Review for Exam 2 (The following problems are all from old exams)

Review for Exam 2 (The following problems are all from old exams). Write down, but do not evaluate, an integral which represents the volume when y = e x, x, is rotated about the y axis. Soln: Divide the solid into equal horizontal slices (along the y-axis) and notice that the volume of each slice is approximatly V π(log y) 2 dy. e Thus the total volume is then V = π(log y) 2 dy. 2. Find the volume of the solid formed by rotating the graph of y = x for x about the x-axis. Soln: π( x )2 dx = π. 3. Let R be the curve y = 3x 2, x 2. (a) Set up (but do not evaluate) an integral to compute the volume of the solid generated by rotating R around the x-axis. (b) Set up an integral to compute the volume of the solid generated by rotating R around the line y =. (c) Set up an integral to compute the volume of the solid generated by rotating R around the y-axis. 2 Soln: (a) The volume of each slice is V πy 2 dx = π(3x 2 ) 2 dx = π9x 4 dx. The total volume is then π9x 4 dx. (b) Here the volume element is V π( + y) 2 dx = π( + 3x 2 ) 2 dx and so V = 2 π( + 3x 2 ) 2 dx. (c) Here the volume element is V πx 2 dy = π( y/3) 2 dy = πy/3dy and so V = 2 πy/3dy. 4. Find the volume when y = e x2, < x <, is revolved about the y-axis The volume element here is V πx 2 dy = π( e log y) 2 dy = π log ydy. The total volume is V = π log ydy. 5. Use integration to derive a formula for the volume of a sphere of radius r. [Hint: Think of the sphere as a surface of revolution.] Soln: The sphere is obtained by revolving the circle y = r r 2 x 2, r x r, about the x-axis. The volume is π( r 2 x 2 ) 2 dx = 4 r 3 πr2. 6. We wish to estimate the volume of a flower vase using only a calculator, a piece of string, and a ruler. We measure the height of the vase to be 6 inches. We then use the string and the ruler to find the circumference of the vase (in inches) at /2 inch intervals starting at the bottom. The circumferences are 5.4, 5.8, 6.5, 7., 8.5, 9.8,.2,.8..6,.6,.8, 9., 6.3. (a) Using the above information, approximate the volume of the vase. Soln: If C(x) is the circumference at height x, the radius is equal to r(x) = C(x)/2π and so the volume of each slice is πr(x) 2 dx = π(c(x)/2π) 2 2. The approximate volume (using a left-hand sum) is V 5.4 π(( 2 2π )2 + ( 5.8 2π )2 + ( 6.5 2π )2 + ( 7. 2π )2 + ( 8.5 2π )2 + ( 9.8 2π )2 + (.2 2π )2 + (.8 2π )2 + (.6 2π )2 + (.6 2π )2 + (.8 2π )2 + ( 9. 2π )2 ) = 4.234 (b) If C(x) represents the circumference (in inches) of the vase at height x inches from the bottom, find a integral which represents the volume of the vase. 6 Soln: At each height, the radius is equal to C(x)/2π. Thus the volume is equal to π( C(x) 2π )2 dx 7. Write down an integral which represents the length of the hyperbola xy = from the point (, ) to the point (2, /2).

