*Its All about Pj Problem Strings -
7 Spaces Of Interest and their associated Basic Sequences; 7 Pj Problems of Interest (PPI) and their Alleles (A)*

1. Define the following:

(a) *Change*

(b) *Rate of change*

(c) *Change* can be categorize broadly into two groups. Indicate the two groups.

(d) Give examples of *change* in each of the two groups you indicated in problem 1(c).
**Ans:**1(a) Consider matter *m* in a state of existence x_{1}. If x_{1} is caused to become x_{2}, another state of existence of *m*, then *m* is said to have *changed*. The *force* (doer) that caused the *change* is a *change agent*. The transitioning from x_{1} to x_{2} is *change in progress* that always involves *time* (the time it took for x_{1} to become x_{2}). This time can be *instantaneous* or *spread-out*.

(b)*Rate of change* is the measured *change* of a dependent variable per unit *change* of an independent variable. There has to be a relation between the dependent variable and the independent variable. In the context of single variable derivatives, the relation is the functional relation f(x). Often the independent variable is time. However, it can be any variable so long as a functional relation has been established between it and the dependent variable.

(c) Physical Change and Chemical Change

(d) The chemical properties of *matter* involved in a physical change remain the same. Two examples of physical changes are the melting of solid water (ice) to liquid water and the change in position of a moving vehicle.

The chemical property of *matter* involved in a chemical change does not remain the same (it changes). For example, the gases hydrogen and oxygen combine chemically to becom liquid water. The digestion of food is another example of chemical change.

2. Given that f(x) = x^{2}. Calculate:

(a) The average rate of change of f(x) over the interval 2 to 2.01

(b) The instantaneous rate of change of f(x) when x = 2.
**Ans:**2(a) The average rate of change is given by the difference quotient:

[f(x + Δx) - f(x)]/Δx = Δy/Δx.
So, Δy/Δx = [(2.01)^{2} - 2^{2}](.01) = 4.0401-4/.01 = .0401/.01 = 4.01

So average rate of change is 4.01.

The instantaneous rate of change of f(x) when x = 2 = f'(2) = 4.

3. Suppose a substance *x* is used in a system when injected into the system. Let A (the amount of *x* remaining in the system t days after its been injected into the system) be a function of time as follows:

A = t^{-3/2} (where t is in days and ≥ 0.5 and A is measured in a suitable unit).

(a) Calculate the derivative of A when t = 1

(b) Interprete the result in 3(a) as a rate of change of A.
**Ans:**3(a) Derivative of A = A'(t) = (-3/2)t^{-5/2}

So, Derivative of A at t = 1 is A'(1) = (-3/2)(1)^{-5/2} = -3/2.

(b) The amount of x in the system is changing at the rate of -3/2 units per day when t = 1. The negative sign indicates that the amount of x is decreasing. So the interpretation can be that the system is utilizing 3/2 units of x per day when t=1.

4. A company's profit P from the sales of x units of its product is given as follows:

P(x) = 0.003x^{3} + 0.01x dollars. Calculate:

(a) The additional profit gained from increasing sales from 100 to 101 units

(b) The marginal profit at a production level of 100 units.
**Ans:**4(a) P(101) - P(100) = 3090.91 - 30001.1 = 91.81 dollars.

(b) Marginal profit at 100 units = P'(100) = 0.003(3)(100)^{2} + 0.01 = 90.01 dollars.

The marginal profit at 100 units is the profit gained by producing 1 extra unit of product when production level is at 100. If the Cost function C(x) or Revenue function R(x) is known, maginal cost at x =a or maginal revenue at x =a is similarly interpreted.

5. Suppose rain causes a river to overflow its banks and w(t) is the amount of water on main street t hours after the rain starts.

(a) If w'(100) = 1/4, by approximately how much will the water level change during the next hal-hour?

(b) Which of the following two conditions are the best news?

(i) w(100) = 2.5, w'(100) = 1.5, w''(t) = -4

(ii) w(100) = 2.5, w'(100) = -1.5, w''(t) = 4
**Ans:**5(a) The interpretation of w'(100) = 1/4 is that after 100 hours, the rain causes the water level on main street to rise 1/4 inch every hour. So during the next half-hour, the water level will rise approximately 1/8 inch.
(b) (ii) is the best news because it indicates that the water level is decreasing.

6. A chemical reaction converts the reactants A and B to product C. [C], the concentration of C as a function of time is given as:

[C] = (a^{2}kt)/(akt + 1) moles/liter.

(a) What is [C] as t → ∞?

(b) If the rate of reaction = d[C]/dt = (a^{2}k)/(akt + 1)^{2}. What information does this derivative representing the rate of reaction give as t → ∞?
**Ans:**6(a) Limit_{t → ∞} [C] = Limit_{t → ∞} (a^{2}kt)/(akt + 1) = (a^{2}k)/(ak + 1/t) = (a^{2}k)/(ak) = a moles/L.

(b) As t → ∞, d[C]/dt = (a^{2}k)/(akt + 1)^{2} → 0. So as t increases, nearly all of the reactants A and B are converted into product C. In practical terms, the reaction ends.

