2419Su2011Exam1.pdf

June 27, 2011

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Notation. Any of the three variants of vector notations may be used, all of the following are the same vector: (1,2,3),h1,2,3i,ı+ 2+ 3k.

Instructions.

• All work must be shown andexactanswers are expected. You are allowed calculators, but you should only use these to check your work not to perform your work. For example,sin(2π/3) =√3/2 will be accepted,sin(π/4) =.70...will not.

• You are allowed a single standard81₂ ×11piece of notebook paper.

• Write your final answer in the answer box provided.

Problem 1 (10 pt). (a) Find a unit vector, inR3, orthogonal to2ı+ 3−k.

There are infinitely many such vectors, you just need(a, b, c)satisfying(a, b, c)(2,3,−1) = 0, e.g.,(−3,2,0),(1,1,5), etc. Now just normalize these to have length 1, i.e,(−3,2,0)/√13,

(1,1,5)/√27, etc.

(b) For what values ofb areh−6, b,2i and hb, b2_{, biorthogonal?}

Here you need(−6, b,2)(b, b2, b) =−6b+b3+ 2b=b(b2−4) =b(b−2)(b+ 2) = 0, so

b= 0,±2are the desired values of b.

(c) Find the vector in the same direction asu= (2,−2,1)of length5. The vector is

5 u kuk = 5

Problem 2 (10 pt). Let u = (2,−1,2), v = (3,3,0), and w = 13

3u−v. Let p be the projection of the vector (2,−1,2) onto w and let q be the vector orthogonal to w so that (2,−1,2) = p+q. In the notation used in class, p = proj_w(2,−1,2) and q= proj⊥_w(2,−1,2).

(a) Findw.

w= 1₃(6,−3,6)−(3,3,0)= 1₃(3,−6,6) = (1,−2,2)

(b) Findp.

p= proj_w(2,−1,2) = w(2,−1,2) kwk2 w=

(1,−2,2)(2,−1,2)

9 (1,−2,2) = 8

9(1,−2,2)

(c) Findq.

q = proj⊥_w(2,−1,2) = (2,−1,2)−8

9(1,−2,2)

(d) Findcos(θ) forθ the angle betweenu andv.

cos(θ) = uv kukkvk =

3 3√18 =

Problem 3 (10 pts). (a) Find the equation for the plane containing the pointsP(1,0,0),

Q(0,2,0), and R(0,0,3). Write your answer in the form ax+by+cz=d.

−−→

P Q×−→P R= det 



ı  k

−1 2 0 −1 0 3 

= (6,3,2)

is a vector normal to the plane. So the plane is given by(6,3,2)(x−1, y−0, z−0) = 6(x−1) + 3y+ 2z= 0, so in standard form this is 6x+ 3y+ 2z= 6.

(b) Find the area of triangle PQR.

Problem 4 (10 pts). Suppose a particle with inital velocity v0 = (−5,4,2) and initial

position r0 = (1,1,1) is moving through a field with force applied to the particle F = ma =m(cos(t),sin(t),−9)where m is the mass of the particle. Find the trajectory, r(t), of the particle.

v(t) = (sin(t) +c1,−cos(t) +c2,−9t+c3)

v(0) = (c1, c2−1, c3) = (−5,4,2)soc1=−5, c2 = 5, c3 = 2

v(t) = (sin(t)−5,−cos(t) + 5,−9t+ 2)

r(t) = (−cos(t)−5t+d1,−sin(t) + 5t+d2,−9/2t2+ 2t+d3)

r(0) = (−1 +d1, d2, d3) = (1,1,1)sod1 = 2, d2 = 1, d3 = 1

Problem 5 (10 pts). Consider the curve

r(t) = (cos(t) +tsin(t),sin(t)−tcos(t), t3).

