Fall 016 Math B Suggested Homework Problems Solutions Antiderivatives Exercise : For all x ], + [, the most general antiderivative of f is given by : ( x ( x F(x = + x + C = 1 x x + x + C. Exercise 4 : For all x ], + [, the most general antiderivative of f is given by : ( x 6 ( x ( x F(x = 6 8 9 + C = x 6 8 6 x x + C. Exercise 8 : For all x [0, + [, the most general antiderivative of f is given by : ( x F(x = x4.4 4.4 + C = x4.4 x + C. Exercise 9 : For all x ], + [, the most general antiderivative of f is given by : F(x = x + C. Exercise 11 : For all x [0, + [, the most general antiderivative of f is given by : ( ( F(x = x/ 4 x4/ + C = x / x4/ + C. Exercise 1 : The most general antiderivative of f is given by : 1 F(x = x ln x + C 1, if x<0, 1 x ln x + C, if x>0. Exercise 1 : For all t ]0, + [, the most general antiderivative of g is given by: G(t = t 1/ + t/ + t/ + C. 1
Exercise 17 : For all θ ] π, π [, the most general antiderivative of h is given by: H(θ = cos(θ + tan(θ + C. Exercise 19 : For all x ], + [, the most general antiderivative of f is given by F(x = x + 4 cosh(x + C. ln Exercise : We can rewrite the expression of f as follows : Then F(x = x + tan 1 (x + C. f (x = x + x + 1 = (x + 1 + x + 1 = + x + 1. Exercise 7 : For all x ], + [, the most general antiderivative of f is given by : f (x = x + e x + C. For all x ], + [, the most general antiderivative of f is given by : f (x = x + ex + Cx + D. Exercise : by : For all x ], + [, the most general antiderivative of f is given f (x = 4 tan 1 (x + C. We plug in 1 into this expression to get f (1 = 4 tan 1 (1 + C = 4 π 4 + C = π + C. Thus the condition f (1 = 0 gives us C = π and for all x ], + [ f (x = 4 tan 1 (x π. Exercise 8 : An antiderivative of f has the following general form t ln ln t + C 1, if t<0, f (t = t ln ln t + C, if t>0. We use the condition f ( 1 = 1 to determine the constant C 1. Thanks to the general form of f, we have f ( 1 = 1 ln ln 1 + C 1 = 1 ln + C 1.
So C 1 = 1 ln. We use the condition f (1 = to determine the constant C. Thanks to the general form of f, we have So C = ln. Thus f (1 = 1 ln ln 1 + C = ln + C. f (t = t 1 ln t, if t<0, ln ln t ln t +, if t>0. ln ln Exercise 47 : An antiderivative of f has the following general form f (x = 1 x + C for all x ]0, + [. For all x ]0, + [, the most general antiderivative of f is given by : f (x = ln x + Cx + D = ln(x + Cx + D We use the conditions f (1 = 0 and f ( = 0 to determine the constants C and D. Thanks to the general form of f, we have f (1 = ln(1 + C 1 + D = C + D, f ( = ln( + C + D. The first condition gives us C = D. We plug this in the second and get : ln( D = 0. So we have finally : C = ln(, D = ln( and f (x = ln(x + ln(x ln( for all x > 0. Exercise 6 : The velocity has the following general form : v(t = sin(t + cos(t + C, for all t > 0. We use the initial condition v(0 = 4 on the velocity to determine C. Thanks to the general form of v, we have v(0 = sin(0 + cos(0 + C = + C. So C = and v(t = sin(t + cos(t +, for all t > 0. The position has the following general form : s(t = cos(t + sin(t + t + D,
