We are tasked with finding $\frac{dz}{dt}$ for the function:

$z = f(x, y) = x^2 + y^3$

where $x = t^2$ and $y = 2t$.

Solution:

To find $\frac{dz}{dt}$, we need to apply the chain rule. The chain rule for a function $z = f(x, y)$, where both $x$ and $y$ are functions of $t$, is given by:

$\frac{dz}{dt} = \frac{\partial z}{\partial x} \frac{dx}{dt} + \frac{\partial z}{\partial y} \frac{dy}{dt}$

Now, let’s compute each term step by step:

1. Partial derivative of $z$ with respect to $x$: $\frac{\partial z}{\partial x} = \frac{\partial}{\partial x}(x^2 + y^3) = 2x$

2. Partial derivative of $z$ with respect to $y$: $\frac{\partial z}{\partial y} = \frac{\partial}{\partial y}(x^2 + y^3) = 3y^2$

3. Derivative of $x = t^2$ with respect to $t$: $\frac{dx}{dt} = \frac{d}{dt}(t^2) = 2t$

4. Derivative of $y = 2t$ with respect to $t$: $\frac{dy}{dt} = \frac{d}{dt}(2t) = 2$

Now, applying the chain rule:

$\frac{dz}{dt} = \left( 2x \right) \cdot \frac{dx}{dt} + \left( 3y^2 \right) \cdot \frac{dy}{dt}$

Substitute $x = t^2$ and $y = 2t$ into the equation:

$\frac{dz}{dt} = 2(t^2) \cdot (2t) + 3(2t)^2 \cdot 2$

Simplify each term:

$\frac{dz}{dt} = 2t^2 \cdot 2t + 3 \cdot 4t^2 \cdot 2$

$\frac{dz}{dt} = 4t^3 + 24t^2$

Thus, the derivative is:

$\frac{dz}{dt} = 4t^3 + 24t^2$

Would you like further details or have any questions?

Here are five related questions to deepen your understanding:

1. How would the chain rule change if $z$ had more variables, such as $z = f(x, y, w)$?
2. What is the interpretation of partial derivatives in multivariable functions?
3. How does implicit differentiation apply to related rates problems?
4. Can you apply this process to different functions of $x$ and $y$ with respect to $t$?
5. What if $x$ and $y$ were functions of two variables, say $t$ and $u$?

Tip: Always apply the chain rule carefully by ensuring that each term is properly differentiated with respect to the desired variable.