Math 161

Sample Third Midterm

Problem 1

Dr. Wilson

1. An object's velocity is measured as a function of time.

The graph will look like

The distance the object falls is the area under the graph over the interval (0, 5).

We will look at three ways to find the distance the object falls.

a) The average of the upper and lower Riemann Sums

We divide the domain into intervals, and find the maximm and the minimum values of the velocity over those intervals

The upper Riemann sum is the area of the boxes which are striped in green, and the lower Riemann sum is the area of the boxes which are striped in red. Since the boxes for the lower sum is contained in the boxes for the upper sum, there is some overlap between the green and red.

The area of the green boxes is





+

32
64
96
128
160
480

The area of the red boxes is





+

0
32
64
96
128
320

The average is

Since the velocity function, in this case, is linear, it doesn't matter how wide we choose our intervals. The exact answer will always be halfway between the upper and lower Riemann sums. In more general cases one would want to take narrower intervals so that both the upper and lower Riemann sums, and, hence, their average, would be closer to the actual real area than if one were to take wider intervals. Since the width of the intervals does not matter with this linear function, we have taken the width to be one so that the are of each box is equal to its height.

The area of the triangle

Since the area of a triangle is

we get

We will find a formula for the linear function and integrate it

It is easy to verify that the follwing function yields all of the correct results

v(t) = 32t

This gives us the following differential equation

Separate the variables

Integrate both sides

To find the value of the constant of integration, C, let t = 0. Then s = 0 and we get

0 = 16(0)2 + C

so

C = 0, and the fromula for the distance is

s = 16t2

When t = 5, we get

s = 16(5)2 = 16(25) = 400

and all three methods give us the same answer.