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12.2E: Exercises for Section 12.2

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Taking Derivatives of Vector-Valued Functions

In questions 1 - 10, compute the derivative of each vector-valued function.

1) r(t)=t3ˆi+3t2ˆj+t36ˆk

Answer
r(t)=3t2ˆi+6tˆj+12t2ˆk

2) r(t)=sin(t)ˆi+cos(t)ˆj+etˆk

3) r(t)=etˆi+sin(3t)ˆj+10tˆk. A sketch of the graph is shown here. Notice the varying periodic nature of the graph.

Graph of the 3D corkscrew path of r(t)=e^(−t) i + sin(3t) j+10 sqrt(t) k.

Answer
r(t)=etˆi+3cos(3t)ˆj+5tˆk

4) r(t)=etˆi+2etˆj+ˆk

5) r(t)=ˆi+ˆj+ˆk

Answer
r(t)=0,0,0=0

6) r(t)=tetˆi+tln(t)ˆj+sin(3t)ˆk

7) r(t)=1t+1,arctan(t),lnt3

Answer
r(t)=1(t+1)2,11+t2,3t

8) r(t)=tan2t,sec2t,sin2t

9) r(t)=3,4sin(3t),tcos(t)

Answer
r(t)=0,12cos(3t),costtsint

10) r(t)=t2ˆi+te2tˆj5e4tˆk

11) a. Describe and sketch the curve represented by the vector-valued function r(t)=6t,6tt2.

b. Locate the highest point on the curve r(t)=6t,6tt2 and give the value of the function at this point.

Answer
b. r(t)=18,9 at t=3

12) Find the parametric equations of the tangent line to the curve r(t)=t,t2,t at t=2.

13) Find the parametric equations of the tangent line to the curve r(t)=et,et,0 at t=0.

Answer
x=1+t,y=1t,z=0

14) Compute the first, second, and third derivatives of r(t)=3tˆi+6ln(t)ˆj+5e3tˆk.

Describing Motion with Vector-Valued Functions

In questions 15 - 17, find the velocity and acceleration at the given times, plot the graph of the position function, and draw in the velocity and acceleration vectors at the corresponding locations on the curve.

15) r(t)=tˆi+t3ˆj, at t=0 and at t=1

Answer
t=0:t=1:r(t)=tˆi+t3ˆj,r(0)=0,r(1)=ˆi+ˆjv(t)=ˆi+3t2ˆj,v(0)=ˆi,v(1)=ˆi+3ˆja(t)=6tˆj,a(0)=0,a(1)=6ˆj
Path along graph of y = x^3 from left-to-right.  Also showing velocity and acceleration vectors at t = 0 and t = 1.

16) r(t)=costˆi+sin3tˆj, at t=0, at t=π4, and at t=π2

17) r(t)=lntˆi+(t2)ˆj, at t=1 and at t=2

Answer
t=1:t=2:r(t)=(lnt)ˆi+(t2)ˆj,r(1)=ˆj,r(2)=(ln2)ˆiv(t)=1tˆi+ˆj,v(1)=ˆi+ˆj,v(2)=12ˆi+ˆja(t)=1t2ˆi,a(1)=ˆi,a(2)=14ˆi
Path along graph of r(t) = ln(t) i + (t - 2) j.  Also showing velocity and acceleration vectors at t = 1 and t = 2.

In questions 18 - 24, find the velocity, speed, and acceleration of a particle with the given position function. Remember that the speed is the magnitude of the velocity represented by v(t) or r(t).

18) r(t)=e2tˆi+sintˆj

19) r(t)=cost3ˆi+sint3ˆj

Answer
v(t)=3t2sint3ˆi+3t2cost3ˆj,speed(t)=v(t)=3t2,a(t)=(6tsint39t4cost3)ˆi+(6tcost39t4sint3)ˆj

20) r(t)=et,et,0

21) r(t)=t+cost,tsint

Answer
v(t)=1sint,1cost,speed(t)=v(t)=32(sint+cost),a(t)=cost,sint

22) r(t)=2t12t+1ˆi+ln(14t2)ˆj

23) r(t)=cos3tˆi+sin3tˆj+0.5tˆk

Answer
v(t)=3sin3tˆi+3cos3tˆj+0.5ˆk,speed(t)=v(t)=9.25 units/sec,a(t)=9cos3tˆi9sin3tˆj

24) r(t)=etˆi+(lnt)ˆj+(sin7t)ˆk

25) Consider the position vector for a particle to be r(t)=tˆi+t2ˆj+t3ˆk. The graph is shown here:

CNX_Calc_Figure_13_02_207.jpeg

a. Find the velocity vector at any time.

b. Find the speed of the particle at time t=2 sec.

c. Find the acceleration at time t=2 sec.

