The surface area of a cylinder is given by:

$S = 2\pi r^2 + 2\pi r h$

where:

• $r$ is the radius of the base,
• $h$ is the height of the cylinder.

To find the derivative of the surface area $S$ with respect to time $t$, we use the chain rule:

$\frac{dS}{dt} = \frac{\partial S}{\partial r} \frac{dr}{dt} + \frac{\partial S}{\partial h} \frac{dh}{dt}$

Step 1: Partial derivatives of $S$

1. $\frac{\partial S}{\partial r} = \frac{\partial}{\partial r}(2\pi r^2 + 2\pi r h) = 4\pi r + 2\pi h$
2. $\frac{\partial S}{\partial h} = \frac{\partial}{\partial h}(2\pi r^2 + 2\pi r h) = 2\pi r$

Step 2: Chain rule

Substitute into the chain rule formula:

$\frac{dS}{dt} = (4\pi r + 2\pi h) \frac{dr}{dt} + 2\pi r \frac{dh}{dt}$

This is the expression for the rate of change of the surface area of a cylinder with respect to time.

Explanation:

• $\frac{dr}{dt}$: Rate of change of the radius over time.
• $\frac{dh}{dt}$: Rate of change of the height over time.

Final Formula:

$\boxed{\frac{dS}{dt} = (4\pi r + 2\pi h) \frac{dr}{dt} + 2\pi r \frac{dh}{dt}}$

Do you want an example or further breakdown?

---

Related Questions:

1. What is the physical interpretation of $\frac{dS}{dt}$ in a real-world scenario?
2. How does $\frac{dS}{dt}$ change if the radius is constant over time?
3. Can you find the second derivative of the surface area $S$ with respect to time?
4. What happens to $\frac{dS}{dt}$ when both $r$ and $h$ are functions of time?
5. How does the derivative change if the cylinder has an open top or bottom?

---

Tip: Always check if the radius or height is constant, as this simplifies the derivative significantly.