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Thread: Four tough related rate problems

  1. #1
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    Angry Four tough related rate problems

    Hey everyone, i am new here. I am in BC Calc at my school and we got a tough related rate problems worksheet over the weekend. I got the first two and the last one, but the these three have stumped me. Last year's problems weren't this hard. Can you help with whatever you can, please?

    I took a picture of the problems. I know it's a lot, but if possible, just help with whatever you know.



    Thanks so much!!!
    Last edited by SuperTyphoon; Sep 12th 2008 at 02:29 PM.
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  2. #2
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    So far, i have some work done.

    3. I have dV/dt = k/r

    V = (4/3)pi*r^3

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

    Thus, 4pi*r^2 (dr/dt) = k/r

    I'm stuck here.


    4. The volume is 8pi in^3

    That's all i have for 4.


    5. I don't' know how to relate the eqautions.
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  3. #3
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    Hi there. I'm Nowegian and don't know what reciprocal means, but if the increase of r (referring to task 3) is constant (r(t) is linear), then you can find the rate of increase of the radius by $\displaystyle \frac{r_{t_1}-r_{t_0}}{t_1-t_0}$. If we define $\displaystyle \Delta t=t_1-t_0$ we then get $\displaystyle \frac{r_{t_0+\Delta t}-r_{t_0}}{\Delta t}$, which is the definition of the derivate. (For linear functions anyway. For other functions let $\displaystyle \Delta t$ go to zero.) You know that $\displaystyle r_{t_1}=2$ and $\displaystyle r_{t_0}=1$ and $\displaystyle t_1-t_0=15$. You can use this to find the rate of increase of the radius, then apply it to problem B.
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  4. #4
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    #5:

    Circumference is $\displaystyle C=2{\pi}r$

    and we know that $\displaystyle \frac{dC}{dt}=6$

    Therefore, $\displaystyle \frac{dC}{dt}=2{\pi}\frac{dr}{dt}$

    $\displaystyle \frac{dr}{dt}=\frac{3}{\pi}$

    The perimeter of the square is P=8r.

    So, $\displaystyle \frac{dP}{dt}=8\frac{dr}{dt}$

    $\displaystyle \frac{dP}{dt}=8(\frac{3}{\pi})=\frac{24}{\pi}$

    For part b, you need the difference between the area of the circle and the square.

    Since the area of the circle is $\displaystyle {\pi}r^{2}=25{\pi}$

    r=5. Then a side length of the square is 10.

    Continue on with this part?.

    Can you use this to solve other ones now?.
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  5. #5
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    Hello, SuperTyphoon!

    5. A circle is inscribed in a square. The circumference of the circle is increasing at a constant
    rate of 6 in/sec. As the circle expands, the square expands to maintain tangency.

    A. Find the rate at which the perimeter of the square is increasing.
    Code:
          *-------*-*-*-------*
          |   *           *   |
          | *               * |
          |*                 *|
          |                   |
          *                   *
       2r *         * - - - - *
          *              r    *
          |                   |
          |*                 *|
          | *               * |
          |   *           *   |
          *-------*-*-*-------*
                     2r

    The radius of the circle is $\displaystyle r.$
    The side of the square is $\displaystyle 2r.$

    The circumference of the circle is: .$\displaystyle C \:=\:2\pi r$

    Differentiate with respect to time: .$\displaystyle \frac{dC}{dt} \:=\:2\pi\!\cdot\!\frac{dr}{dt}$

    We are told that $\displaystyle \frac{dC}{dt} = 6$

    So we have: .$\displaystyle 6 \:=\:2\pi\!\cdot\!\frac{dr}{dt} \quad\Rightarrow\quad\frac{dr}{dt} \:=\:\frac{3}{\pi}$ in/sec. .[1]


    The perimeter of the square is: .$\displaystyle P \:=\:4(2r) \:=\:8r$

    Differentiate with respect to time: .$\displaystyle \frac{dP}{dt} \:=\:8\!\cdot\!\frac{dr}{dt}$ .[2]


    Substitute [1] into [2]: .$\displaystyle \frac{dP}{dt} \:=\:8\left(\frac{3}{\pi}\right) \:=\:\boxed{\frac{24}{\pi}\text{ in/sec}}$



    B. When the area of the circle is 25π inē,
    find the rate on increase of the area between the circle and the square.

    The area of the circle is: .$\displaystyle A_c \:=\:\pi r^2$

    If the area is $\displaystyle 25\pi$ inē, we have: .$\displaystyle 25\pi \:=\:\pi r^2\quad\Rightarrow\quad r \:=\:5$ .[3]


    The area of the square is: .$\displaystyle A_s \:=\:(2r)^2 \:=\:4r^2$


    The difference of the areas is: .$\displaystyle D \:=\:A_s - A_c \:=\:4r^2 - \pi r^2\quad\Rightarrow\quad D \:=\:(4-\pi)r^2$

    Differentiate with respect to time: .$\displaystyle \frac{dD}{dt} \:=\:2(4-\pi)r\!\cdot\!\frac{dr}{dt}$ .[4]


    Substitute [1] and [3] into [4]: .$\displaystyle \frac{dD}{dt} \:=\:2(4-\pi) (5)\left(\frac{3}{\pi}\right) \:=\:\boxed{\frac{30(4-\pi)}{\pi}\text{ in}^2\text{/sec}}$

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  6. #6
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    OMG Thank you guys so much!!! You don't know how much stress this has saved me. I understand it very well the way you write them too! Now i just need number 4, but still, thanks for all that.
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