MATH 222 (Lectures 1,2, and 4) Fall 2015 Practice Midterm 2.1 Solutions. Carolyn Abbott Tejas Bhojraj Zachary Carter Mohamed Abou Dbai Ed Dewey

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1 MATH (Lectures,, and 4) Fall 05 Practice Midterm. Solutions Student ID#: Circle your TA s name from the following list. Carolyn Abbott Tejas Bhojraj Zachary Carter Mohamed Abou Dbai Ed Dewey Jale Dinler Di Fang Bingyang Hu Canberk Irimagzi Chris Janjigian Tao Ju Ahmet Kabakulak Dima Kuzmenko Ethan McCarthy Tung Nguyen Jaeun Park Adrian Tovar Lopez Polly Yu Please inform your TA if you find any errors in the solutions. Score Problem Problem Problem 3 Problem 4 Problem 5 Problem 6 Problem 7 Instructions Write neatly on this exam. If you need extra paper, let us know. On Problems,, and 3 only the answer will be graded. On Problems 4, 5, 6, and 7 you must show your work and we will grade the work and your justification, and not just the final answer. Limited partial credit will be available. Problem 3 is worth 0points. All other problems worth 5 points. No calculators, books, or notes (except for those notes on your 3 inch by 5 inch notecard.) Please simplify any formula involving a trigonometric function and an inverse trigonometric function. For example, please write cos(arcsin x) = p x. Note that we have provided some formulas on the next page to help with this.

2 Formulas T e x = P k=0 xk k! T sin x = P xk+ k=0 ( )k (k+)! T cos x = P xk k=0 ( )k (k)! T x = P k=0 xk T +x = P k=0 ( )k x k T ( + x) b = P k=0 b k xk where b k = b(b )(b ) (b k+) k!

3 . On this page are three True/False statements. On the following page you will be asked to match direction fields to their defining equations. CIRCLE the correct answers below. (a) (b) (c) (d) (e) True False True False True False I II III I II III True or false: (a) The function (x + x 3 ) is o(x 3 ). (b) R 4 e x = e x ( + x + x + x 3 + x 4 ) (c) Let f(x) andg(x) befunctionswhosetaylorseriesexist. Thenforanyn we have T n (f(x)g(x)) = (T n f(x))(t n g(x)).

4 Below are three direction fields. The equations for two of those fields are given below. Match the equation to the appropriate direction field and record your answer on the previous page. (d) dy dx = x(x )( y) (e) dy dx = x + y I. II. Solution: III.

5 (a) True. (b) False. (c) False. (d) III (e) I

6 . (a) Use Euler s method with step size h =0. to approximate y(.) where y(x) isthe solution of dy = y +3x and y(.0) = 0. dx Solution: x k y k m k = y k +3x k y k+ = y k + m k h x 0 = y 0 =0 (0) + 3 () 0 + 3(0.) = 0.3 x =. y =0.3 So we have that y(.) 0.3. (b) Find T (arctan x). You may find it helpful to recall that d dx (arctan x) =. You +x do not need to simplify any values of arctan in your answer. Solution: We compute: f(x) = arctan x f 0 (x) = +x =(+x ) Thus: f() = arctan f 0 () = + = 5 This yields: T (arctan x) =f() + f 0 ()(x ) = arctan() + (x ) 5 (c) Find T of p +x (your final answer must not contain any binomial expression a b ). p Solution: We have T +u = T ( + u) / =+ / u + / u. Substituting u =x yields p / / T +x =+ (x)+ (x). Lastly we need to rewrite the binomial expressions. We have / = (/) =(/) and / = ( ) ( )! = 4 = 8.Thus: T p +x =+(/)(x)+( /8)4x =+x x.

7 3. In the problem below:. Clearly define variables (including units!);. Set up the appropriate di erential equation; and 3. Write down the appropriate initial condition. DO NOT SOLVE THE DIFFERENTIAL EQUATION. Start with a full 50 quart vat containing a 8% (by volume) solution of vinegar, at 0:00am. A solution of 5% vinegar flows in at a rate of quarts per minute. The solution is kept thoroughly mixed and drawn o at a rate of 3 quarts per minute. We are interested in a function describing the total amount of vinegar in the vat at a given time. Variables (pts): Di erential equation (5pts) Initial condition (3pts): Solution: Variables: t = time in minutes from 0:00am and A(t) = amount of vinegar in the vat at time t, in quarts. Di erential equation: da dt =. Initial condition: A(0) = 4 3A 50 t

8 4. Find a solution to each initial value problem. (a) dy dx e x y = e x and y(0) = 0. Solution: Rewriting, we get the separable di erential equation: dy dx = ex ( + y ) After separating variables we get: Z Z dy +y = e x dx arctan(y) =e x + C y = tan(e x + C) Plugging in y(0) = 0 gives: 0 = tan(e 0 + C) and thus C = yielding y = tan(e x ). (b) ( + x ) dy dx +xy = 3( + x ) and y() = 5. Solution: Dividing through by (+x )yields dy dx di erential equation with a(x) = we get m(x) =e ln +x = +x =+x. We then get: y = Z +x ( + x )(3)dx = Z +x 3+3x dx +y x +x = 3. This is a then a linear x and k(x) = 3. Since R x =ln +x + C +x +x = +x [3x + x3 + C]. Plugging in y() = 5 gives: 5 = [3 + + C] and thus C =yielding y = +x [3x + x 3 + ].

