MAT292 - Calculus III - Fall Solution of Term Test 1 - October 6, 2014 DO NOT WRITE ON THE QR CODE AT THE TOP OF THE PAGES.

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1 MAT9 - Calculus III - Fall 04 Solution of Term Test - October 6, 04 Time allotted: 90 minutes. Aids permitted: None. Full Name: Last First Student ID: Instructions DO NOT WRITE ON THE QR CODE AT THE TOP OF THE PAGES. Please have our student card read for inspection, turn off all cellular phones, and read all the instructions carefull. DO NOT start the test until instructed to do so. This test contains 6 paes (includin this title pae). Make sure ou have all of them. You can use paes 4 6 for rouh work or to complete a question (Mark clearl). DO NOT DETACH PAGES 4 6. GOOD LUCK!

2 PART I No explanation is necessar. For questions 8, consider a constant a R and the differential equation. (8 marks) d dt = ( + a)( a).. If a > 0, then the critical point a is stable / semistable / unstable. If a > 0, then the critical point a is stable / semistable / unstable 3. If a < 0, then the critical point a is stable / semistable unstable 4. If a < 0, then the critical point a is stable / semistable / unstable 5. Without solvin the differential equation, sketch the solution for a = with the initial condition () = Pae of 3

3 6. For a =, the solution has an asmptote at = as t + if the initial condition is (a) (4) = 0 (c) (0) = ( 8) = (d) Onl the equilibrium solution can have asmptote at =. 7. For a =, the solution has an asmptote at = as t + if the initial condition is (a) (0 0 ) = 0 (04) = (a) ( 4000) = (c) Onl the equilibrium solution can have asmptote at =. 8. Let a < 0 and let = φ(t) be the solution with initial condition (0) = a. Then the maximum of φ(t) for t 0 is max φ(t) = t [0, ) a/. Pae 3 of 3

4 PART II Justif our answers. 9. Consider the autonomous differential equation = f(), with a critical point c. (8 marks) (a) Assume that f (c) > 0. Graph z = f() for values of near c. z c Is c stable or unstable? Justif our answer. Solution. followin The critical point c is unstable, because in the raph from part (a), we have the If > c, then > 0. This implies that is becomin larer: movin awa from c If < c, then < 0. This implies that is becomin smaller: movin awa from c So solutions with initial condition (t 0 ) = 0 for 0 near c, will move awa from c. Thus the critical point is unstable. Pae 4 of 3

5 (c) Assume that f (c) < 0. Graph z = f() for values of near c. z c (d) Is c stable or unstable? Justif our answer. Solution. followin The critical point c is stable, because in the raph from part (a), we have the If > c, then < 0. This implies that is becomin smaller: movin towards c If < c, then > 0. This implies that is becomin larer: movin towards c So solutions with initial condition (t 0 ) = 0 for 0 near c, will convere to c. Thus the critical point is stable. Pae 5 of 3

6 0. You are a baseball pitcher and ou want to throw a ball from our position (0 marks) to the catcher 8m awa and m below our throwin position. Consider ravit onl. (a) If the pitcher throws the ball horizontall, how fast should he throw it? And how much time will it take 8 x for the ball to reach the catcher? Proof. Solution. ] Usin Newton s nd Law of motion, we have F = m a. Define ( x(t), (t) ) as the position of the ball at time t. Then a = ( x (t), (t) ) and F = (0, m). So we have the differential equations: x (t) = 0 and (t) =. ( ) The solution is x(t) = u 0 t + x 0 and (t) = t + v 0 t + 0, assumin the initial conditions ( x(0), (0) ) = (0, 0) and ( x (0), (0) ) = (u 0, v 0 ). The ball is thrown horizontall, so v 0 = 0. Also, we want the ball to reach the catcher, so the solution must satisf x(t ) = 8 and (T ) =. This implies that u 0 T = 8 T = u 0 = 8 T T = u 0 = 8 T = The pitcher should throw the ball at 8 m/s and it will take s to reach the catcher. Pae 6 of 3

7 Assume that the pitcher is used to cricket: he throws the ball horizontall, the ball bounces once on the round (m below the throwin position), but loses a quarter of its velocit on the bounce. With exactl one bounce, how fast should he throw the ball? 8 x Solution. We need to split the calculations in two parts: before and after the bounce. Before the bounce. We have x(t) = u 0 t and (t) = t. Define t b = time when the ball touches the round. Then (t b ) =, which implies that t b = 4 t b =. This means that x(t b ) = u 0 and the velocit at the time of the bounce is x (t b ) = u 0 and (t b ) = t b =. After the bounce. solution is the same The differential equation after the bounce is the same as before, so its x(t) = u b (t t b ) + x(t b ) and (t) = (t t b) + v b (t t b ) + (t b ), where u b = 3 4 u 0 and v b = 3 4 = 3. We have x(t) = 3 4 u 0(t t b ) + u 0 and (t) = (t t b) + 3 (t tb ). We now need to find T such that x(t ) = 8 and (T ) =, Pae 7 of 3

