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Math Help - speed derivative

  1. #1
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    speed derivative

    A point moves along the curve y=2x^2 + 1 in such a way that the y value decreasing at the rate of 2 units per second. At what rate is x chaning when x = \frac {3} {2}?
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  2. #2
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    Quote Originally Posted by Samantha View Post
    A point moves along the curve y=2x^2 + 1 in such a way that the y value decreasing at the rate of 2 units per second. At what rate is x chaning when x = \frac {3} {2}?
    If y=2x^2 + 1 then differentiate with respect to t:

    <br />
\frac{dy}{dt}=2x \frac{dx}{dt}<br />

    and the rest is easy.

    RonL
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  3. #3
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    decreasing  \frac {1} {3} unit/sec ?
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  4. #4
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    Hello, Samantha!

    A point moves along the curve: y \:=\:2x^2 + 1 in such a way
    that the y value decreasing at the rate of 2 units/sec.
    At what rate is x changing when x = \frac {3} {2} ?

    Differentiate with respect to time: . \frac{dy}{dt} \:=\:4x\cdot\frac{dx}{dt} \quad\Rightarrow\quad \frac{dx}{dt}\;=\;\frac{1}{4x}\cdot\frac{dy}{dt}

    We are given: . \frac{dy}{dt} = -2,\;x = \frac{3}{2}

    Therefore: . \frac{dx}{dt} \;=\;\frac{1}{4\left(\frac{3}{2}\right)}(-2) \;=\;-\frac{1}{3} units/sec

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