ad2014 calvec-industrial-jllf.ps14000302.curvas (1)

1(ALU) 1IM1 ANGEL DAVID ORTIZ RESENDIZ AD2014-CALVEC-INDUSTRIAL-JLLFAssignment CURVAS due 09/07/2014 at 11:54pm CDT

1. (1 pt) From Rogawski ET 2e section 13.1, exercise 27.Use cos(t) and sin(t), with positive coefficients, to parame-

trize the intersection of the surfaces x2 + y2 = 4 and z = 5x2.r(t) = 〈 , , 〉Solution:

The points on the cylinder x2 + y2 = 4 and on z = 5x2 can bewritten in the form:x2 + y2 = 4→ (2cos t,2sin t,z)z = 5x2→

(x,y,5x2

)The points (x,y,z) on the intersection curve must satisfy the

following equations:x = 2cos ty = 2sin tz = 5x2 = 5(2cos t)2

We obtain the vector parametrization:r(t) =

⟨2cos t,2sin t,5(2cos t)2

⟩Answer(s) submitted:

2cost

(correct)Correct Answers:

2*cos(t); 2*sin(t); 5*[2*cos(t)]ˆ2

2. (1 pt) From Rogawski ET 2e section 13.1, exercise 37.Find a parametrization, using cos(t) and sin(t), of the follo-

wing curve:The intersection of the plane y = 7 with the sphere x2 +y2 +z2 =74

r(t) = 〈 , , 〉Solution: Substituting y = 7 in the equation of the sphere

gives:

x2 +(7)2 + z2 = 74⇒ x2 + z2 = 25

This circle in the horizontal plane y = 7 has the parametrizationx =√

25cos t,z =√

25sin t.Therefore, the points on the inter-section of the planey = 7 and the sphere x2 +y2 +z2 = 74, can be written in the form(5cos t,7,5sin t), yielding the following parametrization:

r(t) = 〈5cos t,7,5sin t〉Answer(s) submitted:

5sint


5*cos(t); 7; 5*sin(t)

3. (1 pt) From Rogawski ET 2e section 13.1, exercise 27.Use cos(t) and sin(t), with positive coefficients, to parame-

trize the intersection of the surfaces x2 + y2 = 16 and z = 7x4.r(t) = 〈 , , 〉Solution:

The points on the cylinder x2 + y2 = 16 and on z = 7x4 can bewritten in the form:x2 + y2 = 16→ (4cos t,4sin t,z)z = 7x4→

(x,y,7x4

)The points (x,y,z) on the intersection curve must satisfy the

following equations:x = 4cos ty = 4sin tz = 7x4 = 7(4cos t)4

We obtain the vector parametrization:r(t) =

⟨4cos t,4sin t,7(4cos t)4


4cost


4*cos(t); 4*sin(t); 7*[4*cos(t)]ˆ4

4. (1 pt) From Rogawski ET 2e section 13.1, exercise 37.Find a parametrization, using cos(t) and sin(t), of the follo-

wing curve:The intersection of the plane y = 5 with the sphere x2 +y2 +z2 =125

r(t) = 〈 , , 〉Solution: Substituting y = 5 in the equation of the sphere

gives:

x2 +(5)2 + z2 = 125⇒ x2 + z2 = 100

This circle in the horizontal plane y = 5 has the parametrizationx =√

100cos t,z =√

100sin t.Therefore, the points on the inter-section of the planey = 5 and the sphere x2 + y2 + z2 = 125, can be written in theform (10cos t,5,10sin t), yielding the following parametriza-tion:

r(t) = 〈10cos t,5,10sin t〉Answer(s) submitted:

10cost


10*cos(t); 5; 10*sin(t)

1

5. (1 pt) From Rogawski ET 2e section 13.1, exercise 14.The function r(t) traces a circle. Determine the radius, cen-

ter, and plane containing the circler(t) = 7i+ (7cos(t)) j+ (7sin(t))k

Plane : x =Circle’s Center : ( , , )Radius :

Solution: We have:x(t) = 7,y(t) = 7cos(t),z(t) = 7sin(t)Hence,y(t)2 + z(t)2 = 49cos2(t) + 49sin2(t) = 49(cos2(t) +

sin2(t)) = 49This is the equation of a circle in the vertical plane x = 7.

