# Multivariable Calculus chain rule question

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..

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

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