Find the arc lengths(t) on [0, t].

dr

dt = (−sin(t) + sin(t) +tcos(t),cos(t)−cos(t) +tsin(t),3t

2_{) = (tcos(t), tsin(t),2t)}

So

dr

dt

=pt2_{+ 9t}4_=|t|p_{1 + 9t}2

and hence

arc length= Z t

0 p

u2_{+ 9u}4_du= Z t

0

up1 + 9u2_du

= 1 18

Z 1+9t2

1

√

v dv= 1 27v

3/2

1+9t2

1

= 1 27

Problem 6 (10 pts). Consider the curve r(t) =hsin(t),cos(2t), ti. Decompose accelera-tion d_dt22r att= π₄ into its normal and tangential components. (Make sure that the following are clear,T,N,aT, and aN and writer00(t) =aTT +aNN.)

dr

dt = (cos(t),−2 sin(2t),1) d2r

dt2 = (−sin(t),−4 cos(2t),0) dr

dt(π/4) = √

2 2 ,−2,1

! dr dt(π/4) = r 11 2

T(π/4) = r

2 11

√ 2 2 ,−2,1

!

d2r

dt2(π/4) = − √

2 2 ,0,0

!

aT(π/4) = d 2_r

dt2(π/4)T(π/4) =− r 1 22

d2_r dt2(π/4)

= √ 2 2 dr

dt(π/4)× d2r

dt2(π/4) = 0,− √

2 2 ,−

√ 2 ! dr

dt(π/4)× d2r dt2(π/4)

= r 5 2 aN = dr

dt(π/4)× d2_r

dt2(π/4)

d_dtr(π/4) = r 5 11

N =B×T =

r 2 5 r 2 11 √ 2 2 ,−2,1

!

× 0,− √

2 2 ,−

√ 2

!! = √1

55

Problem7 (10 pts). For the same function,r(t), as in problem 6, compute the curvature at π₄,κ π₄.

The calculations were all done above

κ(π/4) =

dr

dt(π/4)× d2_r

dt2(π/4)

d_dtr

3 =

Problem 8 (10 pts). For the same function,r(t), as in problem 6, find the equation for the osculating plane in the formax+by+cz=d. (Recall this is just the plane determined by the pointr(π/4)and the vectors d_dtr(π/4)(“velocity”), and d_dt2r(π/4)(“acceleration”).)

Again all calculations have already been done. The plane is given by:

dr

dt(π/4)× d2r

dt2(π/4)((x, y, z)−r(π/4)) = 0,− √

2 2 ,−

√ 2

!

(x, y, z)− √

2 2 ,0,

π 4

!! = 0

This becomes

Problem 9 (10pts). For the curve y =x3−x find the equation of the osculating circle at √1

3,− 2 3√3

.

To find N just find the tangent line at √1

3,− 2 3√3

, say ax+by = c, then n = (a, b) is

normal to the tangent line, thus N = ±√ 1

a2_+b2(a, b) (you must figure out which sign to

take). The center of the osculating circle is

1

√

3,− 2 3√3

+ 1

κN.

Since y0(1/√3) = 0, the tangent line at

1

√

3,− 2 3√3

is the horizontal line y =− 2 3√3 and

N = (0,1).

Curvature is

κ= |y

00_|

1 + (y0₎23/2 = 2 √

3

so the radius of the osculating circle is 1

2√3 and center of the osculating circle is

1 √

3,− 2 3√3

+ 1

2√3(0,1) =

1 √

3,− 1 6√3

.

The equation of the osculating circle is

x− √1

3 2

+

y+ 1 6√3

2 = 1

Problem 10 (10pts). Consider the spiralr(θ) =θ, find

(a) the arc length on0≤θ≤π (set up the integral, 2 bonus points if you actually work it out and show your work)

Z π

0 s

r2₊ dr dθ 2 dθ = Z π 0 p

θ2_{+ 1dθ}

To work this setθ= tan(t)so thatdθ= sec2(t)dtand√θ2_{+ 1 =}p

tan2_{(t) + 1 = sec(t)}

(here0≥0 so0≤t≤π/2and thus sec(t)≥0). So

Z p

θ2_{+ 1dθ =} Z

sec3(t)dt= 1

2 sec(t) tan(t) + ln|sec(t) + tan(t)|

+C

Nowθ2+ 1 = sec2(t) sosec(t) =√θ2_{+ 1and we have}

Z p

θ2_{+ 1dθ=} 1 2 θ

p

θ2_{+ 1 + ln}

p

θ2_{+ 1 +θ}

+C

(b) the slope of the tangent line at π

6 √ 3 2 , 1 2 . dy dx = dy/dθ dx/dθ =

sin(θ) +θcos(θ) cos(θ)−θsin(θ)

(x, y) = π 6 √ 3 2 , 1 2

at θ=π/6 so the slope of the tangent line is

1 2 + π 6 √ 3 2 √ 3 2 + π 6 1 2

= 6 +π √