Homework Set 13 Solutions

Homework Set 13 Solutions 5.4.2) Consier the functiong(x) = 10 10e.1x, x 0. a.) Show that g(x) is increasing an concave own for x. g (x) = e.1x an g (x) =.1e.1x. Since g (x) is positive for all x (in particular all x 0, g is increasing for x > 0. Since g (x) is negative for all x, g is concave own for x > 0. b.) Explain why g(x) approaches 10 as x gets large. As x gets large,.1x becomes very big an negative, so e.1x goes to 0, so g goes to 10. c.) Sketch the graph of g(x), x 0. g(x) is a function of the from M(1 e kx ), so its graph looks like: 5.4.4) Suppose y = 5(1 e ). Compute y an show y = 10 2y y = 5( e )( ) = 10e. Also, 10 2y = 10 10(1 e ) = 10 10 + 10e = 10e. 5.4.6) Sketch the graph of the function f(t) = 85e.5t + 15, t 0 f(t) is an exponential ecay function shifter up by 15 units, so it looks like:

5.4.12) A news item is broacast by mass meia to a potential auience of 50,000 people. After t ays, f(t) = 50, 000(1 e.3t ) people will have hear the news. a.) How many people will have hear the news in 10 ays? f(10) = 50, 000(1 e 3 ) = 47510.64, which is 47510, rouning own to only count whole people. b.) At what rate is the news spreaing initially? f (t) = 15000e.3t, so f (0) = 15000e 0 = 15, 000 people per ay. c.) When will 22,500 people have hear the news? We solve for t in f(t) = 50, 000(1 e.3t ) = 22, 500. So, 1 e.3t = 22, 500/50, 000 = 9/20. So e.3t = 1 (9/20) = 11/20. Taking the natural log of both sies, we get.3t =ln11/20, so t = 10ln(11/20) 3 = 1.993. Roun this to 2..) Approximately when will the news be spreaing at the rate of 2500 people per ay? We solve f (t) = 15000e.3t = 2500 for t. So, e.3t = 2500/15000 = 1/6, so again taking the natural log of both sies, we get.3t =ln1/6 = ln6. This gives us t = (10ln6)/3 5.97, so roun to 6. e.) Use equations (3) an (4) to etermine the ifferential equation satisfie by f (t). Equations (3) an (4) are f (t) = k[p f(t)] an f(t) = P (1 e kt ), respectively. We have k =.3 an P = 50, 000, so f(t) satisfies the ifferential equation y = 15000.3y. f.) At what rate will the news be spreaing when half of the potential auience has hear the news?

