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Continuous Functions - Setting up work problems

  1. Apr 5, 2010 #1
    Continuous Functions - Setting up word problems

    1. The problem statement, all variables and given/known data
    Each side of a square is expanding at 5 cm/sec. What is the rate of change when the length of the sides are 10 cm.


    2. Relevant equations

    [tex] A = ab [/tex]

    3. The attempt at a solution

    [tex] a = 5t, b = 5t [/tex]

    and the area is described as [tex] A = ab [/tex]

    so the rate of change in the area should be

    [tex] \frac {dA}{dt} = a \frac {db}{dt} + b \frac {da}{dt} [/tex]

    then [tex] \frac {da}{dt} = 5 [/tex] and [tex] \frac {db}{dt} = 5 [/tex]

    so [tex] \frac {dA}{dt} = 5a + 5b [/tex]

    then at [tex] t = 2 [/tex]

    [tex] \frac {dA}{dt} = 5 *10 + 5 *10 = 50 + 50 = 100 cm^{2}/sec [/tex]

    So the area of the square is changing at a rate of 100 sq cm/sec when the sides are at 10 cm

    is this correct?
     
    Last edited: Apr 5, 2010
  2. jcsd
  3. Apr 5, 2010 #2

    Dick

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    Sure, it's correct. You can also do it directly. A=(5t)^2. Find dA/dt at t=2. Do you get the same thing?
     
  4. Apr 5, 2010 #3
    Indeed I do, I worked it out that way because I was just using the equations as they are given in the book, that's all.

    Thanks though Dick
     
  5. Apr 5, 2010 #4
    Given Information

    The vertical side of a rectangle is expanding at a rate of 1 in/sec, while the horizontal side of the rectangle is contracting at a rate of -1 in/sec. At time [tex] t =1 [/tex] the rectangle is a square with both sides at 2 in in length. How fast is the rate of change in the area of the rectangle at [tex] t = 2 [/tex]?

    Relevant Equations

    [tex] A = ab [/tex]

    Attempt At A Solution

    [tex] \frac {da}{dt} = 1 in/sec [/tex]

    [tex] \frac {db}{dt} = -1 in/sec [/tex]

    [tex] a = 2 + (t-1) [/tex]

    [tex] b = 2 - (t-1) [/tex]

    [tex] \frac {dA}{dt} = a \frac {db}{dt} + b \frac{da}{dt} [/tex]

    [tex] \frac {dA}{dt} = -a + b [/tex]

    so at [tex] t = 2 [/tex]

    [tex] \frac {dA}{dt} = -3 +1 = -2 in^{2}/sec [/tex]

    So the area of the rectangle is decreasing at a rate of -2 sq in/sec at t=2.

    Actually I just found the answers section to my book and found that it is decreasing in size, which I had thought that it was but the answer I get, as shown, is positive, did I do something wrong or am I just suppose to know that it is decreasing and change the sign?

    Thanks!
     
    Last edited: Apr 5, 2010
  6. Apr 5, 2010 #5

    Dick

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    I think you understand these problems pretty well, but I think you've got a sign error. Can you find it? If you take a=2+(t-1) and b=2-(t-1) multiply them and differentiate at t=2, you don't get dA/dt=2, do you?
     
  7. Apr 5, 2010 #6
    Ah yeah, I see it all now, I swapped the da/dt and db/dt values, if I swap back around it comes out -2 in^2/sec, man that was silly
     
  8. Apr 5, 2010 #7

    Dick

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    It's not silly. It's a mistake. You have the basic concept fine.
     
  9. Apr 13, 2010 #8
    Problem

    A cup in the form of a right circular cone with radius [tex] r [/tex] and a height [tex] h [/tex] is being filled with water at the rate of [tex] 5 in^{3}/sec [/tex]. How fast is the level of the water rising when the volume of the water is equal to half the volume of the cup?

    Relevant Equations

    Volume of a right circular cone :

    [tex] V = \frac {\pi r^{2}h}{3} [/tex]

    Attempt At A Solution

    I really haven't gotten that far on this one, I will put down what I think I have so far but even if I am right up to the point that I stop here I really don't know where to go from there.

    Since

    [tex] V = \frac { \pi r^{2}h} {3} [/tex]

    then

    [tex] \frac {dV}{dt} = \frac {3 * [\pi r^{2}h]' - \pi r^{2}h * 0} {3^{2}} [/tex]

    and

    [tex] \pi r^{2}h = h(\pi r^{2}) [/tex]

    [tex] [\pi r^{2}h]' = [h(\pi r^{2})]' [/tex]

    then

    [tex] [h(\pi r^{2})]' = h * [\pi r^{2}]' + \pi r^{2} * 1 [/tex]

    and

    [tex] [\pi r^{2}]' = \pi * 2r + r^{2} * 0 = \pi (2r) = 2r \pi [/tex]

    so

    [tex] [h(\pi r^{2})]' = h * 2r \pi + \pi r^{2} = 2r \pi h + \pi r^{2}[/tex]

    and then finally

    [tex] \frac {dV}{dt} = \frac {3 * (2r \pi h + \pi r^{2})} {3^{2}} [/tex]

    so since there is no specific numbers for me to compute should I have simply cut the volume equation in half and differentiated it from there to get the answer like so:

    [tex] \frac {1}{2}V = \frac { \pi r^{2}h} {(3)(2)} = \frac {\pi r^{2}h}{6} [/tex]

    then

    [tex] \frac {dV}{dt} = \frac {6 * (2r \pi h + \pi r^{2})} {6^{2}} [/tex]

    but if this is true I have no idea where to go from there and I never used the 5 cu in/sec value which would be the value of dV/dt would it not? I am confused....
     
