AP Calculus Testbank (Chapter 6) (Mr. Surowski)

AP Calculus Testbank (Chapter 6) (Mr. Surowski) Part I. Multiple-Choice Questions 1. Suppose that f is an odd differentiable function. Then (A) f(1); (B) f (1) (C) f(1) f( 1) (D) 0 (E). 1 1 xf (x) =. The slope field indicated below most likely depicts the differential equation (A) dy = x + y (B) dy = x + 5 (C) dy = y + 5 (D) dy = x + 5 (E) dy = y + 5

3. You are given differentiable functions f and g. Which of the following is a consequence of the integration by parts formula? (A) f (x)g(x) + f(x)g (x) = f (x)g (x) (B) f (x)g(x) + f (x)g (x) = f(x)g(x) (C) f (x)g(x) + f (x)g (x) = f (x)g(x) (D) f (x)g(x) + f (x)g(x) = f (x)g(x) (E) f (x)g(x) + f(x)g (x) = f(x)g(x). 4. Given the slope field below, what is the most plausible behavior of the solution of the IVP dy = f(x, y), y( 4) = 4? (A) lim x y = (B) lim x y = 0 (C) lim x y = + (D) lim x y = does not exist. (E) lim x y = 5. You are given the differential equation dy = 1 y, where y(0) = 5 5. For which value(s) of x is y = 0? (A) ±5 (B) 10 (C) 50 (D) ±10 (E) 5

6. tan 6 x sec x = (A) tan7 x 7 (B) tan7 x 7 + C + sec3 x 3 (C) tan7 x sec 3 x + C 1 (D) 7 tan 7 x + C (E) 7 tan7 x sec x + C + C 7. 1/ 0 = 1 x (A) π 6 (B) π 3 (C) π 3 (D) π 3 (E) π 3 8. 7xe 3x = (A) 1 + C 4 e3x (B) 6 + C 7 e3x (C) 7 + C 6 e3x (D) 7 e 3x + C (E) 4 e 3x + C

9. e x ( e 3x) = (A) 1 3 e3x + C (B) 1 4 e4x + C (C) 1 4 e5x + C (D) 4 e 4x + C (E) 5 e 5x + C 10. x 5 x = (A) 10 3 (5 x) 3 + C 5x (B) x3 3 + C (C) 10 5x 3 x3 3 + C (D) 10(5 x) 1 + 3 (5 x) 3 + C (E) 10 3 (5 x) 3 + 5 (5 x) 5 + C 11. If dy = x3 + 1 y 7 (A) and y = when x = 1, then when x =, y = (B) 7 8 7 (C) ± 8 (D) ± 3 7 (E) ±

1. ln x 3x = (A) 6 ln x + C (B) 1 ln(ln x) + C 6 (C) 1 3 ln x + C (D) 1 6 ln x + C (E) 1 3 ln x + C 13. sin 5 (x) cos(x) = (A) sin6 x 1 (B) sin6 x 6 (C) sin6 x 3 (D) cos5 x 3 (E) cos5 x 6 + C + C + C + C + C 14. π/ 0 sin(x)e sin x = (A) e (B) e 1 (C) 1 e (D) e + 1 (E) 1 15. π/ 0 sin 5 x cos x = (A) 1 6 (B) 1 6 (C) 0 (D) 6 (E) 6

16. 1 0 sin 1 x = (A) 0 (B) π + (C) π (D) π (E) π 17. If dy = 3y cos x, and y = 8 when x = 0, then y = (A) 8e 3 sin x (B) 8e 3 cos x (C) 8e 3 sin x + 3 (D) 3y cos x + 8 (E) 3y sin x + 8 18. 9 x = (A) sin 1 3x + C (B) ln x + 9 x + C (C) 1 3 sin 1 3x + C (D) sin 1 x 3 + C (E) 1 3 ln x + 9 x + C 19. x sin(x) = (A) x cos(x) + C (B) x 4 cos(x) + C (C) x cos(x) + 1 4 sin(x) + C (D) x cos(x) + 1 cos(x) + C (E) 1 cos(x) + 1 4 sin(x) + C

0. Given the differential equation dz ( dt = z 4 z ), where z(0) = 100 50, what is lim z(t)? t (A) 400 (B) 00 (C) 100 (D) 50 (E) 4

