Related Rates: rate of rise of water in filling pool

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wonky-faint

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A swimming pool is 15 feet wide, 40 feet long, 3 feet deep at one end, and 10 feet deep at the other end. Water is being added to it at a rate of 25 cubic feet per minute.

a) how fast is the water level rising when there is 2000 cubic feet of water in the pool
b) how fast is the water level rising when there is 3000 cubic feet of water in the pool.

This is what I have worked out so far, but I don't know where to go from here:

V=1/2(40)(20)(3+10)=5200 cubic feet

dV/dt=25ft^3 per min

Now I'm not sure if I'm supposed to find the derivative of the volume function and then where to go from there.

I would appreciate not just the answer but how to actually do it because i have to do 20 other related rates problems. thankyou :)
 
Hint: You do need a derivative, but it isn't continuous. While filling up the bottom, you have one result. After the water hits the shallow wall, it is a different story. You need to know how much water is in the pool when it hits that shallow wall. Then you will be able to tell which piece you need for which question.
 
Ah, concept struggling...

One thing you MUST learn is that you do NOT have to see the end before you start. Dive in and let the notation and technique help you.

Question #1 - What is the volume of water in the pool when the water hits the shallow wall? The cross section is a right traingle with its base (40 ft) on top. The width is constant - 15 ft. The height is...I'll leave that and the volume for you.

Hint: What is the difference in depth between the deep end and the shallow end?

Note: Don't give me that! :) You do math and solve word problems every day of your life. Have you ever looked at a clock and tried to decide how much time you have left? Math!!! Just relax and think it through. :wink:
 
so would the height be 7 because (10-3)=7

so the volume would be V=1/2*b*h V=20*7=140

where would the dh/dt go then?

Edit: actually arent you trying to find volume of a triangular PRISM?
 
I am not encouraged by your volume calculation. Where's the third dimension?

Call it a triangular prism if you like, but that is applicable only before the water hits the shallow wall.
 
ya...i dont know what was wrong with me last night...so i woke up early and try to do it again and this is what i got:

with the prism, it would be v=1/2 (40)(15)(7)=2100 cubic ft when it hits the shallow wall

so with that number, i can figure out part b which asks for 3000 ft so i just get the derivative of the volume function so it would be

dV/dt=l*w*dh/dt

so then u just plug in the numbers... (25)=(40)(15)dh/dt

so dh/dt=1/24 cubic ft/min.....

is that right? if it is i still dont get how to figure out part A because it would be also the rate of change of the length...
 
Glad you're back.

You're also backwards.

When the water hits the shallow wall, this is the maximum amount of water that will still be similar in shape to all volumes LESS THAN 2100 ft^3. Try part a with this shape.

For volumes OVER 2100 ft^3, you must use a different formula for volume. It's getting deeper, but it isn't getting longer.

Note: For consitency, imagine the relationship of the two answers. Which one will be faster? Make up your mind before you work it out. It will help you avoid errors.
 
You want \(\displaystyle \frac{dh}{dt}\) given that \(\displaystyle \frac{dV}{dt}=25\)

Do you see what tkh is saying?. For part a, 2100 ft^3 comes to the

bottom of the 3 foot depth, but you want the height when there is 2000 ft^3.

The formula would be for a triangular wedge:

\(\displaystyle V=\frac{1}{2}bh(40)\)...[1]

Express b in terms of h and sub into your formula, perhaps try

\(\displaystyle \frac{b}{h}=\frac{15}{7}\).....[2]

Solve [2] for b, sub into [1], solve for h and differentiate.

Enter in your data and you should be set.

poolyt9.gif
 
Well, I expect you never imagined you would be encouraged in your study of mathematics to work on your art/drawing skills. :)
 
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