The circle is centered at the point (7,0,0) and its radius is√49 = 7Answer(s) submitted:

77007


77007

6. (1 pt) From Rogawski ET 2e section 13.1, exercise 5.Find a vector parametrization of the line through P =

(−7,5,−3) in the direction v = 〈−8,7,−5〉r(t) =

( + t ) i+( + t ) j+( + t ) k

Solution: We use the vector parametrization of the line toobtain:

r(t) = OP + tv = 〈−7,5,−3〉 + t 〈−8,7,−5〉 =〈−7−8t,5+7t,−3−5t〉

or in the form:r(t) = (−7−8t) i+(5+7t) j+(−3−5t)kAnswer(s) submitted:

-7-857-3-5


-7-85

7-3-5

7. (1 pt) From Rogawski ET 2e section 13.1, exercise 30.Determine whether r1 and r2 collide or intersect:

r1(t) =⟨t, t2, t3

⟩r2(t) =

⟨−(5t +7) ,16t2,−216

⟩The two paths

A. do not collideB. collide

The two paths

A. do not intersectB. intersect

Solution:The two parts collide if there exists a value of t such that:⟨t, t2, t3

⟩=⟨−(5t +7) ,16t2,−216

⟩Equating corresponding components we obtain the followingequations:t =−(5t +7)t2 = 16t2

t3 =−216The second equation implies that t = 0, but this value does notsatisfy the other equations. Therefore, the equations have no so-lution, which means that the paths do not collide.

The two paths intersect if there exist values of t and s suchthat:⟨s,s2,s3

⟩=⟨−(5t +7) ,16t2,−216

⟩Or equivalently:s =−(5t +7)s2 = 16t2

s3 =−216The second equation implies that s = ±4t, and the third equa-tion implies s =−6. Hence t =±−1,5We can see that both solutions don’t satisfy the first equation,hence the two paths do not intersect.

Answer(s) submitted:

AA


AA

8. (1 pt) From Rogawski ET 2e section 13.1, exercise 28.Given two paths r1 and r2. such that for each t1 and t2

r1(t1) 6= r2(t2).The two paths

A. intersect and collideB. intersect but do not collideC. do not intersectD. collide but do not intersect

2

Solution:Intersection is a geometric property of the curves. Collision de-pends on the actual parametrization. By definition the two pathsdo not intersect.

Answer(s) submitted:C


C

9. (1 pt) From Rogawski ET 2e section 13.1, exercise 9.

Which of the space curves above describes the vector-valued function: r(t) = 〈cos t,sin t,cos t sin12t〉?Enter number of figure using one of 1 through 6.

Solution:The curve describing the given vector-valued function is the curve in figure 3.

Answer(s) submitted:3


3

10. (1 pt) From Rogawski ET 2e section 13.1, exercise 33.Find a parametrization of the line through the origin whose projection on the xy-plane is a line of slope 5

and on the yz-plane is a line of slope 8 (i.e., ∆z/∆y = 8).Scale your answer so that the smallest coefficient is 1.

r(t) = 〈 t, t, t〉Solution: We denote by (x,y,z) the points on the line.

The projection of the line on the xy-plane is the line through the origin having slope 5,that is the line y = 5x in the xy-plane. The projection of the lineon the yz-plane is the line through the origin with slope 8,that is the line z = 8y. Thus, the points on the desired line satisfythe following equalities:

y = 5x⇒ y = 5x,z = 8 ·5x = 40xz = 8y

We conclude that the points on the line are all the points in the form (x,5x,40x)Using x = t as parameter we obtain the following parametrization:

r(t) = 〈t,5t,40t〉 .Answer(s) submitted:

8cos(5s)58sin(5s)

(score 0.333333343267441)Correct Answers:

1540

11. (1 pt) From Rogawski ET 2e section 13.1, exercise 33.Find a parametrization of the line through the origin whose projection on the xy-plane is a line of slope 2

and on the yz-plane is a line of slope 7 (i.e., ∆z/∆y = 7).Scale your answer so that the smallest coefficient is 1.

r(t) = 〈 t, t, t〉Solution: We denote by (x,y,z) the points on the line.

