Rate of change

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Imagine a trough filled with water (I can't put up a picture).
The water in the tank at time t seconds is given by [tex]V = 12x^2[/tex]. Given that water is flowing into the trough at the rate of [tex]60 cm^3/s[/tex], find the rate at which x is increasing when x = 10.

[tex]\frac{dV}{dx} = 24x[/tex]
[tex]\frac{d?}{ds} = 60 cm^3/s[/tex]
Then what do I do?
 

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  • #2
dextercioby
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Who's "x" and what does your second equation stand for...?

Daniel.
 
  • #3
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If it shouldn't be x, can you tell me what it should be? I don't know what to do for the second equation. I just know its [tex]\frac{d\text{ something}}{dx} = 60 cm^3/s[/tex]. Again, if I'm wrong please tell me what it should be.
 
  • #4
dextercioby
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I don't know about "x",what is its point...But:
[tex] \frac{dV}{dt}=24x \frac{dx}{dt} [/tex]

,uding the chain rule...Now plug in the values and solve for the rate of "x"...

Daniel.
 
  • #5
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Oh now I know why you asked me about the second equation. I always get mixed up and put ds instead of dt :grumpy: .
So I got 1440x. I know thats not the final answer because theres still the x = 10 part. And I'm confused on that part.
 
  • #6
dextercioby
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Well,[itex] \frac{dV}{dt}=60cm^{3}s^{-1} [/itex]...x=10,you must find [itex] \frac{dx}{dt} [/itex]...

Daniel.
 
  • #7
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I thought [tex]\frac{dx}{dt} = 60 cm^3/s[/tex] and I've to find [tex]\frac{dV}{dt}[/tex] which is 1440x?
 
  • #8
dextercioby
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Nope,it's the other way aroung,the volume is increasing at the rate given in the problem...:wink:

Daniel.
 
  • #9
HallsofIvy
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The point is that
[tex]\frac{dV}{dt}= \frac{dV}{dx}\frac{dx}{dt}[/tex]
(the chain rule).

You are given [itex]\frac{dV}{dt}[/itex] and you can (and did) calculate [itex]\frac{dV}{dx}[/itex]. Put them into that equation and solve for [itex]\frac{dx}{dt}[/itex].
 
  • #10
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Oh... Got it! Thank you!
 

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