First, we fin the t such that f(t) = 50, 000(1 e.3t ) = 25, 000. This is the t that gives us 1 e.3t =.5, so e.3t = 1.5 =.5. We take the ln of both sies to get.3t =ln.5 = ln2, so t = (10ln2)/3. We plug this t into f (t) to fin the rate at that time. f (10ln2)/3) = 15000e.3 (10ln2)/3 = 15000e ln2 = 15000e ln(1/2) = 15000 (1/2) = 7500 people per ay. Supplementary Exercises: 10.) Two ifferent bacteria colonies are growing near a pool of stagnant water. Suppose that the first colony initially has 1000 bacteria an oubles every 21 minutes. The secon colony has 710,000 bacteria an oubles every 33 minutes. How much time will elapse before the first colony becomes as large as the secon? We set up two equations f(t) an g(t) that measure the number of bacteria in the first an secon colonies, respectively. Since both colonies are growing exponentially, the equations will satisfy f(t) = 1000e λ1t an g(t) = 710, 000e λ2t for some λ 1 an λ 2. We can solve for λ 1 by using 2000 = 1000e λ1 21, so 2 = e λ1 21, an so ln 2 = λ 1 21. This gives us λ 1 = ln 2 21. We can solve for λ 2 using 1420000 = 710000e λ2 33. This gives us 2 = e λ2 33, so using the same steps as above, we get λ 2 = ln 2 33. Now we nee to fin the time when f(t) = g(t), that is, when 1000e (t ln 2)/21 = 710, 000e (t ln 2)/33 ln 2)/21 e(t. This happens when 710 = = e (t ln 2)/33 e (t ln 2)/21 (t ln 2)/33 = e (11t ln 2 7t ln 2)/231 = e (4t ln 2)/231. Taking the ln of both sies, we get ln 710 = t 4 ln 2 231, 231 ln 710 so t = 4 ln 2 546.989 547 minutes. 21.) A company can sell q = 1000p 2 e.02(p+5) calculators at a price of p ollars per calculator. The current price is $200. If the price is ecrease, will the revenue increase or ecrease? The revenue, since we on t know the cost of prouction, is the number of calculators multiplie by the price per calculator, so R(p) = p q(p) = 1000p 3 e.02(p+5). We want to know whether this function is increasing or ecreasing at p = 200. First, we fin R (p) using the prouct rule. R (p) = 1000(3p 2 )e.02(p+5) + 1000p 3 (.02)e.02(p+5) = 1000p 2 e.02(p+5) [3.02p]. At p = 200, we get R (p) = 1000(200) 2 e.02(205) (3.02 200). The first three terms are always positive, since one is a constant, one is a square, an one is e to a power, so the sign of R (200) is etermine by the sign of 3.02 200 = 3 4 = 1, which is negative. So R(p) is ecreasing at p = 200. This means that R ecreases as p increases, so R increases when p ecreases. 24.) The growth of the yellow nutsege wee is escribe by formula f(t) of type (9) in Section 5.4. A typical wee has length 8 centimeters after 9 ays, length 48 centimeters after 25 ays, an reaches length centimeters at maturity. Fin the formula for f(t). M Since f(t) is of type (9), it has the form f(t) =. We know M is the maximum we can reach, so 1+Be Mkt in our case it is the height at maturity, i.e. centimeters. We also know that f(9) = = 8 an 1+Be k9 that f(25) = = 48. We solve the first equation for B. First, we get 1 + Be 495k = /8, 1+Be k25 so Be 495k = /8 1 = 47/8, so B = (47/8) e 495k. We plug this into our secon equation to solve for k. First, we get 48 = =, so 1 + (47/8)e 880k = /48, so 1+(47/8) e 495k e 1375k 1+(47/8)e 880k (47/8)e 880k = /48 1 = 7/48, an so e 880k = (7/48) (8/47). We take the ln of both sies to get ln 7 ln 282 880k = ln [(7/48)(8/47)] = ln 7/282, so k =. We plug this back into our equation for b to solve 880 ln 7 ln 282 495 for B. B = (47/8) e 495k = (47/8) e our equation is f(t) = = 1+47e (.0042)t 1+47e.23t 880 = (47/8)e 9(ln 282 ln 7) 16 46.98 47. Also, k.0042. So, 6.1.10) Determine k 2 x for k a constant. k 2 x = k 2 x + C, since k 2 is also a constant. 6.1.12) Determine x x 2 x x x 2 x = x 3 x = (1/4)x 4 + C 6.1.14) Determine 1 7x x 1 7x x = 1 1 7 x x = 1 7 ln x + C

