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Math Help - Find the Rate of Change

  1. #1
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    Find the Rate of Change

    Find the rate of change of the distance between the origin and a moving point on the graph of
    y = x^2 + 1 It dx/Dr = 2 cm/second

    My work:

    I proceeded taking the implicit derivative of the given function to find dy/dt.

    dy/dt = 2x* dx/dt

    dy/dt = 2x* (2) and the I plugged x = 0 from the origin value for x.

    dy/dt = 2(0)* (0)

    dy/dt = 0...My Answer.

    Book's Answer:

    2(2x^3+3x)/sqrt(x^4+3x^2+1)

    My Answer is zero and the book's answer is a crazy rational function.

    Can someone answer this question?
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  2. #2
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    Re: Find the Rate of Change

    Quote Originally Posted by nycmath View Post
    Find the rate of change of the distance between the origin and a moving point on the graph of
    y = x^2 + 1 It dx/Dr = 2 cm/second

    My work:

    I proceeded taking the implicit derivative of the given function to find dy/dt.

    dy/dt = 2x* dx/dt

    dy/dt = 2x* (2) and the I plugged x = 0 from the origin value for x.

    dy/dt = 2(0)* (0)

    dy/dt = 0...My Answer.

    Book's Answer:

    2(2x^3+3x)/sqrt(x^4+3x^2+1)

    My Answer is zero and the book's answer is a crazy rational function.

    Can someone answer this question?
    Yes. Someone can answer it.

    -Dan
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  3. #3
    Forum Admin topsquark's Avatar
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    Re: Find the Rate of Change

    Quote Originally Posted by nycmath View Post
    Find the rate of change of the distance between the origin and a moving point on the graph of
    y = x^2 + 1 It dx/Dr = 2 cm/second

    My work:

    I proceeded taking the implicit derivative of the given function to find dy/dt.

    dy/dt = 2x* dx/dt

    dy/dt = 2x* (2) and the I plugged x = 0 from the origin value for x.

    dy/dt = 2(0)* (0)

    dy/dt = 0...My Answer.

    Book's Answer:

    2(2x^3+3x)/sqrt(x^4+3x^2+1)

    My Answer is zero and the book's answer is a crazy rational function.

    Can someone answer this question?
    Sorry, just a little joke.

    The problem is asking about how fast the distance is changing between a point on the graph and the origin. So say we have a point on the graph (x, y). What is the distance between that point and the origin? Once you get that you can take your derivative. (And recall that y = x^2 - 1 before you differentiate.)

    -Dan
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    Re: Find the Rate of Change

    This is what I did:

    (1) I used the distance formula for points to find d.

    d = sqrt{(x-0)^2 + (x^2-1-0)}

    d = sqrt{x^4-2x^2+1}

    (2) I then differentiated implicitly and found dr/dt to be

    [2x(2x^2-1)]/sqrt{x^4-2x^2+1}

    It is not correct.
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  5. #5
    Forum Admin topsquark's Avatar
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    Re: Find the Rate of Change

    Quote Originally Posted by nycmath View Post
    d = sqrt{(x-0)^2 + (x^2-1-0)}
    Your parabola is y = x^2 + 1. You have a minus sign there. And don't forget the (x - 0)^2 term.

    -Dan
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  6. #6
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    Re: Find the Rate of Change

    Almost there...take a look.

    My answer:

    dr/dt = 2(2x^3+ 3)/sqrt(x^4+3x^2+1)

    The numerator should be
    2(2x^3+3x). I am missing an "x" for the term 3x. See it?
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  7. #7
    Forum Admin topsquark's Avatar
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    Re: Find the Rate of Change

    Quote Originally Posted by nycmath View Post
    Almost there...take a look.

    My answer:

    dr/dt = 2(2x^3+ 3)/sqrt(x^4+3x^2+1)

    The numerator should be
    2(2x^3+3x). I am missing an "x" for the term 3x. See it?
    r = \sqrt{x^4 + 3x^2 + 1}

    \frac{dr}{dt} = \frac{1}{2} (x^4 + 3x^2 + 1)^{-1/2} \cdot (4x^3 + 6x) \frac{dx}{dt}

    -Dan
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  8. #8
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    Re: Find the Rate of Change

    I clearly see my error. When taking the derivative of the inside using the chain rule, I mistakenly multiplied 4x^3 and 6x by the value given for dx/dt or 2. Thank you for your help. Math is fun.
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