2 2 Soln: The curve is parameterized by y = /x, x 2. The volume is thus + (y ) 2 dx = 2 + /x 2 dx. 8. Find the length of the parabola y 2 = x from (, ) to (, ). b Soln: The length of a curve y = f(x), a x b is + (f (x)) 2 dx. In our case, y = x and so L = + ( a 2 x )2 dx = + 4x dx. 9. A force of lbs. is required to hold a spring stretched 4 in. beyond its natural length. How much is done in stretching it from its natural length to 6 in. beyond its natural length. Soln: By Hooke s law. F = kx (k is Hooke s constant and x is the displacement from its natural position). Note that = k 4 and so k = 5/2. 6 Thus F = 5/2 x. Notice that work element is W = F dx = 5/2 x dx and so the total work is then W = 5/2 x dx. It takes 2, 74 pounds to compress a coil spring assembly on a New York City subway car from its natural height of 8 inches to a fully compressed height of 5 inches. How much work does it take to compress the assembly the first one inch? Soln: By Hooke s law, F (x) = kx. But since F (3) = 274, we get k = 7238. Thus the total work is F (x)dx = 7238xdx = 369.. An upright cylindrical tank with height 2 ft. and radius 5 ft. is ft. underground (i.e., the top of the tank is ft. below the surface). If the tank is /4 filled with a fluid which weighs 6.2 pounds per cubic foot, write down, but do not evaluate, an integral which represents the work it takes pump out the tank. Soln: The work it takes to move a horizontal slab of fluid is W 6.2 π 5 2 dx (2 x). Thus the total work it takes to pump is then 5 6.2 π 5 2 (2 x) dx. Notice the limits of integration since the tank is only /4 filled. 2. Consider an upright cylindrical tank with radius 3 ft and height ft which is completely filled with a fluid which weights 5 lbs/ft 3. Write down, but do not evaluate, an integral which computes the work exactly. Soln: If x is the distance from the bottom of the tank (x = is the bottom of the tank while x = is the top of the tank), then the distance a slab of water at height x must be lifted is d = x. The volume of a slab of water is V = π 3 2 dx. The amount of work is takes to move a slab of water is W 5 V d = 5 π 3 2 ( x) dx. The total work is then 5 π 3 2 ( x) dx. 3. A cylindrical tank with height ft. and radius 2 ft. is buried upright so that the top of the cylinder is 2 ft. underground. If the cylinder is /2 full with a fluid which weights 52.8 lbs/ft 3, write down, but do not evaluate, an integral which represents the exact amount of work it takes to empty the tank. Soln: Let x be the distance from the bottom of the tank (x = is the bottom of the tank while x = is the top of the tank). Then the distance a slab of water at height x must be moves is d = 2 x. The volume of a slab of water is V = π 2 2 dx. The work is takes to move a 5 slab of water at height x is then W 52.8 d V = 52.8 π 2 2 (2 x) dx. The total work is then W = 52.8 π 2 2 (2 x) dx. Notice the 5 in the upper limit of integration since the tank is only half filled with water. 4. A rectangular tank with the following dimensions length width height ft. 2 ft. 5 ft. is buried 3 ft. underground (i.e. the top of the tank is 3 ft. below the surface). If the tank is /2 filled with a fluid which weights 58.6 lbs./ft 3., find the work it takes to pump all its contents to the surface. Soln: Let x be the distance from the bottom of the tank (x = is the bottom of the tank while x = 5 is the top of the tank). The distance a slab of water at height x must be moves is d = 45 x. The volume of a slab of water is V = 2 dx. The work is takes to move a slab of 7.5 water at height x is then W 58.6 V d = 58.6 2 dx (45 x). The total work is then W = 58.6 2 (45 x) dx. Notice the 7.5 in the upper limit of integration since the tank is half filled. 5. A cubical ( x x ) tank, full of water, is buried 2 feet underground (the top of the tank is 2 feet under). Set up an integral to compute the amount of work required to pump all the water to the surface? (Water weighs 62.4 lbs/ft 3.) You do not have to evaluate the integral.

