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The problem asks to find the rate of change of \( x \) with respect to time when given the relationship between \( x \), \( y \), and \( z \), which are functions of time. The equation given is:

\[ x = y z^2 \]

We are given the following conditions:
- \( y \) is increasing at a rate of 2 units per minute, so \(\frac{dy}{dt} = 2\).
- \( z \) is decreasing at a rate of 3 units per minute, so \(\frac{dz}{dt} = -3\).
- We need to find \(\frac{dx}{dt}\) when \( y = 6 \) and \( z = 7 \).

### Solution:
1. **Differentiate both sides of the equation with respect to \( t \):**

\[
\frac{d}{dt}(x) = \frac{d}{dt}(yz^2)
\]

Using the product rule on the right side:

\[
\frac{dx}{dt} = \frac{dy}{dt} \cdot z^2 + y \cdot \frac{d}{dt}(z^2)
\]

2. **Simplify \(\frac{d}{dt}(z^2)\) using the chain rule:**

\[
\frac{d}{dt}(z^2) = 2z \frac{dz}{dt}
\]

3. **Substitute this into the differentiated equation:**

\[
\frac{dx}{dt} = \frac{dy}{dt} \cdot z^2 + y \cdot (2z \frac{dz}{dt})
\]

4. **Plug in the given values:**
   - \( \frac{dy}{dt} = 2 \)
   - \( \frac{dz}{dt} = -3 \)
   - \( y = 6 \)
   - \( z = 7 \)

\[
\frac{dx}{dt} = 2 \cdot 7^2 + 6 \cdot (2 \cdot 7 \cdot (-3))
\]

5. **Calculate \( 7^2 \):**

\[
7^2 = 49
\]

6. **Substitute and simplify:**

\[
\frac{dx}{dt} = 2 \cdot 49 + 6 \cdot (2 \cdot 7 \cdot (-3))
\]
\[
\frac{dx}{dt} = 98 + 6 \cdot (2 \cdot 7 \cdot (-3))
\]
\[
\frac{dx}{dt} = 98 + 6 \cdot 14 \cdot (-3)
\]
\[
\frac{dx}{dt} = 98 + 6 \cdot (-42)
\]
\[
\frac{dx}{dt} = 98 - 252
\]
\[
\frac{dx}{dt} = -154
\]

Therefore, the rate at which \( x \) is changing with respect to time is \(-154\) units per minute.

Would you like more details, or do you have any questions? Here are some related questions:

1. What is the significance of the negative sign in the rate \(\frac{dx}{dt}\)?
2. How would the rate change if \( z \) were increasing instead of decreasing?
3. Can you explain the product rule in differentiation in more detail?
4. What are some other examples of related rates problems in calculus?
5. How would the result change if \( y \) and \( z \) were both increasing?

**Tip:** When solving related rates problems, it is crucial to correctly identify which variables are changing and at what rates, and then systematically apply differentiation rules like the product rule and chain rule.

Functions x, y, and z are differentiable with respect to time and are related by the equation x = yz^2. If y is increasing at a constant rate of 2 units per minute and z is decreasing at a constant rate of 3 units per minute, at what rate is x changing, with respect to time, when y = 6 and z = 7?

Learn how to solve a related rates problem involving the functions x, y, and z, where x = yz^2. Given y's increase at 2 units/min and z's decrease at 3 units/min, find the rate at which x is changing when y = 6 and z = 7. This calculus problem demonstrates the use of the product rule and chain rule.

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