Answer
a. v(t)=ˆi+2tˆj+3t2ˆk
b. 161 units/sec
Note that speed(t)=v(t)=12+(2t)2+(3t2)2=1+4t2+9t4.
Hence, speed(2)=1+16+9(16)=161 units/sec.
c. Since a(t)=2ˆj+6tˆk, a(2)=2ˆj+12ˆk

26) A particle travels along the path of an ellipse with the equation r(t)=costˆi+2sintˆj+0ˆk. Find the following:

a. Velocity of the particle

Answer
v(t)=sint,2cost,0

b. Speed of the particle at t=π4

c. Acceleration of the particle at t=π4

Answer
a(t)=22,2,0

27) Show that if the speed of a particle traveling along a curve represented by a vector-valued function is constant, then the velocity function is always perpendicular to the acceleration function.

Answer
v(t)=kv(t)·v(t)=k2ddt(v(t)·v(t))=ddt(k2)=0v(t)·v(t)+v(t)·v(t)=02v(t)·v(t)=0v(t)·v(t)=0

The last statement implies that the velocity and acceleration are perpendicular or orthogonal.

28) Given the vector-valued function r(t)=tant,sect,0 (graph is shown here), find the following:

CNX_Calc_Figure_13_02_209.jpeg

a. Velocity

Answer
v(t)=sec2t,secttant

b. Speed

Answer
v(t)=sec4t+sec2ttan2t=(sec2t)(sec2t+tan2t)

c. Acceleration

Answer
a(t)=2sec2ttant,secttan2t+sec3t

29) Find the minimum speed of a particle traveling along the curve r(t)=t+cost,tsint, where t[0,2π). Then also find its maximum speed on this interval.

Answer
Min. speed is 3220.41421 when t=π4.
Max. speed is 3+222.41421 when t=5π4.

For questions 30 - 31, consider a particle that moves on a circular path of radius b according to the function r(t)=bcos(ωt)ˆi+bsin(ωt)ˆj, where ω is the angular velocity, dθdt.

CNX_Calc_Figure_13_02_201.jpeg

30) Show that the speed of the particle is proportional to the angular velocity.

31) Find the velocity function and show that v(t) is always orthogonal to r(t).

Answer
r(t)=bωsin(ωt)ˆi+bωcos(ωt)ˆj. To show orthogonality, note that r(t)r(t)=0.

Smoothness of Vector-Valued Functions

For questions 32 - 40,

a. Determine any values of t at which r is not smooth.

b. Determine the open intervals on which r is smooth.

c. Graph the vector-valued function and describe its behavior at the points where it is not smooth.

32) r(t)=3t,5t21

33) r(t)=t3ˆi+5t2ˆj

Answer
a. r is not smooth at t=0, since r(0)=0.
b. r is smooth on the open intervals (,0) and (0,).
c. There is a cusp when t=0.

34) r(t)=5,2sin(t),cos(t)

35) r(t)=t33t2,7

Answer
a. r is not smooth at t=0 and t=2, since r(0)=0 and r(2)=0.
b. r is smooth on the open intervals (,0), (0,2), and (2,).
c. The motion on the curve reverses along the same path at both t=0 and t=2.

36) r(t)=t2ˆi+t3ˆj5e4tˆk

37) r(t)=ln(t2+4t+5),t334t,5

Answer
a. r is not smooth at t=2, since r(2)=0.
Since the domain of r is (,), this is all we have to remove.
b. r is smooth on the open intervals (,2) and (2,).
c. There is a cusp when t=2.

38) r(t)=(5costcos5t)ˆi+(5sintsin5t)ˆj, for 0t2π

39) r(t)=t3+9t2ˆi+(t2+12t)ˆj+7ˆk

Answer
a. The domain of r is [9,).
And r is not smooth at t=6, since r(6)=0.
The domain of r is (9,), since r is undefined at t=9.
b. r is smooth on the open intervals (9,6) and (6,).
c. There is a cusp when t=6.

40) r(t)=cos3tˆi+sintˆj, for 0t2π

Answer
a. The domain of r is (,).
r(t)=3(cos2t)(sint)ˆi+costˆj. It's domain is also (,).
But note that both components have a factor of cost, so both components will be 0 when cost=0.
Therefore, r is not smooth at t=π2 and at t=3π2, since r(π2)=0 and r(3π2)=0. Note then that r is not smooth for any odd multiple of π2, that is for t=(2n+1)π2, for any integer value n.
b. r is smooth on the open intervals ((2n1)π2,(2n+1)π2), for any integer value n.
c. There is a cusp when t=(2n+1)π2, for any integer value n.

Properties of the Derivative

For questions, 41 - 43, evaluate each expression given that r(t)=tˆi+t2ˆjt4ˆk and s(t)=sin(t)ˆi+etˆj+cos(t)ˆk

41) ddt[r(t2)]

Answer
ddt[r(t2)]=2t,4t3,8t7

42) ddt[t2s(t)]

43) ddt[r(t)s(t)]

Answer
ddt[r(t)s(t)]=sint+2tet4t3cost+tcost+t2et+t4sint

44) Find r(t)r(t)forr(t)=3t5ˆi+5tˆj+2t2ˆk.

