Question

Find g'(t) when g(t) = f(4 sin 2t, 5 cos 2t).

a) g'(t) = 8 cos 2t - 10 sin 2t
b) g'(t) = 8 sin 2t + 10 cos 2t
c) g'(t) = 8 cos 2t + 10 sin 2t
d) g'(t) = 8 sin 2t - 10 cos 2t

Answer 1

To find g'(t) when g(t) = f(4 sin 2t, 5 cos 2t), we need to find the partial derivatives of f with respect to each variable and evaluate them at the given values. Using the chain rule, we can then find g'(t) = 32 + 18sin²(2t).

Step-by-step explanation:

The function g(t) is defined as g(t) = f(4 sin 2t, 5 cos 2t). To find g'(t), we need to find the partial derivatives of f with respect to each variable and then evaluate them at the given values. Let's find the partial derivative of f with respect to the first variable:

∂f/∂x = ∂f/∂(4 sin 2t) = 4(cos 2t)

Next, let's find the partial derivative of f with respect to the second variable:

∂f/∂y = ∂f/∂(5 cos 2t) = 5(-sin 2t)

Now we can apply the chain rule to find g'(t):

g'(t) = (∂f/∂x)(∂x/∂t) + (∂f/∂y)(∂y/∂t)

Plugging in the partial derivatives we found:

g'(t) = 4(cos 2t)(8 cos 2t) + 5(-sin 2t)(-10 sin 2t) = 32 cos²(2t) + 50 sin²(2t)

Simplifying the expression:

g'(t) = 32(1 - sin²(2t)) + 50 sin²(2t) = 32 - 32sin²(2t) + 50sin²(2t) = 32 + 18sin²(2t)

Therefore, the correct answer is g'(t) = 32 + 18sin²(2).