6.1.16) ( 2 x + 2 x)x ( x 2 + 2 x)x = ( 2 x x + 2 xx = 1/2 x + 1/2 x = 4x 1/2 + C 1 + (4/3)x 3/2 + C 2 = 4 x + (4/3)x 3/2 + C 6.1.18) ( 7 3 x)x 3 ( 7 3 x)x = (7/2) x 3 x x 1/3 x = (7/2)( 1/2)x 2 + C 3 1 (3/4)x 4/3 + C 2 = ( 7/4)x 2 (3/4)x 4/3 + C 6.1.24) 7 2e x 7 2e x = (7/2) e x = (7/2)(1/ 2)e + C = ( 7/4)e + C 6.1.26) Determine the value of k that makes 3e t/10 t = ke t/10 + C true. 3e t/10 t = 3 e t/10 t = 3 10e t/10 + C so k = 30. 6.1.34) Determine the value of k that makes ( 1) 3 x = k( 1) 4 + C true. x k( 1)4 + C = 4k( 1) 3 (2) = 8k( 1) 3. Since this must be ( 1) 3, k = 1/8. 6.1.36) Determine the value of k that makes 5 xkln 2 3x + C = k 3k ( 3) =. Since this equals 5 x = k ln 2 3x + C true., 5 = 3k, so k = ( 5/3). 6.1.42) Fin all functions f(x) such that f (x) = e x, f(0) = 1 Since f (x) = e x, we know f is an antierivative of e x, so f(x) = e x x = x 2 + e x + C. Since f(0) = 1, we get 1 = f(0) = 0 2 +e 0 +C = 1+C, so C = 0. This give us a single function f(x) = x 2 +e x. 6.1.46)Fin all functions f(x) such that f (x) = x 2 + x, f(1) = 3 Since f (x) = x 2 + x, we know f is an antierivative of x 2 + x, so f(x) = x 2 + xx = (1/3)x 3 + (2/3)x 3/2 + C. Using f(1) = 3, we get 3 = f(1) = (1/3)(1) + (2/3)(1) + C, so C = 2. This give us a single function f(x) = (1/3)x 3 + (2/3)x 3/2 + 2 6.1.50) Which of the following is x x + 1x? (a) (2/5)(x + 1) 5/2 (2/3)(x + 1) 3/2 + C (b) (1/2)x 2 (2/3)(x + 1) 3/2 + C x (2/5)(x + 1)5/2 (2/3)(x + 1) 3/2 + C = (x + 1) 3/2 (x + 1) 1/2 = (x + 1) 1/2 (x + 1 1) = x + 1 x, so (a) is the right antierivative. 6.1.56) A rock is roppe from the top of a 400-foot cliff. Its velocity at time t secons is v(t) = 32t feet per secon. (a) Fin s(t), the height of the rock above the groun at time t. We know s(t) = v(t)t = 32tt = 16t 2 + C. We also know that s(0) = 400, so we can solve for C. 400 = s(0) = 16(0) + C = C, so C = 400. Thus, s(t) = 16t 2 + 400. (b) How long will the rock take to reach the goun? We solve for the t that gives us s(t) = 0. s(t) = 16t 2 + 400 = 0, so 400 = 16t 2, so t 2 = 25, an so t = ±5. We take the positive solution t = 5. (c) What will be its velocity when it hits the groun? It hits the groun at time t = 5 an its velocity is given by v(t) = 32t, so when it hits the groun its velocity is v(5) = 32(5) = 160 feet per secon. 6.1.60) A flu epiemic hits a town. Let P (t) be the number of persons sick with the flu at time t, where time is measure in ays from the beginning of the epiemic an P (0) = 100. Suppose that after t ays the flu is spreaing at a rate of P (t) = 120t 3t 2 people per ay. Fin the formula for P (t).

To fin P (t), we take the antierivative of its erivative. So, P (t) = 120t 3t 2 t = 60t 2 t 3 + C. Now we use the initial conition P (0) = 100 to solve for C. 100 = P (0) = 60(0 2 ) (0 3 ) + C, so C = 100. Thus, we get a formula P (t) = 60t 2 t 3 + 100. 6.1.64) Since 1987, the rate of prouction of natural gas in the Unite States has been approximately R(t) quarillion British thermal units per year at time t, with t = 0 corresponing to 1987 an R(t) = 17.04e.016t. Fin a formula for the total U.S. prouction of natural gas from 1987 until time t. Since R(t) is the rate of prouction at time t, it is the erivative of the total prouction P (t). So, we can fin P (t) by antiifferntiating R(t). P (t) = 17.04e.016t t = 17.04(1/.016)e.016t + C = 1065e.016t + C. We use P (0) = 0 since no gas has been prouce when we start, so we can solve for C: P (0) = 0 = 1065e 0 +C = 1065 + C, so C = 1065. Thus, P (t) = 1065e.016t 1065.