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Math Help - Rate of change calculus?

  1. #1
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    Rate of change calculus?

    I have two different questions

    my first one is differentiate with respect to t

    y=x^2-3x find dy/dt when x=3 and dx/dt=2

    This is what I did
    dy/dt=2x (dx/dt)-3 I am not sure if this is correct


    my second one is

    xy=4 find dy/dt when x=8 dx/dt=10

    I am not sure how to begin this one
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  2. #2
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    Re: Rate of change calculus?

    Quote Originally Posted by homeylova223 View Post

    my first one is differentiate with respect to t

    y=x^2-3x find dy/dt when x=3 and dx/dt=2

    This is what I did
    dy/dt=2x (dx/dt)-3 I am not sure if this is correct

    dy/dx = dy/dt * dt/dx

    when x=3

    2(3)-3 = dy/dt * 1/2

    dy/dt = 6

    For the second one make xy=4 so y=4/x find dy/dx
    Last edited by pickslides; July 21st 2011 at 08:41 PM.
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  3. #3
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    Re: Rate of change calculus?

    Quote Originally Posted by homeylova223 View Post
    I have two different questions

    my first one is differentiate with respect to t

    y=x^2-3x find dy/dt when x=3 and dx/dt=2

    This is what I did
    dy/dt=2x (dx/dt)-3(dx/dt)


    my second one is

    xy=4 find dy/dt when x=8 dx/dt=10

    I am not sure how to begin this one

    product rule ...

    x(dy/dt) + y(dx/dt) = 0
    ...
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  4. #4
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    Re: Rate of change calculus?

    I used the product rule for xy=4 to get

    x(dy/dt)+y(dx/dt)=0 then I plug x=8 dx/dt=10

    8(dy/dt)+y(10)=0

    How would I proceed now?
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  5. #5
    MHF Contributor Siron's Avatar
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    Re: Rate of change calculus?

    You now y=4/x so y(10)=4/10, substituted in the equation:
    8(dy/dt)+(4/10)=0
    <-> ...
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    Re: Rate of change calculus?

    Quote Originally Posted by homeylova223 View Post
    I used the product rule for xy=4 to get

    x(dy/dt)+y(dx/dt)=0 then I plug x=8 dx/dt=10

    8(dy/dt)+y(10)=0

    How would I proceed now?
    8\frac{dy}{dt}=-10y\Rightarrow\frac{dy}{dt}=-\frac{10}{8}y

    and now use the fact that xy=4 to find "y" as you have "x".
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  7. #7
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    Re: Rate of change calculus?

    Hmm I understand up the part where you get

    dy/dt=-10/8y

    But what would I do knowing this
    I am unsure where to plug it in.
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  8. #8
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    Re: Rate of change calculus?

    Quote Originally Posted by homeylova223 View Post
    Hmm I understand up the part where you get

    dy/dt=-10/8y

    But what would I do knowing this
    I am unsure where to plug it in.
    \frac{dy}{dt}=-\frac{10y}{8}

    Now you need the value of "y" to obtain a value for dy/dt.

    Since xy=4, then 8y=4 since x=8.
    This gives the value for "y" that allows you to find the value for dy/dt.
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