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Thread: rate of change

  1. #1
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    rate of change



    The diagram shows an isosceles triangle $\displaystyle ABC$ with fixed length $\displaystyle AB$ and $\displaystyle AC$ of $\displaystyle 10cm$ each and Angle $\displaystyle ACB = $Angle $\displaystyle ABC = \theta $radians.

    $\displaystyle A$ is a variable point which is at a height $\displaystyle h cm$ directly above the point $\displaystyle O$ while $\displaystyle B$ and $\displaystyle C$ are variable points which move horizontally along the line $\displaystyle l$.

    Given that $\displaystyle A$ descends vertically towards the point $\displaystyle O$ such that the area of triangle $\displaystyle ABC$ is decreasing at a constant rate of $\displaystyle 0.7cm^2/s$, determine, at the instant when $\displaystyle A$ is $\displaystyle 6cm$ above $\displaystyle O$, the rate of change of $\displaystyle \theta$

    $\displaystyle A=\frac{1}{2}(20cos\theta)h$

    $\displaystyle =10hcos\theta$

    $\displaystyle \frac{dA}{d\theta}=-10hsin\theta $

    $\displaystyle when h=6, sin\theta =\frac{6}{10}$

    $\displaystyle \theta=0.6435 rad$

    $\displaystyle \frac{dA}{d\theta}=-10(6)(sin0.6435)$
    $\displaystyle =-36$

    $\displaystyle \frac{d\theta}{dt}=(\frac{dA}{dt})(\frac{d\theta}{ dA})$

    $\displaystyle =(-0.7)(\frac{1}{-36})$

    $\displaystyle =\frac{7}{360}$

    but the answer is -0.025

    And another part of the question asks : Find the rate at which C is moving away from O. I could not start
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  2. #2
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    note that the area of a triangle can be calculated from the side-angle-side formula

    $\displaystyle A = \frac{1}{2}ab\sin{\phi}$

    where a, b are the adjacent sides and $\displaystyle \phi$ is the angle between them

    $\displaystyle A = \frac{1}{2} \cdot 10^2 \sin(\pi - 2\theta)$

    using the sine difference identity ...

    $\displaystyle A = 50 \sin(2\theta)$

    now take the time derivative, sub in the given/calculated values and determine the value of $\displaystyle \frac{d\theta}{dt}$
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  3. #3
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    Quote Originally Posted by skeeter View Post
    note that the area of a triangle can be calculated from the side-angle-side formula

    $\displaystyle A = \frac{1}{2}ab\sin{\phi}$

    where a, b are the adjacent sides and $\displaystyle \phi$ is the angle between them

    $\displaystyle A = \frac{1}{2} \cdot 10^2 \sin(\pi - 2\theta)$

    using the sine difference identity ...

    $\displaystyle A = 50 \sin(2\theta)$

    now take the time derivative, sub in the given/calculated values and determine the value of $\displaystyle \frac{d\theta}{dt}$
    but i cant seem to spot the mistake in my workings
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  4. #4
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    Quote Originally Posted by Punch View Post
    but i cant seem to spot the mistake in my workings
    note that h is changing. you should have dh/dt in your derivative also ... problem is, that value was not given.
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    Quote Originally Posted by skeeter View Post
    note that h is changing. you should have dh/dt in your derivative also ... problem is, that value was not given.
    yes, i should have expressed h in terms of $\displaystyle \theta$, thanks
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  6. #6
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    hello, the question has a second part which i couldn't start, could i get some help with it? i think its linked to the first part.

    The other part of the question asks : Find the rate at which C is moving away from O. I could not start
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  7. #7
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    Did you get the first part done alright?

    The second part requires you to look at the length OC. Call the length OC a new variable (I called it x) and try to solve for $\displaystyle dx/dt$
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