To find $\frac{dz}{dt}$, we need to use the chain rule because $z$ is a function of $x$ and $y$, which in turn are functions of $t$.

Given: $z = x^2y + y^2, \quad x = \log(t), \quad y = e^t$

We will first compute the partial derivatives of $z$ with respect to $x$ and $y$, and then use the chain rule to compute $\frac{dz}{dt}$.

Step 1: Find $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$

$z = x^2y + y^2$

• $\frac{\partial z}{\partial x} = 2xy$
• $\frac{\partial z}{\partial y} = x^2 + 2y$

Step 2: Find $\frac{dx}{dt}$ and $\frac{dy}{dt}$

• $x = \log(t)$, so $\frac{dx}{dt} = \frac{1}{t}$
• $y = e^t$, so $\frac{dy}{dt} = e^t$

Step 3: Apply the chain rule

Using the chain rule:

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

Substitute the values:

$\frac{dz}{dt} = (2xy) \cdot \frac{1}{t} + (x^2 + 2y) \cdot e^t$

Step 4: Evaluate at $t = 1$

• When $t = 1$, $x = \log(1) = 0$ and $y = e^1 = e$.

Substitute these into the equation for $\frac{dz}{dt}$:

$\frac{dz}{dt} = (2(0)(e)) \cdot \frac{1}{1} + (0^2 + 2(e)) \cdot e$

Simplifying:

$\frac{dz}{dt} = 0 + (2e) \cdot e = 2e^2$

So, $\frac{dz}{dt}$ at $t = 1$ is:

$\boxed{2e^2}$

Would you like any further clarification or breakdown of the steps? Here are some related questions you might find useful:

1. How do you apply the chain rule to functions with multiple variables?
2. What are the steps to calculate partial derivatives?
3. Can we use this same method for functions involving more variables?
4. How does the result change if we evaluated at a different value of $t$?
5. What is the role of logarithms in differentiation, especially in relation to $x = \log(t)$?

Tip: Always carefully track each function's dependency when applying the chain rule, especially in problems involving composite functions.