This related rates problem is killing me, help please?

In summary: You need the volume of the full pool to solve this problem, not the volume when it's full. There is no differentiation needed. Just plug in the values for h and V and solve.In summary, a pool is 20 ft wide, 40 ft long, 3 ft deep at the shallow end, and 9 ft deep at its deepest point. If the pool is shaped as shown, and is being filled at a rate of 8 ft^3/min, the water level is rising at a rate of 5 ft^3/min.
  • #1
pitt14
3
0

Homework Statement


A pool is 20 ft wide, 40 ft long, 3 ft deep at the shallow end, and 9 ft deep at its deepest point. If the pool is shaped as shown (I attached a picture), and is being filled at a rate of 8 ft^3/min, how fast is the water level rising when the depth at the deepest point is 5ft?

Homework Equations


I *think* I need the volume of a trapezoidal prism?
V = Bh = (.5*a*b*h)*w
I am completely lost with the rest of this

The Attempt at a Solution


All I can figure out is that V=2700. I need a lot of help with this.
 

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  • #2
pitt14 said:

Homework Statement


A pool is 20 ft wide, 40 ft long, 3 ft deep at the shallow end, and 9 ft deep at its deepest point. If the pool is shaped as shown (I attached a picture), and is being filled at a rate of 8 ft^3/min, how fast is the water level rising when the depth at the deepest point is 5ft?


Homework Equations


I *think* I need the volume of a trapezoidal prism?
V = Bh = (.5*a*b*h)*w
I am completely lost with the rest of this


The Attempt at a Solution


All I can figure out is that V=2700. I need a lot of help with this.
Is 2700 ft3 the volume when the pool is filled? If so, that won't be any use in this problem.

What you need to do is relate the volume V and the height of water h, then differentiate to get dV/dh.

The volume at a given level will be the cross-sectional area times 20, the width of the pool. The lower base of the trapezoid is 12 ft. The upper base will be 12 plus the contributions of the two triangular pieces. The one on the left has an altitude of 6' and a base of 6', so when the water is h ft deep, the contribution to the upper base of the trapezoid will also be h ft, using similar triangles. I'll leave figuring out the contribution of the triangle on the right to you.
 
  • #3
Mark44 said:
Is 2700 ft3 the volume when the pool is filled? If so, that won't be any use in this problem.

What you need to do is relate the volume V and the height of water h, then differentiate to get dV/dh.

The volume at a given level will be the cross-sectional area times 20, the width of the pool. The lower base of the trapezoid is 12 ft. The upper base will be 12 plus the contributions of the two triangular pieces. The one on the left has an altitude of 6' and a base of 6', so when the water is h ft deep, the contribution to the upper base of the trapezoid will also be h ft, using similar triangles. I'll leave figuring out the contribution of the triangle on the right to you.

Thank you. I am still really confused, though. I can't figure out how to set this up. I figured out the perimeter of both triangles in the trapezoid. But what do I use for the volume equation if I don't use the volume of the full pool?
 
  • #4
The volume of the pool when it's full is of no use whatsoever in this problem. As I already said, you need to get an equation for the volume V as a function of h, the height of water at a given time.

The perimeters of the triangles aren't relevant either. The length of the upper base of the trapezoid is 12 + A + B, where A is the length of the base of the triangle on the left and B is the length of the base of the triangle on the right.

When you get V as a function of h, differentiate to get V'(h), and then evaluate at h = 5.
 
  • #5
Mark44 said:
The volume of the pool when it's full is of no use whatsoever in this problem. As I already said, you need to get an equation for the volume V as a function of h, the height of water at a given time.

The perimeters of the triangles aren't relevant either. The length of the upper base of the trapezoid is 12 + A + B, where A is the length of the base of the triangle on the left and B is the length of the base of the triangle on the right.

When you get V as a function of h, differentiate to get V'(h), and then evaluate at h = 5.

This is the part I can't figure out. I don't know how to get V as a function of h. I have been trying this problem for a week now. Can I get some guidance as to how to get this?
 
  • #6
Read what I wrote in my two previous posts.
 

Related to This related rates problem is killing me, help please?

1. What is a related rates problem in science?

A related rates problem in science involves finding the rate at which one quantity changes in relation to another quantity. This is typically done by using derivative equations and solving for the unknown rate.

2. How do I approach solving a related rates problem?

The first step in solving a related rates problem is to clearly define the variables and their rates of change. Then, use the given information to set up an equation that relates the variables. Finally, use derivative equations to solve for the unknown rate of change.

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

Some common mistakes to avoid when solving related rates problems include not clearly defining the variables and their rates of change, using the wrong derivative equations, and not correctly setting up the equation that relates the variables.

4. Are there any tips for solving related rates problems more efficiently?

Yes, there are a few tips that can help make solving related rates problems more efficient. These include drawing a diagram to visualize the problem, using the chain rule when necessary, and avoiding unnecessary calculations by simplifying equations.

5. Can you provide an example of a related rates problem and its solution?

Sure, here is an example: A spherical balloon is being inflated at a rate of 10 cm^3/s. How fast is the radius increasing when the volume is 1000 cm^3? Solution: Let V be the volume and r be the radius. We know that dV/dt = 10 cm^3/s and V = 1000 cm^3. We can use the formula for the volume of a sphere, V = (4/3)πr^3, to relate the variables. Taking the derivative of both sides with respect to time, we get dV/dt = 4πr^2 (dr/dt). Plugging in the known values, we can solve for dr/dt and find that the radius is increasing at a rate of approximately 0.16 cm/s.

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