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Thread: [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 $\displaystyle y = 2x+c$ is a tangent to the curve $\displaystyle y^2 = 4x$

    Attempt:
    NO IDEA
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  2. #2
    Super Member wingless's Avatar
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    Hint: $\displaystyle y=2x+c$ tells us the tangent lines slope is $\displaystyle m=2$. 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:

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

    and you want when it is equal to 2:

    $\displaystyle \frac{dy}{dx} = 2$

    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:

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

    So,
    $\displaystyle y^2 = 4x$

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

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

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

    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 $\displaystyle \frac{dy}{dx} = 2$, then $\displaystyle \frac{dx}{dy} = \frac{1}{2}$

    $\displaystyle y^2 = 4x$

    $\displaystyle x' = \frac{y}{2}$

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