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Thread: 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; Mar 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 $\displaystyle f(x)$ in the first quadrant, you need to realize that the "worst" situation is that the line is tangent to $\displaystyle f(x)$ at point $\displaystyle (2,0)$, the slope corresponding to this situation is just the slope of the tangent line to graph $\displaystyle f(x)$ at point $\displaystyle (2,0)$, which is $\displaystyle f^{\prime}(2) = -2(2)=-4$. Hence for any $\displaystyle m>-4$, at least part of the line will below the curve of $\displaystyle f(x)$. So the range of $\displaystyle m$ for which the line lies above $\displaystyle f(x)$ in the first quadrant is $\displaystyle m\le -4$.

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

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

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

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

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

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

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


    Roy
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