Question

b and c please the first photo is just some background information ! thank you.

Unless otherwise specified, the domain of a function f is assumed to be the set of all real numbers x for which f (x) is a real number. A large vat is initially filled with a saltwater solution. A solution with a higher concentration of salt flows into the vat, and solution flows out of the vat at the same rate. The number of pounds of salt in the vat at timet minutes is modeled by the function A that satisfies the differential equation * = 6 – 0.02 A. At time t = 10 minutes, the vat contains 50 pounds of salt. dA (a) Write an equation for the line tangent to the graph of A at t = 10. Use the tangent line to approximate the number of pounds of salt in the vat at time t = 12 minutes.

Answer 1

02-0.02t Solution O : To show that Alt)= 300 – 250 e s atrafies the differentwal eqni dA = 6-0.02A, with Alok 50 A Here 02-002t Alt)= 300 – 250 € 1: A (10) = 300 - 250 € 0.2-0.02 X10 0.2-0.2 300-250 e = 300 - 250 - 50 and 0.2-0.02t de = (300) – 250 e = 0 - 250 x -0.02 e 0.2-0.027 5 0.2-0.02t alt = 300-A 50 from 0 e de A = 3:0-A the ha - 6-0.02A] Hence Alt) is a solution of gren differential eqh. The given model is dh = -kih. oh - - kdt at t=0, h=16. 2th = -kt + 2116 = c c = 8 285 = kt+8 - kt = 8-255 - 2116 - C kat +8 0 Hence 49... when t=30 h 30k = 8-2145 2-6 Olekt: 8-215 - = 5(257-8) It= 5(255-8)H) = (t+4012 CS sfrenal @ KL - he) = (4+74052 CamScanner LLTILY

Answer 2

b. solution: Given differential equation intins. = 6-0-02A with condition Aco) = 50. Now, da dt 6 -0.02A Integrating weget log (6 -0.02A) atte, e being constant -0.02 log (6 -0.02A) = -0.02t -0.02€ 6 -0.024 2e-0027-0.02C -- with initial condition AC10) = 50 we have 6-0.02850 ne -0.02x10- 0.02e. 5 = -0.2 -0.022 lag 5 = -0.2 -0.020 c=110 + 50 logg) -0.6 - 0.02 A = -0.027- 0.02 (-10-50 log 5 ) 0.02A = 6 - - 0.027+ 2 + log 5 = 6- & 0.2-0.027 = 6 - 580.2 -0.027 =) A - 6 5 0 .2 -0.0 24 0. 2 - 300 - 250 e 0.2 0.2-0.027 300-250 0.2-0.027 0.02 This This shows showe that A- 300-250 e satisfies given differential equation

ton h that satisfies the differential solution. The function h that sat equation dh crh , where het measured in meters, kis a constant 7 dh Vī E-kat => 1 ch =-klat > 285 =-kt te, cbeing constant According to problem n = 16 when t-o Ley 2176 = -K :0. te o becomes 20h =-Ktts Again h=4 when t=30 204 = -k. 30+8 - 4 = -K 13048 ) x - 4 - 5 2ra = - Ķ + +8 in = 14-17) h = (4-6) Lie, het) (GoT h(t) in terms oft