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Math Help - What is being asked?

  1. #1
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    What is being asked?

    I have a 4 part problem that I am unsure what is being asked. I have shown what i have worked out.





    a) Find the differential of y





    b) Calculate dy when x=4 and delta x = 0.01. Interpret.



    dy= \frac {1}{4} * \frac{1}{100} = \frac{1}{400 }= 0.0025

    c) Calculate delta y when x =4 and delta x = 0.01. Interpret.









    d) Illustrate dy and delta y. Interpret.

    So what are they asking for "interpret" and "illustrate"?

    Thanks,
    Rod
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  2. #2
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    Re: What is being asked?

    Quote Originally Posted by friesr View Post
    I have a 4 part problem that I am unsure what is being asked. I have shown what i have worked out.

    a) Find the differential of y


    b) Calculate dy when x=4 and delta x = 0.01. Interpret.
    dy= \frac {1}{4} * \frac{1}{100} = \frac{1}{400 }= 0.0025

    c) Calculate delta y when x =4 and delta x = 0.01. Interpret.




    d) Illustrate dy and delta y. Interpret.

    So what are they asking for "interpret" and "illustrate"?

    To be truthful, I not sure what the expected answer is.

    But \sqrt{4.01} - \sqrt{4} \approx 0.002498 tell you that \sqrt{4.01}\approx 2.002498
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  3. #3
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    Re: What is being asked?

    Quote Originally Posted by friesr View Post
    I have a 4 part problem that I am unsure what is being asked. I have shown what i have worked out.





    a) Find the differential of y





    b) Calculate dy when x=4 and delta x = 0.01. Interpret.



    dy= \frac {1}{4} * \frac{1}{100} = \frac{1}{400 }= 0.0025
    You know that you can think of the derivative as the slope of the tangent line, don't you? If you start from (4, 2) and x increases by .01 then the linear approximation to y(x) says that y will increase by 0.0025.

    c) Calculate delta y when x =4 and delta x = 0.01. Interpret.







    This is saying that if, starting at (4, 2), x increases by .01, then y actually changes by 0.002498.
    Pretty close to the linear approximation, before, 0.0025 isn't it?

    d) Illustrate dy and delta y. Interpret.
    Draw a graph! Draw the graph of y= \sqrt{x} and its tangent line at x= 4. Draw a horizontal line from (4, 2) to (4.01, 2). \Delta y is the distance from (4.01, 2) up to the graph of y= \sqrt{x} and dy is the distance from (4.01, 2) to the tangent line.


    So what are they asking for "interpret" and "illustrate"?

    Thanks,
    Rod
    Last edited by HallsofIvy; November 22nd 2012 at 03:26 PM.
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