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Related Rates height of rising water

  • Thread starter uradnky
  • Start date
30
0
1. Homework Statement
A rectangular swimming pool 16m x 10m is being filled at a rate of .8m^3/min. How fast is height of the water rising?



2. Homework Equations
V = LWH


3. The Attempt at a Solution

H = V/LW
(dh/dt) = LW(dv/dt)
(dh/dt) = (10)(16)(.8)


This question looks so easy and I realize Im probablly making a stupid mistake somewhere, could someone point it out for me please?
 

Answers and Replies

1,631
4
V=abh----(*), where a=16, b=10, h is the height and it is changing so we don't know it.

we need to know

[tex]\frac{dh}{dt}=??[/tex] we also know that

[tex]\frac{dV}{dt}=.8\frac{m^{3}}{min}[/tex]

Now differentiate implicitly (*) and plug in the information you were provided. Does this help?
 
1,631
4
Ok, where u went wrong is that LW should be on denominator, and not on the numerator.

(dh/dt) = (dv/dt)/(LW)
 
30
0
Thanks for the help. One thing I don't understand though..
I rearrange V = LWH to solve for H.
That gives me H = V/LW.
Now when I take the derivate implicitly, wouldnt I need to use the quotient rule to get the derivative of V/LW? That is how I got LW(dv/dt) and not the correct answer of (dv/dt)/(LW)
 
1,631
4
Thanks for the help. One thing I don't understand though..
I rearrange V = LWH to solve for H.
That gives me H = V/LW.
Now when I take the derivate implicitly, wouldnt I need to use the quotient rule to get the derivative of V/LW? That is how I got LW(dv/dt) and not the correct answer of (dv/dt)/(LW)
BUt notice that Length (L) and Width(W) are constants. In general:

[tex]\frac{d(Cf(x))}{dx}=C\frac{d(f(x))}{dx}[/tex] C is a constant
 
30
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So since the denominator is a constant it isn't necessary to use the quotient rule?
 
1,631
4
So since the denominator is a constant it isn't necessary to use the quotient rule?
Well, even if you used the quotient rule, you will get the exact same answer, if you do the calculations right at first place.
 
1,631
4
See what happenes if you really want to apply the quotient rule

[tex](\frac{V}{LW})'=\frac{V'LW-(LW)'V}{(LW)^{2}}=\frac{V'LW}{(LW)^{2}}=\frac{V'}{LW}[/tex]

SO, the answer is the same, that's why i love math, because it is all consistent, no matter which way u decide to pursue.
 
30
0
If I tried this using the quotient rule, would this be the correct way to go about it? (I know it isnt..)
(dh/dt) = (160)(dv/dt)-(v)(0)
 
Last edited:
1,631
4
(I know it isnt..)
(dh/dt) = (160)(dv/dt)-(v)(0)
Well, i think you answered it yourself.

Do you know the quotient rule??????
 
30
0
(dh/dt) = (160)(.8)

= 128 ...this isn't the correct answer?
 
1,631
4
(dh/dt) = (160)(.8)

= 128 ...this isn't the correct answer?
No.

it should read sth like

.8/160= whatever

You need to review the quotient rule!!!
 
V= L*w*h
dh/dt=?
Derivative of V = K h

dv/dt =K dh/dt
.8m^3/min = 160 dh/dt
(.8m^3/min)/160m^2 = dh/dt
Notice we know it is the correct answer because the Units cancel out this is very interesting fact that even if you dont really know what your doing as long as the unit makes sense then your good remeber this for your next test lol!


( the length is not changin the width is not changing only the height is changin as you fill the pool)
 
30
0
kjlhdfkljdshfkljnvckjnd pretty ugly mistake. I guess I haven't done quotient rule in awhile hah. I didn't pay any attention when you showed me how to do it with the quotient rule. Sorry for wasting your time bro, but thanks for all the help.
 

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