Question

how we do thi sone brainly keeps kicking me oput on purpse its trwo parts

Answer 1

Answer:

Part A:

The the level in the pot is rising at a rate of 0.239 inches per minute.

Part B:

The the level in the cone is decreasing at a rate of 0.716 inches per minute.

Step-by-step explanation:

Part A:

For the cylindrical coffepot, we'll have that the volume is:

$V=\pi r^2h$

Now, we have that both the volume and the height of the liquid on the coffeepot change over time (They are a function of time).

This way, we'll have that the volume is:

$V(t)=\pi r^2h(t)$

Now, we can take the derivative as following:

$\begin{gathered} V(t)=\pi r^{2}h(t) \\ \\ \rightarrow\frac{dV(t)}{dt}=\pi r^2\frac{dh(t)}{dt} \end{gathered}$

The derivative (rate of change) of the volume is constant: 12 cubic inches per minute. And in this case, always positive because the coffeepot is being filled. This way,

$\begin{gathered} \frac{dV(t)}{dt}=\pi r^2\frac{dh(t)}{dt}\rightarrow12=\pi r^2\frac{dh(t)}{dt} \\ \\ \rightarrow\frac{dh(t)}{dt}=\frac{12}{\pi r^2} \end{gathered}$

And using the data given, we'll have that:

$\frac{dh(t)}{dt}=0.239\text{ in/min}$

This way, we can conclude that the the level in the pot is rising at a rate of 0.239 inches per minute.

Part B:

For the cone, we'll have that the volume is:

$V=\frac{1}{3}\pi r^2h$

Now, we have that both the volume and the height of the liquid on the coffeepot change over time (They are a function of time).

This way, we'll have that the volume is:

$V(t)=\frac{1}{3}\pi r^2h(t)$

Now, we can take the derivative as following:

$\begin{gathered} V(t)=\frac{1}{3}\pi r^2h(t) \\ \\ \rightarrow\frac{dV(t)}{dt}=\frac{1}{3}\pi r^2\frac{dh(t)}{dt} \end{gathered}$

The derivative (rate of change) of the volume is constant: 12 cubic inches per minute. And in this case, always negative because the coffeepot is being emptied. This way,

$\begin{gathered} \begin{equation*} \frac{dV(t)}{dt}=\frac{1}{3}\pi r^2\frac{dh(t)}{dt} \end{equation*} \\ \\ \rightarrow-12=\frac{1}{3}\pi r^2\frac{dh(t)}{dt}\rightarrow\frac{-12\times3}{\pi r^2}=\frac{dh(t)}{dt} \\ \\ \rightarrow\frac{dh(t)}{dt}=-\frac{36}{\pi r^2} \end{gathered}$

And using the data given, we'll have that:

$\frac{dh(t)}{dt}=-0.716\text{ in/min}$

This way, we can conclude that the the level in the cone is decreasing at a rate of -0.716 inches per minute.