    Last edited: Apr 13, 2010
  10. Apr 13, 2010 #9

    Dick

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    You probably shouldn't be posting all of these in a single thread. People will stop looking at threads as the number of posts gets larger. One thing that is wrong is that you should be differentiating with respect to t. ' has to mean d/dt. That means for example that (pi*r^2)'=2*pi*r*dr/dt. Not just 2*pi*r. So dV/dt=(pi/3)*(2*r*h*dr/dt+r^2*dh/dt). But I really don't see how to answer the question numerically. Are they asking for a relation between dh/dt when the cup is full versus dh/dt when the cup is half full?
     
  11. Apr 13, 2010 #10
    Hmm I see your point, and they are probably looking for dh/dt, that didn't occur to me until you mentioned it. I also had thought that I was missing d/dt(s) I was just really confused. I will try it out and see if I can figure out something but one question, why is r differentiated with respect to t, its not a function of t. Even though it is technically a variable its not changing with time, right? or does that not matter?

    After I get this one done I post any further questions in a new thread!
     
  12. Apr 13, 2010 #11

    Dick

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    The shape of the cone is fixed, right? That means h=k*r for some constant k. Where k is the slope of the r versus h function. So as the cup is filling both h and r are changing. So if dh/dt is nonzero then so is dr/dt. But it still seems like there are too many variables to actually solve dh/dt numerically. There's always the possibility that I'm just not seeing something. That would be the point to going to a new thread.
     
  13. Apr 13, 2010 #12
    hmm ok yeah, so just to check to make sure I understand you are saying that the radius just like the height is going to be changing as the water is being poured in right? I get what you are saying, I am just making sure that's what you meant. I will try to work through it tomorrow and see if i can at least come up with something even if its not numerical. I agree that there are too many variables to solve it that way, even though all of the selected answers in this section are completely numerical.... (this not being one of the selected obviously).

    Thanks for the differentiating tip!
     
  14. Apr 13, 2010 #13

    Dick

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    Sure, that's what I mean. The cone has a fixed slope angle. r and h are related through that slope.
     
  15. Apr 14, 2010 #14
    Ok so going for the same problem, which I have not successfully got to a numeric number I think I did solve for dh/dt, probably as close as I am going to get on this one unless you see something that I don't.

    [tex] V = \frac { \pi r^{2}h} {3} [/tex]

    [tex] \frac {dV}{dt} = \frac {3 * [\pi r^{2}h]' - \pi r^{2}h * 0} {3^{2}} [/tex]

    [tex] [\pi r^{2}h]' = h * [\pi r^{2}]' + \pi r^{2} * \frac {dh}{dt} [/tex]

    [tex] [\pi r^{2}]' = \pi * 2r\frac {dr}{dt} + r^{2} * 0 = \pi 2r\frac {dr}{dt} [/tex]

    [tex] [\pi r^{2}h]' = h * \pi 2r\frac{dr}{dt} + \pi r^{2}\frac{dh}{dt} = \pi 2r\frac{dr}{dt}h + \pi r^{2}\frac{dh}{dt} [/tex]

    so then finally

    [tex] \frac {dV}{dt} = \frac{3 * (\pi 2r\frac{dr}{dt}h + \pi r^{2}\frac{dh}{dt})} {3^{2}} [/tex]

    and distributing


    [tex] \frac {dV}{dt} = \frac{\pi 2r\frac{dr}{dt}h3 + \pi r^{2}\frac{dh}{dt}3} {3^{2}} [/tex]

    then substitute the 5 cu in/sec for dV/dt

    [tex] 5 = \frac{\pi 2r\frac{dr}{dt}h3 + \pi r^{2}\frac{dh}{dt}3} {3^{2}} [/tex]

    [tex] 45 = \pi 2r\frac{dr}{dt}h3 + \pi r^{2}\frac{dh}{dt}3 [/tex]

    [tex] 45 - \pi 2r\frac{dr}{dt}h3 = \pi r^{2}\frac{dh}{dt}3 [/tex]

    [tex] \frac {45 - \pi 2r\frac{dr}{dt}h3} {\pi r^{2} 3} = \frac{dh}{dt} [/tex]

    Simply a couple things

    [tex] \frac{dh}{dt}= \frac {45 - \pi 6r\frac{dr}{dt}h} {3 \pi r^{2}}[/tex]

    Definitely not close to a numerical answer and if I tried to solve for h or r values they would involve each other and never give me exact values. Unless I am just not seeing it.
     
  16. Apr 14, 2010 #15

    LCKurtz

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    One suggestion I would make is when you have

    [tex]V = \frac 1 3 \pi r^2 h[/tex]

    there is no need for the quotient rule.

    [tex]V' = \frac \pi 3 (r^2h)' = \frac \pi 3 (2r r' h + r^2h')[/tex]
     
  17. Apr 14, 2010 #16

    Dick

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    I would use the relation that r=k*h from the start. At least you can get rid of the variable r.
     
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