The projection of the line on the xy-plane is the line through the origin having slope 2,that is the line y = 2x in the xy-plane. The projection of the lineon the yz-plane is the line through the origin with slope 7,that is the line z = 7y. Thus, the points on the desired line satisfythe following equalities:

y = 2x⇒ y = 2x,z = 7 ·2x = 14xz = 7y

We conclude that the points on the line are all the points in the form (x,2x,14x)Using x = t as parameter we obtain the following parametrization:

r(t) = 〈t,2t,14t〉 .Answer(s) submitted:

2

(score 0.333333343267441)Correct Answers:

1214

12. (1 pt) From Rogawski ET 2e section 13.2, exercise 1.Evaluate the limit:lımt→−5

⟨t2,−2t, 1

t

⟩= 〈 , , 〉

Solution: By the theorem on vector-valued limits we have:lımt→−5

⟨t2,−2t, 1

t

⟩=⟨lımt→−5 t2, lımt→−5−2t, lımt→−5

1t

⟩=⟨25,10, 1

−5


2510-1/5


2510-0.2

3

13. (1 pt) From Rogawski ET 2e section 13.2, exercise 5.Evaluate the limit:lımh→0

r(t+h)−r(t)h for r(t) =

⟨t−6,sin t,−3

⟩r(t) = 〈 , , 〉Solution: This limit is the derivative dr

dt . Using componentwise differentiation yields:

lımh→0r(t+h)−r(t)

h = drdt =

⟨ ddt (t−6), d

dt (sin t), ddt (−3)

⟩=⟨− 6

t7 ,cos t,0⟩

Answer(s) submitted:-6tˆ(-7)cost0


-6*tˆ(-7)cos(t)0

14. (1 pt) From Rogawski ET 2e section 13.2, exercise 50.Find the solution r(t) of the differential equation with the given initial condition:

r′(t) = 〈sin5t,sin5t,9t〉 ,r(0) = 〈6,9,8〉r(t) = 〈 , , 〉

Solution: We first integrate the vector r′(t) to find the general solution:

r(t) =Z〈sin5t,sin5t,9t〉 dt

=⟨Z

sin5tdt,Z

sin5tdt,Z

9tdt⟩

=⟨−1

5cos5t,−1

5cos5t,

92

t2⟩

+ c

Substituting the initial condition we obtain:

r(0) =⟨−1

5cos0,−1

5cos0,

92

02⟩

+ c

= 〈6,9,8〉=⟨−1

5,−1

5,0⟩

+ c

Hence,

c = 〈6,9,8〉−⟨−1

5,−1

5,0⟩

=⟨

315

,465

,8⟩

Hence the solution to the differential equation with the given initial condition is:

r(t) =⟨−1

5cos5t,−1

5cos5t,

92

t2⟩

+⟨

315

,465

,8⟩

=⟨

15

(31− cos5t) ,15

(46− cos5t) ,8+92

t2⟩

Answer(s) submitted:(-cos(5t))/5+31/5(-cos(5t))/5+46/5(9tˆ2)/2+8


6.2-[cos(5*t)]/59.2-[cos(5*t)]/58+4.5*tˆ2

4

15. (1 pt) From Rogawski ET 2e section 13.2, exercise 25.Evaluate d

dt r(g(t)) using the Chain Rule:r(t) =

⟨et,e2t,6

⟩,g(t) = 8t +4

ddt r(g(t)) = 〈 , , 〉Solution: We first differentiate the two functions:r′(t) = d

dt

⟨et,e2t,6

⟩=⟨et,2e2t,0

⟩g′(t) = d

dt (8t +4) = 8Using the Chain Rule we get:ddt r(g(t)) = g′(t)r′(g(t)) = 8

⟨e8t+4,2e2(8t+4),0

⟩=⟨

8e8t+4,16e2(8t+4),0⟩

Answer(s) submitted:8 eˆ(8 t+4)16 eˆ(16 t+8)0


8*eˆ(8*t+4)16*eˆ[2*(8*t+4)]0

16. (1 pt) From Rogawski ET 2e section 13.2, exercise 27.Let r(t) =

⟨t2,1− t,4t

⟩. Calculate the derivative of r(t) ·a(t) at t = 9,

assuming that a(9) = 〈−4,−3,2〉 and a′(9) = 〈5,1,7〉ddt r(t) ·a(t)|t=9 =Solution: By the Product Rule for dot products we haveddt r(t) ·a(t) = r(t) ·a′(t)+ r′(t) ·a(t)At t = 9 we haveddt r(t) ·a(t)|t=9 = r(9) ·a′(9)+ r′(9) ·a(9)We compute the derivative r′(9) :r′(t) = d

