Need help on related rates problem

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In summary, to solve for the rate of change of the depth of water in a tank with the shape of an inverted right cone, we can use the Chain Rule and the concept of similar cones. After finding the volume of the water in the tank, we can then find the derivative of the volume with respect to the height to determine the rate of change of the depth.
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Austindick5
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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!
 
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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

$\displaystyle \begin{align*} \frac{\mathrm{d}h}{\mathrm{d}t} &= \frac{\frac{\mathrm{d}V}{\mathrm{d}t}}{\frac{\mathrm{d}V}{\mathrm{d}h}} \\ &= \frac{2\, \textrm{m}^3 / \textrm{min}}{ \frac{900\,\pi}{49}\,\textrm{m}^2} \\ &= \frac{49}{450\,\pi}\,\textrm{m}/ \textrm{min} \end{align*}$
 

FAQ: Need help on related rates problem

1. What is a related rates problem?

A related rates problem is a type of mathematical problem that involves finding the rate at which one variable changes in relation to another variable. These problems often involve multiple variables that are changing over time, and the goal is to determine how the rate of change of one variable affects the rate of change of another.

2. How do you approach a related rates problem?

The first step in solving a related rates problem is to identify the variables involved and the relationships between them. Then, you can use the chain rule to differentiate the equation and solve for the rate of change of the desired variable. It is important to carefully interpret the problem and set up the equation correctly in order to arrive at the correct solution.

3. What are some common mistakes to avoid when solving related rates problems?

One common mistake is to forget to take the derivative when differentiating the equation. Another mistake is to mix up the variables and their rates of change, leading to an incorrect solution. It is also important to be careful with units and make sure they are consistent throughout the problem.

4. Can you give an example of a related rates problem?

Sure, here is an example: A hot air balloon is rising at a rate of 10 feet per second. At the same time, a person on the ground is walking away from the balloon at a rate of 3 feet per second. How fast is the distance between the person and the balloon changing when the person is 40 feet from the initial position?

In this problem, the variables are the distance between the person and the balloon (D), the rate at which the person is walking (dD/dt), and the rate at which the balloon is rising (dh/dt). Using the Pythagorean theorem, we can set up the equation D^2 = (40)^2 + h^2 and differentiate both sides to get 2D(dD/dt) = 2h(dh/dt). Then, we can plug in the given values and solve for dD/dt, which is the rate at which the distance is changing.

5. How can I practice and improve my skills in solving related rates problems?

As with any math problem, practice is key. You can find many related rates problems online or in textbooks to work through. It is also helpful to review the chain rule and make sure you understand how to apply it in different situations. Additionally, seeking out resources such as videos or tutoring sessions can provide additional guidance and practice opportunities.

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