for all t > 0. We use the initial condition s(0 = 0 on the position to determine D. Thanks to the general form of s, we have s(0 = cos(0 + sin(0 + 0 + D = + D. So D = and for all t > 0. s(t = cos(t + sin(t + t +, Exercise 68 : Let s 1 and s be the altitudes of the first and second ball. Let v 1 and v be the velocities of the first and second ball. Both balls have a negative constant acceleration equal to. The velocities have the following general forms : v 1 (t = t + C 1, for all t>0, v (t = t + C, for all t>1. We use the initial conditions v 1 (0 = 48 and v (1 = 4 to determine C 1 and C. Thanks to the general forms of v 1 and v, we have : So C 1 = 48 and C = 6, and v 1 (0 = C 1, v (1 = + C. v 1 (t = t + 48, for all t>0, v (t = t + 6, for all t>1. The positions of the two balls have the following general forms : s 1 (t = + 48t + D 1, for all t>0, s (t = + 6t + D, for all t>1. We use the initial conditions s 1 (0 = 4 and s (1 = 4 to determine D 1 and D. Thanks to the general forms of s 1 and s, we have : s 1 (0 = D 1, s (1 = + 6 + D = 40 + D. So D 1 = 4 and D = 4 40 = 9, and s 1 (t = + 48t + 4, for all t>0, s (t = + 6t + 9, for all t>1. 4
The balls pass each other if there exits t > 1 such that s 1 (t = s (t that is : + 48t + 4 = + 6t + 9. Or 48t + 4 = 6t + 9, that is 8t = 40 and t =. The balls pass each other at t = s. Exercise 78 : (athe acceleration of the rocket verifies : a(t = 60t, for all 0 t, a(t =, for all t 17. Therefore the velocity has the following general form : v(t = 0t + C 1, for all 0 t, v(t = t + C, for all t 17. We use the initial condition v(0 = 0 to determine C 1, and get C 1 = 0. We use the continuity of the velocity at time t = to find C. The first expression of the velocity gives us v( = 0 = 70. The second expression of the velocity gives us v( = + C. Therefore we have C = + 70 and v(t = 0t, for all 0 t, v(t = (t + 70, for all t 17. At t = 17s, the second expression gives us v(17 = 14 + 70 = 178 ft/s. Then the rocket slows linearly to 18 ft/s in s. So for 17 < t <, v(t = We have therefore : 18 + 178 (t 17 178 = (t 17 178. v(t = 0t, for all 0 t, v(t = (t + 70, for all t 17, v(t = (t 17 178, for all 17 t v(t = 18, for all t. The position has then the following general expression : s(t = 10t + D 1, for all 0 t, s(t = 16(t + 70t + D, for all t 17, s(t = 16(t 17 178t + D, for all 17 t s(t = 18t + D 4, for all t.
We use the initial condition s(0 = 0 to determine D 1, and get D 1 = 0. We use the continuity of the position at time t =,time t = 17 and time t = to find D, D and D 4. The first expression of the position gives us s( = 10 = 70. The second expression of the position gives us s( = 70 + D. Therefore we have D = 70 + 70. and for all t 17, s(t = 16(t + 70(t + 70. The second expression of the position gives us s(17 = 16 14 + 70 14 + 70 = 914. Hence D = 178 17 + 914. and 17 t, s(t = 16(t 17 178(t 17 + 914. The third expression of the velocity gives us s( = 44 and thus for all t, s(t = 18(t + 44. 00 Velocity 1,00 Position 00 100 1,000 0 00 100 00 0 10 1 0 0 0 10 1 0 (b The rocket reaches its maximum height when its velocity is zero for the first time. Let t m be the time for which the rocket reach this maximum height. We know that v( = 70 and v(17 = 178. So t m 17, and t m verifies (t m + 70 = 0. We hence obtain t m = 1/ 11.4s. The maximum height that the rocket reaches is then s(t m = 16(t m + 70(t m + 70 1409.1ft. (c The rocket lands when its position reaches zero. Let t l be the landing time of the rocket. We know that s( > 0 so t l > 0 and we have That is t l = 44/18 + 4.6s. 18(t l + 44 = 0. 6