Louis A. Talman
Department of Mathematical & Computer Sciences
Metropolitan State College of Denver
The area of the given region is .
We must first solve the equation to find the limits of integration:
Although the solutions appear to be complex, the imaginary parts are very tiny and result from numerical errors. We need the roots that lie in the interval , so we will ignore the third solution.
The area of that part of the region which lies below the horizontal line is given by .
The area of a cross-section of the solid perpendicular to the -axis at is . Thus, the volue of the solid is :
Numerically, this is
Under the conditions given, the pool is a region in 3-space whose base is and whose cross-section perpendicular to the -axis at has area . The volume is thus :
At 5:30pm, the rate at which the number of people standing in line was changing was approximately people per hour.
The average number of people standing in line during the first four hours that tickets were on sale was . From the data in the table and the trapezoid rule, that is approximately
By the Mean Value Theorem, there must be a point such that and there must be a point such that . Now is defined for all , and this means that must be a continuous function on . By the Intermediate Value Theorem for Continuous Functions, there must be a point where . By similar reasoning, there must be where , and so where . Further, there must be for which , and therefore where .
We conclude that takes on the value at least times in the interval .
Note: It actually suffices to know that is differentiable throughout to make this argument: Derivatives have the Intermediate Value Property even when they are not continuous.
If denotes the number of tickets sold at time , we are given and . Consequently,
To the nearest whole number, 973 tickets have been sold by 3:00pm.
We write , and find that . We are given that cc/min for all , and that when we have cm, cm, and cm/min. Thus , so that
At the given instant, cm/min.
Taking to be the moment when the recovery device goes into action, we have . Thus, has a critical point at , when . Because for , while when , it follows from the First Derivative Test that is maximal when .
If there were 60,000 cc of oil in the slick at the moment , when the recovery device began to operate, then by the Fundamental Theorem of Calculus we must have . From what we saw in part b) above, we must therefore have
By the Fundamental Theorem of Calculus and what we are given, we must have . This means that , , and . From the figure and the information given, we have for and for , while for . Thus is decreasing when and when , while is increasing when . Thus, the particle is farthest to the left when and its position at that instant is .
Because and , (see part a)), the particle moves through once (leftward bound) when . Because , and (see part a)), it moves through again (rightward bound) at some time in the interval . Because and (see part a) again) it moves through still again (now leftward bound) at some time in the interval . The existence of these times is guaranteed, in each case, because the differentiable function must be continuous on , and continuous functions have the intermediate value property. That these three instances are the only instances is guaranteed by the fact that must be monotonic on each of the intervals , , and . We conclude that the particle passes through just three times.
Speed is the magnitude of velocity. On the interval , velocity is—from the graph—increasing, but negative. On this interval, we obtain the magnitude of velocity by reflecting the relevant portion of the curve about the -axis. So speed is decreasing on .
Acceleration is . Thus, acceleration is negative on intervals where is decreasing. From the graph and what we have been given about it, acceleration is negative on and on .
We have , with . Hence . Integrating, we find that . But is given, so . Thus, , or .
, so an equation of the tangent line at the point which corresponds to is .
when , or when , so the -coordinate of the critical point of is . We note that when , but that when , and it follow that has a local maximum at .
From , we have . We note that when but that when . The function therefore has an inflection point at .
Created by Mathematica (March 24, 2010)