Review 2, Math 2415, Ch. 12

Ace

SHORT ANSWER.  Write the word or phrase that best completes each statement or answers the question.

 

If r(t) is the position vector of a particle in the plane at time t, find the indicated vector.

1) Find the velocity vector.                                                                                                                              1)

r(t) = (cot t)i + (csc t)j

 

 

The position vector of a particle is r(t). Find the requested vector.

2) The velocity at t = π for r(t) = 5sec2(t)i – 8tan(t)j + 6t2k                                                                      2)

4

 

 

3) The acceleration at t = 3 for r(t) = (6t – 3t4)i  + (7 – t)j + (6t2 – 3t)k                                                   3)

 

 

Find the unit tangent vector of the given curve.

4) r(t) = (4 + 11t4)i + (5 + 2t4)j + (5 + 10t4)k                                                                                                   4)

 

 

Compute rʹʹ(t).

5) r(t) = (9 ln(6t))i + (8t3)j                                                                                                                                 5)

 

 

Evaluate the integral.

4

 

6)  ∫

0


i –    2t

(5 + t2)2


j + 15

2


tk  dt                                                                                                                 6)

 

 

 

MULTIPLE CHOICE.  Choose the one alternative that best completes the statement or answers the question.

 

 

 

7)  ∫


π/4


 

(8 cos t i + 5 sin t j) dt                                                                                                                                       7)

 

-π/4

A) 8   2 i + 5   2 j                 B) 5   2 i                               C) 0                                       D) 8   2 i

 

 

Solve the problem. Assume the x axis is horizontal, the positive y axis is vertical (opposite g), the ground is horizontal, and only the gravitational force acts on the objects.

8) A projectile is launched from level ground at a launch angle of 26 ° and an initial speed of 48 m/sec.      8) How far away from the launch point does the projectile hit the ground?

A) ≈ 230 m                           B) ≈ 60 m                             C) ≈ 185 m                           D) ≈ 290 m

 

Horizontal range = v^2*sin2@ / g = 48^2*sin 52 / 9.8 = 2304*.788/9.8 = 185  m

 

Answer C .

 

 

Calculate the arc length of the indicated portion of the curve r(t).

9) r(t) = 4ti  +  10 cos  3 t j   +  10 sin  3 t k; 1≤ t ≤ 8                                                                                                  9)

10                        10

A) 225                                   B) 35                                     C) 45                                     D) 175

 

 

To get the arc length, we need to integrate the square root of
(dx/dt)^2 + (dy/dt)^2 + (dz/dt)^2
where r = < x(t), y(t), z(t) >

 

=  (4)^2 + ( 3 sin 10 t / 3)^2  + (-3 cos 10t /3 )^2

 

=  16  + 9  +  9  +  (sin^2 10 t / 3 + cos^210 t / 3)

 

= 16 + 18 + 1 = 35.

 

 

 

 

 

Find the length of the indicated portion of the trajectory.

10) r(t) = (etcos t)i + (etsin t)j + 3etk,  -ln 2 ≤ t ≤ 0


 

10)

 

A)     11

2

WE  DO dr/dt

 

(Sint *e^t +cos t * e^t)I  + (e^ t  – cos t  + sin t *e^ t )*j +  3 e^t * k

 

Length  would be  =  sqrt [  e^2t (sint+cost)^2 + [e^2*t(sint  – cos t ]  +  9 *e^2t .]

 

=  e ^2 t   {  2 *(sin^2t +cos^2t )  +    9  }  =  11 e^2t

 

=  [-ln2,0]  ∫sqrt 11 *e^t

 

=  sqrt11/2.


B)     7

2


C)     7

4


D)     11

4

 

 

 

 

 

 

 

 

 

 

 

 

 

Find the curvature of the curve r(t).

11) r(t) = (7 + 6 cos 9t) i – (4 + 6 sin 9t)j + 10k


 

11)

 

A) κ = 3

2

Dr /dt  = ( 54 sin9t  ) i  + 54 cos 9t j  +

|dr/dt|=  sqrt( 54*54+54*54 )  =   1.41 *54.