7. The time-dependent flow of charge through a cross-sectional area A, of a conductor is given as follows:

Q(t) = t^{3} - 2t^{2} + 6t + 2 (where t is in seconds).
(a) What is the current passing through A at t = 1 sec?
(b) At what time is the current passing through a, the lowest?
**Ans:**7(a) Current passing through A = Q'(t) = 3t^{2} - 4t + 6.

So, Q'(1) = 3(1) - 4 + 6 = 5A (Amperes).

(b) The current is lowest when Q' has a minimum. Use the second derivative test to determine the minimum.

So, Q"(t) = 6t - 4 < 0 when t < 2/3; and > 0 when t > 2/3.

So the current decreases when t < 2/3 and increases when t > 2/3.

Therefore, the current is lowest at 2/3 secs.

8. A viral epidemic hits a metropolis. Public Health officials have determined that the number of persons sick with the virus at time t days from the beiginning of the epidemic can be approximated by the following function:

P(t) = 60t^{2} - t^{3} (0 < t < 40).

(a) At what rate is the epidemic spreading when t = 20?

(b) When is the epidemic spreading at the rate of 900 people per day?

(c) Speculate on your answer in (b).
**Ans:**8(a) P'(t) = the rate at which the epidemic spreads.

So, P'(t) = 120t - 3t^{2} (0 < t < 40).

So, P'(20) = 120(20) - 3(20)^{2} = 1200.

20 days after the start of the epidemic, it is spreading at 1200 people per day.

(b) P'(t) = 900 = 120t - 3t^{2}

So, - 3t^{2} So, (t - 10)(t - 30) = 0. Thus t = 10 or t = 30.

(c) If at t = 30 the spread is 900 people per day and at t = 20 it is 1200 people per day, Then the spread is decreasing perhaps by the actions of Public Health officials, or the virus has run out of steam or both.

9. The distance traveled (in meters) by a particle after t secs is given by the following function:

s(t) = t^{3} - 12t^{2} + 36t (t in seconds).

(a) What does the velocity of the particle at t = 3 say about the direction of the motion of the particle?

(b) When is the particle at rest?

(c) At what times is the particle moving forward?

(d) What is the total distance traveled by the particle after 8 seconds?

(e) What is the acceleration of the particle at t = 3?

(f) When is the particle speeding up; when is the particle slowing down?
**Ans:** 9(a) Velocity of particle = v(t) = s'(t) = 3t^{2} - 24t + 36

v(3) = s'(3) = 27 - 72 + 36 = -9 m/sec. particle is moving in the opposite direction. If forward movement is the postive direction, then particle is moving backward.

Particle is at rest when v(t) = 0. that is 3t^{2} - 24t + 36 = 0

So, 3(t - 2)(t - 6) = 0; implies t = 2 or t = 6.

(c) Forward movement is when v(t) > 0. That is 3t^{2} - 24t + 36 > 0.

So, 3(t - 2)(t - 6) > 0; implies 0 ≤ t < 2 or t > 6

(d) Distance traveled during forward movement forward = [s(2) - s(0)] + [s(8) - s(6)] = 32-0 + 32- 0 = 64 meters.

Distance traveled during backward movement = [s(6) - s(2)] = 0 - 32 = -32 (negative sign indicate movement backward)

So, total distance traveled after 8 seconds = 64 + 32 = 96 meters.

(e) Acceleration = a(t) = v'(t) = 6t - 24. So, a(3) = 18 - 24 = -6 m/s^{2}.

(f) Particle is speeding up when v(t) and a(t) have the same sign. This occurs when 2 < t < 4.

Particle is slowing down when v(t) and a(t) have opposite sign. This occurs when 0 ≤ t < 2 and 4 < t < 6

10. The parametric equations of a projectile are: x = 20t and y = -16t^{2} + 30t. Determine the range of the projectile.
**Ans:**The range of a projectile is the horizontal ground distance the projectile covered when it lands on the ground. At this time the vertical distance y is zero.

So setting y = 0, we have 16t = 30. From which we have t = 15/8. This t is the time that elapsed before the projectile reached the ground, that is, the time it took to cover the range.

So range of projectile = x = 20 x 15/8 = 75/2 = 37.5 ft.

11(a). What is the speed of a pebble 4 secs after it is dropped?

(b) What is the average speed during the 4 secs?

(c) At what instant does the pebble actually posssess this average speed?
**Ans:**11(a) The speed v, of a free falling object on earth where the acceleration due to gravity is constant at 32ft/sec^{2}
= 32t ft/sec (where t is time in secs)

Distance traveled by the object, d = 16t^{2}.

So, at t = 4 secs, v = 32 x 4 = 128ft/sec

(b) Averaage speed = Distance traveled/time.

At t = 1 secs, d = 16ft;
at t = 2 secs, d = 64ft;
at t = 3 secs, d = 144ft;
at t = 4 secs, d = 256ft.

So, Total distance traveled = 256ft
and average speed = 256/4 = 64ft/sec.

(c)
Average speed occured at t = 2 secs.
In general average speed occurs in t/2 secs.