Exam 2 Solutions, Math March 17, ) = 1 2

Transcription

1 Eam Solutions, Math 56 March 7, 6. Use the trapezoidal rule with n = 3 to approimate (Note: The eact value of the integral is ln 5 +. (you do not need to verify this or use it in any way to complete this problem.)) Solution: We have = b a n = 3 = and f() = +. Thus, + = (f() + f() + f(3) + f()) ( ). 5. Which of the following is a direction field for the differential equation = y? (a) (b) (c) (d)

2 (e) Solution: When y =, =, thus on the line y =, we should have horizontal lines. That leaves only (a) as an answer. The graph of the direction field is shown below. 3. The length of the curve y = cos( ) from (, ) to (, cos()) is given by: Solution: We have = sin( ). Thus, Arclength = + ( sin( )) = + sin ( ).

3 . The improper integral e (ln()). Solution: Use the definition of improper integral and making the substitution u = ln with = du. Then t ln t e (ln ) t e (ln ) t u du t [ u ] ln t t ( ln t + ) =. 5. Using Euler s method with step size, which of the following gives an approimation for y(3), where y is the solution of with y() =? y = y Solution: Thus y(3) 5 n n y n n, y n ) ()() = = + () = 3 ()(3) = 6 = () = 5 6. Find n= ( ) n 3 n 5 n. Solution: Note that ( ) n 3 n = n n= This is a geometric series. The first term a is 3 and the ratio r is 5 5 Note that r = < hence our series converges. Now the sum is 5 in our case. 3 a r = = =

4 7. Evaluate the improper integral π tan (). Solution: Note that there is a trouble spot at π, since tan( π ) is undefined. Let us write Note that That gives us π tan () = tan () = π π tan () + sec () tan () t π π π tan (). = tan() + C. t tan () [tan() ] t t π tan(t) t t π = π. Hence π tan () diverges, therefore, π tan () also diverges. 8. The solution to the initial value problem satisfies the implicit equation = e ( + y ), y() = Solution: We can use the separation of variables: + y = e tan (y) = e = e e + C.

5 We have used integration by parts with u =, dv = e in the above integral. The initial condition gives us tan () = e e + C, or C =. Hence our solution satisfies the implicit equation tan (y) = e e. 9. Find the general solution to the differential equation y + 3y = + 5. Solution: We can put the equation in the following form: y + 3 y = + 5. This is a first-order linear differential equation. In order to solve it, we find the integrating factor. I() = e 3 = e 3 ln = e ln 3 = 3. Multplying across by I() we get Hence y = C 3. y y = (y 3 ) = ( + 5) y 3 = ( + 5 ) y 3 = C.. Consider the following sequences: (I) { ( ) n n n } n= (II) Which of the following statements is true? a) All three sequences converge. { ( ) n n n } n= b) Sequences I and II converge but sequence III diverges. (III) { e n n } n= 5

6 c) All three sequences diverge. d) Sequences I and III converge but sequence II diverges. e) Sequence I converges but sequences II and III diverge. { Solution: (I) ( ) n n } is an alternating sequence. It is a theorem that it n n= converges if and only if lim n n ( )n n n =. n n We can use L Hospital s rule to determine the above limit. n lim n n Hence sequence I converges. (II) ln (ln ) =. { } ( ) n n is an alternating sequence as well. As in (I), we consider n n= lim n n ( )n n Hence sequence (II) diverges. n n n n n (III) We write e n lim n n e. We have an indeterminate form so we can use LHospitals rule { e n Thus sequence III = n } n= e lim e =. is divergent. =. To sum up, the solution is: Sequence I converges but sequences II and III diverge. 6

7 . A tank initially contains liters of salt water with kilogram of dissolved salt. A well mied salt water solution containing kilograms of salt per liters is pumped into the tank at 5 liters per minute. The salt water in the tank is kept thoroughly mied and is drained at the rate of 5 liters per minute. (a) Let y = y(t) be the amount of salt in the tank at time t. Give a differential equation relating to y. dt Solution: We note that the rate of change of the amount of salt in the tank, denoted, is given by dt dt = rate in - rate out, where rate in denotes the rate at which the salt is entering the tank (in kg/min) and rate out denotes the rate at which the salt is eiting the tank (also in kg/min). Now, Similarly, rate in = (Conc. of Solution Entering Tank)(Flow Rate In) = kg L 5 L min = kg/min rate out = (Conc. of Solution Eiting Tank)(Flow Rate Out). The difference here is that the concentration of the solution eiting the tank is no longer constant. This concentration in the tank (assuming the tank is evenly mied) is given by C(t) = y(t), where V (t) is the volume in the tank at time t. Since the V (t) rate at which solution is entering the tank is the same at which it is eiting, we have dv/dt = and the volume of solution in the tank is constant at L. So C(t) = y(t) kg/l. Putting this together, we see that Thus, rate out = y(t)kg L dt 5 L min = y(t) kg/min. = rate in - rate out = y(t). 7

