and du . this simplifies to one over the natural log of eight times the anti-derivative of e to the u , e to the u , that 's the u , du . this times this times that is du , du . and this is straightforward , we know what this is going to be . this is going to be equal to , let me just write the one over natural log of eight out here , one over natural log of eight times e to the u , and of course if we 're thinking in terms of just anti-derivative there would be some constant out there . then we would just reverse the substitution . we already know what u is . this is going to be equal to , the anti-derivative of this expression is one over the natural log of eight times e to the , instead of u , we know that u is x to the third times the natural log of two . and of course we could put a plus c there . now , going back to the original problem . we just need to evaluate the anti-derivative of this at each of these points . let 's rewrite that . given what we just figured out , let me copy and paste that . this is just going to be equal to , it 's going to be equal to the anti-derivative evaluated at one minus the anti-derivative evaluated at zero . we do n't have to worry about the constants because those will cancel out . so we are going to get , we are going to get one -- let me evaluate it first at one . you 're going to get one over the natural log of eight times e to the one to the third power , which is just one , times the natural log of two , natural log of two , that 's evaluated at one . then we 're going to have minus it evaluated it at zero . it 's going to be one over the natural log of eight times e to the , well when x is zero this whole thing is going to be zero . well e to the zero is just one , and e to the natural log of two , well that 's just going to be two , we already established that early on , this is just going to be equal to two . we are left with two over the natural log of eight minus one over the natural log of eight , which is just going to be equal to one over the natural log of eight . and we are , and we are done .
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you might be saying , `` okay , look . `` maybe u substitution could be at play here . '' because i have this crazy expression , x to the third times the natural log of two , but what 's the derivative of that ?
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instead of reversing the u-substitution , could n't we adjust the bounds of the original definite integral for u and calculate the definite integral from the expression with u ?
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sal : let 's see if we can calculate the definite integral from zero to one of x squared times two to the x to the third power d x . like always i encourage you to pause this video and see if you can figure this out on your own . i 'm assuming you 've had a go at it . there 's a couple of interesting things here . the first thing , at least that my brain does , it says , `` i 'm used to taking derivatives and anti-derivatives of e to the x , not some other base to the x . '' we know that the derivative with respect to x of e to the x is e to the x , or we could say that the anti-derivative of e to the x is equal to e to the x plus c. since i 'm dealing with something raised to , this particular situation , something raised to a function of x , it seems like i might want to put , i might want to change the base here , but how do i do that ? the way i would do that is re-express two in terms of e. what would be two in terms of e ? two is equal to e , is equal to e raised to the power that you need to raise e to to get to two . what 's the power that you have to raise two to to get to two ? well that 's the natural log of two . once again the natural log of two is the exponent that you have to raise e to to get to two . if you actually raise e to it you 're going to get two . this is what two is . now what is two to the x to the third ? well if we raise both sides of this to the x to the third power , we raise both sides to the x to the third power , two to the x to the third is equal to , if i raise something to an exponent and then raise that to an exponent , it 's going to be equal to e to the x to the third , x to the third , times the natural log of two , times the natural log of two . that already seems pretty interesting . let 's rewrite this , and actually what i 'm going to do , let 's just focus on the indefinite integral first , see if we can figure that out . then we can apply , then we can take , we can evaluate the definite ones . let 's just think about this , let 's think about the indefinite integral of x squared times two to the x to the third power d x. i really want to find the anti-derivative of this . well this is going to be the exact same thing as the integral of , i 'll write my x squared still , but instead of two to the x to the third i 'm going to write all of this business . let me just copy and paste that . we already established that this is the same thing as two to the x to the third power . copy and paste , just like that . then let me close it with a d x. i was able to get it in terms of e as a base . that makes me a little bit more comfortable but it still seems pretty complicated . you might be saying , `` okay , look . `` maybe u substitution could be at play here . '' because i have this crazy expression , x to the third times the natural log of two , but what 's the derivative of that ? well that 's going to be three x squared times the natural log of two , or three times the natural log of two times x squared . that 's just a constant times x squared . we already have a x squared here so maybe we can engineer this a little bit to have the constant there as well . let 's think about that . if we made this , if