3 Soln: Let x be the distance from the bottom of the tank (x = is the bottom of the tank while x = is the top of the tank). The distance a slab of water at height x must be moves is d = 3 x. The volume of a slab of water is V = dx. The work is takes to move a slab of water at height x is then W 62.4 V d = 62.4 dx (3 x). The total work is then W = 62.4 (3 x) dx. 6. An upright cylindrical tank of height 2 feet and radius 4 feet is /3 filled with a fluid which weights 52.6 lbs/ft 3. If the tank is buried 2 feet underground, find the work it takes to empty it. Soln: Let x be the distance from the bottom of the tank (x = is the bottom of the tank while x = 2 is the top of the tank). The distance a slab of water at height x must be moves is d = 22 x. The volume of a slab of water is V = π 4 2 dx. The work is takes to move a slab of 2/3 water at height x is then W 52.6 V d = 52.6 π 4 2 dx (22 x). The total work is then W = 52.6 π 4 2 (22 x) dx. Notice the 2/3 in the upper limit of integration since the tank is /3 filled. 7. A worker on a scaffolding 75 ft. above the ground needs to lift a 5 lb. bucket of cement from the ground to a point 3 ft. above the ground by pulling on a rope weighing 5 lbs./ft. How much work is required? Soln: Let x be the distance from the ground. The weight of the rope and cement at a point x feet off the ground is 5 + 5 (75 x). The 3 total work is then W = [5 + 5 (75 x)]dx. 8. A lb. weight is being lifted to a height ft. off the ground. It is lifted using a rope which weighs 4 lbs./ft. and which is being pulled up by a construction workers standing on a roof 3 ft. off the ground. Find the work done to lift the weight. Soln: Let x be the distance from the ground. The weight of the rope and lb. weight at a point x feet off the ground is + 4 (3 x). The total work is then W = [ + 4 (3 x)]dx. 9. A circular oil slick has the following density function ρ(r) = e r2 kg/m 2, where r (in meters) is the distance from the center of the slick. If the slick extends from r = to r =, write down an integral which gives the mass of the slick exactly. You need not evaluate the integral. Soln: Divide up the oil slick into annuli (concentric circles), we see that the area of an annulus with inner (or outer) radius r is approximately 2 π r dr. The mass of this portion of the oil slick is then M ρ(r) 2 π r dr and so the total mass is then M = ρ(r) 2 π r dr. 2. The density of people living r miles from the center of a city is ρ(r) millions/square mile. Write down an integral (in terms of the density function ρ) which represents the total number of people living within a 5 mile radius of the center of the city. Soln: This is very similar to the previous problem. 2. Scientists are looking to compute the mass of a certain tumor which is 5 mm tall. An MRI analysis reveals the following results. height h (mm) cross sectional area A(h) (mm 2 ) density ρ(h) (g/mm 3 ) 2.6.2 2 3.8.6 3 4.9.9 4 2.28. 5.8.9 (a) Use numerical integration to estimate the mass of the tumor. (b) Write down an integral (in terms of the functions A(h) and ρ(h)) which computes the exact mass of the tumor. (b) Soln: Using the basic identity that mass is density times volume, we get the mass of a slab of the tumor h units from the base and of 5 thickness dh is M A(h) dh ρ(h). The total mass is then M = A(h) ρ(h) dh. (a) Using a RHS to approximate this integral, we get M A() ρ() + + A(5) ρ(5).

4 22. (a) A rectangular swimming pool is 3 feet wide and 4 feet long. The table below shows the depth h(x) (in feet) of the water at 5-ft. intervals from one end of the pool to the other. Estimate the volume of the water in the pool. position x. 5. 5 2 25 3 35 4 depth h(x) 6. 8.2 9. 9.9.5..5.9 2.3 (b) Find a mathematical expression which gives the exact volume of the pool. Soln: Let x = and x = 4 be the two ends of the pool. The volume of a slab at (position x and of thickness dx) of water is V 3 h(x) dx. 4 The total volume is V = 3 h(x) dx. (a) Use a RHS to estimate V 3 h(5) 5 + 3 h() 5 + + 3 h(4) 5. 23. The density (in kg/m 2 ) of a thin 3 m by 5 m rectangular plate is ρ(x) = 4x 2, where x is the distance from the 3 m side. Write down, but do not evaluate, an integral which represents the exact mass of the plate. Soln: Let x = and x = 5 represent the two ends of the plate and divide up the plate into vertical strips each of thickness dx. The mass of 5 each strip is then M ρ(x) 3 dx (density times area) and so the total mass is then M = 4x 2 3 dx. 24. Match the slope field with the corresponding differential equation. (a) dy dx = xy dy (b) dx = x y dy (c) dx = x + y2 (d) dy dx = x + x2

5 Soln: a - d - b - c 25. The slope field below represents which differential equation. (a) y = sin x, (b) y = y, (c) y = y /2y 2, (d) y = x + y, (e) y = x 2, (f) y =, (g) y = y 2, (h) y = x 2 + y 2. Soln: f. 26. The slope field below represents which differential equation. (a) y = sin x, (b) y = y, (c) y = y /2y 2, (d) y = x + y, (e) y = x 2, (f) y =, (g) y = y 2, (h) y = x 2 + y 2. Soln: e. 27. The slope field below represents which differential equation.

