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Math Help - Differentiation

  1. #1
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    Differentiation

    Find the coordinates of the point where the tangent to the curve y = x^2 + 1 at the point (2, 5) meets the normal to the same curve at the point (1, 2).

    How can I solve this problem? Please help me.
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  2. #2
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    Given the curve y = x^2 +1 the slope of the tangent at (2,5) is 4 and the slope of the normal at (1,2) is \frac{-1}{2}.
    Now write the equations of the two lines.
    Find the point of their intersection.
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  3. #3
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    Thank you Plato. Problem resolved
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  4. #4
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    So  y-5 = 4(x-2) or  y = 4x-3 (tangent line).

     y = -\frac{1}{2}x (normal at point).

     4x-3 = -\frac{1}{2}x,  \ x = \frac{2}{3}, y = -\frac{1}{3} .
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  5. #5
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    Quote Originally Posted by Plato View Post
    Given the curve y = x^2 +1 the slope of the tangent at (2,5) is 4 and the slope of the normal at (1,2) is \frac{-1}{2}.
    Now write the equations of the two lines.
    Find the point of their intersection.
    You are probably wondering how Plato got those slopes.

    f(x) = x^2 +1

    Find the derivative of f

    f'(x) = 2x

    Slope at (2, 5) is a simple plug and chug of the x-coordinate, so f'(2) = 4

    Slope at (1, 2) is just f'(1) = 2, so the line normal to this point is perpendicular. Perpendicular slopes are opposite signed reciprocals, so that's how Plato got \frac{-1}{2}
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  6. #6
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    Iíve another problem.

    The curve with equation y = ax^2 + bx + c passes through the point (1, 2). The gradient of the curve is zero at the point (2, 1). Find the values of a, b, c.

    How can I solve this?
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by geton View Post
    I’ve another problem.

    The curve with equation y = ax^2 + bx + c passes through the point (1, 2). The gradient of the curve is zero at the point (2, 1). Find the values of a, b, c.

    How can I solve this?
    let f(x) = ax^2 + bx + c.

    set up three simultaneous equations. since (1,2) and (2,1) are on the curve, we have:
    f(1) = 2 and f(2) = 1. so form two equations from that.

    but there are three unknowns, we need at least three equations. so use the information from the derivative. we are also told that f ' (2) = 0, that's your third equation
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  8. #8
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    New questions need to start a new thread.
    You know that y(1) = 2,\quad y(2) = 1\quad \& \quad y'(2) = 0 .
    Make the substitutions and solve.
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  9. #9
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Plato View Post
    New questions need to start a new thread.
    thanks, i forgot to mention that
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  10. #10
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    Thank you so much Jhevon & Plato.
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