To solve for $\frac{dz}{dt}$ using the Chain Rule, we'll follow these steps based on the given equations and values:

We know that:

$z = \cos(x + 6y)$

Using the Chain Rule for multivariable functions, we have:

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

Given derivatives:

• $\frac{\partial z}{\partial x} = -\sin(x + 6y)$
• $\frac{dx}{dt} = -8t^3$
• $\frac{\partial z}{\partial y} = -6\sin(x + 6y)$
• $\frac{dy}{dt} = \frac{2}{t^2}$

Step-by-step calculation:

1. First term: $\frac{\partial z}{\partial x} \cdot \frac{dx}{dt} = -\sin(x + 6y) \cdot (-8t^3) = 8t^3 \sin(x + 6y)$

2. Second term: $\frac{\partial z}{\partial y} \cdot \frac{dy}{dt} = -6\sin(x + 6y) \cdot \frac{2}{t^2} = -\frac{12}{t^2} \sin(x + 6y)$

3. Combine the terms: $\frac{dz}{dt} = 8t^3 \sin(x + 6y) - \frac{12}{t^2} \sin(x + 6y)$

Factoring out $\sin(x + 6y)$, we get:

$\frac{dz}{dt} = \sin(x + 6y) \left( 8t^3 - \frac{12}{t^2} \right)$

Thus, the derivative $\frac{dz}{dt}$ is:

$\frac{dz}{dt} = \sin(x + 6y) \left( 8t^3 - \frac{12}{t^2} \right)$

Let me know if you'd like further clarification or details!

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Here are 5 follow-up questions related to this:

1. What is the significance of the Chain Rule in multivariable calculus?
2. How does factoring the result simplify the expression?
3. Can the Chain Rule be applied to functions with more than two variables?
4. What would happen if we changed the expressions for $x(t)$ or $y(t)$?
5. How would you interpret this result in terms of rates of change in physics?

Tip: When applying the Chain Rule, always ensure that the partial derivatives and the corresponding rates of change are aligned properly with respect to each variable.