9 5. We have a vat containing a mixture of acid and water. Let: t stand for time in minutes from :00pm A(t) denote the total amount of acid in the vat at time t V (t) denote the total volume of liquid in the vat at time t. Assume that da dt =3 A 0 t and A(0) = 0 and V (t) = t. What is the concentration (%) of acid in the vat after 0 minutes? Solution: We want to know A(0)/V (0). The di erential equation da dt + A 0 t =3is linear with a(t) = 0 t and k(t) = 3. The integrating factor is m(t) =er a(t)dt = e ln 0 t = 0 t Note that the initial value is at t = 0 and 0 0 is positive, so we can replace 0 t by (0 t) near t = 0. We check that this is the right integrating factor by checking whether dm dt we check: dm dt = (0 t) 3 ( ) = (0 t) 3 and m(t)a(t) = (0 t) 0 t = (0 t) 3. We then get A = Z m(t)k(t)dt m(t) Z = (0 t) (0 t) 3 We use the initial condition at t = 0 to get: Our equation is: We compute = (0 t) [3(0 t) + C] = 3(0 t)+c(0 t). 0 = 3(0 0) + C(0 0) = C and thus C = = 3 0 A(t) = 3(0 t) 3 (0 t) 0 = m(t)a(t). So A(0) V (0) = 3(0 0) 3 0 (0 0) (0) = = = 3 00 After ten-minutes, the concentration of acid in the vat is 3% or 3 00.

10 6. Let f(x) =e 5 x.findanumberb such that f(x) T 4 f(x) appleb for all x in the range apple x apple. You must justify your answer. Solution: We want to use the error-bound formula for R 4 f(x). We have f(x) =e 5 x and c = and n = 4. But we need to find M which requires bounding f (5) (x). So we compute: f 0 (x) = e 5 x f () (x) = e 5 x f (3) (x) = 3 e 5 x f (4) (x) = 4 e 5 x f (5) (x) = 5 e 5 x We want to maximize f (5) (x) = 5 e 5 x on the range apple x apple. This will be largest when the exponent 5 x is largest, which will happen when x = and 5 x = 7. Thus f (5) (x) apple 5 e 7 for all apple x apple. We choose M = 5 e 7. We then apply the error bound formula to obtain: R 4 f(x) = f(x) T 4 f(x) apple Mcn+ n! Thus B = 5 e 7 5! is our answer. = (5 e 7 ) n+ 5! for all apple x apple.

11 7. Let f(x) be a function satisfying the di erential equation f 00 (x)+sin(x ) f(x) =0 and also satisfying the initial conditions f(0) = and f 0 (0) = 0. Compute T 4 f(x). Note: it is essential that you use notation correctly in your answer, as part of what we are testing is whether you understand what the notation means. Solution: We write f(x) =a 0 + a x + a x + a 3 x 3 + a 4 x 4 + o(x 4 ). By definition of the Taylor series of f(x), we have that a 0 = f(0) and a = f 0 (0). Thus a 0 = and a = 0. Thus f(x) =+a x + a 3 x 3 + a 4 x 4 + o(x 4 ). We then compute: and also Note that f 0 (x) =a x +3a 3 x +4a 4 x 3 + o(x 3 ), f 00 (x) =a +6a 3 x + a 3 x 4 + o(x ). T sin(u) =u u 3 3! + u5 5! u 7 7! +... By the substitution method with u =x (which is a polynomial), we then get: T sin(x )=(x ) (x ) 3 + =x + o(x 4 ) 3! (NB: we only track terms of degree at most 4 because we are interested in computing T 4 f(x).) We then compute 0=f 00 (x)+sin(x ) f(x) =(a +6a 3 x + a 4 x + o(x )) + (x 8x 6 + o(x 5 )) ( + a x + a 3 x 3 + a 4 x 4 + o(x 4 )) 3! =(a ) + (6a 3 )x + (a 4 + a )x + o(x ) By equating coe cients we see that 0 = a and thus a =. Also a 3 = 0. And a 4 + a = 0, but a = and so a 4 = and thus a 4 =. We conclude that T 3 f(x) =+x x4.

Carolyn Abbott Tejas Bhojraj Zachary Carter Mohamed Abou Dbai Ed Dewey. Jale Dinler Di Fang Bingyang Hu Canberk Irimagzi Chris Janjigian

MATH 222 (Lectures 1,2, and 4) Fall 2015 Midterm 2a Name: Student ID#: Circle your TA s name from the following list. Carolyn Abbott Tejas Bhojraj Zachary Carter Mohamed Abou Dbai Ed Dewey Jale Dinler