APPM 2350 Section Exam points Wednesday October 24, 6:00pm 7:30pm, 2018

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1 APPM 250 Section Eam 2 40 points Wednesda October 24, 6:00pm 7:0pm, 208 ON THE FRONT OF YOUR BLUEBOOK write: () our name, (2) our student ID number, () lecture section/time (4) our instructor s name, (5) a grading table. Tet books, class notes, calculators are NOT permitted. A one-page one-sided crib sheet is allowed. Problem True/False: (20 points) For the following true/false questions, write TRUE (for alwas true) or FALSE (if not alwas true). Your work will not be graded. (a) If f(, ) L as (, ) (a, b) along ever straight line passing through (a, b), then (b) Given f((r, s), (r), z(r, s, t)) then lim f(, ) = L (,) (a,b) r = r + d dr + z z r (c) Given f(,, z) = z the smallest (ie most negative) derivative of f(,, z) at the point (, 2, ) (out of all possible directions) is 7. (d) The vector f(a, b) is normal to the graph of z = f(, ) at the point (a, b, f(a, b)). (a) False. In order for the limit to eist, the limit along ever path through that point must agree, not just the straight line paths. (b) False. r = r + d dr + z z r (c) True. The smallest derivative is f(, 2, ). f(,, z) = zi + zj + k = f(, 2, ) = 6i + j + 2k = f(, 2, ) = = 7 (d) False. The vector f(, ) is normal to the level curves of f(, ). Thus the vector f(, ) is normal to the graph of the level curve f(, ) = f(a, b) at the point (, ) = (a, b). Problem 2 Short Answer Questions: (60 points) For the questions in this problem, show all our work clearl bo our final answer. Partial credit ma be given. (a) Find z/ for z = z(, ) that satisfies the relation ln (z) + sin (z) z2 2 = π. (b) Find a value of the constant c that makes the following function continuous at (0, 0): e f(, ) = ( if (, ) (0, 0) ) c if (, ) = (0, 0) (c) If g(s, t) = f((s, t), (s, t)) where (s, t) = s + t (s, t) = s t, find a simplified epression for g tt in terms of derivatives of f. Assume all partial derivatives involved eist are continuous. (d) At a certain point P on a mountain, a skier looks directl east measures a 0 drop-off (decline), then looks directl north measures a 45 rise (incline). Using these two measurements of the mountain topograph, find a vector that points in the direction of steepest descent from the point P.

2 (a) We use implicit differentiation (ln (z) + sin (z) z2 2 ) = (π) ( z z + cos (z) z + z ) z z = 0 ( ) z + cos (z) z = z cos (z) z z = z z cos (z) + cos (z) z or z = z 2 cos (z) + z cos (z) z 2 (b) Converting to polar coordinates, note that lim f(, ) = lim (,) (0,0) r 0 + e r2 2re r2 = lim r 0 + 6r = lim r 0 + er2 =, r 2 indeterminate 0 0 L Hôpital s Rule so we should set c =. (c) Set = s + t = s t. Then g t = f t + f t = f f g tt = (f ) t (f ) t = [(f ) t + (f ) t ] [(f ) t + (f ) t ] = f f (f f ) = f 2f + f where we have used Clairaut s theorem. (d) If the elevation on the mountain range is given b z = f(, ), the given information sas that = tan(0 ) = = tan(45 ) =,

3 V + - FIGURE. Two-resistor circuit with resistors in parallel. so the gradient at the point the skier measures is,. This points in the direction of steepest ascent, so the vector, points in the direction of steepest descent. Problem : (0 points) The total resistance R of two resistors that are connected in parallel, each with resistances such as in the circuit in Fig., are related according to R = +. (a) If,, R are allowed to var, find an epression for the differential dr in terms of the differentials d d. (b) You have designed a two-resistor circuit like the one in Fig. to have resistances = 0 Ohms = 40 Ohms, but there is alwas some variation in manufacturing the resistors received b our firm will probabl not have these eact values. Will the total resistance R be more sensitive to variation in or to variation in? (c) Find the linear approimation (linearization) L(, ) of the function R(, ) about the resistances (, ) = (0, 40) Ohms. (a) Differentiating the epression, we obtain Solving for dr, dr R 2 = d 2 d 2. dr = ( ) R 2 d + ( ) R 2 d. (b) Inserting = 0 = 40 into the result from part (a), we obtain ( ) R 2 ( ) R 2 dr = d + d, 0 40 so that dr will be more sensitive to d because (R/0) 2 > (R/40) 2. (c) We need to compute the first partials R R. Either from part (a) or direct calculation, we obtain R = ( ) 2 + R = ( + ) 2.

4 Evaluating at (, ) = (0, 40), we have R(0, 40) = (/0 + /40) = (5/40) = 8, R (0, 40) = ( + /4) 2 = (4/5) 2 = 6/25, R (0, 40) = ( + 4) 2 = /25. Then the linear approimation near (, ) = (0, 40) is R(, ) L(, ) = ( 0) + ( 40) 25 = Problem 4: (0 points) You recentl moved to the spherical planet A2PP5M0 whose radius ou have been told is 4 units. You need to find a place to live on the surface of the planet. Thankfull, before building a residence, our realtor told ou about a subterranean radiation source located at (,, ). Since ou do not want to glow in the dark, ou want to live as far from the radiation source as possible. Where on the planet surface should ou build our house? You can assume the center of the planet is at the origin of a stard z coordinate sstem. To obtain full credit, ou must use Calculus III techniques that ou have learned in this course. Let (,, z) be the coordinates of our residence. We want to maimize the distance (squared for ease of calculation) between our residence the radiation source, which we designate as f(,, z) = ( ) 2 + ( + ) 2 + (z ) 2 Since we are required to live on the surface of the planet, the coordinates of our residence are constrained to satisf g(,, z) = z 2 = (4 ) 2 = 48 Set up the Lagrange equations f = 2( ) g = 2 f = λg = 2( ) = 2λ f = 2( + ) g = 2 f = λg = 2( + ) = 2λ f z = 2(z ) g z = 2z f z = λg z = 2(z ) = 2zλ We need to solve the following sstem of simultaneous equations: = λ () + = λ (2) z = λz () z 2 = 48 (4) Note that neither nor nor z can be zero since none of Eqs. ()-() can be satisfied in that case. Hence we can write = λ = + = λ = z z = = + = = = z z = z = z =

5 Eq. (4) then ields = 2 = 48 = 2 = 6 = = ±4 The critical points are (4, 4, 4) ( 4, 4, 4) f(4, 4, 4) = (4 ) 2 + ( 4 + ) 2 + (4 ) 2 = 27 f( 4, 4, 4) = ( 4 ) 2 + (4 + ) 2 + ( 4 ) 2 = 75 Thus, we should place our residence at ( 4, 4, 4).