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Thread: speed derivative

  1. #1
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    speed derivative

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

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

    and the rest is easy.

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

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

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

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

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

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