Suppose that $k^{2} + h^{3} = 9$. Find $\frac{dh}{dt}$ when $k=1$ and $\frac{dk}{dt} = 3$ ans $= \frac{1}{2}$. I'm differentiating with respect to $t$ but I cannot get the answer if you could show me the steps, or how to approach this question, would be very helpful :)
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Use the fact that $$\frac{dh}{dk} = \frac{\frac{dh}{dt}}{\frac {dk}{dt}} \iff \frac{dh}{dt} = \frac{dh}{dk}\cdot \frac{dk}{dt}$$
Integrate the equation with respect to $k$, evaluate $\frac{dh}{dk}$ at $k = 1$, use that value, and use the given value for $\frac {dk}{dt}$, and solve for $$\frac{dh}{dt} = \frac{dh}{dk}\cdot \frac{dk}{dt}$$
amWhy
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Your work is great$\Large\color{green}{✓}$@amWhy – Software May 26 '13 at 04:37
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thanks ofr answering my question although a little TOO simplified! hahahahaha but thanks also your explanation is helpful when used with Adriano's – Unistudent9 May 26 '13 at 12:59
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You're welcome. Sorry for the "oversimplification". ;-) – amWhy May 26 '13 at 13:01
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@amWhy: I agree - you always do very nice write-ups! +1 – Amzoti May 27 '13 at 01:01
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Treating $k$ and $h$ as functions of $t$, we take the derivative of each side of the given equation with respect to $t$ by using implicit differentiation and chain rule to obtain: $$\begin{align*} 2k\frac{dk}{dt}+3h^2\frac{dh}{dt} &= 0 \\ 3h^2\frac{dh}{dt} &= -2k\frac{dk}{dt} \\ \frac{dh}{dt} &= \frac{-2k\frac{dk}{dt}}{3h^2} \\ \end{align*}$$ Note that when $k=1$, we have $1^2+h^3=9 \iff h^3=8 \iff h=2$. Thus, substitution yields: $$\frac{dh}{dt} = \frac{-2(1)(3)}{3(2)^2} = \frac{-2}{2^2} = \boxed{\dfrac{-1}{2}}$$
Adriano
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That makes alot of sense, its so simple thankyou so much for your explanation it is really helpful! – Unistudent9 May 26 '13 at 12:58