Help with AB A.P. Exam Problems?

**mak15** · March 19th 2008, 05:47 PM

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!

**roy_zhang** · March 20th 2008, 11:59 AM

(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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