If z=r^2 tan^2(theta),x=r cos(theta),y=r sin(theta),
find the {partial^2 z}/{partial(x)partial(y)}
staying with parametric functions makes it hard.
So, let
r^2 = x^2+y^2
tanθ = y/x
z = r^2 tan^2θ = (x^2+y^2)(y^2/x^2) = y^2 + y^4/x^2
∂z/∂x = -2y^4/x^3
∂z/∂y = 2y + 4y^3/x^2
∂^2z/∂x∂y = ∂^2z/∂y∂x = -8y^3/x^3 = -8tan^3θ
To find the second partial derivative of z with respect to x and y, we need to differentiate the expression z = r^2tan^2(theta) with respect to x and y separately and then take the partial derivative.
First, let's find the partial derivative of z with respect to x, denoted as dz/dx:
Taking the given expression for x = rcos(theta), we differentiate both sides with respect to x:
dx/dx = d(rcos(theta))/dx
1 = -rsin(theta)*d(theta)/dx
Since x is independent of theta, we can assume that d(theta)/dx = 0
Therefore, -rsin(theta)*d(theta)/dx = 0
Now, let's find the partial derivative of z with respect to y, denoted as dz/dy:
Taking the given expression for y = rsin(theta), we differentiate both sides with respect to y:
dy/dy = d(rsin(theta))/dy
1 = rcos(theta)*d(theta)/dy
Since y is independent of theta, we can assume that d(theta)/dy = 0
Therefore, rcos(theta)*d(theta)/dy = 0
Now that we have found dz/dx = 0 and dz/dy = 0, we can find the second partial derivative of z with respect to x and y, denoted as d^2z/dxdy:
d^2z/dxdy = d(dz/dy)/dx
Since dz/dy = 0, we have:
d^2z/dxdy = d(0)/dx = 0
Therefore, the second partial derivative of z with respect to x and y is equal to 0.