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Math Help - Related Rates

  1. #1
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    Related Rates

    I need help understanding the steps to solving these problems:

    1. A spherical balloon is expanding. If the radius is increasing at the rate of 2 inches per minute, at what rate is the volume increasing when the radius is 5 inches?

    I know to start off, I have to take the derivative of V, which is 4piR^2 * dr/dt, which equals dr/dt=dV/dt/(4piR^2). I'm stuck after that.

    2. A particle moves in a circular orbit x^2 + y^2 = 16. As it passes through the point (2, 2sqrt(3)), its y-coordinate decreases at a rate of 3 units per second. At what rate is the x-coordinate changing?

    3. A triangle, whose h = 2b, has an area A = 1/2bh=1/2b(2b)=b^2 is expanding with time. If dA/dt = 8 cm2/s, what is db/dt when b = 2?

    4. A heap of garbage in the shape of a cube is being compacted. Given that the volume decreases at a rate of 2 cubic meters per minute, find the rate of change of an edge of the cube when the volume is exactly 27 cubic meters.

    There are a lot more, but I think if I knew how to do these, I could figure out the others.

    Thanks!
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  2. #2
    MHF Contributor Calculus26's Avatar
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    for #1
    dr/dt=dV/dt/(4piR^2).
    you want dV/dt = 4pi r^2 dr/dt

    you know dr/dt and r don't you ?

    For #2

    x^2 + y^2 = 16.

    differentiate 2xdx/dt + 2ydy/dt = 0

    you know dy/dt, x , and y don't you?

    once you have #1 and # 2 then the others should be no problem
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  3. #3
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    Mmm...Thank you, but I'm still having a little trouble understanding :/

    For number one, I have another similar problem including the solution, but I'm having trouble seeing the steps:
    A sphere is expanding at the rate of 6 cm^3/sec, the formula for volume of a sphere is: V=4/3pir^3, what is dr/dt when r = 2?
    Here is the solution:
    V=4/3pir^3
    dV/dt=4pir^2*dr/dt
    dr/dt=dV/dt/(4piR^2)
    =3/8cm/s

    I don't understand how the answer became 3/8cm/s from the third step.


    For number two, this is what I did but I'm not sure if it's correct.
    2xdx/dt + 2ydy/dt = 0
    2(2)*dx/dt+2(2sqrt(3))(-3)=0 [-3 since it's decreasing?]
    4*dx/dt-20.78=0
    dx/dt=5.2

    Did I solve it correctly?
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  4. #4
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    I tried to solve number one and this is what I got:

    dV/dt = 4pi r^2 dr/dt
    4pi(5^2)(2)=200pi

    And I understand the problem and the solution on the previous post. Sorry, my posts are very confusing.
    Last edited by iyppxstahh; February 10th 2010 at 02:05 AM.
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  5. #5
    MHF Contributor Calculus26's Avatar
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    dx/dt=5.2
    is correct for #1

    dV/dt = 4pi r^2 dr/dt = 200pi is correct


    dr/dt=dV/dt/(4piR^2)
    =3/8cm/s
    dV/dt = 6 r= 2

    dr/dt=dV/dt/(4piR^2) = 6/(4pi *4) = 3/(8pi)
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  6. #6
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    Alright, thanks so much. You've helped me immensely on this lesson.
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  7. #7
    MHF Contributor Calculus26's Avatar
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    I think you have the ability but I think you need to slow down and realize the problems are more straightforward than you think.

    Calculus always takes more patience than intelligence.


    If interested I have several related rates problems with animations and solutions on my web site at Optimization and Related Rates.
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  8. #8
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    I think you're completely correct about the patients.
    Thank you, and again for the link - I started Optimization today!
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