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Math Help - [SOLVED] Differentiation Question

  1. #1
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    [SOLVED] Differentiation Question

    Question:
    The equation of a curve is y = \frac{6}{5 - 2x}
    (i) Calculate the gradient of the curve at the point where x = 1
    (ii) A point with coordinates (x, y) moves along the curve in such a way that the rate of increase of y has a constant value of 0.02 units per second. Find the rate of increase of x when x = 1.


    Answers:

    (i) y = 6(5 - 2x)^{-1}
    \frac{dy}{dx} = 6 \times -1(5 - 2x)^{-1-1} \times (-2)
    \frac{dy}{dx} = 12(5 - 2x)^{-2}

    Is this right?

    (ii) How can I get the rate of increase of x?
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    Hello ^^

    Quote Originally Posted by looi76 View Post
    Question:
    The equation of a curve is y = \frac{6}{5 - 2x}
    (i) Calculate the gradient of the curve at the point where x = 1
    (ii) A point with coordinates (x, y) moves along the curve in such a way that the rate of increase of y has a constant value of 0.02 units per second. Find the rate of increase of x when x = 1.


    Answers:

    (i) y = 6(5 - 2x)^{-1}
    \frac{dy}{dx} = 6 \times -1(5 - 2x)^{-1-1} \times (-2)
    \frac{dy}{dx} = 12(5 - 2x)^{-2}

    Is this right?
    Yes it is
    Remember, you're asked the gradient where x=1

    (ii) How can I get the rate of increase of x?
    The rate of increase is 0.02 means that \frac{d^2y}{dx^2}=0.02 (second derivative)

    The second derivative represents the speed of variation of y, so if you're asked a rate of increasing or decreasing, use this
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