6 (a) y = sin x, (b) y = y, (c) y = y /2y 2, (d) y = x + y, (e) y = x 2, (f) y =, (g) y = y 2, (h) y = x 2 + y 2. Soln: d. 28. The slope field below represents which differential equation. (a) y = sin x, (b) y = y, (c) y = y /2y 2, (d) y = x + y, (e) y = x 2, (f) y =, (g) y = y 2, (h) y = x 2 + y 2. Soln: b. 29. The slope field below represents which differential equation. (a) y = sin x, (b) y = y, (c) y = y /2y 2, (d) y = x + y, (e) y = x 2, (f) y =, (g) y = y 2, (h) y = x 2 + y 2. Soln: a. 3. Solve the following differential equation y = y(y + ), y() = 2. Soln: Separate the variables and use partial fractions to get [ y ]dy = dt. Now integrate to get log y log y + = t + c. y + Use y y laws of loge to get log = t + c. Exponentiate to get y + y + = Cet. A little algebra says y(t) = Ce t. The initial conditions says that Cet 2 = C/( C) and so C = 2/3. The final solution is then y(t) = 2/3et 2/3e t. 3. Solve the following differential equations. (a) y + 2y = e t Soln: Use an integrating factor for this one. The integrating factor is µ = e 2t and so, multiplying both sides of the DE by µ we get (µy) = e t and so µy = e t + c. This says y(t) = + ce 2t. (b) P = P P 2 dp Soln: Use separation of variables to write this as = dt. Now integrate using partial fractions to get P (t) = Ce t P ( P ) + Ce t. 32. Solve the following differential equation: x 2 y + 6xy = 2x Soln: rewrite the equation as y + (6/x)y = 2/x. The integrating factor for this equation is µ = x 6. Thus (µy) = x 6 (2/x) = 2x 5 and so µy = 3 x6 + C. From here we get y(x) = 3 + C x 6.

7 33. Find the most general solution to the following differential equations. (a) y = 2y( y) dy Soln: Separate variables to get y( y) (b) P = P log P 2. 2t Ce = 2dt. Now integrate, using partial fractions and some algebra, to get y(t) = Ce 2t. dp Soln: Separate variables and get = 2dt. Now integrate, using u = log P, to get log log P = 2t + C. This means that log P = Ce 2t P log P and so P (t) = e Ce 2t. 34. Find the general solution to the following differential equations. (a) dx = 2x x2 dt dx Soln: Separate variables to get x(2 x) = dt. Now integrate, using partial fractions to obtain 2 log x log(2 x) = t + c. Now use a little 2 2Ce2t + Ce 2t. (b) ( + t)y + 2y = ( + t) 2 algebra to get the final solution as x(t) = Soln: First rewrite the differential equation as y + 2 + t y = + t and multiply both sides by the integrating factor µ = ( + t)2 to get (µy) = ( + t) 3 ( + t)4 ( + t)2 C. Now integrate and obtain µy = + C and do a little algebra to get the final solution as y(t) = + 4 4 ( + t) 2. 35. For the initial value problem v = v 2, v() =, (a) use Euler s method with 2 steps to estimate v(). Soln: t =, v =. t = t + h = /2, v = v + h( v 2 ) = /2. t 2 = t + h =, v 2 = v + h( v 2 ) = 7/8. So v() 7/8 =.875 (b) compute the exact value of v(). Soln: Separate variables and use partial fractions to get ln( + v + v ) = 2t + C. Thus v v = Ce2t. From here, + v = ( v)ce 2t = Ce 2t vce 2t and so v = Ce2t 2t e 2 e. Using the initial condition v() = says C = and so v = + Ce2t e 2t. So v() = + e 2 + =.76594. 