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Math Help - TLA Implicit Differentiation

  1. #1
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    TLA Implicit Differentiation

    Hello,

    I have a problem here that I am somewhat unsure about:

    "Point (-1, 1) lies on the curve y^3 - \frac{x}{y} + \frac{y}{x} = 1. Use tangent line approximation to approximate the y-coordinate of the point where x = -1.01."

    (A) 1.02
    (B) 0.978
    (C) 0.965
    (D) 1.101
    (E) 0.98

    Here is my go at it:

    First, use implicit differentiation to find the derivative:

    3y^2 \cdot \frac{dy}{dx} - \left(\frac{1}{y} - \frac{x}{y^2} \cdot \frac{dy}{dx}\right) + \left(\frac{dy}{dx} \cdot \frac{1}{x^2} - \frac{y}{x^2}\right) = 0

    \frac{dy}{dx} = \frac{\left(\frac{y}{x^2} + \frac{1}{y}\right)}{\left(3y^2 + \frac{x}{y^2} + \frac{1}{x^2}}\right)

    Substitute (-1, 1) and solve for the tangent line:

    \frac{dy}{dx} = \frac{ \left( \frac{(1)}{(-1)^2} + \frac{1}{(1)}\right) }{ \left( 3(1)^2 + \frac{(-1)}{(1)^2} + \frac{1}{(-1)^2} \right) } = \frac{2}{3}

    Tangent line linear formula:

    y = \frac{2}{3}x + \frac{5}{3}

    Use TLA approximation formula with a = -1.01:

    L(x) = f(x) + f'(a)(x - a) = 1 + \frac{2}{3}(-1.01)(0.01) = 1.009933 \approx 1.01

    So my guess would probably be A, or 1.02. Am I doing this right? (I could have made a mistake differentiating).

    Thanks.
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  2. #2
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    Re: TLA Implicit Differentiation

    Quote Originally Posted by Biff View Post
    3y^2 \cdot \frac{dy}{dx} - \left(\frac{1}{y} - \frac{x}{y^2} \cdot \frac{dy}{dx}\right) + \left(\frac{dy}{dx} \cdot \frac{1}{x^2} - \frac{y}{x^2}\right) = 0
    The last expression in parentheses should be \left(\frac{dy}{dx} \cdot \frac{1}{x} - \frac{y}{x^2}\right). I get y' = 2 and the answer is (E).
    Thanks from Biff
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  3. #3
    MHF Contributor MarkFL's Avatar
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    Re: TLA Implicit Differentiation

    When I differentiate, I get:

    3y^2\frac{dy}{dx}-\frac{y-x\frac{dy}{dx}}{y^2}+\frac{x\frac{dy}{dx}-y}{x^2}=0

    3y^2\frac{dy}{dx}-\frac{1}{y}+\frac{x}{y^2}\cdot\frac{dy}{dx}+\frac{  1}{x}\cdot\frac{dy}{dx}-\frac{y}{x^2}=0

    \left(3y^2+\frac{x}{y^2}+\frac{1}{x} \right)\frac{dy}{dx}=\frac{1}{y}+\frac{y}{x^2}

    \frac{3xy^4+x^2+y^2}{xy^2}\cdot\frac{dy}{dx}=\frac  {x^2+y^2}{x^2y}

    \frac{dy}{dx}=\frac{y(x^2+y^2)}{x(3xy^4+x^2+y^2)}

    At the point (-1,1), we have the slope:

    \frac{dy}{dx}\left|_{(-1,1)}=\frac{1((-1)^2+1^2)}{(-1)(3(-1)1^4+(-1)^2+1^2)}=\frac{2}{1}=2

    Hence:

    y-1=2(x+1)

    y=2x+3

    y(-1.01)=0.98
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  4. #4
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    Re: TLA Implicit Differentiation

    Quote Originally Posted by emakarov View Post
    The last expression in parentheses should be \left(\frac{dy}{dx} \cdot \frac{1}{x} - \frac{y}{x^2}\right). I get y' = 2 and the answer is (E).
    That makes better sense, thanks.
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  5. #5
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    Re: TLA Implicit Differentiation

    Quote Originally Posted by MarkFL2 View Post
    Hence:

    y-1=2(x+1)

    y=2x+3

    y(-1.01)=0.98
    Thanks Mark, I thought it might have been a differentiating mistake. Also, I see how the TLA equation isn't necessary.
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