1. Ma 262 Purdue

MA 262Solutions Manual Practice Problems: Ma 26200 - Linear Algebra and Differen. From Purdue University.

1 Solutions to Section 1.1 True-False Review: 1. A derivative must involve some derivative of the function y = f (x), not necessarily the first derivative. The initial conditions accompanying a differential equation consist of the values of y, y 0,.

If we define positive velocity to be oriented downward, then dv = g, dt where g is the acceleration due to gravity. We can justify this mathematically by starting from a(t) = g, and integrating twice to get 1 v(t) = gt + c, and then s(t) = gt2 + ct + d, which is a quadratic equation. The restoring force is directed in the direction opposite to the displacement from the equilibrium position.

According to Newton’s Law of Cooling, the rate of cooling is proportional to the difference between the object’s temperature and the medium’s temperature. Since that difference is greater for the object at 100 F than the object at 90 F, the object whose temperature is 100 F has a greater rate of cooling. The temperature of the object is given by T (t) = Tm + ce kt, where Tm is the temperature of the medium, and c and k are constants. Since e kt 6= 0, we see that T (t) 6= Tm for all times t. The temperature of the object approaches the temperature of the surrounding medium, but never equals it.

Since the temperature of the coffee is falling, the temperature difference between the coffee and the room is higher initially, during the first hour, than it is later, when the temperature of the coffee has already decreased. The slopes of the two curves are negative reciprocals of each other.

If the original family of parallel lines have slopes k for k 6= 0, then the family of orthogonal trajectories are parallel lines with slope k1. If the original family of parallel lines are vertical (resp. Horizontal), then the family of orthogonal trajectories are horizontal (resp. Vertical) parallel lines. The family of orthogonal trajectories for a family of circles centered at the origin is the family of lines passing through the origin. Problems: d2 y 1.

Starting from the differential equation 2 = g, where g is the acceleration of gravity and y is the unknown dt position function, we integrate twice to obtain the general equations for the velocity and the position of the object: dy gt2 = gt + c1 and y(t) = + c1 t + c2, dt 2 where c1, c2 are constants of integration. Now we impose the initial conditions: y(0) = 0 implies that c2 = 0, 2 and dy dt (0) = 0 implies that c1 = 0. Hence, the solution to the initial-value problem is y(t) = gt2. 2 The object hits the ground at the time t0 for which y(t0 ) = 100. Hence 100 = s, where we have taken g = 9.8 ms 2. Gt20 2, so that t0 = q 200 g ⇡ 4.52 d2 y 2. Starting from the differential equation 2 = g, where g is the acceleration of gravity and y is the unknown dt position function, we integrate twice to obtain the general equations for the velocity and the position of the ball, respectively: 1 dy = gt + c and y(t) = gt2 + ct + d, dt 2 where c, d are constants of integration.

Setting y = 0 to be at the top of the boy’s head (and positive direction downward), we know that y(0) = 0. Since the object hits the ground 8 seconds later, we have that y(8) = 5 (since the ground lies at the position y = 5). From the values of y(0) and y(8), we find that d = 0 5 32g and 5 = 32g + 8c. Therefore, c =.

8 (a) The ball reaches its maximum height at the moment when y 0 (t) = 0. That is, gt + c = 0. Therefore, t= 32g 5 c = ⇡ 3.98 s. G 8g (b) To find the maximum height of the tennis ball, we compute y(3.98) ⇡ 253.51 feet. So the ball is 253.51 feet above the top of the boy’s head, which is 258.51 feet above the ground.

Starting from the differential equation 2 = g, where g is the acceleration of gravity and y is the unknown dt position function, we integrate twice to obtain the general equations for the velocity and the position of the rocket, respectively: 1 dy = gt + c and y(t) = gt2 + ct + d, dt 2 where c, d are constants of integration. Setting y = 0 to be at ground level, we know that y(0) = 0. (a) The rocket reaches maximum height at the moment when y 0 (t) = 0.

