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Math Help - Slope of Tangent

  1. #1
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    Slope of Tangent

    The gravitational potential energy of the MIR space station, in joules, at a distance of x kilometers from the centre of the Earth, is described by the equation
    E = -\frac{(5.38)(10)^{19}}{x}. Determine the slope of the tangent to the graph at the given value of x.

    x = (6.5)(10)^6 m

    My answer:


    x = (6.5)(10)^6 m = 6500 km

    m = \frac{E(6500 + 0.001) - E(6500)}{0.001}

    m = \frac{(-\frac{(5.38)(10)^{19}}{(6500 + 0.001)}) -(-\frac{(5.38)(10)^{19}}{(6500)})}{0.001}


    m = 1.273(10)^{12}

    Is this correct?
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  2. #2
    CSM
    CSM is offline
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    Yes your anwers is correct.

    Your formula can be rewritten as:
    E = -(5.38)(10)^{19}* \frac{1}{x}
    Which has derivative:
    E' = (5.38)(10)^{19}* \frac{1}{x^2}

    Proof see: Calculus with polynomials - Wikipedia, the free encyclopedia
    With your x, this gives 1.27337278 \times 10^{12}
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