36. Solve the following initial value problem: x 2 y + 4xy = x 4, y() = 2 Soln: Rewrite the differential equation as y + 4 x y = x2. Multiply both sides by the integrating factor µ = x 4 to get (µy) = x 6 and integrate to get µy = x7 7 3 x + C. Finally we get y = 37. Solve the following differential equations. (a) y = y(y + 2) Soln: Separate variables and use partial fractions to get y(t) = (b) y + 3 x y = e 2x x 3 7 + C. Use the initial condition to compute C = 3/7. x4 2Ce2t Ce 2t. Soln: Multiply both sides of the equation by the integrating factor µ = x 3 to get (µy) = e 2x. Now integrate and obtain µy = 2 e 2x + C. Finally y(x) = 2 e 2x x 3 + C x 3. 38. Solve the following differential equations: (a) y = y ln y 2 Soln: This is very similar to one we have already solved. (b) y + 4 x y = e 2x x 4 Soln: Use the integrating factor µ = x 4 to obtain the solution y(x) = 2 e 2x x 4 + C x 4. 39. An RL circuit is an electric circuit consisting of a resistor R (measured in ohms) and an inductor L (measured in Henrys) all hooked up to a voltage source E(t) which possibly changes in time. The current I(t) at time t satisfies the differential equation LI + RI = E.

8 (a) Solve this differential equation with L =, R = 2, and E =. (b) Solve this differential equation with L =, R = 2, and E = sin t. Soln: All of the above differential equations can be solved using integrating factors. 4. The velocity (in feet per second) of a falling object subject to air resistance is governed by the following differential equation v = 32.4v 3/2. (a) Compute the limiting velocity of this object. Soln: The limiting velocity occurs when v = (i.e., there is no change in the velocity). Thus the limiting velocity is the solution v to the equation 32.4v 3/2 which is v = (32/.4) 2/3 8.5664 (b) Using Euler s method with a step size of.25, estimate the velocity of the object after second. Assume the object is dropped and not thrown, i.e., t =, v =. Soln: Let f(t, v) = 32.4v 3/2. Note that t =, v =, and h =.25. Then t = t + h = +.25 =.25, v = v + f(t, v )h = + f(, ).25 = 32.25 = 8. t 2 = t + h =.25 +.25 =.5, v 2 = v + f(t, v )h = 8 + f(.5, 8).25 = 3.7373 t 3 = t 2 + h =.5 +.25 =.75, v 3 = v 2 + f(t 2, v 2 )h = 6.6457 Thus v() 7.8544 t 4 = t 3 + h =.75 +.25 =., v 4 = v 3 + f(t 3, v 3 )h = 7.8544 4. Consider the following initial value problem: ( + x)y = y 2, y() =. (a) Use Euler s method with step size equal to.25 to estimate y(). Soln: Let f(x, y) = y 2 /( + x), x =, y =, and h =.25. Then x = x + h =.25, y = y + f(x, y )h =.25 x 2 = x + h =.5, y 2 = y + f(x, y )h =.5625 x 3 = x 2 + h =.75, y 3 = y 2 + f(x 2, y 2 )h =.9694 x 4 = x 3 + h =., y 4 = y 3 + f(x 3, y 3 )h = 2.52348 Thus y() 2.52348. (b) Solve the above initial value problem and use your answer to compute y() exactly. Soln: Use separation of variables to get dy y 2 = dx + x. Integrate and use some algebra to obtain y =. Use the initial condition c + log( + x) y() = to get that c =. Thus the solution to the IVP is y(x) =. So y() = + log( + x) + log 2 3.25889. 42. A painting attributed to the Dutch artist Johannes Vermeer (632-675), which should contain more than 96.2% of its original carbon-4, contains 99.5% instead. About how old is the forgery? Note that the half-life of carbon-4 is 578 years and that the rate of change of the amount of a radioactive substance is proportional to the amount present. Soln: If Q(t) is the amount of isotope at time t, the Q = rq and so Q(t) = Q e rt, where Q is the initial amount. Using the fact that Q(578) =.5Q, we get the equation.5q = Q e 578r. Thus r =.99 and so Q(t) = Q e.99t. Now we find the t so that Q(t) = Q e.99t =.995Q which turns out to be t 4 years. 