If $u = x^3 + y^3$ where $x = a \cos(t)$, $y = b \sin(t)$, choose the value of $\frac{du}{dt}$.

Steps to Solve

1. Find $\frac{du}{dx}$ Differentiate $u = x^3 + y^3$ with respect to $x$, treating $y$ as a function of $x$.

   $\frac{du}{dx} = 3x^2$

2. Find $\frac{du}{dy}$ Differentiate $u = x^3 + y^3$ with respect to $y$, treating $x$ as a function of $y$.

   $\frac{du}{dy} = 3y^2$

3. Find $\frac{dx}{dt}$ Differentiate $x = a \cos(t)$ with respect to $t$.

   $\frac{dx}{dt} = -a \sin(t)$

4. Find $\frac{dy}{dt}$ Differentiate $y = b \sin(t)$ with respect to $t$.

   $\frac{dy}{dt} = b \cos(t)$

5. Apply the chain rule Using the chain rule, we have: $\frac{du}{dt} = \frac{du}{dx} \cdot \frac{dx}{dt} + \frac{du}{dy} \cdot \frac{dy}{dt}$

6. Substitute the derivatives Substitute the derivatives we found in the previous steps: $\frac{du}{dt} = (3x^2)(-a \sin(t)) + (3y^2)(b \cos(t))$

7. Substitute $x$ and $y$ Substitute $x = a \cos(t)$ and $y = b \sin(t)$ into the equation: $\frac{du}{dt} = 3(a \cos(t))^2(-a \sin(t)) + 3(b \sin(t))^2(b \cos(t))$

8. Simplify Simplify the expression: $\frac{du}{dt} = -3a^3 \cos^2(t) \sin(t) + 3b^3 \sin^2(t) \cos(t)$ $\frac{du}{dt} = 3b^3 \sin^2(t) \cos(t) - 3a^3 \cos^2(t) \sin(t)$