If , and , then the curves have an intersection in the first quadrant where .
We have , so the equation of the line tangent to the curve at is ,
or . This is , or .
The curve intersects the -axis at , and the line crosses the -axis at . Therefore the area of the region is
The curve crosses the -axis at and at . Thus, the volume generated when the region is revolved about the -axis is
(Note: I mistakenly found the area generated by the region instead of the region in an earlier version of these solutions. Thanks to Heather Kelly for alerting me to the mistake.)
The circle of radius 2 has equation in the first quadrant and extends over the interval there. The circle of radius 1 has equation in the first quadrant and extends over the interval there. Using the fact that the two semi-circles intersect at , the required expression is therefore .
The circle of radius 2 has equation in the first quadrant and extends over the interval there. The first-quadrant portion of the circle of radius 1 that lies inside the larger circle has equation and extends over the interval . The required expression is therefore .
Note: A phrase seems to have been left out of the distributed version of this question. I assume that the blank space between the symbol " " and the symbol " " should contain the phrase "that represents the area of", or some phrase of similar meaning.
The segment connecting the origin to the first-quadrant intersection point of the two circles, given as lies in the ray . The area of the sector of the large circle that lies below this segment but above the positive -axis is a part of and has area . The remainder of the region after this sector is removed has bounding curve , where . Hence, its area is . The required expression is therefore .
Average radius is .
The required midpoint Riemann sum is
The integral gives the volume, in cubic millimeters, of the segment of the blood vessel that extends from mm to mm.
We have . Consequently, by Rolle's Theorem, there is a number where . We also have , so, once again by Rolle's Theorem, there is a number where . But then , so, yet again by Rolle's Theorem, there must be a number between and , such that . Noting that , we see that .
The velocity vector is . At , this is , and this vector has length , which is speed at .
The tangent line is horizontal when . By part b, above, the tangent line can be horizontal only when . Because both exponentials are always positive, there are no horizontal tangent lines.
The tangent line is vertical precisely when the tangent vector is vertical. By part a, the tangent line can be vertical only when . This is equivalent to , or . (It should be noted that we saw in part c., above, that never takes on the value zero; if were zero, this would complicate our analysis.)
is the area of a rectangle of base 1 and height 2 plus the area of a triangle of base 1 and height 2, or . By the Fundamental Theorem of Calculus, , and when , this is . In the interval , the FTC tells us that . Thus in that interval, and .
The average rate of change of on is . (See part a for the calculation of ; for , we have , which is the negative of the area of a triangle of base 2 and height 4, or )
By the Fundamental Theorem of Calculus, . Thus, on the interval , takes on its average value, at two different points: , and
An inflection point occurs where the monotonicity of the derivative changes from increasing to decreasing or vice versa. There are two such points for , (We know by the Fundamental Theorem of Calculus.) They are and .
The Taylor series about for has general term , where . Through the first four terms, we have . The general term is .
We apply the Ratio Test: . This limit is less than when , so the radius of convergence for this series is .
We have , and we may integrate the series term-by-term (as long as ). Thus, . The general term is .
When , the general term of the series for becomes . As , this does not converge to zero, so the series of which it is the -th term cannot converge.
Converted by Mathematica (May 14, 2003)