we defined this as u , if we said u is equal to x to the third times the natural log of two , what is du going to be ? du is going to be , it 's going to be , well natural log of two is just a constant so it 's going to be three x squared times the natural log of two . we could actually just change the order we 're multiplying a little bit . we could say that this is the same thing as x squared times three natural log of two , which is the same thing just using logarithm properties , as x squared times the natural log of two to the third power . three natural log of two is the same thing as the natural log of two to the third power . this is equal to x squared times the natural log of eight . let 's see , if this is u , where is du ? oh , and of course we ca n't forget the dx . this is a dx right over here , dx , dx , dx . where is the du ? well we have a dx . let me circle things . you have a dx here , you have a dx there . you have an x squared here , you have an x squared here . so really all we need is , all we need here is the natural log of eight . ideally we would have the natural log of eight right over here , and we could put it there as long as we also , we could multiply by the natural log of eight as long as we also divide by a natural log of eight . we can do it like right over here , we could divide by natural log of eight . but we know that the anti-derivative of some constant times a function is the same thing as a constant times the anti-derivative of that function . we could just take that on the outside . it 's one over the natural log of eight . let 's write this in terms of u and du . this simplifies to one over the natural log of eight times the anti-derivative of e to the u , e to the u , that 's the u , du . this times this times that is du , du . and this is straightforward , we know what this is going to be . this is going to be equal to , let me just write the one over natural log of eight out here , one over natural log of eight times e to the u , and of course if we 're thinking in terms of just anti-derivative there would be some constant out there . then we would just reverse the substitution . we already know what u is . this is going to be equal to , the anti-derivative of this expression is one over the natural log of eight times e to the , instead of u , we know that u is x to the third times the natural log of two . and of course we could put a plus c there . now , going back to the original problem . we just need to evaluate the anti-derivative of this at each of these points . let 's rewrite that . given what we just figured out , let me copy and paste that . this is just going to be equal to , it 's going to be equal to the anti-derivative evaluated at one minus the anti-derivative evaluated at zero . we do n't have to worry about the constants because those will cancel out . so we are going to get , we are going to get one -- let me evaluate it first at one . you 're going to get one over the natural log of eight times e to the one to the third power , which is just one , times the natural log of two , natural log of two , that 's evaluated at one . then we 're going to have minus it evaluated it at zero . it 's going to be one over the natural log of eight times e to the , well when x is zero this whole thing is going to be zero . well e to the zero is just one , and e to the natural log of two , well that 's just going to be two , we already established that early on , this is just going to be equal to two . we are left with two over the natural log of eight minus one over the natural log of eight , which is just going to be equal to one over the natural log of eight . and we are , and we are done .
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then we can apply , then we can take , we can evaluate the definite ones . let 's just think about this , let 's think about the indefinite integral of x squared times two to the x to the third power d x. i really want to find the anti-derivative of this . well this is going to be the exact same thing as the integral of , i 'll write my x squared still , but instead of two to the x to the third i 'm going to write all of this business .
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shouldnt you take the derivative of ln2 after the derivative of x^3 ?
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sal : let 's see if we can calculate the definite integral from zero to one of x squared times two to the x to the third power d x . like always i encourage you to pause this video and see if you can figure this out on your own . i 'm assuming you 've had a go at it . there 's a couple of interesting things here . the first thing , at least that my brain does , it says , `` i 'm used to taking derivatives and anti-derivatives of e to the x , not some other base to the x . '' we know that the derivative with respect to x of e to the x is e to the x , or we could say that the anti-derivative of e to the x is equal to e to the x plus c. since i 'm dealing with something raised to , this particular situation , something raised to a function of x , it seems like i might want to put , i might want to change the base here , but how do i do that ? the way i would do that is re-express two in terms of e. what would be two in terms of e ? two is equal to e , is equal to e raised to the power that you need to raise e to to get to two . what 's the power that you have to raise two to to get to two ? well that 's the natural log of two . once again the natural log of two is the exponent that you have to raise e to to get to two . if you actually raise e to it you 're going to get two . this is what two is . now what is two to the x to the third ? well if we raise both sides of this to the x to the third power , we raise both sides to the x to the