Part II. Free-Response Questions 1. Compute the following indefinite integrals: (a) (1 x) 5 (1 x)6 = + C 6 (b) x 1 x = 7 (1 x)7/ + 4 5 (1 x)5/ 3 (1 x)3/ + C x (c) x 1 = x 1 + C (d) e x sin(e x ) = cos(e x ) + C (e) ln ex = 3 x3/ + C (sin (f) x 1 ) d(sin x) = 1 3 sin3 x sin x + C x (g) x 4 + 1 = 1 tan 1 x + C (h) x sec ( x ) = 1 tan x + C (i) sec x tan x = 1 sec x + C (j) = 4 1 x x 1 x (k) = sin 1 x + C x 1 x (l) x ln x = x x ln x 4 + C (m) xe x = 1 + C ex (n) xe x = xe x e x + C (o) e x sin x = 1 ex (sin x cos x) + C (p) x 3 3 1 x = 3 4 x (1 x ) 4/3 9 56 (1 x ) 7/3 + C

. Compute the following definite integrals: (a) (b) (c) (d) e e π/4 0 π/ 0 π π x ln x = ln(ln x) e e = ln x 1 + x = 1 4 tan 1 x π/4 = 1 0 x sin x = ( x cos x + x sin x + cos x) cos x = π 4 + 3 sin x 3 4 + 3 sin x = 0 π π/ 0 = π 3. Suppose that f is an even differentiable function. Compute This can t be computed explicitly, but we can simplify a bit. Using integration by parts with u = x, dv = f (x), we get du =, v = f(x), and so 1 1 xf (x) = xf(x) 1 1 1 1 1 1 1 f(x) = f(1) f(x). 0 4. Below is sketched the slope field for the differential equation dy = f(x, y). Sketch a possible solution of the above differential equation satisfying the initial value y( 5) =. y -5-4 -3 - -1 1 3 4 5 Equation 1: y'=a(y+3) 3 1-1 - -3 x xf (x).

5. Given the slope field indicated below, sketch solutions of the IVPs (i) dy = f(x, y), y( 1) = 4, x 1 (ii) dy = f(x, y), y( 1) = 1, x 1 4 y x -1 1 3 - -4 Equation 1: y'=ax/(x+3) 6. Sketch the slope field describing the solutions of the ODE dy = x y. 4 y x -6-5 -4-3 - -1 1 3 4 5 6 - Equation 1: y'=x y -4 7. Solve the initial value problem dy y(π) = 0. y = sin x cos x 1. = cos x + sin x, 8. Solve the initial value problem d y = cos x + sin x, y(π) = 0, y (π) = 0. y = cos x sin x x + π

9. Solve the initial value problem dy y = 100e x = y, y(0) = 100. 10. Solve the initial value problem d y = y, y(0) = 1, y (0) = 0. y = cos x 11. The (continuous) growth of money is modelled by the IVP dy dt = ry, y(0) = y 0, where y 0 is an initial amount (of money). (i) Solve the IVP, given that there is initially $10,000 in the bank, and that the interest rate is 6.%. y = 10000e.06t (ii) How much is in the bank after 1 years? y(.5) = 10000e (.06) (.5) $11, 676.58 (iii) How long will it take for the money to double in value? Solving 0000 = y(t) = 10000e.06t leads to 0.06t = ln t 11.18 years. 1. Suppose that we have the initial value problem dy dt = ky, y(0) = y 0 and that we know that y(150) = 1 y 0. Find k. We have that y(t) = y 0 e kt, and that 1 y 0 = y(150) = y 0 e 150k. This gives 150k = ln, and so k 4.6 10 3. 13. (Carbon-14 Dating). The initial value problem of importance in Carbon-14 dating is dy dt = ky, y(0) = y 0. It is typically assumed that the half-life of the radioactive nuclei of carbon 14 is roughly 5,700 years. (i) Using the above, show that k = (ln )/5700. We are given that y(5700) = 1 y 0. This implies that 1 y 0 = y 0 e 5700k k = (ln )/5700.

(ii) After how many years of radioactive decay will the original carbon 14 have decayed by 0%? We are given that y 0 e kt = y(t) = 0.8y 0 t = (ln 0.8)/( k) = (ln 0.8)/( (ln )/5700) 1, 835 years. 14. Newton s Law of Cooling says that the IVP governing the temperature T of a heated object immersed in an environment at a temperature of T s is dt dt = k(t T s), T (0) = T 0, (where k is a positive constant depending on the various media). (i) Show by differentiation that the solution of the above IVP is given by T (t) = T s + (T 0 T s )e kt. We have that dt = dt k(t 0 T s )e kt = k(t T s ). Also, T (0) = T s +(T 0 T s )e 0 = T 0, so the initial condition is satisfied. (ii) Compute lim t T (t) = T s. (iii) Does your answer to part (ii) make sense? Of course this makes sense, as we expect the temperature of a heated object eventually to cool to the temperature of the environment.

