MATH165 Homework 7. Solutions March 23. 2010 Problem 1. 3.7.9 The stone is thrown with speed of 10m/s, and height is given by h = 10t 0.83t 2. (a) To find the velocity after 3 seconds, we first need to find the expression for velocity, v, which is given by v(t) = dh = 10 1.66t. Hence velocity at t = 3s, is given by v(3) = 10 (1.66 3) = 5.02m/sec. (b) When h = 25m, we have 25 = 10t 0.83t 2. 0.83t 2 10t + 25 = 0. Solving for t we have that t = 3.54or 8.51. The value t 1 = 3.54 corresponds to the time stone takes to rise 25m on the way up, and t 2 = 8.51, corresponds to time when the stone is 25m high on the way down. So, velocity of the stone when the stone has risen 25m is given by v(t 1 ) = 10 (1.66 3.54) = 17 = 4.12m/sec. Problem 2. 3.7.15 Surface area of the balloon is given by S = 4πr 2. To find the rate of increase of surface area w.r.t radius, we need to find ds. Differentiating w.r.t r we have that dr ds = 8πr. dr When r = 1ft, we have ds = 8π1 = dr 8πft2 /ft. Similarly, when r = 2ft, we have ds = dr 16πft2 /ft. And when r = 3ft, ds = dr 24πft2 /ft. We can observe that as radius is increasing, rate of increase of surface area is also increasing in a linear fashion. 1
Problem 3. 3.7.23 Suppose the bacteria population triples every hour. Let the initial population be n 0. So, we have f(1) = 3n 0, f(2) = 3 2 n 0, and so on. Hence we end up with general formula given by n(t) = 3 t n 0. Rate of growth is found by differentiating the above equation. And dn(t) = 3 t ln3 n 0. Now initial population is given by n 0 = 400. Hence rate of growth is given by 3 t ln3 400bacteria/ hr. Now put t = 2.5hrs to get n (2.5) = 3 2.5 ln3 400 = 6850bacteria/hr. Problem 4. 3.7.29 Cost of prodction as function of x is given by C(x) = 1200+12x 0.1x 2 + 0.0005x 3. So marginal cost function is given by C (x) = 12 0.2x+0.0015x 2 dollars per yard. Now when x = 200, we have C (200) = 12 0.2(200) + 0.0015(200) 2 = 32dollars/yard. And this the rate at which production cost is increasing when x = 200. It predicts the cost of production of 201st yard of fabric. Now cost of production of 201st yard is given by C(201) C(200) = 3632.20 3600 = 32.20 dollars, which is approximately equal to C (200). Problem 5. 3.8.9 Half life is 30 yrs.(i.e. 100 mg decays to 50 mg in 30 yrs.).initial mass (m 0 ) is 100mg. Now, we know that mass at any time t is given by m(t) = m(0)e kt. 50 = 100e k30. Solving for k we obtain that ln0.5 = 30k. Hence k = ln0.5 30 = 0.023 The mass remaining after t = 100yrs is given by m(100) = 100e 0.023 100 = 100e 2.3 10mg. 2
To find the amount of time after which 1mg will remain can be solved by 1 = 100e 0.023t. On solving for t we have that t 200yrs. Problem 6. 3.8.13 Temperature of the surroundings, T s = 75 F. And initial temperature of turkey is 185 F. Using newton s law of cooling we have dt = k(t 75) where T (t) is the temperature of the turkey after t minutes. Suppose y = T 75, then y(0) = 185 75 = 110 F. Hence we have the form given by dy = ky, and y(0) = 110 F.Hence the solution is given by y(t) = y(0)e kt = 110e kt. Now we know that T (30) = 150 F, so y(30) = 150 75 = 75 F. Hence we end up with 75 = 110e k30. Solving for k we obtain k = 0.012. Hence y(45) = 110e 0.012 45 = 62.10. Now T (45) = 62.10 + 75 = 137.10 F. Now when T = 100 F, we have that y = 25. So, we have that 25 = 110e 0.012t. Solving ln 0.227 for t we obtain that t = 116min. Hence turkey cools down to 0.012 100 F in 116 mins. Problem 7. 3.8.19 a.)if 3000 dollars is invested at 5 percent, then the value of investement is given by A 0 (1 + r n )nt where A 0 is the initial amount(here it is 3000), r is the rate of interest,n is the number of compounding periods in a year, nt is number of compounding periods in t years. If interest is compounded annually, we have A = 3000(1 + 5 100 )5 = 3000(1.05) 5 = 3828.24 dollars. 3
Now if interest is componded semiannually, we have that A = 3000(1 +.05 2 )10 = 3840.25 dollars. If interest is compounded monthly, then A = 3000(1 +.05 12 )60 = 3850.08 dollars. If interest is compounded weekly, then A = 3000(1 +.05 48 )5 48 = 3851.61 dollars. Now if interest is compounded daily, then A = 3000(1 +.05 365 )5 365 = 3852.01 dollars. In case of continuos compounding, A(5) = 3000e 0.05 5 = 3852.08 dollars. b.) Now da = ra 0 e rt = 3000 0.05 e.05t = 0.05 A(t), where A(t) = 3000 e 0.05t. Problem 8. 3.9.21 Refer to the figure 1. for more explanation. We know that dx = 35km/h, dy = 25km/h. And we know that z 2 = (x + y) 2 + 100 2, using Pythogorean theorem. Differentiating the equation we get that2z dz dx = 2(x + y)( ). Now at 4 : 00p.m. we know that x = 4(35) = 140km, y = 4(25) = 100km. Hence at 4p.m z = (140 + 100) 2 + 100 2 = 260km. + dy So dz = x+y z ( dx + dy ) = 140+100 260 (35 + 25) = 720 13 55.4km/h. Problem 9. 3.9.25 Refer to figure 2 for more explanation. The volume of the trapezoid is given by 1 2 (b 1 + b 2 )h l. where b 1 and b 2 are the parallel sides, 4
Figure 1: Figure for Problem 21 and h is the height, l is the length of the tank. Using similar triangles we have that h = a a = h. 0.5 0.25 2 Volume of the trapezoid of height h is given by (10) 1 [0.3+(0.3+2a)]h, after simplication 2 and getting rid of a, we have V = 3h + 5h 2. So we have dv = dv dh dh 0.2 = (3 + 10h) dh = 0.2. dh 3+10h When h = 0.3, we have that dh = 0.2 = 10cm/min. 3+10(0.3) 3 Problem 10. 3.9.33 When R 1 = 80Ω, R 2 = 100Ω, we have R = R 1 R 2 R 1 +R 2 = 100 80 180 = 400 9. Differentiating w.r.t time t we have that dr = R 2 ( 1 dr 1 + 1 dr 2 ). R1 2 R2 2 When R 1 = 80Ω, R 2 = 100Ω, dr = ( 400 9 )2 [ 1 80 2 (0.3) + 1 100 2 (0.2)] = 107 810 = 0.13Ω/second. 5
Figure 2: Figure for Problem 25 6