third power , two to the x to the third is equal to , if i raise something to an exponent and then raise that to an exponent , it 's going to be equal to e to the x to the third , x to the third , times the natural log of two , times the natural log of two . that already seems pretty interesting . let 's rewrite this , and actually what i 'm going to do , let 's just focus on the indefinite integral first , see if we can figure that out . then we can apply , then we can take , we can evaluate the definite ones . let 's just think about this , let 's think about the indefinite integral of x squared times two to the x to the third power d x. i really want to find the anti-derivative of this . well this is going to be the exact same thing as the integral of , i 'll write my x squared still , but instead of two to the x to the third i 'm going to write all of this business . let me just copy and paste that . we already established that this is the same thing as two to the x to the third power . copy and paste , just like that . then let me close it with a d x. i was able to get it in terms of e as a base . that makes me a little bit more comfortable but it still seems pretty complicated . you might be saying , `` okay , look . `` maybe u substitution could be at play here . '' because i have this crazy expression , x to the third times the natural log of two , but what 's the derivative of that ? well that 's going to be three x squared times the natural log of two , or three times the natural log of two times x squared . that 's just a constant times x squared . we already have a x squared here so maybe we can engineer this a little bit to have the constant there as well . let 's think about that . if we made this , if we defined this as u , if we said u is equal to x to the third times the natural log of two , what is du going to be ? du is going to be , it 's going to be , well natural log of two is just a constant so it 's going to be three x squared times the natural log of two . we could actually just change the order we 're multiplying a little bit . we could say that this is the same thing as x squared times three natural log of two , which is the same thing just using logarithm properties , as x squared times the natural log of two to the third power . three natural log of two is the same thing as the natural log of two to the third power . this is equal to x squared times the natural log of eight . let 's see , if this is u , where is du ? oh , and of course we ca n't forget the dx . this is a dx right over here , dx , dx , dx . where is the du ? well we have a dx . let me circle things . you have a dx here , you have a dx there . you have an x squared here , you have an x squared here . so really all we need is , all we need here is the natural log of eight . ideally we would have the natural log of eight right over here , and we could put it there as long as we also , we could multiply by the natural log of eight as long as we also divide by a natural log of eight . we can do it like right over here , we could divide by natural log of eight . but we know that the anti-derivative of some constant times a function is the same thing as a constant times the anti-derivative of that function . we could just take that on the outside . it 's one over the natural log of eight . let 's write this in terms of u and du . this simplifies to one over the natural log of eight times the anti-derivative of e to the u , e to the u , that 's the u , du . this times this times that is du , du . and this is straightforward , we know what this is going to be . this is going to be equal to , let me just write the one over natural log of eight out here , one over natural log of eight times e to the u , and of course if we 're thinking in terms of just anti-derivative there would be some constant out there . then we would just reverse the substitution . we already know what u is . this is going to be equal to , the anti-derivative of this expression is one over the natural log of eight times e to the , instead of u , we know that u is x to the third times the natural log of two . and of course we could put a plus c there . now , going back to the original problem . we just need to evaluate the anti-derivative of this at each of these points . let 's rewrite that . given what we just figured out , let me copy and paste that . this is just going to be equal to , it 's going to be equal to the anti-derivative evaluated at one minus the anti-derivative evaluated at zero . we do n't have to worry about the constants because those will cancel out . so we are going to get , we are going to get one -- let me evaluate it first at one . you 're going to get one over the natural log of eight times e to the one to the third power , which is just one , times the natural log of two , natural log of two , that 's evaluated at one . then we 're going to have minus it evaluated it at zero . it 's going to be one over the natural log of eight times e to the , well when x is zero this whole thing is going to be zero . well e to the zero is just one , and e to the natural log of two , well that 's just going to be two , we already established that early on , this is just going to be equal to two . we are left with two over the natural log of eight minus one over the natural log of eight , which is just going to be equal to one over the natural log of eight . and we are , and we are done .
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this simplifies to one over the natural log of eight times the anti-derivative of e to the u , e to the u , that 's the u , du . this times this times that is du , du . and this is straightforward , we know what this is going to be .
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why did n't sal use the product rule for du ?
