Suppose z = f(x, y) is a di erentiable function. We could view z as a function o

Question

Suppose z = f(x, y) is a dierentiable function. We could view z as a function of rectangular coordinates (x,y) or, if we set x = r cos(theta) and y = r cos(theta), we could view z as a function of polar coordinates. Use the chain rule to show  (dz/dr)^2+(1/r^2)(dz/dtheta)^2 = (dz/dx)^2 + (dz/dy)^2 (d^2z/dr^2)+(1/r)(dz/dr)+ (1/r^2)(d^2z/dtheta^2) = (d^2z/dx^2) + (d^2z/dy^2) Your best friend for these problems will be the identity cos^2(theta) + sin2(theta) = 1

Explanation / Answer

z= f(x,y) ; x = r cos theta and y = r sin theta ; dz/dr = dz/dx * dx/dr + dz/dy*dy/dr = dz/dx* cos theta + dz/dy* sin theta ; dz/dtheta = dz/dx*dx/dtheta + dz/dy*dy/d theta = -dz/dx* r sin teta + dz/dy* r cos theta ; (dz/dr)^2 = (dz/dx)^2 * (cos theta )^2 + (dz/dy)^2 * (sin theta)^2 + 2*dz/dx*dz/dy * sin theta * cos theta ; ---(1)----------> (1/r^2)(dz/dtheta)^2 = (dz/dx)^2 * sin theta)^2 + (dz/dy)^2*( cos theta)^2 - 2*dz/dx*dz/dy* sin theta * cos theta ; -----(2) -------> Form (1) and (2) (dz/dr)^2 + (1/r^2)dz/dtheta)^2 = (dz/dx)^2 + (dz/dy)^2; since (cos theta)^2 + (sin theta)^2 = 1 ; d^2z/dr^2 = d^2z/dx^2 * (cos theta)^2 + (d^2z/dy^2)*( sin theta)^2 -------(A) ; and d^2z/dtheta^2 =r^2*(d^2z/dx^2)*(sin theta)^2 - (dz/dx)*r*(cos theta) + r^2*(d^2z/dy^2)*(cos theta)^2 - (dz/dy)*r*(sin theta); so (1/r^2)*d^2z/dtheta^2 = (d^2z/dx^2)*(sin theta)^2 + (d^2z/dy^2)*(cos theta)^2 - 1/r*[ (dz/dx)*r*(cos theta) + (dz/dy)*(sin theta)] ; (1/r^2)*d^2z/dtheta^2 = (d^2z/dx^2)*(sin theta)^2 + (d^2z/dy^2)*(cos theta)^2 - 1/r* (dz/dr) so (1/r^2)*d^2z/dtheta^2 +1/r* (dz/dr) = (d^2z/dx^2)*(sin theta)^2 + (d^2z/dy^2)*(cos theta)^2 -------(B) ; (A)+(B) = d^2z/dr^2 + (1/r^2)*d^2z/dtheta^2 +1/r* (dz/dr) = (d^2z/dx^2) + (d^2z/dy^2)

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