Use the chain rule to find 
∂z
∂s
 and 
∂z
∂t
.
z = x5y3,   x = s cos(t),   y = s sin(t)

To find \( \frac{\partial z}{\partial s} \) and \( \frac{\partial z}{\partial t} \) using the chain rule, we will use the given relationships:
- \( z = x^5 y^3 \)
- \( x = s \cos(t) \)
- \( y = s \sin(t) \)

We need to express \( \frac{\partial z}{\partial s} \) and \( \frac{\partial z}{\partial t} \) in terms of partial derivatives of \( z \), \( x \), and \( y \), applying the chain rule.

### 1. For \( \frac{\partial z}{\partial s} \):

Using the chain rule:
\[
\frac{\partial z}{\partial s} = \frac{\partial z}{\partial x} \cdot \frac{\partial x}{\partial s} + \frac{\partial z}{\partial y} \cdot \frac{\partial y}{\partial s}
\]

First, calculate the partial derivatives:

- \( \frac{\partial z}{\partial x} = 5x^4 y^3 \)
- \( \frac{\partial z}{\partial y} = 3x^5 y^2 \)

Next, the partial derivatives of \( x \) and \( y \) with respect to \( s \):
- \( \frac{\partial x}{\partial s} = \cos(t) \)
- \( \frac{\partial y}{\partial s} = \sin(t) \)

Now substitute these into the chain rule:
\[
\frac{\partial z}{\partial s} = 5x^4 y^3 \cdot \cos(t) + 3x^5 y^2 \cdot \sin(t)
\]

### 2. For \( \frac{\partial z}{\partial t} \):

Again, applying the chain rule:
\[
\frac{\partial z}{\partial t} = \frac{\partial z}{\partial x} \cdot \frac{\partial x}{\partial t} + \frac{\partial z}{\partial y} \cdot \frac{\partial y}{\partial t}
\]

The partial derivatives of \( x \) and \( y \) with respect to \( t \) are:
- \( \frac{\partial x}{\partial t} = -s \sin(t) \)
- \( \frac{\partial y}{\partial t} = s \cos(t) \)

Substitute into the chain rule:
\[
\frac{\partial z}{\partial t} = 5x^4 y^3 \cdot (-s \sin(t)) + 3x^5 y^2 \cdot s \cos(t)
\]

Thus, the two partial derivatives are:
\[
\frac{\partial z}{\partial s} = 5x^4 y^3 \cos(t) + 3x^5 y^2 \sin(t)
\]
\[
\frac{\partial z}{\partial t} = -5x^4 y^3 s \sin(t) + 3x^5 y^2 s \cos(t)
\]

These can be left in terms of \( x \) and \( y \), or we can substitute \( x = s \cos(t) \) and \( y = s \sin(t) \) if you prefer fully expanded results.

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Would you like further details on any of the steps?

### Related Questions:
1. How would you apply the chain rule if \( x \) and \( y \) were functions of both \( s \) and \( t \)?
2. What happens if we change the exponents in \( z = x^5 y^3 \)?
3. Can the same process be used for implicit differentiation?
4. What is the geometric interpretation of partial derivatives in this context?
5. How does the chain rule apply to multivariable functions with more than two variables?

#### Tip:
When using the chain rule, always check how many variables each function depends on to ensure you're applying the correct number of terms.

Use the chain rule to find ∂z/∂s and ∂z/∂t. z = x^5 y^3, x = s cos(t), y = s sin(t)

Learn how to apply the chain rule to find the partial derivatives ∂z/∂s and ∂z/∂t for the function z = x^5 y^3, with x = s cos(t) and y = s sin(t). This problem demonstrates the use of partial derivatives and the chain rule in multivariable calculus. Suitable for college-level students.

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