Mathematical Modeling I Test 1 with Instructor's Notes


Mathematical Modeling 2001

Test #1.

1. Suppose that beginning on your birthday in the year 2000, you receive money at the rate of $10 / day, which you immediately invest into an account paying annual 8% interest starting from the day you receive it. After earning interest at rate r for t years, an initial amount P of money will grow to value P e ^ ( r t ). Note that an 8% rate is 8 per hundred, or .08.

2. Suppose you are driving at 60 mph down the highway and realize that you have to stop and take 10 seconds to secure part of your baggage in the back seat. You brake to a stop, changing velocity at a constant rate and requiring 10 seconds to reach a complete stop. You take 10 seconds to secure the load, then accelerated uniformly back to 60 mph, requiring another 20 seconds to do so. How much time did you just lose? ** For the slowdown and the speedup velocity changes at a uniform rate so your average velocity is half the 60 mph. At half the velocity you only go half as far, so half the time required to slow down then to speed up is wasted. You lose half of the 30 seconds spent slowing down and speeding up, plus the 10 seconds you were stopped. **

3. A tree has a number of bugs on it equal to the number of days until your next birthday.It is in the middle of a long line of bugless trees. Every hour there is a 20% transition of bugs from each tree to each of its neighbors.

4. Using Pascal's Triangle determine the probabilities of ending up 10, 8, 6, 4, 2 and 0 steps away from your original position on a random walk of 10 steps. Explain what the whole analysis has to do with coin flips, and what Pascal's Triangle has to do with coin flips. ** n=10 row is 1 10 45 120 210 252 210 120 45 10 1. This means that there is 1 way to take 0 ‘rights’ resulting in distance 10, 10 ways to take 1 ‘right’ resulting in distance 8, 45 ways to take 2 ‘rights’ resulting in distance 6, etc.. Following this reasoning there are 2 ways to end up 10 units away, 20 ways to end up 8 units away, 90 ways to end up 6 units away, 240 ways to end up 4 units away, 420 ways to end up 2 units away and 252 ways to end up where you started. The probabilities are therefore 2/1024, 20/1024, 90/1024, etc..

5. Set up a transition matrix for a sane-demented model in which each transition sees 13% of the Sane become Demented and 19% of the Demented become Sane.

6. The plants in a certain garden are arranged in four concentric circles, i.e., circles with a common center. Plants are spaced at 1-foot intervals, and the radius of each circle is 10 feet greater than the next one inside it. The first circle has a 10-foot radius. In any given time interval, 20% of the bugs in one circle will move to the next circle out from the center, and 20% to the next circle in toward the center. Note that this refers to the total number of bugs in each circle, not on each plant. Each plant initially has 50 bugs.

The bugs on the innermost circle have no circle inside them so they don't move inward. 20% of the bugs on the outer circle get lost.

 

Diameters 10, 20, 30, 40 ft imply circumferences 31.4, 62.8, 94.2, 125.6 ft. At 1 tree per foot and 50 bugs per tree we have 1670, 3340, 5010, 6580 bugs in the four circles.

After 1 transition we have 670 bugs moving each way from circle #2, 1000 each way from circle #3, 1320 each way from circle #4; also 330 out from circle #1. So we end up with 2000 bugs in center circle, 3000 in second circle, 5010 in third circle, 5900 on fourth circle (check numbers).

This is 62 bugs per tree on 1st circle, 50 per tree on 2nd circle, 52 per tree on 3rd circle, 46 per tree on 4th circle (check numbers).

7. A container has total height 100 cm, and is in the shape of a cone. The diameter of the cone is at every point 1/10 the height of that point above the apex of the cone, which rests on a table. There is a hole .3 cm in diameter at a point 10 cm above the apex, and water exits freely from the hole. The cone is initially full to the brim. If y represents the depth of water above the hole, then the exit velocity of the water from the hole is v = `sqrt( 2 * 980 * y), where y is in cm and v is in cm / sec.

1 sec of water stream is cylinder with c.s. area .07 cm^2, length 420 cm, so volume is about 30 cm^3.

