answers1 - Mathematics 234, Fall 2004 Lecture 3 (Wilson)...

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Unformatted text preview: Mathematics 234, Fall 2004 Lecture 3 (Wilson) First Midterm Exam October 4, 2004 ANSWERS Problem 1 In class we saw pictures of the helix (corkscrew) given by the position vector m) : sin(t)?+ tj+ cos(t)lZ. Calculate the arclength of one turn of this helix, from t = 0 to t : 271 ANSWER: The velocity vector 17(t) is cos(t)T+f—sin(t)E, so (17‘ is (/cos2(t) + 1 + sin2(t) : V? at any value of t. Thus the arclength is 271' fidt : 2\/§7r. 0 Problem 2 Let F(t) : 3sin(t)?+ 3cos(t)f+ 4tk describe the motion of an object along a curve in space. Find as functions of t: (a) The velocity 17(t) The acceleration 5(2?) The unit tangent vector ) ) d) The principal unit normal vector 1W1?) ) The curvature H(t) ) The tangential (scalar) component of acceleration (1T (g) The normal (scalar) component of acceleration aN ANSWERS: Differentiating, 27(t) : 3 cos(t)?— 3 sin(t)f+ 4i? and (7(1?) 2 —3 sin(t)7— 3 cos(t)f+ 01?. To find T, we need a vector of unit length in the direction of 17(25). The magnitude of 17(25) is |27(t)| = (/9 cos2(t) + 9sin2(t) + 16 = m = 5. Thus “(t)_1_, _3 H 3. #4 _ — Eva) — ECOS(t)Z — 6811105)]... 6 IE. There are several ways to proceed to JV. One way is to go ahead and find the components of acceleration, CLT and cm, and since we need them anyway I will do that. For the tangential component we have am) : cm) for) : —3sin(t) >< geese) + (—3cos(t)) >< (—gsin(t)) + 0 x g = 0. Since (7(2?) 2 aTT(t) + aNZWt) : (MIN/(t), where = 1, we have is a unit vector in the direction of (7(1?) and (ii) cm is the magnitude of (fit). Hence cm 2 9sin2(t) + 9sin2(t) 2 \/§ : 3, and = écflt) : —sin(t)i’— cos(t)f+ 01;. This leaves only Mt) to be determined. Again there are several ways to find it: Given what we have already computed, it may be easiest to compute GN 3 H“) : W2 Z % which happens not to depend on 15. Problem 3 For each of the four descriptions, fill in A or B or C or D to indicate which graph below it corresponds to. (i) 2; : y sin($) ANSWER: Notice that figure C has a sine—wave cross—section in one direction, like .2 : sin(x), while in the other horizontal direction it has cross sections that are straight lines, like 2 =(some constant)y. C is the answer here. (ii) 7" = 1 + cos(6), any value of z ANSWER: From the variables 7“, 6, and z, we can infer this is in cylindrical coordinates. Since .2 does not appear in the equation, if any point is on the surface so is the entire vertical line through that point. That alone is enough to pick out figure D. When you also notice that a horizontal cross—section will be 7" = 1 + cos6 in polar coordinates, a cardioid, you have confirming evidence. (iii) z:3—$2—y ANSWER: This surface slopes downward as y increases, and upward in the —y direction, and follows the downward opening parabola z = 3 — x2 as you move in the ::x directions along the :r—axis, and surface B certainly fits this description. (iv) p : sin(q§), any value of 6 ANSWER: This must be in spherical coordinates, and since 6 does not appear in the equation it will be completely symmetric around the z axis. That picks out surface A. In addition, for any fixed 6, the cross—section will be p = sin gb: 7“ : sinH gave a circle in polar coordinates, so looking out from the z—aXis in any direction we should see a circle: This confirms the ” donut” shape in figure A. x is . xfifigfifigflufi .\ . l m N ¢ \ xx . 2 a. ___.______________________ m W, EEEESEE m x, .l e§§w§§§x a fe §§§§3. .‘AVVSR‘KS KW / fir. . an) a); 0 Ham 2 1 8 4% 3“ + 2y). 2 ll} ( = 2:1: cos Problem 4 find: 2 sin(x2 + 2y 1193731) For ANSWER: Remembering the Chain rule7 % : cos(x2 + 2y) >< 2x + 2y) X 2 : 2(:0s(x2 + 2y). 2 : cos(:1: fl 3y ANSWER: Similarly7 {if 8:162 ) C ( ANSWER: We had % = 2x cos(ar2 + 2y): we differentiate that with respect to 36, using the product and chain rules, we get % : 2 cos(x2 + 2y) — 4:152 sin(x2 + 2y). 32 f (d) agar ANSWER: Since this says to take the derivative first with respect to ac and then with respect to y, we take the derivative of % = 236 cos(ac2 + 23/) with respect to 3/. Using the chain rule we get 883/28]; : —4x sin(x2 + 2y). 82 (e) f amay ANSWER: We can cite the theorem that says if the partials are all continuous then the mixed second partials will be equal, so the answer should be the same as for Or we can directly compute the derivative of 2cos(ar2 + 2y) with respect to 1:. Either way, we get a2 - aw]; : —4:E s1n(a:2 —I— 2y). 