43. Snow is falling at a constant rate of in/hour and is melting at a rate equal to 3/4 of the amount already on the ground per hour. If y(t) is the amount of snow on the ground (in inches) and y() = 2, (a) Write a differential equation satisfied by y(t). Soln: y = 3 4 y (b) Solve this differential equation. Soln: Use an integrating factor (or separation of variables - either one works) to get y(t) = 2 3 (2 + e 3/4t ). (c) What is the equilibrium level of snow. Soln: You can get this in one of two ways. First you can find the y so that y =, i.e. the solution to the equation 3/4y, i.e. y = 4/3. Second, you can compute the limit lim y(t) = lim 2 t t 3 (2 + e 3/4t ) = 4/3.

9 44. A drug is being injected in the bloodstream at a constant rate of 2 ml/hr and is absorbed by the body at a rate equal to /2 the amount present. (a) If a(t) is the amount of the drug present after t hours, write down an differential equation which a(t) satisfies. Soln: a = 2 2 a (b) If a() = ml, find a(t). Soln: Use separation of variables - or integrating factors - to solve the differential equation as a(t) = 4 3e t/2. (c) Find the equilibrium amount of the drug in the bloodstream. Soln: The equilibrium level is lim a(t) = lim 4 t t 3e t/2 = 4 45. At 4: AM. snow begins to fall at a constant rate of 4 inches per hour. The snow melts at a rate equal to /2 the current height. (a) What is the height of the snow at 9: PM? (b) What is the limiting height of the snow? Soln: This problem is very similar to one we have already seen. 46. A familiar model for population growth is dp dt = kp. Another similiar model, called the doomsday model is given by dp +ɛ = kp dt where ɛ is a very small constant. Consider the doomsday model, with k = 3 and ɛ = 3, dp dt = 3P 4 3 (a) Find P (t) if P () = 728 and t is measured in years. 728 Soln: Use separation of variables to get P (t) = ( + 2t) 3. (b) What happens to P (t) as the end of the first month approaches (t = 2 )? Soln: From the solution to the differential equation, P (t) as t /2 47. The velocity v(t) of a falling object with air resistance taken into account satisfies the differential equation dv dt = 2.7v 3 2 feet per second. () If v() =, use Euler s Method with 2 steps to estimate v(). (2) Compute lim v(t). t Soln: This problem is very similar to one we have already seen. 48. The velocity (in feet per second) of a falling object subject to air resistance is governed by the following differential equation v = 32.4v 3/2. (a) Compute the limiting velocity of this object. (b) Using Euler s method with a step size of.25, estimate the velocity of the object after second. Assume the object is dropped and not thrown. Soln: This problem is very similar to one we have already seen. 49. Consider the Gompertz equation which is used to model certain populations. P = P lnp. (a) For the initial value P () = 2, estimate, using Euler s method with step size equal to, the value of P (2).

