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



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moongazer
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I only got to the part:
dy/dt = 2at/3y^2

Yttrium
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Just derive again, so you can solve the d^2y/dt^2. :))

Yttrium
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it's like solve for y''

amriju
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let it remain as 3y^2=2at..differntiate again....using multiplication rule....

moongazer
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it says the answer is:
2a/9y^2
but I am getting a different answer

moongazer
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is "a" here constant?

Yttrium
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Quotient rule, remember?

amriju
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yo constnt

Yttrium
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for easier calculation, you can factor out 2a/3 and just derive (t/y^2)

moongazer
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could you show me the first step for the second derivative?
maybe I did something wrong

moongazer
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i'll try again :)

amriju
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may be u need to substitute y as (at^2)^2/3

moongazer
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I got:
(6ay^4  8a^2 t^2 y)/9y^6
I think I am doing something wrong.

moongazer
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i'll try again :)

Yttrium
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\[y' = \frac{ 2at }{ 3y^2 }\], 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
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can you follow @moongazer ?

moongazer
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That's what I got. I think there is some typo in my answer sheet.
Thanks :)

myininaya
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Well you could simplify

Yttrium
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@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
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@myininaya could you simply it further?
@Yttrium yes, I understood your solution

myininaya
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Could I? Yes. Can you?

myininaya
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Eye ball the numerator and the denominator. You should see they share a common factor.

Yttrium
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For our class, the definition of simplest form is that it is factored out completely. Just factor out some stuff and do cancellation.

moongazer
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2a(3y^3  4at^2) / 9y^5

moongazer
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that's what I got

myininaya
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You can also choose to write at^2 as y^3

myininaya
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Recall our initial equation: y^3=at^2

myininaya
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I see at^2

myininaya
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Replace it y^3
then combine like terms

Yttrium
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@moongazer , do you what myininaya told you?

myininaya
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@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
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WOW! Thank you very much to both of you.
I failed to notice that until you said it.
Thanks :)
I now got the answer. :)