dt

⟨t2,1− t,4t

⟩= 〈2t,−1,4〉 ⇒ r′(9) = 〈18,−1,4〉

Also, r(9) =⟨92,1−9,4 ·9

⟩= 〈81,−8,36〉. Substituting the vectors in the equation above, we obtain:

ddt r(t) ·a(t)|t=9 = 〈81,−8,36〉 · 〈5,1,7〉+ 〈18,−1,4〉 · 〈−4,−3,2〉= (405−8+252)+(−72+3+8) = 588The derivative of r(t) ·a(t) at t = 9 is 588 .Answer(s) submitted:

588


588

17. (1 pt) From Rogawski ET 2e section 13.2, exercise 17.Use the appropriate Product Rule to evaluate the derivative, wherer1(t) =

⟨10t,7,−t6

⟩,r2(t) = 〈−4,et,−8〉

ddt (r1(t) · r2(t)) =Solution: By the Product Rule for dot products we have:ddt r1 · r2 = r1 · r′2 + r′1 · r2We compute the derivatives of r1 and r2:r′1 = d

dt

⟨10t,7,−t6

⟩=⟨10,0,−6t5

⟩r′2 = d

dt 〈−4,et,−8〉= 〈0,et,0〉Then,ddt r1(t) · r2(t) =

⟨10t,7,−t6

⟩· 〈0,et,0〉+

⟨10,0,−6t5

⟩· 〈−4,et,−8〉

= 7et +48t5−40Answer(s) submitted:

48tˆ5+7eˆt-40

5


7*eˆt+48*tˆ5-40

18. (1 pt) From Rogawski ET 2e section 13.2, exercise 50.Find the solution r(t) of the differential equation with the given initial condition:

r′(t) = 〈sin6t,sin3t,9t〉 ,r(0) = 〈7,4,4〉r(t) = 〈 , , 〉

Solution: We first integrate the vector r′(t) to find the general solution:

r(t) =Z〈sin6t,sin3t,9t〉 dt

=⟨Z

sin6tdt,Z

sin3tdt,Z

9tdt⟩

=⟨−1

6cos6t,−1

3cos3t,

92

t2⟩

+ c

Substituting the initial condition we obtain:

r(0) =⟨−1

6cos0,−1

3cos0,

92

02⟩

+ c

= 〈7,4,4〉=⟨−1

6,−1

3,0⟩

+ c

Hence,

c = 〈7,4,4〉−⟨−1

6,−1

3,0⟩

=⟨

436

,133

,4⟩

Hence the solution to the differential equation with the given initial condition is:

r(t) =⟨−1

6cos6t,−1

3cos3t,

92

t2⟩

+⟨

436

,133

,4⟩

=⟨

16

(43− cos6t) ,13

(13− cos3t) ,4+92

t2⟩


(-cos(6t))/6+43/6(-cos(3t))/3+13/3(9tˆ2)/2+4


7.16667-[cos(6*t)]/64.33333-[cos(3*t)]/34+4.5*tˆ2

19. (1 pt) From Rogawski ET 2e section 13.2, exercise 46.Evaluate the integral: Z t

0(7si+21s2j+2k)ds

Answer : i+ j+ kSolution: We first compute the integral of each component:

Z t

07sds =

72

s2∣∣∣∣t0=

72

t2

Z t

021s2 ds =

213

s3∣∣∣∣t0= 7t3

6

Z t

02ds = 2s

∣∣∣∣t0= 2t

Hence,

Z t

0(7si+21s2j+2k)ds =

(Z t

07sds

)i+(Z t

021s2 ds

)j+(Z t

02ds)

k

=(

72

t2)

i+(7t3) j+(2t)k


(7tˆ2)/221tˆ3/(3)2t


3.5*tˆ27*tˆ32*t

20. (1 pt) From Rogawski ET 2e section 13.4, exercise 29.For a plane curve r(t) = 〈x(t),y(t)〉,

κ(t) =|x′(t)y′′(t)− x′′(t)y′(t)|

(x′(t)2 + y′(t)2)3/2 .

Use this equation to compute the curvature at the given point.

r(t) = 〈2t4,−4t4〉, t = 2.