Answer
r(t)r(t)=900t7+16t

45) Given r(t)=tˆi+3tˆj+t2ˆk and u(t)=4tˆi+t2ˆj+t3ˆk, find ddt(r(t)×u(t)).

46) Evaluate ddt[u(t)×u(t)] given u(t)=t2ˆi2tˆj+ˆk.

Answer
ddt[u(t)×u(t)]=0ˆi+2ˆj+4tˆk

47) Given r(t)=tˆi+2sintˆj+2costˆk and u(t)=1tˆi+2sintˆj+2costˆk, find the following:

a. r(t)×u(t)

Answer
r(t)×u(t)=0,2(cost)(1tt),2(sint)(t1t)

b. ddt(r(t)×u(t))

Answer
ddt(r(t)×u(t))=0,2(sint)(t1t)2(cost)(1+1t2),2(sint)(1+1t2)+2(cost)(t1t)

c. Now, use the product rule for the derivative of the cross product of two vectors and show this result is the same as the answer for the preceding problem.

Unit Tangent Vectors

For questions 48 - 51, find a unit tangent vector at the indicated value of t.

48) r(t)=3t3ˆi+2t2ˆj+1tˆk;t=1

49) r(t)=tˆi+sin(2t)ˆj+cos(3t)ˆk;t=π3

Answer
r(π3)=1,1,0 is a tangent vector, so a unit tangent vector would be:
121,1,0=22,22,0

50) r(t)=cos(2t)ˆi+2sintˆj+t2ˆk;t=π2

51) r(t)=3etˆi+2e3tˆj+4e2tˆk;t=ln(2)

Answer
r(ln(2))=6,34,32 is a tangent vector, so a unit tangent vector would be:
11060.56256,34,32=241696916969,121696967876,1281696916969

For questions 52 - 58, find the unit tangent vector T(t) for the following parameterized curves.

52) r(t)=tˆi+3tˆj+t2ˆk

53) r(t)=6ˆi+cos(3t)ˆj+3sin(4t)ˆk,0t<2π

Answer
T(t)=19sin2(3t)+144cos2(4t)0,3sin(3t),12cos(4t)

54) r(t)=tcost,tsint

55) r(t)=t+1,2t+1,2t+2

Answer
T(t)=131,2,2

56) r(t)=costˆi+sintˆj+sintˆk,0t<2π. Two views of this curve are presented here:

CNX_Calc_Figure_13_02_203.jpeg

57) r(t)=t,1t. The graph is shown here:

CNX_Calc_Figure_13_02_210.jpeg

Answer
T(t)=t2t4+1,1t4+1

58) r(t)=3cos(4t)ˆi+3sin(4t)ˆj+5tˆk,1t2

Answer
T(t)=1213sin(4t)ˆi+1213cos(4t)ˆj+513ˆk

59) A particle travels along the path of a helix with the equation r(t)=cos(t)ˆi+sin(t)ˆj+tˆk. See the graph presented here:

CNX_Calc_Figure_13_02_208.jpeg

Find the following:

a. Velocity of the particle at any time

Answer
v(t)=sint,cost,1

b. Speed of the particle at any time

c. Acceleration of the particle at any time

Answer
a(t)=costˆisintˆj+0ˆk

d. Find the unit tangent vector for the helix.

Integration of Vector-Valued Functions

Evaluate the following integrals:

60) (etˆi+sintˆj+12t1ˆk)dt

61) 10r(t)dt, where r(t)=3t,1t+1,et

Answer
34ˆi+ln(2)ˆj+(11e)ˆk

62) Evaluate 30tˆi+t2ˆjdt.

Answer
13(10321)

63) The acceleration function, initial velocity, and initial position of a particle are

a(t)=5costˆi5sintˆj,v(0)=9ˆi+2ˆj,andr(0)=5ˆi

Find v(t) and r(t).

Answer
v(t)=(95sint)ˆi+(3+5cost)ˆj
r(t)=(9t+5cost)ˆi+(3t+5sint)ˆj

64) Find the antiderivative of r(t)=cos(2t)ˆi2sintˆj+11+t2ˆk that satisfies the initial condition r(0)=3ˆi2ˆj+ˆk.

65) An object starts from rest at point P(1,2,0) and moves with an acceleration of a(t)=ˆj+2ˆk, where a(t) is measured in feet per second per second. Find the location of the object after t=2 sec.

Contributors:

  • Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org.

  • Paul Seeburger (Monroe Community College) created questions 11 - 19.

 


12.2E: Exercises for Section 12.2 is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by LibreTexts.

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