15. Assume, as in Problem 14 above, that Newton s Law of Cooling has solution given by T (t) = T s + (T 0 T s )e kt. We apply this model to a bowl of hot noodle soup, where the soup was originally at 90 C and cooled to 60 C after 10 minutes in a room whose ambient temperature was 0 C. (i) Find the value of k in the above model. We re given the solution T (t) = T s +(T 0 T s )e kt = 0+70e kt. Also, we re given that 60 = T (10) = 0 + 70e 10k k = (ln(7/4))/10 0.056. (ii) What is the temperature of the soup after half an hour? Using k 0.056 we infer that T (30) = 0 + 70e 30k 0 + 70e 30 0.056 33.05 C. (iii) After how many minutes will the soup cool to 30 C?Here, we re trying to solve the equation 30 = 0 + 70e 0.056t for t. This leads to t = (ln 7)/0.056 34.75 minutes (or close to 35 minutes). (Note that this is consistent with the result of part (b).) 16. Assume, as in Problem 14 above, that Newton s Law of Cooling has solution given by T (t) = T s + (T 0 T s )e kt. We apply this model to a bowl of hot noodle soup, where the soup was originally at 90 C and cooled to 60 C after 10 minutes in a room, and then cooled to 50 C after another 10 minutes. Determine the ambient temperature of the room. 1 We have the two equations in the two unknowns T s and k: We have 60 = T s + (90 T s )e 10k and 50 = T s + (90 T s )e 0k. e 10k = 60 T s 90 T s, e 0k = 50 T s 90 T s. Since e 0k =)e 10k ), we see that This simplifies slightly to 50 T s 90 T s = (60 T s) (90 T s ). (90 T s )(50 T s ) = (60 T s ). 1 This problem is a bit more difficult than Problem 15, above.

The above ultimately reduces to a linear equation in T s, with solution T s = 45 C. 17. (Mr. S on roller skates.) Believe it or not, Mr. S is pretty good at roller skating! One of his favorite activities is skating as fast as he can and then stop, seeing how far he can coast. Once he starts coasting, the differential equation that governs his motion is m dv = kv, where v is his velocity in km/hr, where m is dt Mr. S s mass (in kg), and where k > 0 is a constant representing friction. Assume that k = 60, 000, that m = 80 kg, and that at the instant he starts to coast, Mr. S is moving at 30 km/hr. (a) Determine Mr. S s velocity, in km/hr, as a function of time. The solution of the initial-value problem is v(t) = v 0 e kt/m = 30e 750t. (b) Determine how far Mr. S will coast (in meters). Since dt = v = 30e 750t, we have that the total distance traveled is x = 0 ( ) dt dt = 0 30e 750t dt = 1 5 e 750t 0 = 1 5 km. Therefore, Mr. S will coast for 40 meters.

18. The Logistic Differential Equation is given by where k is a positive constant. dp dt = k M P (M P ), (i) Show by differentiation that the solution of the above differential equation is given by P = M 1 + Ae kt. This is routine but a bit tedious. The details have been omitted. (ii) Compute lim t P. = M 19. Suppose that the spread of measles in a given school is predicted by the logistic function P (t) = 00 1 + 199e t, where t is the number of days after a student comes into contact with an infected student. (i) Write down the corresponding logistic differential equation. This is dp =.005P (00 P ). dt (ii) Compute P (0) and explain what this means. P (0) = 1, which means that initially one person is infected with measles. (iii) After how many days will half of the students have become infected? We solve the equation 100 = 00 1 + 199e t, which reduces to e t = 1, so t = ln 199 5.3days. 199

(iv) After how many days will 90% of the students have become infected? In this case we are to solve which leads to 180 = 00 1 + 199e t, t = ln 1990 7.6 days. 0. Sketch the slope field describing the solutions of the logistic differential equation dp = P (5 P ), P 0. dt 1. Consider the differential equation dy = 3 x. y (a) Let y = f(x) be the particular solution to the given differential equation for 1 < x < 5 such that the line y = is tangent to the graph of f. Find the x-coordinate of the point of tangency, and determine whether f has a local maximum, local minimum, or neither at this point. Justify your answer. The line whose equation is y = is horizontal; the only value of x for which dy = 0 is x = 3. Next, we have, using implicit differentiation and the quotient rule, that y = y (3 x)y y.