κ(2) =Solution:We quickly compute

Function Formula at t = 2x′(t) 2 ·4t3 64x′′(t) 2 ·4 ·3t2 96y′(t) −4 ·4t3 −128y′′(t) −

(4 ·4 ·3t2

)−192

And so

κ(2) =|(64)(−192)− (96)(−128)|

((64)2 +(−128)2)(3/2) .


0


abs((64)*(-192)-(96)*(-128))/(((64))ˆ2+((-128))ˆ2)ˆ(3/2)

21. (1 pt) From Rogawski ET 2e section 13.4, exercise 17.Find the curvature of the plane curve

y =−4t2

at the point t = 1.κ(1) =Solution:By the curvature of a graph of y = f (t) in the plane, we have:

κ(t) =| f ′′(t)|

(1+( f ′(t))2)3/2 .

7

Here f (t) =−4t2, so f ′(t) =−4 ·2t and f ′′(t) =−8. Hence

κ(t) =8(

1+((−8)t)2)1,5 .

We evaluate this at t = 1 to obtain

κ(1) =−8 ·10

(1+64 ·12)3/2 ≈ 0,0152658.


8/(65ˆ(3/2))


0.0152658

22. (1 pt) From Rogawski ET 2e section 13.4, exercise 10.Calculate κ(t) when

r(t) = 〈3t−1,−5,6t〉κ(t) = .Solution:By the formula for curvature we have

κ(t) =||r′(t)× r′′(t)||||r′(t)||3

.

We now findr′(t) = 〈−3t−2,0,6〉

thenr′′(t) = 〈6t−3,0,0〉

and so their cross product is

r′(t)× r′′(t) =

∣∣∣∣∣∣i j k

−3t−2 0 66t−3 0 0

∣∣∣∣∣∣=

∣∣∣∣ 0 60 0

∣∣∣∣ i− ∣∣∣∣ −3t−2 66t−3 0

∣∣∣∣ j+ ∣∣∣∣ −3t−2 06t−3 0

∣∣∣∣k= 〈0,36t−3,0〉.

We compute the necessary lengths:

||r′(t)× r′′(t)|| = 36|t|−3

||r′(t)|| =√

(−3t−2)2 +02 +62 =√

9t−4 +36

leaving

κ(t) =36|t|−3

(9t−4 +36)3/2

which can be simplified to36|t|3

(9+36t4)3/2 .


((36t)/(tˆ4))/((sqrt(((-3/(tˆ2))ˆ2)+36))ˆ3)

(incorrect)Correct Answers:

36*|t|ˆ(-3)/([9*tˆ(-4)+36]ˆ1.5)

8

23. (1 pt) From Rogawski ET 2e section 13.4, exercise 3.Calculate r′(t) and T(t), where

r(t) = 〈4t−4,−3t,8+5t〉.r′(t) = 〈 , , 〉.

T(t) = 〈 , , 〉.Solution:We first find

r′(t) = 〈4,−3,5〉and so

||r′(t)||=√

42 +(−3)2 +52 =√

50.

The unit tangent vector is therefore:

T(t) =r′(t)||r′(t)||

=1√50〈4,−3,5〉=

⟨4√50

,−3√

50,

5√50

⟩.

We see that the unit tangent is constant, since the curve is a straight line.Answer(s) submitted:

4-35(2sqrt(2))/5(-3sqrt(2))/10sqrt(2)/2


4-350.565685-0.4242640.707107

24. (1 pt) From Rogawski ET 2e section 13.4, exercise 1.Calculate r′(t), T(t), and T(4) where

r(t) = 〈4t2, t〉.r′(t) = 〈 , 〉.

T(t) = 〈 , 〉.T(4) = 〈 , 〉.

Solution:We differentiate r(t) to obtain:

r′(t) = 〈4 ·2t,1〉and so

||r′(t)||=√

(8t)2 +1.

We now find the unit tangent vector:

T(t) =r′(t)||r′(t)||

=1√

(8t)2 +1〈4 ·2t,1〉.

For t = 4 we obtain the vector:

T(4) =1√

1025〈32,1〉=

⟨32√1025

,1√

1025

⟩.