That is, gt + c = 0. Therefore, the c time that the rocket achieves its maximum height is t =.

At this time, y(t) = 90 (the negative sign g accounts for the fact that the positive direction is chosen to be downward). Hence, ✓ ◆ ◆2 ◆ ✓ ✓ c 1 c2 c c2 c2 c +c = g = =. 90 = y g 2 g g 2g g 2g p Solving this for c, we find that c = ± 180g. However, since c represents the initial velocity of the rocket, and the (relative to the fact that the positive direction is downward), we choose p initial velocity is negative c= 180g ⇡ 42.02 ms 1, and thus the initial speed at which the rocket must be launched for optimal viewing is approximately 42.02 ms 1. 3 c ⇡ g (b) The time that the rocket reaches its maximum height is t = 42.02 = 4.28 s. Starting from the differential equation 2 = g, where g is the acceleration of gravity and y is the unknown dt position function, we integrate twice to obtain the general equations for the velocity and the position of the rocket, respectively: 1 dy = gt + c and y(t) = gt2 + ct + d, dt 2 where c, d are constants of integration. Setting y = 0 to be at the level of the platform (with positive direction downward), we know that y(0) = 0.

Ma 262 Purdue

(a) The rocket reaches maximum height at the moment when y 0 (t) = 0. That is, gt + c = 0.

Therefore, the c time that the rocket achieves its maximum height is t =. At this time, y(t) = 85 (this is 85 m above g the platform, or 90 m above the ground). Hence, 85 = y ✓ c g ◆ = 1 g 2 ✓ c g ◆2 +c ✓ c g ◆ = c2 2g c2 = g c2. 2g p Solving this for c, we find that c = ± 170g. However, since c represents the initial velocity of the rocket, and the (relative to the fact that the positive direction is downward), we choose p initial velocity is negative 170g ⇡ 40.84 ms 1, and thus the initial speed at which the rocket must be launched for optimal c= viewing is approximately 40.84 ms 1.

40.84 c ⇡ = 4.16 s. (b) The time that the rocket reaches its maximum height is t = g 9.81 5. If y(t) denotes the displacement of the object from its initial position at time t, the motion of the object can be described by the initial-value problem d2 y dy = g, y(0) = 0, (0) = dt2 dt 2, where g is the acceleration of gravity and y is the unknown position function. We integrate this differential equation twice to obtain the general equations for the velocity and the position of the object: dy = gt + c1 dt and y(t) = gt2 + c1 t + c2. 2 Now we impose the initial conditions: since y(0) = 0, we have c2 = 0. Moreover, since 2 c1 = 2. Hence the solution to the initial-value problem is y(t) = 2 Consequently, h = g(10) 2 2 10 =) h = 10(5g gt 2 dy (0) = dt 2, we have 2t.

We are given that y(10) = h. 2) ⇡ 470 m where we have taken g = 9.8 ms 2. If y(t) denotes the displacement of the object from its initial position at time t, the motion of the object can be described by the initial-value problem d2 y = g, dt2 y(0) = 0, dy (0) = v0. Dt 4 We integrate the differential equation twice to obtain the velocity and position functions, respectively: dy = gt + c1 dt and y(t) = gt2 + c1 t + c2. 2 Now we impose the initial conditions. Since y(0) = 0, we have c2 = 0. Moreover, since c1 = v0.

Hence the solution to the initial-value problem is y(t) = Consequently, h = gt20 + v0 t0. Solving for v0 yields v0 = 7.

From y(t) = A cos (!t 2t0 gt2 + v0 t. We are given that y(t0 ) = h. ), we obtain dy = dt Hence, gt20 2h dy (0) = v0, we have dt A! Sin (!t d2 y +!2 y = dt2 ) and A! 2 cos (!t d2 y = dt2 A!

2 cos (!t ) + A! Substituting y(0) = a, we obtain a = A cos( ) = A cos( ).

Also, from dy dt (0) = 0, we obtain 0 = A! Since A 6= 0 and!