math_273_(spring_2009)_practice_test_1_solutions

math_273_(spring_2009)_practice_test_1_solutions - MATH 273...

Info iconThis preview shows pages 1–7. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 6
Background image of page 7
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: MATH 273 PRACTICE TEST 1 SON/{TIM NAME: 1. (10 points). Suppose that z z 2(m, y) is a function defined implicitly by the equation 23 + x2 - y2 = 1. Use implicit differentiation to find % and 5’5 3:??? '— ; (0V: "FE—#24) o ‘93: , , 2 @X 9 522+X) 9%: 3421;? -r2< 3:? ~23] =0 (9:: - 25' 7’2] 3&‘+¥ Answer: 3—: =~E§fi,g—; = 2. (10 points). Find an equation of the tangent plane to the surface given by the graph of f(:c,y) = 2x2 +4y2 — 5 at the point P(1, %,0). than?) a 2 my 9y 2» 5122-0 VF: “"9 73)’1>' A”; VF/c1 : “taro 3:74 3to Answer: 4(x— 1)+2(y~ %)—z:0. 5":0) :0 3. (10 points). Compute all first—order partial derivatives of f(r,s,t) = (1 — 1'2 — s2 — t2)e’”t. q) ~'t$l" __}j~£ if: " “e + (1421-52'5') (*SHe “D13 th-f é ~25 ifs": “'23 e + [zwfziszwtfl-Gfik 6% J‘ch ,«29 IL ,H : ~Zf€ + [fridlc‘rfzj’wae 4. (10 points). Prove that ' 2 my 23+y3 does not exist. Hint: find two different direction of approach (x,y) a (0,0) leading to different ”limits”. ) G77/%d c4 «169.9 {:0 I 013 1) wwdw 45914.7 M=y x“ , X311" 2’ .limmwmm 5 2,4‘0 =3 72% gum/’3‘ ac; “:07! {NHL 5.(10 points). Find the equation of the plane passing through the points (3, -1, 2), (8, 2, 4) and (-1, -2, -3). cc 7W0 M0611; 1k 0% ézm; “(o-=43, '1 7»> / I L $4:<2§-3/ 2r(—1)'V—2>:4573,2> Q3: < 3+1), 2+1) fl~('3)>:<!,9,¥> C A7 V ._ E ) I? .7 %( X z), 3 F 3) Z : ‘1‘.- /}J.~p ? v 1 Answer: 13(36 — 3) — 17(y + 1) — 7(2 — 2) = O 6.(10 points). Find the maximal rate of change of ‘f(m,y) = sin(3$ — 4y) at P(§, %) and the direction in which it occurs. vafi; < 003(91 -v:,)~3)m (SK‘YJ)‘('Y)> M P -. 2H 4:; : (31’7> Answer: 5, %i — gj. 7.(10 points). The curve is given by r(t) = 2x/ii + tzj + tk. Find the unit (length must be 1) tangent vector at the point P corresponding to the value t = 1. Then find the equation of the tangent line to the curve at P. Answer: figi+fisj+71gkx=2+t,y=1+2t,z=1+t. 8. (10 points). In the two-dimensional space, the trajectory of a particle is given by . l 2. 1'“): “591$ J + Vet, where g z 10m/s2, v0 = 2\/§i + 2j. a) Find the time when the particle hits the ground; b) Find the distance d between the origin and the point of impact. (To do this, you would need to determine X—coordinate of the impact point). at) {36—— W 0% I'M/0610!; 7, ' ML m, 3H“):0 <2) igéx-Pafigz-O 4&6“: +*'('2)>0 Answer: a) Z’ b) 4455. 4‘) 1: K[+¢) 5156/: : at"? 9. (10 points). Find the length of the curve r(t) = ti+ “Tfitz j + §t3k, 0 S t g 1. Hint: to integrate, represent the quantity under the square root as a complete square. This should be easy. ETH: < 4, Fit, 46> WI“) \= \Jfilt‘r-t" : V@++Z)Ic 7.6+?— 1— ’L L :: flfialfi: M++M+= (H - 1,1 “17‘s” . Ii 0 Answer: 0914:. ...
View Full Document

Page1 / 7

math_273_(spring_2009)_practice_test_1_solutions - MATH 273...

This preview shows document pages 1 - 7. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online