2 (f) ANSWER: Differentiating % : 2 cos(:v2+2y) with resepect to y we get 227’; = —4 sin(:v2+2y). Problem 5 (a) Evaluate lim $2 cos(2y). ($7y)—>(2,7T) ANSWER: This function is constructed by multiplying two functions (362 and cos(2y)) which are each continuous at all input values, so it is continuous at all points (1:, 3/). Hence we can evaluate the limit by “plugging in77 and ( lian ) $2 cos(2y) : 22 cos(27r) : 4. J},y H 771' (b) Show that lim L does not exist. ($7y)—>(0,0) V $2 + y2 ANSWER: If we approach (0, 0) along the y—axis, where 36 = 0, the function is constantly 0 for all values of y 7E 0. Hence the limit is 0 as we approach the origin along the y—axis. If we instead approach along the x—axis, where y : 0, the function values are For 2 :1: > 0 this evaluates to 1, so the limit coming in along the postive :r—axis is 1, vxfhich is not the same as the 0 we got along the y—aXis, and so the limit must not exist. In fact the limit coming in along the negative m—aXis is actually —1, since for :v < 0 we have V? : : —x, so we even get different limits coming to zero along the $—aXlS from opposite sides, and so we did not actually need the y—aXis result. Problem 6 Let f(a:, y) : $2 69. Let P be the point (1, 0). (a) Find the gradient Vf at the point P. ANSWER: The gradient at any point (36,31) is Vf(yc, y) = 236 eyfl— 362 69f, so when ac : 1 and y = 0 we have Vf(1,0) : 27+f (b) Find the directional derivative of f at P in the direction of an arbitrary vector 17 : via—112 f. ANSWER: First find a unit vector in the direction of 17: Let 73 fir? = 1 A v17+ 1 mi. Now the directional derivative Dgf(1,0) : Vf(1, 0) - U m Um; W ‘ (c) Find the directional derivative of f at P in the direction from (1, 0) to (47 4). ANSWER: In this case the vector 17 2 374—41“, where v1 : 3 and v2 : 4, is in the correct direction. Then MD? + 21% 2 V9 + 16 : 5 and the answer to (b) becomes “2+4 : 2. (d) In what direction is the directional derivative of f at P largest? What is the directional derivative in that direction? ANSWER: The direction of the gradient, 2T+ f, gives the direction in which the direc— tional derivative is greatest. In the form of a unit vector this is %?+ if. As we have seen, the directional derivative in that direction will be the magnitude of the gradient, x/E. Problem 7 Suppose w : sin(:1:y) +3: Sin(y), where :1: : u2+v2 and y : 2u—l—v — 2. Using the Chain rule: (a) 2—: = (y 008(xy) + Sin(y)) >< 2U + (:16 COS(:Uy) + x cos(y)) X 2. (b) 66—1: = (y 008(xy) + Sin(y)) X 22} + (x cos(xy) + :U COS(y)) X 1. Problem 8 EDGE} 983 The north—south direction is measured as 869 y and the east—west direction as x. The 500 we Suppose you are walking over some hills. altitude in feet is given by , my a: , = 1000 — 100 — — - fix 9) 8m 3000 100 If you are at the point where a: : 100 and y : —100, and you start walking toward the center where :1: : O and y : 0, will you begin by walking uphill, downhill, or horizontally? Be sure to show how you determine your answer. ANSWER: We need to find the directional derivative for the given altitude function f (:10, y) in the direction of a vector from (100, —100) toward (0,0). A unit vector in that direction i —%2_’+ if The partial derivatives of f are % : —§/—0 cos (fl) — {iv—0 and % : 6 (a) — At (100—100) we get = — 2a and 2—5 = —13—0 cos(—%) + 1. Thus the directional derivative we seek is —%(% cos(—%) — 2) + %(—13—0cos(—%) —|— 1) which simplifies to 1 20 _fi (3 cos(—13—0) — 3). With a calculator you can approximate that as +6.748975979, but all we need to know is whether it is positive or negative, and we can tell that without a calculator: m : 3% is just a 3 bit larger than 7r, so the point on the unit circle determined by —% is just above the x—axis and nearly at :10 = —1. Thus cos(—E) is 3 20 10 negative (nearly —1) so both the terms in g cos(—§) — 3 are negative, so their combination must be < O, and multiplying by —% will give us a positive number. Hence the directional derivative is positive and our path leads uphill. ...
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This note was uploaded on 05/15/2008 for the course MATH 234 taught by Professor Dickey during the Fall '08 term at Wisconsin.

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answers1 - Mathematics 234, Fall 2004 Lecture 3 (Wilson)...

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