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Math Help - find te rate of volume

  1. #1
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    find te rate of volume

    Hello
    i have two questions about calculus. I hope you can help me to solve it. Thanks

    A container formed by a cylinder at the bottom surrounded by a heisphere. The radius decreases at the rate of 3 cm/sec and trhe height increases at the rate of 2 cm/sec. Find the rate at which the volume of the contaner changes when the radius is 5 and the height is 10 cm.
    I try to solve by using : 2r+ h= 8?

    Let U(x,y) be a differentiable and let (r,θ) be the polar coordinates ( that is, x=rcos θ, y=rsinθ)
    Compute Ur and Uθ and show that lldedivative of Ull^2= (Ur)^2 + (Uθ)^2 1/r^2
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  2. #2
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    These are both exercises in using the chain rule for function of two variables.
    Quote Originally Posted by rai2003 View Post
    Hello
    i have two questions about calculus. I hope you can help me to solve it. Thanks

    A container formed by a cylinder at the bottom surrounded by a heisphere. The radius decreases at the rate of 3 cm/sec and trhe height increases at the rate of 2 cm/sec. Find the rate at which the volume of the contaner changes when the radius is 5 and the height is 10 cm.
    I try to solve by using : 2r+ h= 8?
    I take it that "a cylinder at the bottom surrounded by a hemisphere" means that the cylinder has a hemisphere on top of it. The volume of a cylinder of radius r and height h is \pi r^2h and the volume of a hemisphere of radius r is \frac{2}{3}\pi r^3. The volume of the figure is the sum of those two: V= \pi r^2h+ \frac{2}{3}\pi r^3. Find dV/dt in terms of dr/dt and dh/dt. \frac{dV}{dt}= \frac{\partial V}{\partial r}\frac{dr}{dt}+ \frac{\partial V}{\partial h}\frac{dh}{dt}.

    Let U(x,y) be a differentiable and let (r,θ) be the polar coordinates ( that is, x=rcos θ, y=rsinθ)
    Compute Ur and Uθ and show that lldedivative of Ull^2= (Ur)^2 + (Uθ)^2 1/r^2
    Use the chain rule:
    \frac{\partial U}{\partial r}= \frac{\partial U}{\partial x}\frac{\partial x}{\partial r}+ \frac{\partial U}{\partial y}\frac{\partial y}{\partial r} and
    \frac{\partial U}{\partial \theta}= \frac{\partial U}{\partial x}\frac{\partial x}{\partial \theta}+ \frac{\partial U}{\partial y}\frac{\partial y}{\partial \theta}
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  3. #3
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    Thanks for replying
    In the first question how can we dr/dt? so i try to get dr, dh and i replcae it by r=5 and h=10. Since i got dr and dh i will multiply it with like dr*(-3) and dh(2) . I got negative answer. I do not know i got the right answer or not.
    Last edited by rai2003; December 1st 2010 at 07:33 PM.
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  4. #4
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    Quote Originally Posted by rai2003 View Post
    A container formed by a cylinder at the bottom surrounded by a heisphere.
    No that description does not explain the shape clearly enough (for me to understand what it is).

    CB
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  5. #5
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    Quote Originally Posted by rai2003 View Post
    A container formed by a cylinder at the bottom surrounded by a heisphere.
    "Surrounded"? I think you mean that the cylinder has a hemisphere on top.
    The volume of a cylinder of radius r and height h is \pi r^2h and the volume of a hemisphere of radius r is \frac{2}{3}\pi r^3. The volume together is \pi r^2h+ \frac{2}{3}\pi r^3/

    The radius decreases at the rate of 3 cm/sec and trhe height increases at the rate of 2 cm/sec. Find the rate at which the volume of the contaner changes when the radius is 5 and the height is 10 cm.
    i try to get dr, dh and i replcae it by r=5 and h=10. Since i got dr and dh i will multiply it with like dr*(-3) and dh(2) . I got negative answer. I do not know i got the right answer or not.
    What do you mean you tried to get "dr" and "dh"? You are told that dr/dh= -3 and that dh/dt= 2. You are asked to find dV/dt. Use the chain rule.
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