Austindick5 said:
I'm having problems with these related rates problems. Can you please show me how you solve the problem along with an explanation.
Here is the problem:
1. A water tank has the shape of an inverted right (non oblique) cone with a diameter of 20 meters and a depth of 14 meters. Water is flowing into the tank at 2 cubic meters per minute. How fast is the depth of the water increasing at the instant the water is 8 meters deep.
Any help will be much appreciated. Thanks!
I'll assume that the cone has its base on the ground (not the apex).
As we are told water is flowing into the tank at 2m^3/min, that means $\displaystyle \begin{align*} \frac{\mathrm{d}V}{\mathrm{d}t} = 2 \end{align*}$.
You want to find out how fast the depth of the water is increasing at the instant the water is 8m deep. So if we define "h" as the depth/height, then you are wanting to evaluate $\displaystyle \begin{align*} \frac{\mathrm{d}h}{\mathrm{d}t} \end{align*}$ when h = 8.
We can relate the rates with the Chain Rule as follows:
$\displaystyle \begin{align*} \frac{\mathrm{d}V}{\mathrm{d}h} \cdot \frac{\mathrm{d}h}{\mathrm{d}t} &= \frac{\mathrm{d}V}{\mathrm{d}t} \\ \frac{\mathrm{d}h}{\mathrm{d}t} &= \frac{\frac{\mathrm{d}V}{\mathrm{d}t}}{\frac{\mathrm{d}V}{\mathrm{d}h}} \end{align*}$
So we need to be able to work out $\displaystyle \begin{align*} \frac{\mathrm{d}V}{\mathrm{d}h} \end{align*}$.
When full, the volume of the cone is
$\displaystyle \begin{align*} V_{\textrm{Full}} &= \frac{1}{3}\,\pi\,R^2\,H \\ &= \frac{1}{3}\,\pi \cdot 10^2 \cdot 14 \textrm{ m}^3 \\ &= \frac{1400\,\pi}{3}\textrm{ m}^3 \end{align*}$
As the tank fills up, the "empty" part of the cone is a similar cone to the full cone. This empty cone has a height of $\displaystyle \begin{align*} \left( 14 - h \right) \end{align*}$ m and a radius of $\displaystyle \begin{align*} r \end{align*}$ m. As it is similar to the full cone, that means (if we use "k" as the scaling factor)
$\displaystyle \begin{align*} k\,r &= 10 \\ k\left( 14 - h \right) &= 14 \\ \\ k &= \frac{10}{r} \\ k &= \frac{14}{14 - h} \\ \\ \frac{10}{r} &= \frac{14}{14 - h} \\ 10\left( 14 - h\right) &= 14\,r \\ r &= \frac{5\left( 14 - h \right) }{7} \end{align*}$
So the volume of the empty cone is
$\displaystyle \begin{align*} V_{\textrm{Empty}} &= \frac{1}{3}\,\pi\,R^2\,H \\ &= \frac{1}{3}\,\pi\,\left[ \frac{5\left( 14 - h \right)}{7} \right] ^2 \left( 14 - h \right) \textrm{ m}^3 \\ &= \frac{25\,\pi}{147}\left( 14 - h \right) ^3\,\textrm{ m}^3 \end{align*}$
and thus the volume of the water in the tank is
$\displaystyle \begin{align*} V_{\textrm{Water}} &= V_{\textrm{Full}} - V_{\textrm{Empty}} \\ &= \left[ \frac{1400\,\pi}{3} - \frac{25\,\pi}{147}\left( 14 - h \right) ^3 \right] \textrm{ m}^3 \end{align*}$
So now we can find $\displaystyle \begin{align*} \frac{\mathrm{d}V}{\mathrm{d}H} \end{align*}$...
$\displaystyle \begin{align*} V &= \left[ \frac{1400\,\pi}{3} - \frac{25\,\pi}{147} \left( 14 - h \right) ^3 \right] \textrm{m}^3 \\ \frac{\mathrm{d}V}{\mathrm{d}h} &= \left[ 0 -\frac{25\,\pi}{147} \left( -1 \right) \left( 3 \right)\left( 14 - h \right) ^2 \right] \textrm{m}^3 / \textrm{m} \\ &= \frac{25\,\pi}{49} \left( 14 - h \right) ^2 \,\textrm{m}^2 \end{align*}$
and when h = 8
$\displaystyle \begin{align*} \frac{\mathrm{d}V}{\mathrm{d}h} &= \frac{25\,\pi}{49}\left( 14 - 8 \right) ^2 \,\textrm{m}^2 \\ &= \frac{25\,\pi}{49} \left( 6 \right) ^2 \,\textrm{m}^2 \\ &= \frac{900\,\pi}{49} \,\textrm{m}^2 \end{align*}$
So finally, at the point in time where h = 8
$\displaystyle \begin{align*} \frac{\mathrm{d}h}{\mathrm{d}t} &= \frac{\frac{\mathrm{d}V}{\mathrm{d}t}}{\frac{\mathrm{d}V}{\mathrm{d}h}} \\ &= \frac{2\, \textrm{m}^3 / \textrm{min}}{ \frac{900\,\pi}{49}\,\textrm{m}^2} \\ &= \frac{49}{450\,\pi}\,\textrm{m}/ \textrm{min} \end{align*}$
