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moongazer
 3 years ago
If y^3 = at^2 , then (d^2 y)/dt^2 ?
moongazer
 3 years ago
If y^3 = at^2 , then (d^2 y)/dt^2 ?

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moongazer
 3 years ago
Best ResponseYou've already chosen the best response.1I only got to the part: dy/dt = 2at/3y^2

Yttrium
 3 years ago
Best ResponseYou've already chosen the best response.1Just derive again, so you can solve the d^2y/dt^2. :))

Yttrium
 3 years ago
Best ResponseYou've already chosen the best response.1it's like solve for y''

amriju
 3 years ago
Best ResponseYou've already chosen the best response.0let it remain as 3y^2=2at..differntiate again....using multiplication rule....

moongazer
 3 years ago
Best ResponseYou've already chosen the best response.1it says the answer is: 2a/9y^2 but I am getting a different answer

moongazer
 3 years ago
Best ResponseYou've already chosen the best response.1is "a" here constant?

Yttrium
 3 years ago
Best ResponseYou've already chosen the best response.1Quotient rule, remember?

Yttrium
 3 years ago
Best ResponseYou've already chosen the best response.1for easier calculation, you can factor out 2a/3 and just derive (t/y^2)

moongazer
 3 years ago
Best ResponseYou've already chosen the best response.1could you show me the first step for the second derivative? maybe I did something wrong

amriju
 3 years ago
Best ResponseYou've already chosen the best response.0may be u need to substitute y as (at^2)^2/3

moongazer
 3 years ago
Best ResponseYou've already chosen the best response.1I got: (6ay^4  8a^2 t^2 y)/9y^6 I think I am doing something wrong.

Yttrium
 3 years ago
Best ResponseYou've already chosen the best response.1\[y' = \frac{ 2at }{ 3y^2 }\], right? Therefore, \[y' = \frac{ 2a }{ 3 } (\frac{ t }{ y^2 })\] \[y'' = \frac{ 2a }{ 3 } [\frac{ y^2  2tyy') }{ y^4 }]\] \[y'' = \frac{ 2a }{ 3 } \frac{ y^2  2ty(\frac{ 2a }{ 3 })(\frac{ t }{ y^2 }) }{ y^4 }\] Simplifying this equation we will get. \[y'' = \frac{ 2a }{ 3 } [ \frac{ 3y^4 4at^2y }{ 3y^6 }]\]

Yttrium
 3 years ago
Best ResponseYou've already chosen the best response.1can you follow @moongazer ?

moongazer
 3 years ago
Best ResponseYou've already chosen the best response.1That's what I got. I think there is some typo in my answer sheet. Thanks :)

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.1Well you could simplify

Yttrium
 3 years ago
Best ResponseYou've already chosen the best response.1@moongazer, it isn't the answer in simplest form. Maybe you can continue this and let us verify if we get the same answers.

moongazer
 3 years ago
Best ResponseYou've already chosen the best response.1@myininaya could you simply it further? @Yttrium yes, I understood your solution

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.1Could I? Yes. Can you?

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.1Eye ball the numerator and the denominator. You should see they share a common factor.

Yttrium
 3 years ago
Best ResponseYou've already chosen the best response.1For our class, the definition of simplest form is that it is factored out completely. Just factor out some stuff and do cancellation.

moongazer
 3 years ago
Best ResponseYou've already chosen the best response.12a(3y^3  4at^2) / 9y^5

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.1You can also choose to write at^2 as y^3

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.1Recall our initial equation: y^3=at^2

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.1Replace it y^3 then combine like terms

Yttrium
 3 years ago
Best ResponseYou've already chosen the best response.1@moongazer , do you what myininaya told you?

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.1@moongazer I'm leaving for tonight but I think Yttrium is still here to help. Goodnight you guys. I think you guys got it from here for sure. :)

moongazer
 3 years ago
Best ResponseYou've already chosen the best response.1WOW! Thank you very much to both of you. I failed to notice that until you said it. Thanks :) I now got the answer. :)
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