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sal : let 's see if we can calculate the definite integral from zero to one of x squared times two to the x to the third power d x . like always i encourage you to pause this video and see if you can figure this out on your own . i 'm assuming you 've had a go at it . there 's a couple of interesting things here . the first thing , at least that my brain does , it says , `` i 'm used to taking derivatives and anti-derivatives of e to the x , not some other base to the x . '' we know that the derivative with respect to x of e to the x is e to the x , or we could say that the anti-derivative of e to the x is equal to e to the x plus c. since i 'm dealing with something raised to , this particular situation , something raised to a function of x , it seems like i might want to put , i might want to change the base here , but how do i do that ? the way i would do that is re-express two in terms of e. what would be two in terms of e ? two is equal to e , is equal to e raised to the power that you need to raise e to to get to two . what 's the power that you have to raise two to to get to two ? well that 's the natural log of two . once again the natural log of two is the exponent that you have to raise e to to get to two . if you actually raise e to it you 're going to get two . this is what two is . now what is two to the x to the third ? well if we raise both sides of this to the x to the third power , we raise both sides to the x to the third power , two to the x to the third is equal to , if i raise something to an exponent and then raise that to an exponent , it 's going to be equal to e to the x to the third , x to the third , times the natural log of two , times the natural log of two . that already seems pretty interesting . let 's rewrite this , and actually what i 'm going to do , let 's just focus on the indefinite integral first , see if we can figure that out . then we can apply , then we can take , we can evaluate the definite ones . let 's just think about this , let 's think about the indefinite integral of x squared times two to the x to the third power d x. i really want to find the anti-derivative of this . well this is going to be the exact same thing as the integral of , i 'll write my x squared still , but instead of two to the x to the third i 'm going to write all of this business . let me just copy and paste that . we already established that this is the same thing as two to the x to the third power . copy and paste , just like that . then let me close it with a d x. i was able to get it in terms of e as a base . that makes me a little bit more comfortable but it still seems pretty complicated . you might be saying , `` okay , look . `` maybe u substitution could be at play here . '' because i have this crazy expression , x to the third times the natural log of two , but what 's the derivative of that ? well that 's going to be three x squared times the natural log of two , or three times the natural log of two times x squared . that 's just a constant times x squared . we already have a x squared here so maybe we can engineer this a little bit to have the constant there as well . let 's think about that . if we made this , if we defined this as u , if we said u is equal to x to the third times the natural log of two , what is du going to be ? du is going to be , it 's going to be , well natural log of two is just a constant so it 's going to be three x squared times the natural log of two . we could actually just change the order we 're multiplying a little bit . we could say that this is the same thing as x squared times three natural log of two , which is the same thing just using logarithm properties , as x squared times the natural log of two to the third power . three natural log of two is the same thing as the natural log of two to the third power . this is equal to x squared times the natural log of eight . let 's see , if this is u , where is du ? oh , and of course we ca n't forget the dx . this is a dx right over here , dx , dx , dx . where is the du ? well we have a dx . let me circle things . you have a dx here , you have a dx there . you have an x squared here , you have an x squared here . so really all we need is , all we need here is the natural log of eight . ideally we would have the natural log of eight right over here , and we could put it there as long as we also , we could multiply by the natural log of eight as long as we also divide by a natural log of eight . we can do it like right over here , we could divide by natural log of eight . but we know that the anti-derivative of some constant times a function is the same thing as a constant times the anti-derivative of that function . we could just take that on the outside . it 's one over the natural log of eight . let 's write this in terms of u and du . this simplifies to one over the natural log of eight times the anti-derivative of e to the u , e to the u , that 's the u , du . this times this times that is du , du . and this is straightforward , we know what this is going to be . this is going to be equal to , let me just write the one over natural log of eight out here , one over natural log of eight times e to the u , and of course if we 're thinking in terms of just anti-derivative there would be some constant out there . then we would just reverse the substitution . we already know what u is . this is going to be equal to , the anti-derivative of this expression is one over the natural log of eight times e to the , instead of u , we know that u is x to the third times the natural log of two . and of course we could put a plus c there . now , going back to the original problem . we just need to evaluate the anti-derivative of this at each of these points . let 's rewrite that . given what we just figured out , let me copy and paste that . this is just going to be equal to , it 's going to be equal to the anti-derivative evaluated at one minus the anti-derivative evaluated at zero . we do n't have to worry about the constants because those will cancel out . so we are going to get , we are going to get one -- let me evaluate it first at one . you 're going to get one over the natural log of eight times e to the one to the third power , which is just one , times the natural log of two , natural log of two , that 's evaluated at one . then we 're going to have minus it evaluated it at zero . it 's going to be one over the natural log of eight times e to the , well when x is zero this whole thing is going to be zero . well e to the zero is just one , and e to the natural log of two , well that 's just going to be two , we already established that early on , this is just going to be equal to two . we are left with two over the natural log of eight minus one over the natural log of eight , which is just going to be equal to one over the natural log of eight . and we are , and we are done .
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you might be saying , `` okay , look . `` maybe u substitution could be at play here . '' because i have this crazy expression , x to the third times the natural log of two , but what 's the derivative of that ?
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is there no simpler substitution that could be made ?