Soln: Let f(t, P ) = P log P, t =, P = 2, and h =. Then t = t + h =, P = P + f(t, P )h =.6376 t 2 = t + h = 2, P 2 = P + f(t, P )h =.9334 (b) Solve the Gompertz equation with initial condition P () = 2. dp Soln: Separate variables to get P log P = dt. Now integrate and obtain log log P = t + c. Exponentiate and get log P = Ce t and again to get P = e Ce t. Use the initial condition to calculate C as log 2. 5. A gallon tank of pure water is being filled with a water and cobalt solution of 2 grams/gallon at a constant rate of 35 gallons/hour and the well mixed solution is being emptied at the same rate. (a) Set up (but do not solve) a differential equation for S(t) = the amount of cobalt (in grams) in the tank at time t. Soln: The differential equation is S S = 35 2 35. The first term in the rate at which cobalt is entering the tank (rate of solution concentration = rate of cobalt) while the second term is the rate at which the cobalt is leaving. (b) Set up (but do not solve) a differential equation for S(t) if the solution leaves the tank at 5 gallons/hour. Soln: In this situation, the volume of the tank at time t is + 2t and so the differential equation becomes S = 35 2 5 5. A number of fish P (t) at time t satisfy the differential equation dp dt = k P where k is a positive constant. Find a general form for the solution P (t). Soln: Use separation of variables to get P (t) = 4 ( kt + C)2. S + 2t. 52. After giving a final exam, a popular calculus professor is found dead in her office at : A.M. The temperature of the body at the time that it is found is 8.2 F ; an hour later the temperature of the body was 78.4 F. The professor was a little eccentric and always kept her office at 6 F. (a) Write a differential equation for db dt, where B(t) is the temperature of the body at time t. Recall that Newton s Law of Cooling says that the rate at which temperature of a body changes is proportional to the difference between its temperature and the temperature of its surroundings. Soln: B = k(6 B). (b) What time was the professor murdered? Soln: By separation of variables, the solution to the above differential equation is B(t) = 6 + Ce kt. B() = 8.2 and so C = 2.2. Thus B(t) = 6 + 2.2e kt. Also, B() = 78.4 and so k =.465. The complete solution to the differential equation is then B(t) = 6 + 2.2e.465t. To find the time of death, we need to find the t so that B(t) = 98.6, i.e., solve the equation 98.6 = 6 + 2.2e.465t for the variable t. The solution is t = 4.2349. Thus the time of death is roughly 5 : 3AM. 53. (a) Mr. Titewad invests $, in an account which pays 4.2 % per year (compounded continuously). He withdraws $,5 per year from this account. Let A(t) be a amount in this account after t years and set up, but do not solve, an initial value problem involving A(t). Soln: A =.42A 5, A() =. 54. An investor puts $, in an account which pays 5.2 % per year (compounded continuously). He also withdraws $, per year (continuously). When will his money run out? Soln: If A(t) is the amount of money in the account at time t (is years), then A(t) satisfies the initial value problem A =.52A, A() =. By separation of variables, the solution is A(t) = 9238 9238e.52t. The money will run out when A(t) =, or equivalently when 9238 9238e.52t =. This occurs when t = 4.7387. 55. Suppose that h(t) is a function which satisfies the following differential equation h = 2.5 h. Compute lim t h(t).

