Question

Water is leaking out of an inverted conical tank at a rate of 6,500 cm3/min at the same time that water is being pumped into the tank at a constant rate. The tank has height 6 m and the diameter at the top is 4 m. If the water level is rising at a rate of 20 cm/min when the height of the water is 2 m, find the rate at which water is being pumped into the tank. (Round your answer to the nearest integer.)

Answer 1

So the rate at which water is being bumped is \(c\). It's constant with respect to time.

Answer 2

First find the dimensions of this tank.

Answer 3

Suppose \(V(t)\) is the volume and \(h(t)\) is the height.\[
h'(t)=20 cm/min
\]

Answer 4

\[
V'(t) = c - 6,500 cm^3/min
\]

Answer 5

The hard part is finding the function \[
V(h(t))
\]

Answer 6

That is, finding the volume in terms of the height.

Answer 7

\[
V'(t)=V'(h)h'(t)
\]

Answer 8

\[
c−6,500cm^3/min= \frac{dV}{dh}(2)\times 20cm/min
\]\[
c=\frac{dV}{dh}(2)\times 20cm/min+6,500cm^3/min
\]

Answer 9

\[
V=\frac 13\pi r^2h=\frac 13\pi (2)^2h=\frac 43\pi h
\]\[
V'(h) = \frac 43 \pi
\]Weird, I wasn't expecting it to be a constant.

Answer 10

Anyway I predict \[
c=\frac 43 \pi\times 20+6500=\frac{80\pi}3+6500 \approx 6583
\]

Answer 11

This would mean \[
V'(t) =\frac{80\pi}{3}\approx 83
\]

Answer 12

It's strange that we didn't need to use the \(h=200cm\) info.

Answer 13

@primeralph @Directrix Guys, if you see any error in my logic, please let me know. I'm going to sleep.

Answer 14

Okay I did make a small error here guys.
I set the radius to be constant at \(2\) when it really should be a function of \(h\).

Answer 15

\[
\frac h6 = \frac r2\implies r=\frac h3\implies r^2=h^2/9
\]

Answer 16

\[
V=\frac 13\pi r^2h=\frac 13\pi \left(\frac h3\right)^2h=\frac {4\pi}{27} h^3
\]\[
V'(h) = \frac {4\pi}{9} h^2
\]

Answer 17

\(h=200cm\) So \[
V'(h)=\frac{160,000\pi}{9}
\]

Answer 18

These are huge numbers.

Answer 19

Wait, I put in a \(4\) that shouldn't be there. IT should be \[
V'(h) = \frac{40,000\pi }{9}
\]