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Math Help - Couple of Calculus Problems

  1. #1
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    Couple of Calculus Problems

    1. If x = sin(t) and y = cos^2(t) , then d^2y/dx^2 at t = pi/2 is?

    I got
    dy/dx = cos(t)/-2cos(t)sin(t)
    dy/dx = -1/2sin(t)
    d^2y/dx^2 = -2cos(t)/(2sin(t))^2

    at t = pi/2
    d^2y/dx^2 = -2cos(pi/2)/(2sin(pi/2))^2
    =0/4=0

    However, the correct answer is -2




    2. If f(x) = |(x^2-12)(x^2+4)|, how many numbers in the interval [-2,3] satisfy the conclusion of the Mean Value Theorem?

    f is always negative on the interval so I got rid of the absolute value signs and multiplied it by -1.
    Then I differentiated and got:
    16x-4x^3 = f(3)-f(-2)/3+2
    16x-4x^3 = 39-64/5
    4x^3-16x+5=0
    x=.321, -2.141, 1.820

    interval is [-2,3] so c=.321, 1.820

    The thing is the answer is three numbers...


    Any help is greatly appreciated
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  2. #2
    MHF Contributor
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    \frac{dy}{dt}\;=\;\frac{dy}{dx}\cdot \frac{dx}{dt}\;\implies\;\frac{dy}{dx}\;=\;\frac{\  frac{dy}{dt}}{\frac{dx}{dt}}

    \frac{dx}{dt}\;=\;\cos(t)

    \frac{dy}{dt}\;=\;-2\sin(t)\cos(t)\;=\;-sin(2t)

    \frac{dy}{dx}\;=\;\frac{-2\sin(t)\cos(t)}{\cos(t)}\;=\;-2\sin(t)

    You seem to be upside down. Please examine the formulation more carefully and give it another go.

    \frac{d^{2}y}{dx^{2}}\;=\;\frac{d}{dx}\cdot \frac{dy}{dx}\;=\;\frac{d}{dt}\frac{dy}{dx}\cdot \frac{dt}{dx}

    Let's see what you get.
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  3. #3
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    Quote Originally Posted by TKHunny View Post
    \frac{dy}{dt}\;=\;\frac{dy}{dx}\cdot \frac{dx}{dt}\;\implies\;\frac{dy}{dx}\;=\;\frac{\  frac{dy}{dt}}{\frac{dx}{dt}}

    \frac{dx}{dt}\;=\;\cos(t)

    \frac{dy}{dt}\;=\;-2\sin(t)\cos(t)\;=\;-sin(2t)

    \frac{dy}{dx}\;=\;\frac{-2\sin(t)\cos(t)}{\cos(t)}\;=\;-2\sin(t)

    You seem to be upside down. Please examine the formulation more carefully and give it another go.

    \frac{d^{2}y}{dx^{2}}\;=\;\frac{d}{dx}\cdot \frac{dy}{dx}\;=\;\frac{d}{dt}\frac{dy}{dx}\cdot \frac{dt}{dx}

    Let's see what you get.
    Wow, I feel kind of dumb for making such a stupid error, thanks for catching it though.
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