Calculus I question: Given st + t^2 = 4, find t"

[gotmilk4]

The question:

Given that s*t + t^2 = 4, find t''. That is, find (d^2t)/(ds^2).

The given answer (that I couldn't get) was:

8/(s+2t)^3

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I couldn't manipulate the equation such that s was a function of t, so I attempted implicit differentiation with respect to s:

d/ds (s*t +t^2 = 4)
s*t' + (1)t + 2t*t' = 0
t'*(s + 2t) = -t
t' = -t / (s + 2t) = -t*(s + 2t)^(-1)

Then I differentiated again:

t'' = - [ t*(-1)*(s + 2t)^(-2) * (1 + 2t') + t'*(s + 2t)^(-1) ]
t'' = [ t*(1 + 2t') / (s + 2t)^2 ] - [ t'/(s + 2t) ]
t'' = [ t +2t*t' - s*t' - 2t*t' ] / (s + 2t)^2
t'' = [ t + 4t*t' - s*t' ] / (s + 2t)^2
t'' = (t - s*t') / (s + 2t)^2

Substituting t' = -t / (s + 2t):

t'' = (t - s*t') / (s + 2t)^2
...(algebra)...

t'' = 2t^2 / (s + 2t)^3

That's as far as I got (but every time I do this problem I get a different answer - this is the most recent and closest answer). Does anyone see my errors or something I overlooked? Or did I just completely screw the problem up? Thanks.

 

[soroban]

Re: Calculus I question

Hello, gotmilk4!

Given that: \(\displaystyle \,st\,+\,t^2\:=\:4\), find \(\displaystyle t''\)

The given answer was: \(\displaystyle \,\frac{8}{(s\,+\,2t)^3}\)

\(\displaystyle \L\;\;\;t''\;=\;\frac{t\,-\, s\cdot t'}{(s\,+\,2t)^2\;\;\) . . . This is correct!

All the work you showed was correct . . . good work!
Your error must be in the algebra after substituting.
Too bad you didn't show us your steps . . .


Substituting: \(\displaystyle \,t'\:=\:\frac{-t}{s\,+\,2t}\)

\(\displaystyle \L\;\:t''\;=\;\frac{t\,-\,s\left(\frac{-t}{s\,+\,2t}\right)}{(s\,+\,2t)^2}\)


Multiply top and bottom by \(\displaystyle (s\,+\,2t):\)

\(\displaystyle \L\;\;t''\;=\;\frac{t(s\,+\,2t)\,+\,st}{(s\,+\,2t)^3}\;=\;\frac{st\,+\,2t^2\,+\,st}{(s\,+\,2t)^3}\;=\;\frac{2st\,+\,2t^2}{(s\,+\,2t)^3}\;=\;\frac{2(st\,+\,t^2)}{(s\,+\,2t)^3}\)


Better sit down . . . brace yourself for this!

At the very beginning of the problem, it said: \(\displaystyle \L\,st\,+\,t^2\:=\:4\) . . . right?

Therefore: \(\displaystyle \L\:y''\;=\;\frac{2(4)}{(s\,+\,2t)^3}\;=\;\frac{8}{(s\,+\,2t)^3}\;\;\) . . . ta-DAA!