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Looking for help with this calculus 3 questions:

If $u=f(x,y)$ where $ x = e^{4s}\cos 2t $ and $ y = e^{4s}\sin 2t $, show that $$\left(\frac{\partial u}{\partial x}\right)^2 + \left(\frac{\partial u}{\partial y}\right)^2 = g(s,t) \left(\frac{\partial u}{\partial s}\right)^2 + h(s,t) \left(\frac{\partial u}{\partial t}\right)^2 $$

where $g(s,t)=\textrm{?}$, $h(s,t)= \textrm{?}$

So I know that the way to begin is by computing $\partial u/\partial s$ and $\partial u/\partial t$ using the chain rule, squaring them, then solving for $g$ and $h$.

so, I get

$$ \frac{\partial u}{\partial s} = \frac{\partial u}{\partial x}\frac{\partial x}{\partial s} + \frac{\partial u}{\partial y}\frac{\partial y}{\partial s} = \frac{\partial u}{\partial x}(4e^{4s}\cos 2t) + \frac{\partial u}{\partial y}(4e^{4s}\sin 2t) $$

$$ \frac{\partial u}{\partial t} = \frac{\partial u}{\partial x}\frac{\partial x}{\partial t} + \frac{\partial u}{\partial y}\frac{\partial y}{\partial t} = \frac{\partial u}{\partial x}(-2e^{4s}\sin 2t) + \frac{\partial u}{\partial y}(2e^{4s}\cos 2t) $$

I know I'm supposed to square the terms or something but I'm not really sure how to proceed..

Thank you in advance!

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  • $\begingroup$ Use dollar signs and Latex/Mathjax syntax to typeset your equations. $\endgroup$ – MrYouMath Oct 30 '17 at 20:08
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From

$$ \frac{\partial u}{\partial s} = 4e^{4s}\cos 2t\frac{\partial u}{\partial x} + 4e^{4s}\sin 2t\frac{\partial u}{\partial y} $$

$$ \frac{\partial u}{\partial t} = -2e^{4s}\sin 2t\frac{\partial u}{\partial x} + 2e^{4s}\cos 2t\frac{\partial u}{\partial y} $$

we can solve for $\partial u/\partial x$ and $\partial u/\partial y$ by treating them as variables in a system of equations. You can either solve for one in terms of the other, or try to eliminate coefficients.

I'm going to eliminate $\partial u/\partial y$ by multiplying the first equation by $\cos 2t$ and the second equation by $2\sin 2t$

$$ \cos 2t\frac{\partial u}{\partial s} = 4e^{4s}\cos^2 2t\frac{\partial u}{\partial x} + 4e^{4s}\sin 2t\cos 2t\frac{\partial u}{\partial y} $$

$$ 2\sin 2t\frac{\partial u}{\partial t} = -4e^{4s}\sin^2 2t\frac{\partial u}{\partial x} + 4e^{4s}\sin 2t\cos 2t\frac{\partial u}{\partial y} $$

Subtract the two equations to get $$ \cos 2t\frac{\partial u}{\partial s} - 2\sin 2t\frac{\partial u}{\partial t} = 8e^{4s}\frac{\partial u}{\partial x} $$

Or

$$ \frac{\partial u}{\partial x} = \frac{1}{8}e^{-4s}\cos 2t\frac{\partial u}{\partial s} - \frac{1}{4}e^{-4s}\sin 2t\frac{\partial u}{\partial t} $$

You can do the remaining work to get $$ \frac{\partial u}{\partial y} = \frac{1}{8}e^{-4s}\sin 2t\frac{\partial u}{\partial s} + \frac{1}{4}e^{-4s}\cos 2t\frac{\partial u}{\partial s} $$

and plug both into the initial equation

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