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If $u=f(x,y)$, where $x=e^{2s}\cos(5t)$ and $y=e^{2s}\sin(5t)$, then $\Big(\frac{\partial u}{\partial x}\Big)^2+\Big(\frac{\partial u}{\partial y}\Big)^2=g(s,t)\Big(\frac{\partial u}{\partial s}\Big)^2+h(s,t)\Big(\frac{\partial u}{\partial t}\Big)^2$. What are $g(s,t)$ and $h(s,t)$?

So I got that $u=f\left(e^{2s}\cos(5t),\,e^{2s}\sin(5t)\right)$ and that $\frac{\partial u}{\partial s}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial s}$.

This gave me $\frac{\partial u}{\partial s}=\frac{\partial f}{\partial x}\left(2e^{2s}cos\left(5t\right)\right)+\frac{\partial f}{\partial y}\left(2e^{2s}sin\left(5t\right)\right)$.

Of course, I can also do the same thing with $\frac{\partial u}{\partial t}$, but I am not entirely sure on how to move on from this. How exactly will I go about squaring these partial derivatives and separating them to get $g(s, t)$ and $h(s, t)$? Or am I actually supposed to find the second partial derivative?

Any help or guidance would be highly appreciated!

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1 Answer 1

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The notation here is not very clear, so I have slightly changed it:

Let $X$ and $Y$ be the functions $X(s,t)=e^{2s}\cos(5t)$ and $Y(s,t)=e^{2s}\sin(5t)$, and $u$ be the function

$$u(s,t)=f[X(s,t),Y(s,t)]$$

What you have been asked to find is the $g$ and $h$ in:

$$f_x^2+f_y^2=g(s,t)u_s^2+h(s,t)u_t^2 \tag{1}$$ where I have suppressed the arguments of all the derivatives.

Here

$$u_s(s,t)=f_xX_s+f_yY_s\qquad \text{and}\qquad u_t(s,t)=f_xX_t+f_yY_t$$

where $$X_s(s,t)=2e^{2s}\cos(5t)=2X\qquad \text{and}\qquad Y_s(s,t)=2e^{2s}\sin(5t)=2Y$$ while $$X_t(s,t)=-5e^{2s}\sin(5t)=-5Y\qquad \text{and}\qquad Y_t(s,t)=5e^{2s}\cos(5t)=5X$$

So

$$u_s(s,t)=2(f_xX+f_yY)\qquad \text{and}\qquad u_t(s,t)=-5(f_xY-f_yX)$$

You can solve this system of equations to get

$$f_x=\frac{5Xu_s-2Yu_t}{10(X^2+Y^2)}\qquad \text{and}\qquad \frac{2Xu_t+5Yu_s}{10( Y^{2}+X^{2})}$$

Then

$$f_x^2+f_y^2=\frac{25u_s^2+4u_t^2}{100(X^2+Y^2)}=\frac{25u_s^2+4u_t^2}{100e^{4s}}$$

and so $g(s,t)=\frac{1}{4e^{4s}}$ and $h(s,t)=\frac{1}{25e^{4s}}$.

I suspect there is probably a quicker way...

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