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Math Help - [SOLVED] Differentiation - Equation of curve

  1. #1
    Member looi76's Avatar
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    Question [SOLVED] Differentiation - Equation of curve

    Question:
    A curve is such that dy/dx = 2x − 5. Given that the point (3, 8) lies on the curve, find the equation of the curve.

    Attempt:
    dy/dx = 2x-5
    x = 3
    2(3)-5 = 13

    y - y1 = m (x - x1)
    y - 8 = 13 (x - 3)
    y - 8 = 13x - 39
    y = 13x - 39 +8
    y = 13x -31
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  2. #2
    Super Member angel.white's Avatar
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    Quote Originally Posted by looi76 View Post
    Question:
    A curve is such that dy/dx = 2x − 5. Given that the point (3, 8) lies on the curve, find the equation of the curve.

    Attempt:
    dy/dx = 2x-5
    x = 3
    2(3)-5 = 13

    y - y1 = m (x - x1)
    y - 8 = 13 (x - 3)
    y - 8 = 13x - 39
    y = 13x - 39 +8
    y = 13x -31
    (3,8) lies on the initial curve, you have to integrate/anti-differentiate in order to find the formula for this.

    \frac{dy}{dx} = 2x^2-5

    y = \frac{2}3x^3-5x+C

    Now you only need to determine the value of your constant, which can be done by plugging in your x and y values.
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  3. #3
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    earboth's Avatar
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    Quote Originally Posted by looi76 View Post
    Question:
    A curve is such that dy/dx = 2x − 5. Given that the point (3, 8) lies on the curve, find the equation of the curve.

    Attempt:
    dy/dx = 2x-5
    x = 3
    2(3)-5 = 13

    y - y1 = m (x - x1)
    y - 8 = 13 (x - 3)
    y - 8 = 13x - 39
    y = 13x - 39 +8
    y = 13x -31
    Hello,

    you have calculated correctly the equation of the tangent of the curve.

    Use

    \int \frac{dy}{dx} dx=y=\int(2x^2-5)dx = \frac23 x^3 - 5x + C

    Since (3, 8) lies on the curve the coordinates must satisfy the equation:

    8 = \frac23 (3)^3 - 5 \cdot 3 + C~\iff~C = 5

    Therefore the equation of the curve is:

    y = \frac23 x^3 - 5x + 5
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