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 amount pumped into the tank during is given by . Integrating numerically, we obtain
The rate of change of the volume of oil in the tank at time is . At this is
The level of oil in the tank is decreasing when because the volume of oil in the tank has negative rate of change then.
We are given that there were gallons of oil in the tank when . Thus, the volume in the tank at time is, by the Fundamental Theorem of Calculus, . Integrating numerically with , we obtain
Here is a plot of the rate at which the volume of oil in the tank changes.
This rate is positive, and the volume increases, from until about . The rate is negative, and the volume decreases, from about until a little after Thereafter, volume increases. This means that there is a minimum somewhere near Solving numerically, we locate the relevant critical point:
The amount of oil in the tank is minimal when .
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 .
If , then acceleration is . When , we have .
Speed is , and so . When , this is ), which is positive. Hence is increasing when , because its derivative is positive there.
The particle changes direction where derivative of position with respect to time, which is velocity, changes sign. This happens only when , or when .
The total distance traveled (as opposed to the total displacement) by the particle over the interval is . Taking into account the sign change at , this is . This reduces to .
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 .
If , then . Thus,
We have , so, . Substituting in the integral on the left, we transform it to . Thus , or . So .
Converted by Mathematica (May 14, 2003)