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Math Help - [SOLVED] Differentiation Help!

  1. #1
    Member looi76's Avatar
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    [SOLVED] Differentiation Help!

    Question:Find the value of the constant c for which the line y = 2x+c is a tangent to the curve y^2 = 4x

    Attempt:
    NO IDEA
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  2. #2
    Super Member wingless's Avatar
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    Hint: y=2x+c tells us the tangent lines slope is m=2. So y' is 2 at the tangent point.
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  3. #3
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    Well think about it. You are told that you are to find a line that is TANGENT to another line.

    This means when the derivative of the second equation equals the slope of the first one:

    y^2 = 4x \rightarrow y = 2\sqrt{x}

    and you want when it is equal to 2:

    \frac{dy}{dx} = 2

    AND also you know that they both are equal at those x values. So you should be able to get two equations and solve them for C.
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  4. #4
    Super Member wingless's Avatar
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    Quote Originally Posted by TrevorP View Post
    Well think about it. You are told that you are to find a line that is TANGENT to another line.

    This means when the derivative of the second equation equals the slope of the first one:

    y^2 = 4x \rightarrow y = 2\sqrt{x}
    Not true.
    Remember: \sqrt{x^2} is NOT x, its |x|!

    So,
    y^2 = 4x

    \sqrt{y^2} =\sqrt{4x}

    |y| = \sqrt{4x} = 2\sqrt{x}

    This gives us two functions of y:
    y= 2\sqrt{x} and y= -2\sqrt{4x}.

    Now differentiate both and set both derivatives y' = 2. Only one of them will satisfy it and you'll get one value for x. This is the x coordinate of the tangent point. Find its y coordinate too and you'll get the tangent point. Plug this point in the line equation to get c.


    A better approach is here:

    If \frac{dy}{dx} = 2, then \frac{dx}{dy} = \frac{1}{2}

    y^2 = 4x

    x' = \frac{y}{2}

    Now set x'=2 to get the tangent points y coordinate. This is much easier than working with two functions.
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