(9/1/08)Math 10B. Lecture Examples.Section 11.5. Growth and Decay†Example 1 Match problems (I) through (IV) below to differential equations (a) th rough (d) and tothe slope fields in Figures 1 through 4.(I) The thickness of the ice on a lake grows at a rate that is proportional to the reciprocal of itsthickness. Find the thickness y = y(t) as a function of the time t.(II) A population grows at a rate proportional to its size. Find the p opulation y = y(t) as afunction of the time t.(III) A hot potato is taken out of the oven at time t = 0 into a kitchen that is at 20◦Celsius.The rate of change of the potato’s te mperature is proportional to the differenc e between its temperatureand that of the kitchen. Find th e temperature y = y(t) of the potato as a function of t.(IV) Find a function y = y(t) whose rate of change with respect to t is −2t.(a)dydt= 0.2y (b)dydt=20y(c)dydt= −2t (d)dydt= −2(y − 20)t5 10y100200300400t500 1000y50100150200FIGURE 1 FIGURE 2t1 2 3 4 5 6y1020304050t1 2 3y102030405060FIGURE 3 FIGURE 4Answer: Problem I goes with equation (b) and the slope field in Figure 2. • Problem II goes with equation (a) andFigure 1. • Problem III goes with equation (d) and Figure 4. • Problem IV goes with equation (c) and Figure 3.†Lecture notes to accompany Section 11.5 of Calculus by Hughes-Hallett et a l.1Math 10B. Lecture Examples. ( 9/1/08) Section 11.5, p. 2Example 2 (a) Solve the differential equation (a)dydt= 0.2y from Example 1 (a population) withthe initial condition y(0) = 50 and draw its graph with the corresponding slope field.(b) What happens to the solution as t → ∞?Answer: (a) y = 50e0.2t• Figure A2. (b) The population y = 50e0.2ttends to ∞ as t → ∞.t5 10y100200300400Figure A2Example 3 Solve differential equation (b)dydt=20yfrom Example 1 (ice thickness) with theinitial condition y(0) = 50 and draw its graph with the c orre sponding slope field.Answer: y =√40t + 2500 • Figure A3t500 1000y50100150200Figure A3Example 4 Solve differential equation (c)dydt= −2t from Example 1 (a function) with the initialcondition y(0) = 50 and draw its graph with the corre sponding slope field.Answer: y = 50 − t2• Figure A4t1 2 3 4 5 6y102030405060Figure A4Section 11.5, p . 3 Math 10B. Lecture Examples. (9/1/08)Example 5 (a) Solve differential equation (d)dydt= −2(y− 20) from Example 1 (the potato) withthe initial condition y(0) = 50 and draw the graph of its solution with the correspondingslope field. (b) What happens to the solution as t → ∞ and why is this plausible?Answer: (a) y = 20 + 30e−2t• Figure A5.(b) y tends to 20 as t → ∞. (This is plausible because y is the potato’s temperature, which tends to the roomtemperature 20◦C as t → ∞.)t1 2 3y102030405060Figure A5Example 6 Match problems (I), (II), and (III) with differential equations (a), (b), and (c) and th eslope fields in Figures 5 through 7.(I) v is the downward velocity, measured in meters per second, of a ball that is falling under theforce of grav ity but with no air resistance or other forces on it. The time t is measured in seconds andthe acceleration due to gravity is 9.8 meters per second2.(II) v is the downward velocity, measured in mete rs per second, of a suitcase that is falling underthe force of gravity with air resistance that is proportional to its velocity. The suitcase’s d ownwardacceleration is zero when its downward velocity is 49 meters per second. (This is called the equilibriumvelocity because it is the velocity at which the upward force of air resistance equals the downwardforce of gravity. It is also referred to as the terminal velocity because it is the limit of the suitcase’svelocity as t → ∞.)(III) v is the downward veloc ity, measured in meters per second, of a rock that is falling underthe force of gravity with air resistance that is proportional to its velocity. Its eq u ilibrium velocity is 98meters pe r second.(a)dvdt= 9.8 (b)dvdt= 9.8 −110v (c)dvdt= 9.8 −15v.t10 20 30v50100t10 20 30v50100t10 20 30v50100FIGURE 5 FIGURE 6 FIGURE 7Answer: Problem I goes with differential equation (a) and Figure 7. • Problem II goes with equation (c) andFigure 5. • Problem III goes with equation (b) and Figure 6.Math 10B. Lecture Examples. ( 9/1/08) Section 11.5, p. 4Example 7 Find th e solution of differential equation (a)dvdt= 9.8 in Example 6 with th e initialcondition v(0) = 0. Draw its graph with the corresponding slope field.Answer: v = 9.8t • Figure A7t10 20 30v50100v = 9.8tFigure A7Example 8 Solve the initial-value problem,dvdt= 9.8 −110v, v(0) = 0 for differential equation (b)in Example 6. Then draw the graph of the solution with the corresponding slope fie ld.Answer: v = 98 − 98e−t/10(which tends t o 98 a s t → ∞) • Figure A8t10 20 30v50100v = 98 − 98e−t/10Figure A8Example 9 Solve the initial-value problem,dvdt= 9.8 −15v, v(0) = 0 for differential equation (c) inExample 6 and draw the graph of the solution with the corresponding slope field.Answer: v = 49 − 49e−t/5(which tends to 49 as t → ∞) • Figure A9t10 20 30v50100v = 49 − 49e−t/5Figure A9Section 11.5, p . 5 Math 10B. Lecture Examples. (9/1/08)Example 10 Match problems (I) and (II) below with differential equations (a) and (b) and t h e slopefields in Figures 8 and 9.(I) v is the horizontal velocity, measured in miles per hour, of a model car whose acceleration,due to its faltering en gine, oscillates between 0 and 90 miles per hour2. There are no other forces on thecar and time is measured in hours.(II) v is the horizontal velocity, measured in feet per minute, of a motor boat that has its engineturned off and is slowing down because of water and air resistance. The resistance is proportional to t h esquare of the boat’s velocity, and there are no other forces on it. Time is measured in minutes.(a)dvdt= −18v2, (b)dvdt= 45 + 45 cos(15t)t0.5 1v10203040t0.5 1v10203040FIGURE 8 FIGURE 9Answer: Problem I goes with differential equation (b) and Figure 9. • Problem II goes with equation (a) andFigure 8.Example 11 Find the solution of equation (a)dvdt= −18v2from Example 10 with the initialcondition v(0) = 40. Draw its graph with the corresponding slope field.Answer: v =405t + 1• Figure A11 (Notice that v → 0 as t → ∞.)t0.5 1v10203040v =405t + 1Figure A11Math 10B. Lecture Examples. ( 9/1/08) Section 11.5, p. 6Example 12 Solve equation (b)dvdt= 45 + 45 cos(15t) from Example 10 with the initial conditionv(0) = 0 and draw the