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Thread: Rates of change

  1. #1
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    Rates of change

    Sand being poured from a conveyor belt forms a cone with height h and a semi-vertical angle of 60 degrees. Let the volume of the sand is $\displaystyle \pi h^3$.

    Suppose that the sand is being poured at a constant rate of 0.3m^3/min. Find the rate of increase of the base area.
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  2. #2
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    I can suggest

    $\displaystyle \frac{dV}{dh} = 3\pi h^2$

    $\displaystyle \frac{dV}{dt} = 0.3$

    $\displaystyle \frac{dA}{dr} = 2\pi r$
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  3. #3
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    Was $\displaystyle V= \pi h^3$ given in the problem?
    The volume of a cone, of height h and base radius r, is given by $\displaystyle V= \frac{1}{3}\pi r^2 h$. Here, you are told that the "semi-vertical angle" is 60 degrees so using $\displaystyle \frac{r}{h}= tan(60)= \sqrt{3}$. $\displaystyle r= \sqrt{3}h$, $\displaystyle r^2= 3h^2$ and $\displaystyle V= \pi h^3$. That is where that formula came from.

    To change to a "rate of change" formula, differentiate both sides with respect to t, using the chain rule.

    $\displaystyle \frac{dV}{dt}= \frac{d\left(\pi h^3\right)}{dt}= 3\pi h^2 \frac{dh}{dt}$
    Using the fact that $\displaystyle \frac{dV}{dt}= .3$, you can solve that for $\displaystyle \frac{dh}{dt}$, as a function of h.

    The area of the base is, of course, $\displaystyle A= \pi r^2= 3\pi h^2$ so $\displaystyle \frac{dA}{dt}= 6\pi h\frac{dh}{dt}$.
    Last edited by HallsofIvy; Jul 15th 2009 at 07:47 AM.
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  4. #4
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    Quote Originally Posted by HallsofIvy View Post
    Was $\displaystyle V= \pi h^3$ given in the problem?
    The volume of a cone, of height h and base radius r, is given by $\displaystyle V= \frac{1}{3}\pi r^2 h$. Here, you are told that the "semi-vertical angle" is 60 degrees so using $\displaystyle \frac{r}{h}= tan(60)= \sqrt{3}$. $\displaystyle r= \sqrt{3}h$, $\displaystyle r^2= 3h^2 and $V= \pi h^3[/tex]. That is where that formula came from.

    To change to a "rate of change" formula, differentiate both sides with respect to t, using the chain rule.

    $\displaystyle \frac{dV}{dt}= \frac{d\left(\pi h^3\right)}{dt}= 3\pi h^2 \frac{dh}{dt}$
    Using the fact that $\displaystyle \frac{dV}{dt}= .3$, you can solve that for $\displaystyle \frac{dh}{dt}$, as a function of h.

    The area of the base is, of course, $\displaystyle A= \pi r^2= 3\pi h^2$ so $\displaystyle \frac{dA}{dt}= 6\pi h\frac{dh}{dt}$.
    Yes but then, what- I get a figure which still has the pronumeral "h"
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  5. #5
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    Hello requal
    Quote Originally Posted by requal View Post
    Yes but then, what- I get a figure which still has the pronumeral "h"
    Unless you are given a specific value of $\displaystyle h$, that's all you will be able to do: express the rate of change of $\displaystyle A$ as a function of $\displaystyle h$.

    Grandad
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  6. #6
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    just wondering, can't I just diffentiate the area in terms of volume(da/dv) so I can use da/dt=da/dv*dv/dt?
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  7. #7
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    Hello requal
    Quote Originally Posted by requal View Post
    just wondering, can't I just diffentiate the area in terms of volume(da/dv) so I can use da/dt=da/dv*dv/dt?
    You could do, by eliminating $\displaystyle h$ between the equations $\displaystyle V= \pi h^3$ and $\displaystyle A = 3\pi h^2$, but you'd then end up with an expression for $\displaystyle \frac{dA}{dt}$ in terms of $\displaystyle V$ (or $\displaystyle A$, or both).

    Grandad
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