SOLUTION: Hey I've tried this 3 times and I keep getting stuck halfway through it. Could someone please take me through the steps. z = f(x,y) of x and y, which has partial derivatives: dz

Algebra ->  College  -> Linear Algebra -> SOLUTION: Hey I've tried this 3 times and I keep getting stuck halfway through it. Could someone please take me through the steps. z = f(x,y) of x and y, which has partial derivatives: dz      Log On


   

Question 298073: Hey I've tried this 3 times and I keep getting stuck halfway through it. Could someone please take me through the steps.
z = f(x,y) of x and y, which has partial derivatives: dz/dx = 2xy and dz/dy = x^2
given the relations: r = sqrt(x^2+y^2) and theta = arctan (y/2)
x = r cos (theta) and y = r sin (theta)
find dz/d(theta) (expressed in terms of r and theta)
Answer by user_dude2008(1862) About Me  (Show Source):
You can put this solution on YOUR website!
dz/dx = 2xy
z=x^2y+C


dz/dy = x^2
z=x^2y+C

z = f(x,y) = x^2y+C


z=x^2y+C
z=(r*cos(theta))^2*r*sin(theta)+C

z=r^2*cos^2(theta)*r*sin(theta)+C

z=r^3*cos^2(theta)*sin(theta)+C

dz/d(theta) = r^3[-2cos(theta)sin^2(theta)+cos^3(theta)]