Math, Solutions to Homework 2

Solutions to Homework 2 Math. 280, Spring 2008 Problem 1. A cup of coffee in a room at temperature of 70◦ F is being cooled according to Newton’s law of cooling: dT = −k(T − 70), T (0) = 200. (time is measured in minutes) dt The solution to (1) is T (t) = 70 + (200 − 70) exp(−kt). The coefficient k can be computed from (2) using the fact that T (3) = 175. Thus, 1/3  1/3  21 175 − 70 = 175 = 70 + 130 exp(−3k) =⇒ exp(−k) = 130 26 (1) (2) (3) The temperature of the coffee will reach 112◦ F at t = t1 that satisfies the equation t 112 = T (t1 ) = 70 + 130 exp(−kt1 ) = 70 + 130 [exp(−k)] 1 , with exp(−k) given in (3) Thus, 112 − 70 = 130 " 21 26 1/3 #t1 ,   21 3 log 65   ≈ 15.87 minutes. or t1 = 21 log 26 Problem 2. Let N (t) denote the amount of C-14 at time t, with time t = 0 corresponding to the moment when the original amount N0 of C-14 was present. Time evolution of N (t) is governed by the equation dN = −λN, N (0) = N0 . (4) dt The amount of C-14 decays in time, thus λ > 0. From (4), N (t) = N0 exp(λt) and, since the half-life of C-14 is approximately 5568 years, N0 /2 = N0 exp(5568λ). This gives us λ = − ln 2/5568 ≈ −0.00012448. Therefore, the expression for N (t) for all time is given by   t ln 2 N (t) = N0 exp − t = N0 2−( 5568 ) . (5) 5568 At present time t, 14.5% is left, or equivalently from (5), t 5568 ln 0.145 t ln 2 =⇒ t = − ≈ 15511.753 years. 0.145N0 = N0 2−( 5568 ) =⇒ ln 0.145 = − 5568 ln 2 Problem 3. For the equation   dy x (1 + x2 )dy + (xy + x3 + x)dx = 0 or y = −x, + dx 1 + x2   √ R x dx = 1 + x2 so that an integrating factor is exp 2 1+x i p
− 1 1 d h p y 1 + x2 = −x 1 + x2 =⇒ y = − (1 + x2 ) + c 1 + x2 2 . dx 3 100 50 -10 10 20 -50 -100
− 1 1 Figure 1. Graphs of y = − (1 + x2 ) + c 1 + x2 2 for various choices of c. 3 1 (6) (7) 2 Problem 4. (a) For the equation  3y 2 − t2 dy t (8) + 4 = 0, y(1) = 1, y5 dt 2y let M = t/(2y 4 ) and N = (3y 2 − t2 )/y 5 so that My = −2t/y 5 = Nt . From φt (t, y) = M = t/(2y 4 ) we obtain φ(t, y) = t2 /(4y 4 ) + h(y). Thus, using φy = N = (3y 2 − t2 )/y 5 , we have h′ (y) = 3/y 3 , and h(y) = −3/(2y 2 ). The general solution is t2 /(4y 4 ) − 3/(2y 2 ) = c. If y(1) = 1 then c = −5/4 and the solution to initial-value problem (8) is  3 5 t2 − 2 =− . 4y 4 2y 4 (9) 2 1 1 0.5 -1 -0.5 0.5 1 -0.5 -1 Figure 2. Graph of the integral curve (9) -1 1 2 3 4 5 -1 -2 Figure 3. Other integral curves of t2 /(4y 4 ) − 3/(2y 2 ) = c The solutions of the initial value problem (8) and some other integral (dashed) curves of the differential equation  2  3y − t2 dy t + 4 = 0 are shown in Figure 3. 5 y dt 2y Remark 1. Please observe that the graph in Figure 2 does not represent a function !!! Which branch of the graph represents the solution satisfying the initial condition y(1) = 1. Remark 2. Note that if one multiplies both sides of equation (8) by y 5 , the resulting equation dy 1 (3y 2 − t2 ) + ty = 0 dt 2 5 is not exact !!! In other words 1/y is an integrating factor for (10). (See also Problem 5, below) (b) (10) The initial-value problem y dy = 2 , y(1) = 5, or dt 2y − t is linear in t (as a function of y). Its integrating factor is 1/y, thus   d t = 2 =⇒ dy y dt t − = 2y, dy y y(1) = 5, (11) t = 2y 2 + cy. If y(1) = 5 then c = −49/5 and t = 2y 2 − (49/5)y. The solutions of the initial value problem (11) and some other integral (dashed) curves of the differential equation dy y = 2 , are shown in Figure 5. dt 2y − t Remark 3. Please observe that the graph in Figure 4 does not represent a function !!! Which branch of the graph represents the solution satisfying the initial condition y(1) = 5. 