To find $\frac{dz}{dt}$, we need to apply the chain rule. Here's the step-by-step process:

We are given:

• $z = \cos\left(\frac{y}{x}\right)$
• $x = 4t$
• $y = 5 - t^2$

We need to compute $\frac{dz}{dt}$, so we apply the chain rule for multivariable functions:

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

Step 1: Calculate partial derivatives of $z$

Since $z = \cos\left(\frac{y}{x}\right)$, we need to differentiate $z$ with respect to both $x$ and $y$.

Partial derivative of $z$ with respect to $x$:

$\frac{\partial z}{\partial x} = \frac{d}{dx} \left( \cos\left( \frac{y}{x} \right) \right)$ We use the chain rule here: $\frac{\partial z}{\partial x} = -\sin\left( \frac{y}{x} \right) \cdot \frac{d}{dx}\left( \frac{y}{x} \right)$ Next, we compute the derivative of $\frac{y}{x}$ with respect to $x$: $\frac{d}{dx}\left( \frac{y}{x} \right) = \frac{-y}{x^2}$ So: $\frac{\partial z}{\partial x} = -\sin\left( \frac{y}{x} \right) \cdot \left( \frac{-y}{x^2} \right)$ $\frac{\partial z}{\partial x} = \frac{y}{x^2} \sin\left( \frac{y}{x} \right)$

Partial derivative of $z$ with respect to $y$:

$\frac{\partial z}{\partial y} = \frac{d}{dy} \left( \cos\left( \frac{y}{x} \right) \right)$ Using the chain rule: $\frac{\partial z}{\partial y} = -\sin\left( \frac{y}{x} \right) \cdot \frac{1}{x}$ $\frac{\partial z}{\partial y} = -\frac{1}{x} \sin\left( \frac{y}{x} \right)$

Step 2: Compute the derivatives of $x$ and $y$ with respect to $t$

• $\frac{dx}{dt} = \frac{d}{dt}(4t) = 4$
• $\frac{dy}{dt} = \frac{d}{dt}(5 - t^2) = -2t$

Step 3: Substitute into the chain rule formula

Now we substitute everything into the chain rule formula:

$\frac{dz}{dt} = \left( \frac{y}{x^2} \sin\left( \frac{y}{x} \right) \right) \cdot 4 + \left( -\frac{1}{x} \sin\left( \frac{y}{x} \right) \right) \cdot (-2t)$

Simplifying:

$\frac{dz}{dt} = \frac{4y}{x^2} \sin\left( \frac{y}{x} \right) + \frac{2t}{x} \sin\left( \frac{y}{x} \right)$

Step 4: Final expression

Now substitute $x = 4t$ and $y = 5 - t^2$ into this expression:

$\frac{dz}{dt} = \frac{4(5 - t^2)}{(4t)^2} \sin\left( \frac{5 - t^2}{4t} \right) + \frac{2t}{4t} \sin\left( \frac{5 - t^2}{4t} \right)$ $\frac{dz}{dt} = \frac{4(5 - t^2)}{16t^2} \sin\left( \frac{5 - t^2}{4t} \right) + \frac{1}{2} \sin\left( \frac{5 - t^2}{4t} \right)$

This is the final expression for $\frac{dz}{dt}$.

Let me know if you'd like further clarification or if you have any additional questions!

Further exploration:

1. How would the result change if the function $z$ were $z = \cos\left(\frac{x}{y}\right)$?
2. How do partial derivatives with respect to more than two variables work in this case?
3. What is the significance of the chain rule in understanding how multiple variables interact over time?
4. How would you compute the rate of change of a function like $z = \sin(xy)$ with respect to time?
5. Can we apply this method to more complex functions, like exponential or logarithmic functions involving multiple variables?

Tip: When applying the chain rule, always take care to carefully differentiate each nested function step-by-step!