Applied Calculus I. Review Solutions. Qaisar Latif. October 25, 2016

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1 Applied Calculus I Review Solutions Qaisar Latif October 25, 2016

2 2 Homework 1 Problem 1. The number of people living with HIV infections increased worldwide approximately exponentially from 2.5 million in 1985 to 37.8 million in (HIV is a virus that causes AIDS). 1. Give a formula for the number of HIV infections, H, (in millions) as a function of years, t, since 1985 and graph the obtained function. 2. What was the yearly continuous percentage change in the number of HIV infections between 1985 and 2003? Solution. 1. The formula for number of HIV infections is given by H(t) = H 0 e rt. Using the information: 2.5 million infections in 1985 we obtain 2.5 = H(0) = H 0 e r 0 = H 0. Using the information: 37.8 million infections in 2003 (2003 corresponds to t = 18), we obtain 37.8 = H(18) = H 0 e r 18 = 2.5e 18r. Rearranging the terms, we obtain Taking natural logarithm on both sides e 18r = = r = ln(15.12). It follows that r = ln(15.12). 18 The HIV function then takes the form H(t) = 2.5e ln(15.12) 18 t. The graph of the obtained function is given below.

3 exp 1 log(15.12) t The yearly percentage increase is given by ) (e r 1) 100% = (e ln(15.12) % 16.29%. Problem 2. A company s cost of producing q liters of a chemical is C(q) dollars; this quatity can be sold for R(q) dollars. Suppose that C(2000) = 5930 and R(2000) = What is the profit at a production level of 2000? 2. If MC(2000) = 2.1 and MR(2000) = 2.5 what is the approximate in profit if q is increased from 2000 to 2001? Should the company increase or decrease production from q = 2000? 3. If MC(2000) = 4.77 and MR(2000) = 4.32, should the company increase or decrease production from q = 2000? Solution. 1. The profit at the production level of 2000 is given by P (2000) = R(2000) C(2000) = = Notice that R(2001) R(2000) + MR(2000) = =

4 4 Moreover C(2001) C(2000) + MC(2000) = = Finally, P (2001) = R(2001) C(2001) = Since the revenue and the profit both are increased, therefore, the company should increase the production from 2000 to Notice that Therefore the profit will be reduced. MR(4.32) MC(4.77) = = 0.54 < 0. Problem 3. The average adult takes about 12 breath per minute. As a patient inhales, the volume of air in the lungs increases. As the patient exhales, the volume of air in the lung decreases. For t in seconds since start of the breathing cycle, the volume of air inhaled or exhaled since t = 0 is given, in hundreds of cubic centimeters, by A(t) = 2 cos( 2π 5 t) How long is one breathing cycle? 2. Find A (1) and explain what it means. Solution. 1. The breathing cycle is precisely the period of the function A(t) which is First notice that A (t) = 4π 5 sin ( 2π 5 t ) It follows that A (1) = 4π 5 sin ( 2π 5 ) > 0.

5 5 Since A (1) > 0, therefore, the inhaling is taking place at a rate of cm 3 /sec = 239 cm 3 /sec. A graph for the given function is given below x 3 (1 - x) Problem 4. With a yearly inflation rate of 5%, prices are given by P (t) = P 0 (1.05) t, Where P 0 is the price in dollars when t = 0 and t is time in years. Suppose that P 0 = 1. How fast (in cents per year) are prices rising when t = 10? Solution. First notice that with the setting P 0 = 1, the function P (t) is of the form P (t) = (1.05) t. To see the rate at which the function is increasing we look at the value of the derivative at a certain point, in this case t = 10. It follows that P (t) = (1.05) t ln(1.05), P (10) = (1.05) 10 ln(1.05) The price is increasing at a rate of 8 cents per year.

6 6 Problem 5. According to the US Census, the world population P, in billions, was approximately P (t) = 6.8e 0.012t where t is in years since January 1, At what rate was the world s population increasing on that date? Give your answer in millions of people every year. Solution. To find the rate of increase of the population, notice first that P (t) = e 0.012t = e 0.012t. The rate of increase in population in 2009 is given by P (0) = e = The population is increasing at a rate of 81.6 millions per year (since the given population was in billions per year). Problem 6. Find and classify the critical points of f(x) = x 3 (1 x) 4 as local maxima and minima. Solution. To find the critical points we compute the first derivative of the given function, which is given by f (x) = x 2 (1 x) 3 (3 7x). Therefore, the critical points are x = 1, x = 3 7, x = 0. To classify the critical points as local maxima or minima, we first compute the second derivative of f which is f (x) = 6( 1 + x) 2 x(1 6x + 7x 2 ). 1. If we choose x sufficiently close to 0, then we see that 1 + x 1 ( 1 + x) x + 7x 2 1 Thus, f (x) < 0, when x < 0 and therefore the function f is concave down on close to left side of 0. Notice also that f (x) > 0 when X > 0 and therefore the function f is concave up. Since the concavity is changing at 0, it is an inflection point.