8 (Units are still kg/min.) We also note that the tank initially contains kg of dissolved salt, giving us the initial condition y() =. So our differential equation becomes dt = y(t), y() =. (b) Give a formula for the amount of salt in the tank at time t. Solution: We may rewrite the equation above as dt = y(t), y() = and we see that this is an eample of a separable differential equation. Separating variables and integrating both sides with respect to t we see y(t) dt = y(t) dt dt = ln y(t) = t + C dt ln y(t) = t + C ( C = C) y(t) = e t+ C y(t) = ±e Ce t = Ae t (A = ±e C) y(t) = Ae t. We now plug in the initial condition to solve for the constant A. y() = = Ae = A A = Thus our formula for the amount of salt in the tank at time t is given by y(t) = e /t. 8

9 Remark: This differential equation is also an eample of a linear differential equation. We can see this by rewriting the equation as dt + y =. This means there are actually two different ways to approach this problem; we could have also used the method for solving linear differential equations. (c) Find the limit of y(t) as t goes to. Solution: lim y(t) t t e /t t = = kg. e (/)t. Find the arc length of the curve given by from = to =. f() = e + e Solution: Let L be the arc length of the curve given above from = to =. By the arc length formula, we know L = + (f ()). We first need to compute the derivative f (): f () = e + e ( ) = e e = (e e ). 9

10 Net, we will simplify the integrand, ( ) + (f ()) = + (e e ) ( ) = + (e + e ) e + + e = (e + e = ) = e + e. Plugging this back into our formula for arc length, we obtain L = + (f ()) e + e = [ ] e e = = e e = e e. e e

11 3. For each of the following improper integrals, determine whether the integral converges or diverges. To receive full credit for this problem, you must justify each answer and state clearly whether each integral converges or diverges. (a) + 3/ Solution: The p-test for improper integrals states that + converges if p > and diverges if p. p In this case, p = 3 < and this integral DIVERGES. Alternatively, you could try to evaluate this integral directly. + t 3/ t 3/ t t 3/ [ ] / t t [ t / ] t = Since this limit is not finite, the integral diverges. (b) + 3/ Solution: This integral CONVERGES by the p-test, since p = 3 >.

12 You can also show the integral converges by evaluating it directly: + t 3/ t 3/ [ ] / t t [ t / + ] t [ ] t t = (c) + 3/ + Solution: CONVERGES The Comparison Test for Improper Integrals states that if f and g are continuous functions with g() f() for all a. i) If a ii) If a f() is convergent, then g() is convergent. a g() is divergent, then f() is divergent. a Now, for all, we have 3/ + 3/ 3/ +. 3/ We also can see that both g() = and f() = are both continuous 3/ + 3/ functions on the interval [, ] (actually these are both continuous functions at all points ecept = ). Since converges (we showed this in part b), by the 3/ comparison test, so does +. 3/ + (d) + 3/

13 Solution: DIVERGES This actually is an improper integral for two reasons: i) the interval is infinite, ii) f() = is discontinuous at = 6. To find when f() is discontinuous in the 3/ interval [, ] we set the denominator equal to and solve for : So, 3/ = ( ) 3/ / = = or / = / = / = = = / = 6 3/ + 6 3/ Note that BOTH of these integrals must converge for the entire integral to be convergent. We will compute the first integral directly: 6 t 3/ t 6 3/ t t 6 3/ ( Use the rationalizing substitution u = /, du = / 3/, so = 3/ du = u 3 du.) t 6 t 6 t t / 8u 3 u 3 u du 8 u du [ 8 ln u ]t/ t 6 8 ln t/ + 8 ln t 6 = Here, we are using the fact that if t 6, t / approaches from the right and lim ln =. Since this integral diverges, the entire integral must diverge. + 3

14 Remark: We can alternatively show that the second integral 6 3/ diverges (you still only need to show that one of the two diverges). Choosing a test point in the interval (6, ), say = 8, we see that f(8) = (8 3/ 8/) = (7 8/) = ( 7/) = /7. Consequently, f() < for all on the entire interval (6, ) (otherwise, the intermediate value theorem says we must have f() = for some value of in this interval, and this never happens). Now, 6 Furthermore, when 6 3/ = 6 3/ 3/ 3/ 3/ Now, 6 = diverges and so, by the comparison test, so does 6. Thus, 6 3/ 6 3/ = also diverges. This also 6 3/ + is enough to show 3/ diverges.