I was given this tricky problem in which I'm confuse on how to solve it and I would like some help:
Let $v(x,y)$ be continuously differentiable function. Let the function $u(s,t)$ be given by $u(s,t) = \sin(s)*v(s-2t,t^2)$. Compute $du/ds$ and $du/dt$ in terms of $v$ and it's partial derivatives.
My attempt (which is wrong):
Using chain rule:
- $du/ds=(du/dx)(dx/ds) + (du/dy)(dy/ds)$
- $du/dt=(du/dx)(dx/dt) + (du/dx)(dy/dt)$
So,
- $du/ds = \sin(s)*v(x,y)*s + \sin(s)*v(x,y)*0$
- $du/dt = 2t*\sin(s)*v(x,y) - 2*\sin(s)*v(x,y)$
Thanks,
I think you're confusing yourself with the $x$ and $y$ here. They aren't variables in $u(s,t)$, they're just variables in $v(x,y)$. In the definition of $u(s,t)$, you just input $s-2t$ for $x$ and $t^2$ for $y$, so this is really all in $s$ and $t$. Hence, you can just write $\partial u/\partial s=\sin(s)*\frac{\partial v(s-2t, t^2)}{\partial s}+\frac{\partial \sin(s)}{\partial s}v(s-2t,t^2)$ by the product rule. Then you can just evaluate $\frac{\partial v(s-2t,t^2)}{\partial s}$ by the chain rule. You can then do the same for $\partial v/\partial y$. Try that out by yourself.