7 7 2. If you put directly 1 (the critical point x = 1) into the f (x), then we see that f (1) = 0, therefore, the second derivative criterion is not valid here. On the other hand if we choose x very close to 1, then we see that x 1 > 0 1 6x + 7x 2 2 > 0 Thus, with this choice of x, f (x) > 0, therefore, the function is concave up around 1. It then follows that it has local minimum at Notice that f (3/7) = 6 = ( ) ( It follows that f has a local maximum at x = 3/7. The graph of the function f is given below. ( ) ) < x 3 (1 - x) Problem For a, a positive constant, find all critical points of f(x) = x a x. 2. What value of a gives a critical point 5? Does f has a local maxima or local minima at this critical point? Solution. 1. To find the critical points, we consider the first derivative of f, which is, f (x) = 1 a 2 x,

8 8 we consider the equation f (x) = 0 1 a 2 x = 0. It follows that the critical point is given by Notice that Therefore, x = a 2 equivalently x = f ( a 2 f (x) = 4 a 4x 3/2. ) = a2 a 4 a 2 4 = a2 4 aa 2 = a2 4 < 0. ( a 2) 2 = a 2 4. Therefore the critical point is a local maximum for any value of a. The graph of the function f(x) = x 2 x is given below which has local maximum at x = x - 2 x Problem 8. If you have 100 feet of fencing and want to enclose a rectangular area up against a long, straight wall, what is the largest area you can close?

9 9 Solution. One side of the fencing is a wall, so we need to cover only three remaining sides. Let x and y be the lengths of the sides of a rectangular fencing, and let x be the length of the side of the rectangular area which on the side of the wall, then notice that 2x + y = 100 y = 100 2x. The area of the rectangular is given by the following function A = xy = x(100 2x) = 100x 2x 2. Since we want the maximize the area therefore, we look at the critical points of A(x), but first, A (x) = 100 4x. Notice that A (x) = 0 at x = 25. Since the second derivative of A is a negative number, therefore the maximum area is given by x = 25 and Maximum area = A(25) = = 1250 ft 2. y x x Problem 9. For the functions f(x) =e x sin x defined on 0 x 2π g(x) = t 1 + t 2 defined on R, determine the critical points of f and g. Which of these critical points are local extrema? Determine the region where the graph is concave up or concave down. What are the absolute maxima and minima on their domains?

10 10 Solution. 1. Consider the function First we compute For the critical points we consider and obtain f(x) = e x sin(x). f (x) = e x (cos(x) sin(x)). f (x) = 0, e x (cos(x) sin(x)) = 0, cos(x) sin(x) = 0. Therefore the critical points of f between [0, 2π] are Let s also notice that π 4, 7π 4 f (x) = 2e x cos(x), and also that { +ve at x = 7π f(x) = 4 ve at x = π 4 Therefore, f attains local minima at 7π/4, it attains local maxima at π/4. To compute the absolute maximum and minimum, we compute the following four values ( f(π/4) = e π π ) 4 sin ( 4 ) f(7π/4) = e 7π 7π 4 sin = f(0) = e 0 sin (0) = 0 f(2π) = e 2π sin (2π) = 0 Therefore, the absolute maximum is attained at π/4 and it is and the absolute minimum is attained at 7π/4 and it is The graph of the function f is given below exp(-x) sin(x)

11 11 2. Consider the function Notice that Therefore the critical points are Notice also that g(t) = g (t) = g (t) = t 1 + t 2 1 t2 (t 2 + 1) 2. t = 1, 1. 2t(t 3) (1 + t 2 ) 3. There is a local maximum at t = 1 because g (1) = 4 which is negative; and there is a local minimum at t = 1 because g ( 1) = 4 which is positive. Notice that the when t < 0 then both t and t 3 are negative, therefore g (t) is positive, which means the function g is concave up for t < 0. Similarly, for 0 < t < 3, t is positive but t 3 is negative which means the functions g (t) is negative, therefore, g is concave down; and finally for t > 3 both t and t 3 are positive which means g (t) is positive and therefore the function g is concave up. The graph of the function is given below. It is a small exercise to check that the local maximum and minimum are in fact global/absolute maximum and minimum t 1+t Problem 10. The demand for tickets to an amusement park is given by p = q, where p is the price of a ticket in dollars and q is the number of people attending at that price. 1. What price generates an attendance of 3000 people? What is the total revenue at that price? What is the total revenue if the price is $20?

