MATH165 Homework 7. Solutions

Transcription

1 MATH165 Homework 7. Solutions March Problem 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 = t. 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 t 2 10t + 25 = 0. Solving for t we have that t = 3.54or 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 ( ) = 17 = 4.12m/sec. Problem 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

2 Problem 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) = ln3 400 = 6850bacteria/hr. Problem Cost of prodction as function of x is given by C(x) = x 0.1x x 3. So marginal cost function is given by C (x) = x x 2 dollars per yard. Now when x = 200, we have C (200) = (200) (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) = = dollars, which is approximately equal to C (200). Problem 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 = ln = The mass remaining after t = 100yrs is given by m(100) = 100e = 100e mg. 2

3 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 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) = = 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) = = 75 F. Hence we end up with 75 = 110e k30. Solving for k we obtain k = Hence y(45) = 110e = Now T (45) = = F. Now when T = 100 F, we have that y = 25. So, we have that 25 = 110e 0.012t. Solving ln for t we obtain that t = 116min. Hence turkey cools down to F in 116 mins. Problem 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( )5 = 3000(1.05) 5 = dollars. 3

4 Now if interest is componded semiannually, we have that A = 3000( )10 = dollars. If interest is compounded monthly, then A = 3000( )60 = dollars. If interest is compounded weekly, then A = 3000( )5 48 = dollars. Now if interest is compounded daily, then A = 3000( )5 365 = dollars. In case of continuos compounding, A(5) = 3000e = dollars. b.) Now da = ra 0 e rt = e.05t = 0.05 A(t), where A(t) = 3000 e 0.05t. Problem 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) , 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 = ( ) = 260km. + dy So dz = x+y z ( dx + dy ) = ( ) = km/h. Problem 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

5 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 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 When R 1 = 80Ω, R 2 = 100Ω, we have R = R 1 R 2 R 1 +R 2 = = Differentiating w.r.t time t we have that dr = R 2 ( 1 dr dr 2 ). R1 2 R2 2 When R 1 = 80Ω, R 2 = 100Ω, dr = ( )2 [ (0.3) (0.2)] = = 0.13Ω/second. 5

6 Figure 2: Figure for Problem 25 6