If the above limit exists, it will be when h is not changing any more, i.e., h =. Thus the limiting h will be the solution to the equation = 2.5 h which is h = 6. 56. A water tank initially contains 2 cubic feet of water. Water runs into the tank at a constant rate of 2 cubic feet per minute, but a valve allows the water to flow out of the tank at a rate equal to 8 % of the volume of the water in the tank per minute. Find the amount of water in the tank after one hour. Soln: If V (t) is the amount of water in the tank at time t, then V satisfies the initial value problem V = 2.8V, V () = 2. Using separation of variables, the solution to this IVP is V (t) =. + 88.889e.7999t. Note that V () = 85.357. 57. Eulercillin is a drug given to anxious students during exams. The drug is injected into the bloodstream at a rate of 2 cc per hour and is absorbed at a rate equal to twice the amount present. If y(t) is the amount of Eulercillin present after t hours, (a) write down the differential equation that y satisfies. Soln: y = 2 2y. (b) If the initial amount of the drug present is.7 cc, find the amount present at any time t and determine what happens to y as t. Soln: Using separation of variables, the initial value problem y = 2 2y, y() =.7 has the solution y(t) =.3e 2t. Also, as t, y(t). 58. A water tank with a maximum capacity of gallons is initially filled with 5 gallons of pure water. A salt water mixture with 5 grams of salt per gallon of water is flowing in at a rate of 2 gallons per minute and the well mixed solution is leaving at a rate of gallon per minute. (a) Write down a formula for the volume V (t) of water in the tank at time t. Solution: V (t) = 5 + t (5 to start off and a net gain of gallon per minute). (b) How long will it take for the tank to overflow? Solution: The tank will overflow when V (t) = which is when t = 5. (c) Write down a differential equation satisfied by y(t), the amount of salt in the tank at time t Solution: The the change in y is the rate in minus the rate out. This gives us y = 5 2 y(t)/v (t) Note here we are using (twice) the fact that the rate of change of salt is the concentration of salt in solution times the rate of solution. Plugging in for V (t) = 5 + t, we get y = y/(5 + t) (d) Solve the differential equation. Soln: The differential equation can be handled by integrating factors. But first, we need to put the equation in standard form The integrating factor µ = exp( y + y/(5 + t) =. 5 + t ) = elog(5+t) = 5 + t. Using the usual facts about integrating factors, we get (µy) = (5+t). Integrating we get µy = 5(5+t) 2 +C and so y(t) = 5(5+t)+C/(5+t). To compute C, we recall that y() = ( pure water ) and so = 25 + C/5 and so C = 25. Thus the solution to the d.e. is y(t) = 5(5 + t) 25/(5 + t). (e) How much salt is in the tank at the moment it overflows? Solution: The tank will overflow when t = 5. Thus the amount of salt is then y(5) = 375. 59. A tank is being filled with water at a constant rate of 4 gallons/minute and is leaking at a rate equal to /3 of the amount present. If y(t) is the amount of water present after t minutes, (a) Write down a differential equation which y satisfies. (b) If the initial amount of water is 2 gallons, find the amount present at any time t. Soln: This problem is very similar to previous problems. 6. A six foot tall tank is completely filled with water which is leaking out from a small hole in the bottom. Let y(t) be the height (in feet) of the water in the tank at time t (in minutes). From a principle of engineering

2 (called Torricelli s Law), y(t) satisfies the differential equation y = y. (a) Find a formula for the height of the water at any time t. Solution: Separate the variables to get y /2 dy = dt. Now integrate to get 2 y = t + c. A little algebra gives us y(t) = ( t 2 + c)2. To find the constant c note that 6 = y() = c 2 and so c = 6. From this we get y(t) = ( t 2 + 6) 2. (b) How long will it take for the tank to empty completely. Solution: The tank will be empty when y(t) =. This occurs when t 2 + 6 = which is when t = 2 6. (c) If, in the previous problem, water is being pumped into the tank at a constant rate of 2 feet per minute, the differential equation becomes y = 2 y. If the tank is initially empty, use Euler s method with two steps to estimate the height of the water after minute. Soln: This problem is similar to previous problems. 6. A 2 gallon capacity tank contains gallons of pure water. A salt water solution containing /4 pound of salt per gallon, flows in at a rate of 2 gallons per minute and the well-mixed solution leaves at a rate of 5 gallons per minute. Write down (but do not solve) a differential equation (along with an initial condition) satisfied by Q(t), the amount (in pounds) of salt in the tank at time t (in minutes). Soln: Q = Q 2 5, Q() =. 4 + 5t 62. A team of archeologists find a piece of clothing and after extensive Carbon-4 dating, they find this cloth has 25% of the Carbon-4 of a new piece of cloth. If the half-life of Carbon-4 is 573 years, find the age of the cloth. (Note here that we are assuming the rate of change of Carbon-4 is proportional to the amount present). 63. Archeologists find a sample of human hair which they determine has 87 % of the Carbon-4 of living hair. If the amount of Carbon-5 left after t years satisfies the differential equation Q = kq, and the half-life of Carbon-4 is 573 years, find the age of the hair sample.