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sal : let 's see if we can calculate the definite integral from zero to one of x squared times two to the x to the third power d x . like always i encourage you to pause this video and see if you can figure this out on your own . i 'm assuming you 've had a go at it . there 's a couple of interesting things here . the first thing , at least that my brain does , it says , `` i 'm used to taking derivatives and anti-derivatives of e to the x , not some other base to the x . '' we know that the derivative with respect to x of e to the x is e to the x , or we could say that the anti-derivative of e to the x is equal to e to the x plus c. since i 'm dealing with something raised to , this particular situation , something raised to a function of x , it seems like i might want to put , i might want to change the base here , but how do i do that ? the way i would do that is re-express two in terms of e. what would be two in terms of e ? two is equal to e , is equal to e raised to the power that you need to raise e to to get to two . what 's the power that you have to raise two to to get to two ? well that 's the natural log of two . once again the natural log of two is the exponent that you have to raise e to to get to two . if you actually raise e to it you 're going to get two . this is what two is . now what is two to the x to the third ? well if we raise both sides of this to the x to the third power , we raise both sides to the x to the third power , two to the x to the third is equal to , if i raise something to an exponent and then raise that to an exponent , it 's going to be equal to e to the x to the third , x to the third , times the natural log of two , times the natural log of two . that already seems pretty interesting . let 's rewrite this , and actually what i 'm going to do , let 's just focus on the indefinite integral first , see if we can figure that out . then we can apply , then we can take , we can evaluate the definite ones . let 's just think about this , let 's think about the indefinite integral of x squared times two to the x to the third power d x. i really want to find the anti-derivative of this . well this is going to be the exact same thing as the integral of , i 'll write my x squared still , but instead of two to the x to the third i 'm going to write all of this business . let me just copy and paste that . we already established that this is the same thing as two to the x to the third power . copy and paste , just like that . then let me close it with a d x. i was able to get it in terms of e as a base . that makes me a little bit more comfortable but it still seems pretty complicated . you might be saying , `` okay , look . `` maybe u substitution could be at play here . '' because i have this crazy expression , x to the third times the natural log of two , but what 's the derivative of that ? well that 's going to be three x squared times the natural log of two , or three times the natural log of two times x squared . that 's just a constant times x squared . we already have a x squared here so maybe we can engineer this a little bit to have the constant there as well . let 's think about that . if we made this , if we defined this as u , if we said u is equal to x to the third times the natural log of two , what is du going to be ? du is going to be , it 's going to be , well natural log of two is just a constant so it 's going to be three x squared times the natural log of two . we could actually just change the order we 're multiplying a little bit . we could say that this is the same thing as x squared times three natural log of two , which is the same thing just using logarithm properties , as x squared times the natural log of two to the third power . three natural log of two is the same thing as the natural log of two to the third power . this is equal to x squared times the natural log of eight . let 's see , if this is u , where is du ? oh , and of course we ca n't forget the dx . this is a dx right over here , dx , dx , dx . where is the du ? well we have a dx . let me circle things . you have a dx here , you have a dx there . you have an x squared here , you have an x squared here . so really all we need is , all we need here is the natural log of eight . ideally we would have the natural log of eight right over here , and we could put it there as long as we also , we could multiply by the natural log of eight as long as we also divide by a natural log of eight . we can do it like right over here , we could divide by natural log of eight . but we know that the anti-derivative of some constant times a function is the same thing as a constant times the anti-derivative of that function . we could just take that on the outside . it 's one over the natural log of eight . let 's write this in terms of u and du . this simplifies to one over the natural log of eight times the anti-derivative of e to the u , e to the u , that 's the u , du . this times this times that is du , du . and this is straightforward , we know what this is going to be . this is going to be equal to , let me just write the one over natural log of eight out here , one over natural log of eight times e to the u , and of course if we 're thinking in terms of just anti-derivative there would be some constant out there . then we would just reverse the substitution . we already know what u is . this is going to be equal to , the anti-derivative of this expression is one over the natural log of eight times e to the , instead of u , we know that u is x to the third times the natural log of two . and of course we could put a plus c there . now , going back to the original problem . we just need to evaluate the anti-derivative of this at each of these points . let 's rewrite that . given what we just figured out , let me copy and paste that . this is just going to be equal to , it 's going to be equal to the anti-derivative evaluated at one minus the anti-derivative evaluated at zero . we do n't have to worry about the constants because those will cancel out . so we are going to get , we are going to get one -- let me evaluate it first at one . you 're going to get one over the natural log of eight times e to the one to the third power , which is just one , times the natural log of two , natural log of two , that 's evaluated at one . then we 're going to have minus it evaluated it at zero . it 's going to be one over the natural log of eight times e to the , well when x is zero this whole thing is going to be zero . well e to the zero is just one , and e to the natural log of two , well that 's just going to be two , we already established that early on , this is just going to be equal to two . we are left with two over the natural log of eight minus one over the natural log of eight , which is just going to be equal to one over the natural log of eight . and we are , and we are done .
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you might be saying , `` okay , look . `` maybe u substitution could be at play here . '' because i have this crazy expression , x to the third times the natural log of two , but what 's the derivative of that ?
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is there no simpler substitution that could be made ?
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