12 12 2. Write the revenue function as a function of attendance, q, at the amusement park. 3. What attendance maximizes revenue? 4. What price should be charged to maximize the revenue? 5. What is the maximum revenue? Can we determine the corresponding profit? Solution. 1. To find out what price generates an attendance of 3000, we plug in q = 3000 in the price function as follows p(3000) = = 10. If the price is $20, the number of participants are given by 20 = q, which provides Therefore, the total revenue is q = The revenue function is given by R(2500) = = R(q) = ( q) q = 70q 0.02q To find out what attendance maximizes the revenue, we find the critical points of the function R(q). First notice that R (q) = q. Therefore the critical point of R(q) is q = Since the second derivative of R is a positive number, therefore, the maximum is attained at the critical point q = 1750, in other words, an attendance of 1750 maximizes the revenue. 4. The price for each ticket that maximizes the revenue is 5. The maximum revenue is p(1750) = = $35. R(1750) = = $ It is not possible to find the profit because the cost is not given.

13 13 Problem 11. At a price of $8 per ticket, a musical theater group can fill every seat in the theater, which has a capacity of For every additional dollar charged, the number of people buying tickets decreases by 75. What ticket price maximizes revenue? Solution. Notice that if t is the additional dollar charged per ticket, then The revenue function is given by p(t) = 8 + t q(t) = t. R(t) = (8 + t)( t) = t 75t 2. For the critical points of R(q), consider first R (t) = t. Therefore, the critical point of R is 6. The second derivative of R is a negative number, therefore a maximum is obtained at 6. The price that maximizes the revenue, is p(6) = = $14. Problem 12. If t is in years since 1990, one model for the population of the world, P, in billions, is 40 P (t) = e. 0.08t 1. What does this model predict for the maximum sustainable population of the world? 2. Graph P against t. 3. According to this model, when will the earth s population reach 20 billion? 39.9 billion? Solution. 1. The maximum sustainable population of the world is 40 billion, which is obtained by taking t. 2. The graph of the function is given below

14 exp(-0.08 t) Let t 0 be the time when the population reaches 20 billion, we must have 20 = P (t 0 ) = e 0.08t 0 A rearrangement of the terms, for finding the value of t 0 provides t 0 = 1 ( ) ln The population, according to the model, reaches 20 billion by Let t 1 be the time when the population reaches 39.9 billion, then we must have 39.9 = P (t 0 ) = e 0.08t 1 A rearrangement of the terms, for finding the value of t 1 provides t 1 = 1 ( ) ln The population, according to the model, reaches 39.9 billion in Problem 13. A rumor spreads among a group of 400 people. The number of people, N(t), who have heard the rumor by time t in hours since the rumor started to spread can be approximated by a function of the form 1. Find N(0) and interpret it. N(t) = e 0.4t. 2. How many people will have heard the rumor after 2 hours? After 10 hours?

15 15 3. Graph N(t). 4. Approximate how long it will take until half the people have heard the rumor? Virtually everyone? 5. Approximate when is the rumor spreading fastest? Solution. 1. Notice that 400 N(0) = = e The rumor spreads with one person. 2. For the number of people who have heard the rumor by 2 hours are N(2) = e Three people have heard the rumor by 2 hours. For the number of people who have heard the rumor by 2 hours are N(10) = e It follows that 49 people have heard the rumor by 10 hours. 3. The graph of N(t) is given below exp(-0.4 t) Let t 0 be the time until half of the people (200) have heard the rumor, then 200 = N(t 0 ) = e 0.4 t 0. A rearrangement of terms solving for t 0 provides t 0 = 1 ( ) ln

16 16 Therefore, half of the people will have heard the rumor in 6 days, 5 hours, 43 minutes. Let t 1 be the time until virtually everyone (399.1) have heard the rumor, then = N(t 1 ) = e 0.4 t 0. A rearrangement of terms solving for t 0 provides t 0 = 1 ( ( )) ln Therefore, virtually everyone will have heard the rumor in 12 days, 14 hours, 5 minutes.