Taylor's questions at Yahoo Answers regarding related rates

[MarkFL]

Here are the questions:

Find the rate at which water is being pumped into the tank.?


1.) Water is leaking out of an inverted conical tank at a rate of 10,000 cubic centimeters per minute at the same time that water is being pumped into the tank at a constant rate. The tank has 6m (600 cm) and the diameter of the top is 4m (400 cm). If the water level is rising at a rate of 20 centimeters per minute when the height of the water is 2m (200 cm), find the rate at which water is being pumped into the tank.

2.) At noon, ship A is 150 km due west of ship B. Ship A is sailing east at 35km/h and ship B is sailing North at 25km/h. How fast is the distance between the ships changing at 4:00 pm?

I have posted a link there to this topic so the OP can see my work.

 

[MarkFL]

Hello Taylor,

1.) Let all linear measures be in centimeters and time be measured in minutes.

The statement:

"Water is leaking out of an inverted conical tank at a rate of 10,000 cm3/min at the same time that water is being pumped into the tank at a constant rate."

Allows us to write:

\(\displaystyle \frac{dV}{dt}=R-10000\)

Where $V$ is the volume of the water in the tank at time $t$ and $R$ is the rate at which water is being pumped into the tank. $R$ is the quantity we are asked to find.

Using the volume of a cone, we know the volume of the water in the tank may be given by:

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

where $r$ is the radius of the surface of the water, and $h$ is the depth of the water. We know that at any given time or volume of water, the ratio of the radius of the water at the surface to its depth will remain constant, and in fact will be in the same proportions as the tank itself.

Because we are given information regarding the time rate of change of the depth and the depth itself, we need to place the radius with a function of the depth. Hence, we may use:

\(\displaystyle \frac{r}{h}=\frac{2}{6}=\frac{1}{3}\implies r=\frac{h}{3}\)

And so the volume as a function of $h$ is:

\(\displaystyle V=\frac{1}{3}\pi \left(\frac{h}{3} \right)^2h=\frac{\pi}{27}h^3\)

Now, differentiating with respect to $t$, we obtain:

\(\displaystyle \frac{dV}{dt}=\frac{\pi}{9}h^2\frac{dh}{dt}\)

Equating the two expressions for \(\displaystyle \frac{dV}{dt}\), we find:

\(\displaystyle \frac{\pi}{9}h^2\frac{dh}{dt}=R-10000\)

Solving for $R$, we get:

\(\displaystyle R=\frac{\pi}{9}h^2\frac{dh}{dt}+10000\)

Now, using the given data (making sure all of our units match):

\(\displaystyle h=200\text{ cm},\,\frac{dh}{dt}=20\,\frac{\text{cm}}{\text{min}}\)

We have:

\(\displaystyle R=\frac{\pi}{9}\left(200\text{ cm} \right)^2\left(20\,\frac{\text{cm}}{\text{min}} \right)+10000\,\frac{\text{cm}^3}{\text{min}}\)

\(\displaystyle R=\frac{10000}{9}\left(80\pi+9 \right)\,\frac{\text{cm}^3}{\text{min}}\)

2.) Let's orient our coordinate axes such that ship A is at the origin at noon, and so ship B is at $(150,0)$. Thus, we may describe the position of each ship parametrically as follows:

Ship A:

\(\displaystyle x=35t\)

\(\displaystyle y=0\)

Ship B:

\(\displaystyle x=150\)

\(\displaystyle y=25t\)

If we let $D$ be the distance between the ships at time $t$, we may use the distance formula to write:

\(\displaystyle D^2(t)=(35t-150)^2+(25t)^2=50\left(37t^2-210t+450 \right)\)

Implicitly differentiating with respect to $t$, we find:

\(\displaystyle 2D(t)D'(t)=50(74t-210)=100(37t-105)\)

Hence:

\(\displaystyle D'(t)=\frac{50(37t-105)}{D(t)}=\frac{\sqrt{50}(37t-105)}{\sqrt{37t^2-210t+450}}\)

At 4:00 pm, we have $t=4$, and so we find:

\(\displaystyle D'(4)=\frac{\sqrt{50}(37(4)-105)}{\sqrt{37(4)^2-210(4)+450}}=\frac{215}{\sqrt{101}}\,\frac{\text{km}}{\text{hr}}\)

Thus, at 4:00 pm the distance between the ships is increasing at a rate of about 21.3933 kph.