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Math Help - Related Rates problem

  1. #1
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    Related Rates problem

    A point moves along the curve y= 2x^2 - 1 in such a way that the y value is decreasing at the rate of 2 units per second. At what rate is x changing when x = -3/2 ?

    Not really looking to solve just an explanation on the process of how to get to the solution. Thanks!!
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  2. #2
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    Quote Originally Posted by Hibijibi View Post
    A point moves along the curve y= 2x^2 - 1 in such a way that the y value is decreasing at the rate of 2 units per second. At what rate is x changing when x = -3/2 ?

    Not really looking to solve just an explanation on the process of how to get to the solution. Thanks!!

    Well this is pretty general so here it goes

    y=f(x) taking the derivative we get

    \frac{dy}{dx}=f'(x) abusing notation we multiply by dx to get

    dy=f'(x)dx plug in what we know and solve for dx.

    Hint: member dx and dy represent the rates of change of x and y.

    Good luck.
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  3. #3
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    Quote Originally Posted by Hibijibi View Post
    A point moves along the curve y= 2x^2 - 1 in such a way that the y value is decreasing at the rate of 2 units per second. At what rate is x changing when x = -3/2 ?

    Not really looking to solve just an explanation on the process of how to get to the solution. Thanks!!
    what you do is that you take the derivative of your function, your new function is
    \frac{dy}{dx}=4x, then
    dy=4xdx
    you know what dy is, it is -2, the rate of change of y with respect to x. x is -3/2, and dx is what you are looking for, the rate of x.
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    Quote Originally Posted by Mr. Engineer View Post
    what you do is that you take the derivative of your function, your new function is
    \frac{dy}{dx}=4xdx
    you know what \frac{dy}{dx} is, it is -2, the rate of change of y with respect to x. x is -3/2, and dx is what you are looking for, the rate of x.
    Ah, thank you! Very clear answer. Also I got 1/3, and how do I know if it's increasing or decreasing?
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  5. #5
    Newbie Mr. Engineer's Avatar
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    no prob

    no problem
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    Based on what how would I know whether that value is an increasing value or decreasing?
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    Quote Originally Posted by Hibijibi View Post
    Ah, thank you! Very clear answer. Also I got 1/3, and how do I know if it's increasing or decreasing?
    since you said in your problem the value of the rate of y was -2, the graph of the function would be decreasing.
    you could try the first derivative test to see at which intervals the function would be increasing or decreasing, but from the values you have, as x gets bigger, y gets smaller
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    A spherical balloon filled with helium at the rate of 2 in^2 / sec. Find the rate of change of the radius at the moment when the radius is 1 inch. Any advice? (I'm working on the final exam review sheet my professor gave me so lots of questions ^_^)
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  9. #9
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    2nd

    using properties from parabolas, the parabola opens up and its vertex is in the center. at -3/2, you have a decreasing slope, until you hit the vertex at (0,-1) where the slop is zero, then it becomes positive. just another way to look at it
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  10. #10
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    Quote Originally Posted by Hibijibi View Post
    A spherical balloon filled with helium at the rate of 2 in^2 / sec. Find the rate of change of the radius at the moment when the radius is 1 inch. Any advice? (I'm working on the final exam review sheet my professor gave me so lots of questions ^_^)
    for this and other problems with related rates, make sure you know the formulas. for a sphere V=\frac{4}{3}\pi r^3
    when you differentiate the function you get
    \frac{dV}{dt}=4\pi r^2\frac{dr}{dt}
    you know that \frac{dV}{dt} is 2in^2/sec, volume with respect to time. radius is 1, just solve for dr/dt, the rate of change of the radius with respect to time. with these problems, its just taking your derivative, realize what is changing and plug in what you know
    make sure you use the thank button
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    A maximization problem. A square is to be cut from each corner of a piece of paper which is 8 cm by 10 cm, and the sides are to be folded up to create an open box. What should the side of the square be for maximum volume? What's the basic process for maximizing problems?
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  12. #12
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    Quote Originally Posted by Hibijibi View Post
    A maximization problem. A square is to be cut from each corner of a piece of paper which is 8 cm by 10 cm, and the sides are to be folded up to create an open box. What should the side of the square be for maximum volume? What's the basic process for maximizing problems?
    find a way to relate length width and height...if u draw a picture you will see that the length is 10-x and the width is 8-x for the base of the box. the height is simply x, so V=(10-x)(8-x)(x). you can multiply, then differentiate the function. you set it equal to zero and you will get values for x. you must use the 1st or sometimes 2nd derivative test to see whether the values you got are maximums or minimums.
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  13. #13
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    good night

    i wish i can help you longer, but i have to get up early tomorrow, good night and good luck on your exam
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  14. #14
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    Quote Originally Posted by Mr. Engineer View Post
    i wish i can help you longer, but i have to get up early tomorrow, good night and good luck on your exam
    My Exam still isn't for another 2 days, and thank you. Also I think it would be 10-2x and 8-2x correct? Since each corner on both sides you're taking out an x?
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