the height and radius of a cylinder are both changing. the radius of the cylinder is decreasing at a…

Explanation:

Step1: Differentiate volume formula

Given $V=\pi r^{2}h$. Since $V$ is constant ($V = 335$), $\frac{dV}{dt}=0$. Differentiate $V$ with respect to time $t$ using the product - rule $(uv)^\prime = u^\prime v+uv^\prime$. Here $u = \pi r^{2}$ and $v = h$. So $\frac{dV}{dt}=2\pi r h\frac{dr}{dt}+\pi r^{2}\frac{dh}{dt}=0$.

Step2: Find $h$ when $r = 9$

We know $V=\pi r^{2}h = 335$. When $r = 9$, we can solve for $h$: $h=\frac{335}{\pi r^{2}}=\frac{335}{\pi\times9^{2}}=\frac{335}{81\pi}$.

Step3: Substitute values into derivative equation

We know $\frac{dr}{dt}=- 9$ (negative because $r$ is decreasing), $r = 9$, and $h=\frac{335}{81\pi}$. Substitute into $2\pi r h\frac{dr}{dt}+\pi r^{2}\frac{dh}{dt}=0$. [ \begin{align*} 2\pi\times9\times\frac{335}{81\pi}\times(-9)+\pi\times9^{2}\times\frac{dh}{dt}&=0\ 2\times9\times\frac{335}{81}\times(-9)+81\pi\times\frac{dh}{dt}&=0\ \frac{2\times9\times335\times(-9)}{81}+81\pi\times\frac{dh}{dt}&=0\

• 670+81\pi\times\frac{dh}{dt}&=0\ 81\pi\times\frac{dh}{dt}&=670\ \frac{dh}{dt}&=\frac{670}{81\pi} \end{align*} ]

Step4: Calculate the value

$\frac{dh}{dt}=\frac{670}{81\pi}\approx2.637$

Answer:

$2.637$