The volume of a right circular cylinder of radius $$ r $$ and height $$ h $$ is $$ V=\pi r^{2} h $$. (a) Assume that $$ r $$ and $$ h $$ are functions of $$ t $$. Find $$ V^{\prime}(t) $$. $$ \begin{array}{ll} ext { (b) Suppose that } r=e^{6 t} ext { and } h=e^{-6 t} ext {. Use part (a) to find } V^{\prime}(t) ext {. } \ ext { (c) Does the volume of the cylinder of part (b) increase or decrease as } t ext { increases? } \ \begin{array}{ll} ext { A) Find } V^{\prime}(t) ext {. Choose the correct answer below. } V^{\prime}(t)=2 \pi r(t) h(t) r^{\prime}(t)+\pi(r(t))^{2} h^{\prime}(t) & V^{\prime}(t)=\pi(r(t))^{2} h^{\prime}(t) \ ext { C. } V^{\prime}(t)=2 \pi r(t) h(t) r^{\prime}(t) & V^{\prime}(t)=2 \pi r(t) h(t) h^{\prime}(t)+\pi(r(t))^{2} r^{\prime}(t)\end{array}\end{array} $$.

Question

The volume of a right circular cylinder of radius $$ r $$ and height $$ h $$ is $$ V=\pi r^{2} h $$. (a) Assume that $$ r $$ and $$ h $$ are functions of $$ t $$. Find $$ V^{\prime}(t) $$. $$ \begin{array}{ll}	ext { (b) Suppose that } r=e^{6 t} 	ext { and } h=e^{-6 t} 	ext {. Use part (a) to find } V^{\prime}(t) 	ext {. } \ 	ext { (c) Does the volume of the cylinder of part (b) increase or decrease as } t 	ext { increases? } \ \begin{array}{ll}	ext { A) Find } V^{\prime}(t) 	ext {. Choose the correct answer below. } V^{\prime}(t)=2 \pi r(t) h(t) r^{\prime}(t)+\pi(r(t))^{2} h^{\prime}(t) & V^{\prime}(t)=\pi(r(t))^{2} h^{\prime}(t) \ 	ext { C. } V^{\prime}(t)=2 \pi r(t) h(t) r^{\prime}(t) & V^{\prime}(t)=2 \pi r(t) h(t) h^{\prime}(t)+\pi(r(t))^{2} r^{\prime}(t)\end{array}\end{array}> $$.

UpStudy AI · Accepted Answer

To solve the problem, we will follow the steps outlined in the prompt.

### Part (a): Find $$ V^{\prime}(t) $$

The volume of a right circular cylinder is given by the formula:

$$
V = \pi r^2 h
$$

Since $$ r $$ and $$ h $$ are functions of $$ t $$, we will use the product rule and the chain rule to differentiate $$ V $$ with respect to $$ t $$.

Using the product rule, we have:

$$
V^{\prime}(t) = \frac{d}{dt}(\pi r^2 h) = \pi \left( \frac{d}{dt}(r^2) h + r^2 \frac{dh}{dt} \right)
$$

Now, applying the chain rule to $$ r^2 $$:

$$
\frac{d}{dt}(r^2) = 2r \frac{dr}{dt}
$$

Substituting this back into the equation for $$ V^{\prime}(t) $$:

$$
V^{\prime}(t) = \pi \left( 2r \frac{dr}{dt} h + r^2 \frac{dh}{dt} \right)
$$

Thus, we can express $$ V^{\prime}(t) $$ as:

$$
V^{\prime}(t) = 2\pi r(t) h(t) r^{\prime}(t) + \pi (r(t))^2 h^{\prime}(t)
$$

### Part (b): Use $$ r = e^{6t} $$ and $$ h = e^{-6t} $$ to find $$ V^{\prime}(t) $$

Now we will substitute $$ r(t) = e^{6t} $$ and $$ h(t) = e^{-6t} $$ into the expression for $$ V^{\prime}(t) $$.

First, we need to find $$ r^{\prime}(t) $$ and $$ h^{\prime}(t) $$:

$$
r^{\prime}(t) = \frac{d}{dt}(e^{6t}) = 6e^{6t}
$$
$$
h^{\prime}(t) = \frac{d}{dt}(e^{-6t}) = -6e^{-6t}
$$

Now substituting these into the expression for $$ V^{\prime}(t) $$:

$$
V^{\prime}(t) = 2\pi (e^{6t})(e^{-6t})(6e^{6t}) + \pi (e^{6t})^2 (-6e^{-6t})
$$

Simplifying each term:

1. The first term:
   $$
   2\pi (e^{6t})(e^{-6t})(6e^{6t}) = 2\pi (6e^{6t}) = 12\pi e^{6t}
   $$

2. The second term:
   $$
   \pi (e^{6t})^2 (-6e^{-6t}) = -6\pi e^{6t}
   $$

Combining these results:

$$
V^{\prime}(t) = 12\pi e^{6t} - 6\pi e^{6t} = 6\pi e^{6t}
$$

### Part (c): Does the volume of the cylinder of part (b) increase or decrease as $$ t $$ increases?

Since $$ V^{\prime}(t) = 6\pi e^{6t} $$ is positive for all $$ t $$ (as $$ e^{6t} > 0 $$ for all $$ t $$), this indicates that the volume of the cylinder is increasing as $$ t $$ increases.

### Summary of Results
- (a) $$ V^{\prime}(t) = 2\pi r(t) h(t) r^{\prime}(t) + \pi (r(t))^2 h^{\prime}(t) $$
- (b) $$ V^{\prime}(t) = 6\pi e^{6t} $$
- (c) The volume of the cylinder increases as $$ t $$ increases.
$$ V^{\prime}(t) = 6\pi e^{6t} $$, so the volume increases as $$ t $$ increases.

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