At the point in question, x = 3, y =, and dy = 0. Therefore, y (3, ) = > 0, and so a local minimum occurs at 4 this point. (b) Let y = g(x) be the particular solution to the given differential equation for < x < 8, with the initial condition g(6) = 4. Find y = g(x). Here we must actually solve the differential equation. We separate variables and integrate both sides: y dy = (3 x) which leads to y x = 3x + C. Since y = 4 when x = 6, we get 8 = 18 18 + C and so C = 8. Writing y as a function of x gives y = ± 16 + 6x x ; however, as y is negative (at x = 6), we infer that, in fact, y = g(x) = 16 + 6x x.. Consider the differential equation dy = x (y 1). (a) On the axes provided, sketch a slope field for the given differential equation at the twelve points indicated. y 3 1 1 1 x

(b) While the slope field in part (a) is drawn at only twelve points, it is defined at every point in the xy-plane. Describe all points in the xy-plane for which the slopes are positive. The expression dy = x (y 1) > 0 at all values (x, y) where x 0 and y > 1. (c) Find the particular solution y = f(x) to the given differential equation with the initial condition f(0) = 3. Separating the variables and integrating leads to dy y 1 = x ln y 1 = x3 3 + C. Write this as y 1 = Ke x3 /3 ; since y = 3 when x = 0, we get = K. Therefore, the solution is y = e x3 /3 + 1. 3. Consider the differential equation dy = 3x e y. (a) Find a solution y = f(x) to the differential equation satisfying f(0) = 1. Separating the variables and integrating: e y dy = 3x e y = x 3 + C. Since y = 1 when x = 0, get e = C. Therefore ey = x 3 + e so that y = f(x) = 1 ln(x3 + e). (b) Find the domain and range of the function f found in part (a). The domain consists of all real numbers x satisfying x 3 + e > 0 x > 3 e/; the range consists of all real numbers y. The graph of the solution is shown below:

4. The ( function f is differentiable for all real numbers. The point 3, 1 ) is on the graph of y = f(x), and the slope at each point 4 (x, y) on the graph is given by dy = y (6 x). ( (a) Find d y and evaluate it at the point 3, 1 ). 4 First, we have d y dy = y (6 x) y. Next, since dy ( 3, 1 ) = 4 0, we see that d y = ( ) 1 = 1 4 8. (Note that this implies that the function f has a local maximum at the point ( 3, 1 4).) (b) Find y = f(x) by solving the differential equation dy = y (6 x) with the initial condition f(3) = 1 4. Separating the variables and integrating yields dy y dy = (6 x) 1 y = 6x x + C. Since y = 1 when x = 3 we obtain 4 = 18 9 + C, which 4 implies that C = 13. Therefore, we obtain

y = f(x) = 1 x 6x + 13. 5. Let f be the function satisfying f (x) = x f(x) for all real numbers x, where f(3) = 5. (a) Find f (3). f (x) = f(x) + xf (x) f(x) f (3) = 5 + 3 15 5 = 5 + 9 = 19. (b) Write an expression for y = f(x) by solving the differential equation dy = x y with the initial condition f(3) = 5. Separating the variables and integrating: dy = y x dy y = x + C; since y = 5 when x = 3, we infer that 10 = 9 + C, and so C = 11. This implies that y = x + 11 and so y = f(x) = 1 16 (x + 11).

6. Consider the differential equation dy = x y. (a) On the axes provided, sketch a slope field for the given differential equation at the twelve points indicated. y 1 1 1 x 1 (b) Let y = f(x) be the particular solution to the differential equation with the initial condition f(1) = 1. Write an equation for the line tangent to the graph of f at (1, 1) and use it to approximate f(1.1). We have that dy (1, 1) =, and so the line tangent to the graph of f at (1, 1) is y + 1 = (x 1), which can be re-written as y = x 3. Using this to approximate f(1.1) we have f(1.1) (x 3) =. 3 = 0.8. x=1.1 (c) Find the particular solution y = f(x) to the given differential equation with the initial condition f(1) = 1. Separating the variables and integrating:

y dy = x dy y = x + C. Since y = 1 when x = 1 we infer quickly that C = 3 and we have the solution y = f(x) = 3 x. (Note the necessity of the minus sign when taking the square root!) 7. You are given the initial value problem dy = x y, y(0) = 0. (i) Use Euler s method to approximate y(0.) (use = 0.1). From y(0.1) y(0) + dy (0, 0)(0.1) = 0, we then have y(0.) y(0.1) + dy (0.1, 0)(0.1) = 0 + 0.01 = 0.01. (ii) Show by differentiation that the solution of this IVP is y(x) = e x + x 1. From y = e x + x 1 get y = e x + 1 = x (e x + x 1) = x y. (iii) Compare the relative error in the estimation of y(0.). (Use the definition Relative Error = exact value approximate value exact value.) The exact value of y(0.) = e 0. + 0. 1 = 0.019 (accurate to three decimal places), whereas the approximate value was shown above to be y(0.) 0.01. Therefore, Relative Error = 0.019 0.010 0.019 = 0.47, i.e., we have incurred a 47% error in the approximation.