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Thread: Couple of Calculus Problems

  1. #1
    Junior Member
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    Couple of Calculus Problems

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

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

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

    However, the correct answer is -2




    2. If $\displaystyle f(x) = |(x^2-12)(x^2+4)|$, 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:
    $\displaystyle 16x-4x^3 = f(3)-f(-2)/3+2$
    $\displaystyle 16x-4x^3 = 39-64/5$
    $\displaystyle 4x^3-16x+5=0$
    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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    $\displaystyle \frac{dy}{dt}\;=\;\frac{dy}{dx}\cdot \frac{dx}{dt}\;\implies\;\frac{dy}{dx}\;=\;\frac{\ frac{dy}{dt}}{\frac{dx}{dt}}$

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

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

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

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

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

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

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

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

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

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

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

    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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