Now   T’(t)‘ =  54 sin 9t +54 cos 9t  /  √2*54

=  <1/√2  (sin9t) , 1/√2   (cos 9t), 0>

T ‘(t) =<9/√2 * -cos 9t ,  9/ /√2 * sin 9t , 0>

| T ‘(t)|=  √  81/2  (cos^2 9t+  sin ^2 t) =   9 / √2

K  = | T ‘(t)| / r’(t)|| 9/√2  /   √2 *54 = 1/6

Answer = D

 B) κ = 6                                C) κ =  1

36


D) κ = 1

6

 

 

 

 

 

 

 

 

Find the curvature of the space curve.

12) r(t) = (t + 8)i + 2j + (ln(sec t) + 1)k

A) κ = sin t                           B) κ = cos t                          C) κ = sec t                          D) κ = csc t

 

 

r(t) = <t+8, 2, ln(sect )+ 1>

 

r’(t) = <1,0, tant>

 

|r’(t)|=√1 + tan ^2t   =  sec^2 t  =  sec t .

 

 

T’(t)=.  <cost ,0 , sin t>

 

| T’(t)|= √ cos^2t+sin^2t  =  1

 

So  k =  1  /  sec t   = cos t

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 


 

12)

 

 

 

 

Find the unit tangent vector T and the principal unit normal vector N.

13) r(t) = (ln(cos t) + 3)i   + 10j  + (10 + t )k, -π/2 < t < π/2

A) T = (-sin t)i  + (cos t)k; N = (-cos t)i  + (sin t)k

B) T = (-sin t)i  + (cos t)k; N = (-cos t)i  – (sin t)k

C) T = (sin t)i  + (cos t)k; N = (cos t)i  – (sin t)k

D) T = (sin t)i  – (cos t)k; N = (cos t)i  – (sin t)k;

 

 

 

 

 

dr/dt  = < -tant ,  0, 1>

 

d2r/dt2 = <  sec^2t ,  0 , 0 >

 

 

r’ =  < -tant ,  0, 1>

 

|r’|=  sqrt (  1+ tan ^2t) =  sec t

 

Tangent =  r ‘ | |r ‘ |

 

T  =   sec t  < -tant , 0 , 1>

 

=   <- sin t , 0 , cos t >

 

 

T ‘  = < -cos t , 0 , -sint >

 

|T| = sqrt ( sin^2 t+ cos ^2 t)  = 1

 

N  =  1/1< -cos t  , o , -sin t>

 

 

 

Answer  is    B

 

 

 

 

 

 

 

 

 

 

 

 

 


 

13)

 

 

 

 

FInd the tangential and normal components of the acceleration.

14) r(t) = (t + 1)i + (ln(cos t) – 2)j + 3k

A) a = (-sec t tan t)T – (sec t)N                                       B) a = (sec t tan t)T – (sec t)N

C) a = (sec t tan t)T + (sec t)N                                        D) a = (-sec t tan t)T + (sec t)N

 

 

 

 

 

 

 

 

 

Dr /dt  =  i  +   -tan t  * j   +   k

 

D2r/dt2= < 0, -sec^2t ,0>

 

|r'(t)| =   1 ^2  +  tan^2  =  sqrt(  sec^2 t )  =sec t

 

r'(t) X r”(t)  =  < 0 ,  (tant)/(cos^(2)t), 0>

 

 

A Tangential=   < 0, tant *sec t,0>

 

A normal = 1/ sec t  <  sec ^2 t >    =   sec T  N

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

So ,

 

C  =

a = (sec t tan t)T + (sec t)N       is the  answer.

 

 

 


 

 

14)

 

 

 

SHORT ANSWER.  Write the word or phrase that best completes each statement or answers the question.

 

Solve the problem.

15) Find the unit tangent and normal vectors for the curve y = t2, x = t at the point t = 1.

