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Math Help - rate of change

  1. #1
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    rate of change



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

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

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

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

    =10hcos\theta

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

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

     \theta=0.6435 rad

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

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

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

     =\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

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

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

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

    using the sine difference identity ...

    A = 50 \sin(2\theta)

    now take the time derivative, sub in the given/calculated values and determine the value of \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

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

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

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

    using the sine difference identity ...

    A = 50 \sin(2\theta)

    now take the time derivative, sub in the given/calculated values and determine the value of \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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  5. #5
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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 \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 dx/dt
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