Use chain rule to find the derivative d/dt (v) w = z ^ 3 * e ^ (x ^ 2 + y ^ 2) along the path = cost, y=sint, z=sect13:21

To find the derivative d/dt (v) w = z ^ 3 * e ^ (x ^ 2 + y ^ 2) along the path x = cos(t), y = sin(t), z = sec(t), we can use the chain rule.

Step 1: Find the partial derivatives of v with respect to x, y, and z.

∂v/∂x = 3z^3 * e^(x^2 + y^2) ∂v/∂y = 2y * z^3 * e^(x^2 + y^2) ∂v/∂z = 3z^2 * e^(x^2 + y^2)

Step 2: Find the derivatives of x, y, and z with respect to t.

dx/dt = -sin(t) dy/dt = cos(t) dz/dt = sec(t) * tan(t)

Step 3: Apply the chain rule to find d/dt (v).

d/dt (v) = ∂v/∂x * dx/dt + ∂v/∂y * dy/dt + ∂v/∂z * dz/dt

Substituting the partial derivatives and derivatives of x, y, and z, we get:

d/dt (v) = (3z^3 * e^(x^2 + y^2)) * (-sin(t)) + (2y * z^3 * e^(x^2 + y^2)) * cos(t) + (3z^2 * e^(x^2 + y^2)) * (sec(t) * tan(t))

Finally, substitute x = cos(t), y = sin(t), and z = sec(t) into the expression to get the derivative along the given path.