 

<x(t),y(t)> = <t,t^2,0>

 

 

R’(t) =<1, 2t,0 >

 

|R’(t)|= √(1 + 4t^2)

 

 

 

 

 

T(t) =  unit tangent vector =  1/√(1 + 4t^2) i + 2 *t//√(1 + 4t^2)

 

at  t  = 1

=  √5/ 5   i+ 2√5 / 5   j

 

 

Now ,

 

T’(t) =  -4 t/(4t^2+1)^3/2 i   +  2 //(4t^2+1)^3/2 j

 

Now  |T(t)| =  √  16 / (4t^2+1) ^3 +  4/(4  t^2+1)^3

 

Now    N(t)  = 4 t/(4t^2+1)^3/2 i   +  2 //(4t^2+1)^3/2 j   /   √  16 / (4t^2+1) ^3 +  4/(4  t^2+1)^3

 

At  t  =  1

 

=   -4 /(5)^3/2 i +  2 / 5^3/2  j /  √[1/ (5^3) {  20 }]

 

=   1/5 * 2 √5 i –  √5/5j

 

 

 

 

 

16) Find the curvature of x = 2t -3, y = t2 + 2t + 5 at the point where t = 0.

 

Curvature   dy/dx  =    2  /  2 *t  + 2  + 5  =  2 /2  = 1

 

 

 

 

16)

 

 

 

17)


 

 

Evaluate the integral  ∫

0


 

π/2


 

 

tsin t2 i dt .

 

 

=  -1/2  * cos  t^2

 

=

 

 

 


17)

 

 

 

 

18) A particle moves in space. Its position vector at time t is r(t) = ti + j sin t + k cos t. Find the velocity and acceleration vectors at time t = 0.

 

Dr/dt  =  1* i  +  -cos t * j  +  k  * sin t .

 

So   velocity   at  t = 0

=   <1 , -1  , 0>

 

Acceleration  =   < 0 ,  sin t  ,  -cos t >  =  < 0,0,-1>

 


18)

 

 

 

MULTIPLE CHOICE.  Choose the one alternative that best completes the statement or answers the question.

 

Solve the problem. Unless stated otherwise, assume that the projectile flight is ideal, that the launch angle is measured from the horizontal, and that the projectile is launched from the origin over a horizontal surface

 

19) A projectile is fired at a speed of 920 m/sec at an angle of 30°.  How long will it take to get 23 km downrange?  Round your answer to the nearest whole number.

A) 27 sec                                                                              B) 31 sec

C) 29 sec                                                                             D) It will never get that far downrange.

 

 

Horizontal speed

 

V cos 30  m/sec.

 

For  23 km  it will take = 23 *1000 / 920 * cos 30 =  29 sec


19)

 

 

 

 

20) A spring gun at ground level fires a tennis ball at an angle of 35°. The ball lands 24 m away. What was the ballʹs initial speed? Round your answer to the nearest tenth.

A) 15.8 m/sec                       B) 5.1 m/sec                         C) 250.3 m/sec                    D) 20.2 m/sec

 

 

24 = v^2 * sin 7 0 / 9.8

 

So  v^2  =  24 *9.8 / sin 70 = 15.8 m/sec
20)

 

Answer Key

Testname: REV2M2415

 

 

 

1) v = (-csc2 t)i – (cot t csc t)j

 

2) v π

4


 

= 20i – 16j + 3πk

 

3) a(3) = -324i + 12k

4) T = 11 i +  2 j + 2 k

15      15      3

5) rʹʹ(t) = – 9 i + 48tj

t2

6) -4i – 16 j + 40k

105

7) D

8) C

9) B

10) A

11) D

12) B

13) B

14) C

15) T =     5 i + 2   5 j;      N = 2   5 i –   5 j

5           5                        5           5

16)     1   

4   2

17)   2 –  2  j

4

18) v(0) = <0, 1, 0>

a(0) = <0, 0, -1>

19) C

20) A

 

http://en.wikipedia.org/wiki/Calculus