8 which implies (T t b) + 3 (T tb ) = (T t b) 3 (T tb ) + = 0 3 ± 9 4 T t b = T t b = 3 ± So we have two solutions T = t b + or T = t b +. The initial speed of the ball u 0 which is the solution of ( 3 u 0 4 (T t b) + ) = 8u 0 = = 36 7 or u 0 = = 7 m/s. m/s. Pae 8 of 3

9 . (a) Find the eneral solution of the differential equation (8 marks) ( cos()x 3 ) (x) = 3x sin() + cos(x). (Hint. Solution. and You can leave the solution in implicit form) This equation is exact: ( 3x sin() + cos(x) ) + ( cos()x 3) (x) = 0, }{{}}{{} M(x,) N(x,) M = 3x cos() = N x. This means that we can solve it b findin ψ(x, ) such that ψ x = M ψ = M(x, ) dx = x 3 sin() + sin(x) + h(). We now find h() usin ψ = N: ψ = x 3 cos() + h () = x 3 cos() = N so We can then take h () = h() = + C. ψ(x, ) = x 3 sin() + sin(x) +, and the eneral solution is iven b x 3 sin() + sin(x) + = C. Pae 9 of 3

10 The differential equation ( ) x cos()x (x) = 3x sin() is not exact. Find an interatin factor µ(x, ) to make this equation exact. Justif our answer. Solution #. b µ(x, ) = x, we et which is exact: This differential equation is ver similar to the one is part (a). If we multipl it ( cos()x 3 ) (x) = 3x sin(), }{{}}{{} N(x,) M(x,) M = 3x cos() = N x. Solution #. Then If we multipl the DE b xµ(x, ), we et ( ) 3x sin()µ + }{{} x cos()x µ (x) = 0, M }{{} N so we want to choose µ such that M = 3x cos()µ 3x sin()µ N x = x µ + x µ x x cos()µ cos()x µ x, 3x cos()µ 3x sin()µ = x µ + x µ x x cos()µ cos()x µ x If we choose µ = µ(x), then we et 3x cos()µ = x µ + x µ x x cos()µ cos()x µ x x cos() (xµ x µ) = x (xµ x µ) (x cos() x ) (xµ x µ) = 0 So we can choose µ(x) which satisfies The interatin factor is µ(x) = x. xµ x µ = 0 µ µ x = x µ = x Pae 0 of 3

11 . Consider the followin initial value problem: (8 marks) = + (0) = (a) Usin Euler s Method with h =, approximate the solution at t =. Solution. First we write the differential equation as = ( + ). Euler s method ields: 0 = = + f(0, ) = + = 3 = 3 + ( f, 3 ) = = = 39 6 Find the solution of the initial value problem and compute the error of the approximation in (a) at t =. Solution. The DE is separable: The solution satisfies Usin partial fractions, we write ( + ) =. ( + ) d = dt. ( ) d = t + + k ln ln + = t + k + = ce t + = ce t We can find c usin the initial condition: = c. Pae of 3

12 So we can find explicitl: + = e t = ( + )e t ( ) e t = e t = e t. e t This ives () = e. e So the error of the previous approximation is e error = () = e (c) If we need to obtain an error 50 times smaller, which step size h should we choose? Solution. Since Euler s method has an error of the order of h, to obtain an error 50 times smaller, we need h to be 50 times smaller. So we need h = 50 = 00. Pae of 3

13 3. Consider functions p(t) and (t) continuous for t (a, b) and consider the initial (8 marks) value problem + p(t) = (t) for t (a, b) (t 0 ) = 0, where a < t 0 < b. Let φ(t) and ψ(t) be two solutions of this initial value problem. Show that φ(t) = ψ(t) for t (a, b). Hint. Split the proof in three steps: (a) (c) Define F (t) = φ(t) ψ(t). Show that F (t) is a solution of the initial value problem F + p(t)f = 0 for t (a, b) F (t 0 ) = 0. Solve this differential equation and find F (t). Conclusion. Solution. (a) First define F (t) = φ(t) ψ(t). Then F + p(t)f = φ (t) ψ (t) + p(t) ( φ(t) ψ(t) ) = φ (t) + p(t)φ(t) ( ψ (t) + p(t)ψ(t) ) = (t) (t) = 0. and F (t 0 ) = φ(t 0 ) ψ(t 0 ) = 0 0 = 0. The equation is separable: we can write it as F df = ln F = p(t) dt p(t) dt + k F = c e p(t) dt We can use the initial condition to find c = 0 and we obtain F (t) = 0. (c) This implies that φ(t) = ψ(t). Pae 3 of 3 The end.

Autonomous Equations / Stability of Equilibrium Solutions. y = f (y).

Autonomous Equations / Stabilit of Equilibrium Solutions First order autonomous equations, Equilibrium solutions, Stabilit, Longterm behavior of solutions, direction fields, Population dnamics and logistic