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Math Help - Help with AB A.P. Exam Problems?

  1. #1
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    Question Help with AB A.P. Exam Problems?

    Dear all, I am a high school student. Recently, I took a exercise from a book but there are questions confused me. Can somebody help me out? I have three questions as follow, Please help! :

    Question 1:


    Thank you so much!
    Last edited by mak15; March 19th 2008 at 05:58 PM.
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  2. #2
    Junior Member roy_zhang's Avatar
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    (a) In oder to find the range of the slope for which the line lies above f(x) in the first quadrant, you need to realize that the "worst" situation is that the line is tangent to f(x) at point (2,0), the slope corresponding to this situation is just the slope of the tangent line to graph f(x) at point (2,0), which is f^{\prime}(2) = -2(2)=-4. Hence for any m>-4, at least part of the line will below the curve of f(x). So the range of m for which the line lies above f(x) in the first quadrant is m\le -4.

    (b) Given the line passes point (0,12), we can find that m=-6, so our line equation becomes y=-6(x-2). Realize that since m=-6<-4, we know that the line must lie above the curve of f(x). Note that the area of R is just the area bounded by y=-6(x-2) and f(x)=4-x^2 from x=0 to x=2. So the area can be expressed by the following integral:

    A(R) = \int\limits_0^2 [(-6(x-2))-(4-x^2)]dx=\frac{20}{3}

    (c) This part is similar to part (b) except we don't know the exact value of m, again let's express the area bounded by y=m(x-2) and f(x)=4-x^2 from x=0 to x=2 through the following integral:

    A(m) = \int\limits_0^2 [(m(x-2))-(4-x^2)]dx= \int\limits_0^2 (mx-2m-4+x^2)dx=-2m-\frac{16}{3}

    (d) Using the expression of A(m) obtained from part (c), we can easily answer this part. In order to find the rate of change of A(m), let's take derivative with respect to time t to both sides of A(m)=-2m-\frac{16}{3}. We have

    \frac{dA(m)}{dt}=-2\frac{dm}{dt}=-2(-2)=4

    Here we use the given \frac{dm}{dt}=-2, so the area is changing at the rate of 4 square units per second and since it is positive, we can conclude that the area is increasing.


    Roy
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