Suppose that z = f(x, y) has continuous second order partial derivatives...

Question

Suppose that z = f(x, y) has continuous second order partial derivatives

and x = 4rt and y = r^3 - t^3. Find (∂^2)(z) / (∂r∂t)

Answer 1

Second order partial derivatives of f(x,y) are as follows:

f_xx = (f_x)_x = ∂(∂f/∂x)⁄∂x = ∂²f/∂x²

f_yy = (f_y)_y= ∂(∂f/∂y)⁄∂y = ∂²f/∂y²

f_xy = (fx)_y= ∂(∂f/∂x)⁄∂y = ∂²f/∂y∂x

f_yx = (fx)_y= ∂(∂f/∂y)⁄∂x = ∂²f/∂x∂y

If f_xy and f_yx are continuous, then f_xy = f_yx

We know here that z = f(x,y) and x=4rt and y=r³-t³

To find ∂²/∂r∂t we will

1. Differentiate z with respect to t first (as per our rules of second order derivatives above)
2. Differentiate ∂z/∂dt with respect to r.

Because z depends on x and y, which depend on r and t, we will use chain rule.

1. Find ∂z/∂dt

∂z/∂t = f_x∂x/∂t + f_y∂y/∂t

∂x/∂t = 4r and ∂y/∂t = -3t² (remember we treat r as a constant when differentiating with respect to t)

so,

∂z/∂t = 4rf_x - 3t²f_y

2. Differentiate ∂z/∂t with respect to r.

∂(4rf_x - 3t²f_y)/∂r

Term 1:

∂(4rf_x)/∂r = 4f_x + 4r∂f_x/∂r (product rule)

f_x depends on x and y so we need to do the chain rule next:

∂f_x/∂r = f_xxx_r + f_xyy_r

x_r = 4t and y_r = 3r²

So,

∂f_x/∂r = 4tf_xx + 3r²f_xy

∂(4rf_x)/∂r = 4f_x + 16rtf_xx +12r³f_xy

Term 2:

∂(-3t²f_y)/∂r = -3t²∂f_y/∂r

∂f_y/∂r = f_yxx_r + f_yyy_r = 4tf_yx + 3r²f_yy

So,

-3t²(4tf_yx + 3r²f_yy) = -12t³f_yx - 9r²t²f_yy

Finally, combine our terms

∂²z/∂r∂t = 4f_x + 16rtf_xx + 12r³f_xy - 12t³f_xy - 9r²t²f_yy

Because our second partial derivatives are continuous, we know that f_xy = f_yx and we can combine like terms. And our final solution is:

∂²z/∂r∂t = 4f_x + 16rtf_xx + 12f_xy(r³-t³) - 9r²t²f_yy