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Math Help - Calculus problem

  1. #1
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    Calculus problem

    At what point on the curve y = x^4 does the normal have a slope of 16?
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  2. #2
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    Does "normal" mean perpendicular to the tangent?
    If that's the case:
    slope_normal(x) * slope_tangent(x) = -1
    16 * slope_tangent(x) = -1

    where slope_tangent(x) = y'(x)

    16 * y'(x) = -1

    Solve for x.

    -O
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  3. #3
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by mathamatics112 View Post
    At what point on the curve y = x^4 does the normal have a slope of 16?
    If the slope of the normal line at that point is 16, then the slope of the tangent line at that point is -\frac{1}{16}. So what you need to do is find where \frac{\,d}{\,dx}\left[x^4\right]=-\frac{1}{16}

    Can you take it from here?
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  4. #4
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    I am stuck again :(

    Ok so I am at this point where dx/dy (x^4) = -1/16.

    from here I got 4x^3 = -1/16

    i multipled by 4 * 16 to get -1/64 and got the third root of this to get me -1/4 which is my x. How do i get the y vaule? i tried subbing it in into x^4 and i got the wrong answer
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  5. #5
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    (-1/4)^4=1/256 should be the answer. No?

    -O
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