3 5 4 3 2 1 -10 -5 -1 4 2 -10 5 -5 10 5 10 -2 Figure 4. Graph of the integral curve t = 2y 2 − (49/5)y Figure 5. Other integral curves of t = 2y 2 + cy Problem 5. Multiply the equation (t2 + ty) dy 1 − y2 = 0 by µ(t, y) = 2 . The result is dt t y   1 1 dy y + − 2 = 0. y t dt t (12) (13) With M = −y/t2 and N = 1/y + 1/t we have My = −1/t2 = Nt ; therefore equation (13) is exact. From φt = −y/t2 we obtain φ(t, y) = y/t + h(y) But φy = N ; thus h′ (y) = 1/y, and h(y) = log |y|. The solution of (12) is y where c ∈ R is a constant. φ(t, y) = + log |y| = c, t Problem 6. Under the change of the independent variable, t → y, the original equation are transformed into dv = −mg − kv 2 , dt dv = mg − kv 2 , m dt m =⇒ k > 0 for the “down” motion dv = −mg − kv 2 , dy dv mv = mg − kv 2 , dy mv k > 0 for the “up” motion k > 0 for the “up” motion (14) k > 0 for the “down” motion Separating variables and integrating the differential equation corresponding to the “up” motion we have   Z Z m 2ky mv 2 2 = − dy =⇒ ln(mg + kv ) = −y + c =⇒ mg + kv = c exp − 1 2 mg + kv 2 2k m   2ky mg =⇒ v 2 = c3 exp − − m k (15) Using y(0) = 0 and v(0) = v0 we have that v = v0 when y = 0 (the initial condition) so that v02 = c3 − mg/k and c3 = v02 + mg/k. Thus,     2 2ky mg kv0 + mg 2 v = exp − − (16) k m k Setting v = 0 in the left hand side of (16) and solving for y we see that the maximum height is   2 m kv0 + mg h= . (17) ln 2k mg Now, separating variables and integrating the differential equation corresponding to the “down” motion (see (14)) we have   Z Z mv m 2ky 2 2 = dy =⇒ − ln |mg − kv | = y + d1 =⇒ mg − kv = d2 exp − mg − kv 2 2k m    2ky mg 1 − d3 exp − =⇒ v 2 = k m In this case v = 0 when y = 0 (the initial condition) so d3 = 1 and    2ky mg 1 − exp − . v2 = k m (18) (19) 4 Setting y = h from (17) into (19) and solving for v we obtain that the impact velocity is v0 < v0 . vi = s kv02 1+ mg (20) Remark 4. By taking y → ∞ in the right hand side of (19) one obtains the limiting velocity, vf inal , for the problem m (dv/dt) = mg − kv 2 s r r    mg mg v02 mg 2ky mg 2 vf inal = lim = vi . (see, (20)) 1 − exp − = =⇒ vf inal = > y→∞ k m k k k v02 + mg k See also (29), where the same result can be obtained by taking t → ∞ in (29) and using lim tanh(x) = 1. x→∞ (b) It is interesting to compare the results of Problem 6 with the same problem but without air resistance. When k = 0 the change variables from t to y is not necessary, however, for comparison with the case k 6= 0, I proceed in an exactly the same way as before. The corresponding equations are dv dv = −mg, for the “up” motion mv m = −mg, for the “up” motion dy dt =⇒ =⇒ (21) dv dv m mv = mg, for the “down” motion = mg, for the “down” motion dt dy v2 v2 v2 = −gy + c1 , v|y=0 = v0 for the “up” motion = −gy + 0 , for the “up” motion 2 2 2 =⇒ (22) v2 v2 = gy + c2 v|y=0 = 0 for the “down” motion = gy for the “down” motion 2 2 Thus, the maximum height is v02 /2g. Observe that the maximum height given in (17) is less than the maximum height in the case without air resistance. Indeed,        2  m v2 m m kv02 kv0 + mg kv 2 since for x > 0 = 0 = ln ln 1 + 0 < ln(1 + x) < x 2k mg 2k mg 2k mg 2g Also the impact velocity in the case without air resistance is equal to the initial velocity v0 . indeed, substituting v02 /2g into v 2 /g = gy (expression for v for the “down” part of the motion, see, (22)) we obtain vi = v0 (No dissipation of the energy) Additional remarks to Problem 6 Finally, I show how one can integrate the equations dv = −mg − kv 2 , v(0) = v0 , k > 0 for the “up” motion (23) m dt dv = mg − kv 2 , v(0) = 0, k > 0 for the “down” motion (24) m dt without changing the variable t to y. The integrations are slightly more involved. For the “up” motion we have, after separation of variables   ! r r   r m 1 mg mg kg k dv  = −dt =⇒ r = − t + c =⇒ v(t) = tan c1 − t (25) arctan v/ mg k k m k m mg + v2 k k Applying the initial conditions v(0) = v0 we have r   r mg mg tan c1 =⇒ c1 = arctan v0 / v0 = k k =⇒ v(t) = r " #  r  r mg mg kg − tan arctan v0 / t . k k m From (26) the time tup needed for the projectile to attain its maximum height (v(tup ) = 0) is   r r m mg tup = arctan v0 / kg k For the “down” motion   dv m  = dt mg k − v2 k =⇒   r 1 k mg −1 r = t + d1 v/ tanh k m mg k =⇒ v(t) = (26) (27) r mg tanh d1 + k r kg t m ! (28) 5 Applying the initial condition v(0) = 0 one obtains d1 = 0 and ! r r mg kg tanh t v(t) = k m (29) The time tdown needed for the projectile to fall from the height (17) can also be computed; from (20) ! ! r r √ r mg kg kg v0 v0 k s = = tanh tanh t =⇒ p t k m m mg + kv02 kv02 1+ mg s ! √ mk v0 k −1 p =⇒ tdown = tanh g mg + kv02 (30) One can also show that tup < tdown . (31) Furthermore, when air resistance is neglected tup = tdown = v0 . g (No dissipation of the energy) Problem 7. (a) The equation for velocity at any time and position y is 2gR2 , (32) R+y where√R is the radius of the earth, g is the acceleration due to gravity, and v0 is the initial velocity of the projectile (in our case, v0 < 2gR). Now the maximum distance, y = Ymax , is reached by the projectile when v = 0. In other words, we have the following equation for Ymax : 2gR2 0 = V02 − 2gR + . (33) R + Ymax Its solution is Rv02 Ymax = . 2gR − v02 1 (b) The solution to the equation y ′′ = −g with the initial conditions y(0) = 0 and y ′ (0) = v(0) = v0 > 0 is y(t) = − gt2 + v0 t 2 (with v(t) = −gt + v0 ). As in part (a), the maximum distance, ymax , is reached when v = 0; this corresponds to t = v0 /g. Thus, the maximum distance ymax is      2 1 v2 v0 v0 v0 =− g = 0. + v0 ymax = y g 2 g g 2g √ Finally, since v0 < 2gR, we have v02 v2 Rv02 = > 0 = ymax . Ymax = 2 2 2gR − v0 2g − (v0 /R) 2g v 2 = v02 − 2gR + (c) In this case (32) becomes v= dy = dt s 2gR2 R+y , y(0) = 0,  Note the plus sign in front of which is a separable equation. The solution of (34) is p 2 (R + y)3/2 = Rt 2g + c. 3 3/2 With y(0) = 0 we have c = (2/3) R . Thus  √ 2/3 3 2g y(t) = Rt + R3/2 − R. 2 (d) s 2  2gR  R+y (34) (35) Solving (35) for t we obtain   2 [y(t) + R]3/2 − R3/2 . t= √ 3 2gR With R = 3963 miles, y(TM ) = 238, 855 miles, and g = 32/5280 = 0.0061 mile/s2 , we get from (36) TM = 181853 seconds = 50.52 hours. (36)