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Math Help - Writing the equation of a tangent line

  1. #1
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    Writing the equation of a tangent line

    I'm reviewing for the AP exam and would love it if someone could walk me through this problem- I know you find the derivative of the curve at the given point... but I get lost from there
    The slope of a function f at any point (x,y) is y/(2x)^2. The point (2,1) is on the graph of f.
    A. Write an equation of the tangent line to the graph f at x=2.
    B. Use the tangent line in part (a) to approximate f(2.5).
    C. Solve the differential equation dy/dx= y/(2x)^2 with the initial condition f(2)=1.
    D. Use the solution in part c and find f(2.5)
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  2. #2
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    Re: Writing the equation of a tangent line

    Quote Originally Posted by qubez View Post
    I'm reviewing for the AP exam and would love it if someone could walk me through this problem- I know you find the derivative of the curve at the given point... but I get lost from there
    The slope of a function f at any point (x,y) is y/(2x)^2. The point (2,1) is on the graph of f.
    A. Write an equation of the tangent line to the graph f at x=2.
    B. Use the tangent line in part (a) to approximate f(2.5).
    C. Solve the differential equation dy/dx= y/(2x)^2 with the initial condition f(2)=1.
    D. Use the solution in part c and find f(2.5)
    A. point (2,1) , use \frac{dy}{dx} to evaluate the slope at the given point ...

    slope = \frac{1}{(2 \cdot 2)^2} = \frac{1}{16}

    write the tangent line equation using the point-slope form ...

    y - 1 = \frac{1}{16}(x - 2)

    B. use the tangent line to approximate the function value at x = 2.5 ...

    f(x) \approx y = \frac{1}{16}(x - 2) + 1

    f(2.5) \approx y = \frac{1}{16}(2.5 - 2) + 1 = \frac{33}{32}

    C. separate variables ...

    \frac{dy}{y} = \frac{dx}{4x^2}

    integrate ...

    \ln|y| = -\frac{1}{4x} + C_1

    change to an exponential to isolate y ...

    y = C_2 e^{-\frac{1}{4x}}

    utilize the given initial condition to solve for the constant of integration ...

    1 = C_2 e^{-\frac{1}{8}}

    C_2 = e^{\frac{1}{8}}

    simplify ...

    y = f(x) = e^{\frac{x-2}{8x}}

    D. f(2.5) = e^{\frac{1}{8}}
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