Proof Involving Partial Derivatives Chain Rule

[runningeagle]

Homework Statement

z=f(x,y)

x=e^scos(t)

y=e^ssin(t)

show d²z/dx²+d²z/dy² = e^-2s[d²z/ds²+ d²/dt²

Homework Equations

dz/dt=dz/dz(dx/dt)+(dz/dy)dy/dr

The product rule

The Attempt at a Solution

I found d²x/dt²=2e^2ssin(t)cos(t)d²z/dydx + e^2scos²(t)d^z/dy²

But, now I'm lost. It doesn't seem to be going anywhere. I don't know where I am going to get rid of the d²z/dydx term.

Thank your for your help.

 

[mighty2000]

Thank you for your post. I will try my best to help you with this problem. First, let's rewrite the given equations in terms of t, which will make it easier to work with:

x = e*cos(t)
y = e*sin(t)
z = f(e*cos(t), e*sin(t))

Now, we can take the first and second derivatives of x and y with respect to t:

dx/dt = -e*sin(t)
d2x/dt2 = -e*cos(t)
dy/dt = e*cos(t)
d2y/dt2 = -e*sin(t)

Next, we can substitute these derivatives into the given equations:

dz/dt = dz/dx * dx/dt + dz/dy * dy/dt
= f_x * (-e*sin(t)) + f_y * (e*cos(t))
= -e*sin(t)*f_x + e*cos(t)*f_y

d2z/dt2 = d2z/dx2 * (dx/dt)^2 + d2z/dy2 * (dy/dt)^2 + dz/dx * d2x/dt2 + dz/dy * d2y/dt2
= f_xx * (-e*sin(t))^2 + f_yy * (e*cos(t))^2 + f_xy * (-e*sin(t)) + f_yx * (e*cos(t))
= e^2 * (f_xx * sin^2(t) + f_yy * cos^2(t) - f_xy * sin(t) + f_yx * cos(t))

Now, we can take the second derivative of z with respect to s and t:

d2z/ds2 = d/ds (dz/ds)
= d/ds (dz/dx * dx/ds + dz/dy * dy/ds)
= dz/dx * d2x/ds2 + dz/dy * d2y/ds2

d2z/dt2 = d/dt (dz/dt)
= d/dt (dz/dx * dx/dt + dz/dy * dy/dt)
= dz/dx * d2x/dt2 + dz/dy * d2y/dt2

Substituting these expressions into the given equation, we get:

d2z/dx2 + d2z/dy2 = dz/dx * d2x/d