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Math Help - Max/Min and Related Rates Question...

  1. #1
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    Max/Min and Related Rates Question...

    #1 If a and b are lengths of two sides of a triangle, and theta the measure of hte included angle, the area A of hte triangle is A = (1/2)absin(theta). How is dA/dt related to da/dt, db/dt, and dtheta/dt?

    My work:
    A= (1/2)absin(theta)
    dA/dt = ((1/2)(ab))'sin(theta) + (1/2)ab(sin(theta))'
    dA/dt = ((1/2)(dA/dt)b + (1/2)a(db/dt))sin(theta) + (1/2)abcos(theta)(theta)'

    I think there's something wrong either in the power rule or the (theta)'.

    #2 Why do we use the formula
    D^2 = x^2 + y^2 in the following question?...
    A point moves smoothly along the curve y = x^(3/2) in the first quadrant in such a way that its distance from the origin increases at the constant rate of 11 units per second. Find dx/dt when x = 3.
    I know the answer is dx/dt = 4 units per second. I just want to know why we use that formula, and other potential scenerios of when we might use it.
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  2. #2
    Member arpitagarwal82's Avatar
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    For question 1 , it is not mentioned that if a, and \theta.
    So letrs assume that all three are dependent upon time.
    <br />
dA/dt = d(1/2ab\sin\theta)/dt

    Now apply multiplication rule of differentiation.

    dA/dt = 1/2ab * d\sin\theta/dt + 1/2\sin\theta * d(ab)/dt
    =  1/2ab *\cos\theta d\theta/dt + 1/2 \sin\theta * (a*db/dt + b*da/dt)

    Simplify and answer will be in terms of da/dt, db/dt and d\theta/dt
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  3. #3
    Member arpitagarwal82's Avatar
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    second answer:
    according to question rate change of distance from origin is 11 unit per second

    =>d( \sqrt(x^2 - y^2))/dt = 11
    => d\sqrt(x^2 -  (x^(3/2))^2)/dt = 11^2 Since point lies on given curve y = x^(3/2)

    solve for dx/dt
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