Radhika Gupta Math 2210-005, Fall 2014 November 5 Midterm 2 INSTRUCTIONS: 1. Calculators are NOT allowed. 2. REMEMBER : Don’t spend too much time on problems which are worth very few points. 3. Don’t get stuck on a problem, KEEP MOVING ON and in the end come back to the problems you could not figure out on first try. 4. Show proper work to get full points. If your answer is wrong you still have a chance of getting partial credit for the work. Maximum Points 75 points Number of Pages : 6 Question No. Marks 2 3 4 5 ‘Total BEST OF LUCK!! 1 c) z :i c . c) — I—’ . - - c2.) Lt ‘ ‘ C - ‘ ti— ‘Z - 4- (—c_ - ,— U - — — +- -t 1 - II H < — Hi Radhika Gupta 2 Math 22 10-005, Fall 2014 November 5 Chain Rule [10 points] Let f(x.p) sin(iy),x = 2 + t s 2 and y = using the chain rule. Express your answer in st. Find terms of s. t. (No need to simplify too much.) Lx= N ?j) / ± 3 Max Mm -t Problem [20 points] (a) [5 points] Find the local maximum arid minimum value of f(x, y) xy on its domain. D 1 <0 :j ‘X-D, JO -yiVn. 0(0) t )1 ) , - i) (ç)I <0 [os) - ve fl (( ? ci C ‘1’ 0 C I; C 0 - A 0 H ,- — C ) -C_c _) - II Thr-’1 q--± IL rt • 0 Cc 0 ii C 0 A. (J CM +cJ’ - CD CD CD -I . - CM 0CD 0 — CD CD CD N II .- 0 0- c 0 CD CD cl CD CDCD oaq CD CD- CD o o aq CDCD CD o —. o IA CD CD 01 0 CD CD z 0 1’.z CD 01 CD \ Ii 4—, ‘0 - Cc ) If —. sz H ID C?’ — — I ITh K (f. ‘N -5 flN -5 I) \__, 1 \-_-- .—.z ) II E 0 S - 1J ‘—; rj :- — — -S - — 0 Hc — A i ; - c-C 1 I > c [ ii L—__, •S cJ )\ Radhika Gupta 4 Math 2210-005, Fall 2014 November 5 Normal line at a point on a surface [15 points] A fly sat at the point (1,2,—i) on a grape in the shape of an ellipsoid 8x 2 + 4y 2 + 8z 2 —32 = 0. Assume the origin is at the center of the grape. The grape was not sweet enough so it decided to fly away. It took off along the normal line to the grape. (a) Draw a cartoon of the ellipsoid with the point (1,2, —1) on it and the normal line at (1,2, —1). (b) What is the equation of the line along which the fly flies off? CA) Yio’fl9’d 2- rx .+ knvo --iab _ 2- c7f2 - ie1 I) = ( 1 2,) <2, 29- (j j--Q Df I 5 + -I —/t 4- r- c:j sJ V( n 1. Jj 2 ) N — Ii N 2 CD CD Ci CD CD CD II CD 0 CD 0- 0 CD CD CD — CD CD . ‘ 0 CD CD CD 00- I’) CD CD CD- cc: CD q5L CD 0 0—, 0 0 CD c1 C I-. aq CD II CD 0 CD C CD z CD LZ 1) CD