At 90 cm above hole, we’re 100 cm above apex so diameter of cylinder is 1/10 * 100 cm = 10 cm and radius is 5 cm. So c.s. area of the slightly beveled cylinder we lose in 1 sec is about `pi r^2 = 78 cm^2. If water descends by distance `dy, then volume lost is about 78 cm^2 * `dy. This must be 30 cm^3, so

78 cm^2 * `dy = 30 cm^3, and we easily find `dy = 30 cm^3 / 78 cm^2 = .39 cm. Thus water descends at .39 cm/sec. **

8. A random walker starts from the origin of an x-y coordinate system and takes three random steps in the x direction, then three random steps in the y direction. What distances from the original point are possible, and what is the probability of each? Note: don't forget about the Pythagorean Theorem.

 

** To get to (1,1) need to get x=1 then y=1. Each has probability 3/8 so prob of (1,1) is 3/8 * 3/8 = 9/64. Dist. Is `sqrt(1^2+1^2)=`sqrt(2). Same true for (1,-1), (-1,1), (-1,-1). Thus prob of dist `sqrt(2) is 4 * 9/64 = 9/16.

To get to (3,1) need to get x=3 then y=1. Probability 1/8 then 3/8 so prob of (3,1) is 1/8 * 3/8 = 3/64. Dist. Is `sqrt(1^2+3^2)=`sqrt(10). Same true for (3,-1), (-1,3), (-3,-1) etc.—8 points in all. Thus prob of dist `sqrt(10) is 8 * 3/64 = 3/8.

To get to (3,3) need to get x=3 then y=3. Each has probability 1/8 so prob of (1,1) is 1/8 * 1/8 = 1/64. Dist. Is `sqrt(3^2+3^2)=`sqrt(18). Same true for (3,-3), (-3,3), (-3,-3). Thus prob of dist `sqrt(18) is 4 * 1/64 = 1/16.

Total prob. Is 9/16 + 3/8 + 1/16 = 1. **

 

 

9. A mass is suspended by a rubber-band system, and when it is x cm from its equilibrium position its velocity changes a rate v ' = dv / dt = -40 cm / sec^2 * `sqrt( | x | ).

 

** note shoulda been –40 cm/sec^2 * `sqrt |x| ) * ( x / | x| ). **

The mass is initially at rest at a distance of 4 cm from equilibrium.

 

** If x = 4 then v’ = -40 cm/s * `sqrt(4) = -80 cm/s^2.

In .1 sec, v changes by v’ `dt = -80 cm/s^2 * .1 sec = -8 cm/s.

Now v is –8 cm/s. For first .1 sec v averages (0+-8 cm/s) / 2 = -4 cm/s.

Displacement during first .1 sec is thus –4 cm/s * .1 sec = -.4 cm.

Now position is 4 cm + -.4 cm = 3.6 cm, velocity is –8 cm/s. **

 

** rate of vel change is v ‘ = - 40 cm/s^2 * `sqrt(3.6 cm) = -76 cm/s^2.

Change in vel is –76 cm/s^2 * .1 s = -7.6 cm/s.

New velocity is thus –8 cm/s + -7.6 cm/s = -15.6 cm/s.

Ave. vel. for this .1 sec is (-8 cm/s + -15.6 cm/s) = -11.8 cm/s.

Displacement during this .1 sec is –11.8 cm/s * .1 sec = -1.18 cm.

New position is thus 3.6 cm + -1.18 cm = 2.42 cm, moving at –15.6 cm/s. **

Repeat this process until the mass has reached its maximum displacement on the other side of the equilibrium position.

10. Use the program Kinmodel in the simulations folder of the comm folder to obtain 20 randomly chosen x kinetic energies. Find the mean and standard deviation, and determine the percents that lie within 1 standard deviation, and within 2 standard deviations, of the mean. How well do your percents agree with the prediction of a normal-curve model?

 

** mean is in the 400’s, std dev 60 or so, about 60%-70% within 1 std dev, almost all within 2 std dev. **