Answer(s) submitted:8t1(8t)/(sqrt((8t)ˆ2+1))1/(sqrt((8t)ˆ2+1))

9

32/(5sqrt(41))1/(5sqrt(41))


4*2*t18*t/[sqrt((8*t)ˆ2+1)]1/[sqrt((8*t)ˆ2+1)]0.9995120.0312348

25. (1 pt) From Rogawski ET 2e section 13.3, exercise 12.Find the speed: r(t) = 〈cosh(t) ,cosh(t) ,−t〉 at t = 4.v(4) =

Solution:The velocity vector is r′(t) = 〈sinh(t) ,sinh(t) ,−1〉.

At t = 4 the velocity is r′(4) = 〈27,2899,27,2899,−1〉, hence the speed isv(4) = ||r′(4)||=

√(27,2899)2 +(27,2899)2 +(−1)2 = 38,6067


sqrt(sinhˆ2(4)+sinhˆ2(4)+1)


38.6067

26. (1 pt) From Rogawski ET 2e section 13.3, exercise 6.Compute the length of the curve r(t) = 5ti+8tj+(t2 +2)k over the interval 0≤ t ≤ 4

Hint: use the formula Z √t2 +a2 dt =

12

t√

t2 +a2 +12

a2 ln(

t +√

t2 +a2)

+C

L =Solution: The derivative of r(t) is r′(t) = 5i+8j+2tk.Using the Arc Length Formula we get:

L =Z 4

0

∣∣|r′(t)∣∣ |dt =Z 4

0

√52 +82 +(2t)2 dt =

Z 4

0

√4t2 +89dt

We substitute u = 2t, du = 2dt and use the given integration formula. This gives:

L =12

Z 8

0

√u2 +89du =

14

u√

u2 +89+14·89ln

(u+√

u2 +89)∣∣∣∣8

0=

14

8√

82 +89+894

ln(

8+√

82 +89)− 89

4ln√

89

=42

√153+

894

ln(

8+√

153)− 89

4ln√

89 =

42

√153+

894

ln8+√

153√89

≈ 41,86


41.8647


41.8647

10

27. (1 pt) From Rogawski ET 2e section 13.3, exercise 1.Compute the length of the curve r(t) = 〈−9t,4t−1,−2t−3〉 over the interval 0≤ t ≤ 8L =Solution: We have x(t) =−9t,y(t) = 4t−1,z(t) =−2t−3 hence

x′(t) =−9, y′(t) = 4, z′(t) =−2

We use the Arc Length Formula to obtain:

L =Z 8

0‖(r′(t)‖dt =

Z 8

0

√x′(t)2 + y′(t)2 + z′(t)2 dt

=Z 8

0

√(−9)2 +42 +(−2)2 dt = 8

√101


80.399


80.399

28. (1 pt) From Rogawski ET 2e section 13.3, exercise 24.Find an arc length parametrization of r(t) = 〈et sin t,et cos t,1et〉r1(s) = 〈 , , 〉

Solution:An arc length parametrization is r1(s) = r(φ(s)) where t = φ(s) is the inverse of the arc length function. We compute the arc lengthfunction:s(t) =

R t0 ||r′(u)du

Differentiating r(t) and computing the norm of r′(t) gives:r′(t) = 〈et sin t + et cos t,et cos t− et sin t,1et〉= et 〈sin t + cos t,cos t− sin t,1〉||r′(t)||= et

√(sin t + cos t)2 +(cos t− sin t)2 +12 = et

√2(sin2 t + cos2 t)+1 =

√3et

Thus,s(t) =

R t0

√3eudu =

√3 eu|t0 =

√3(et −1)

We find the inverse function of s(t) by solving s =√

3(et −1) for t. We obtain:s√3

= et −1et = 1+ s√

3

t = φ(s) = ln(

1+ s√3

)An arc length parametrization for r1(s) = r(φ(s)) is:⟨

eln(1+0,57735s) sin(ln(1+0,57735s)) ,eln(1+0,57735s) cos(ln(1+0,57735s)) ,eln(1+0,57735s)⟩

= (1+0,57735s)〈sin(ln(1+0,57735s)) ,cos(ln(1+0,57735s)) ,1〉



(1+0.57735*s)*sin(ln(1+0.57735*s))(1+0.57735*s)*cos(ln(1+0.57735*s))1+0.57735*s

11

29. (1 pt) From Rogawski ET 2e section 13.3, exercise 10.Find the speed at the given value of t:r(t) =

⟨et−7,−3,7t−1

⟩, t = 7

v(7) =Solution: The velocity vector is

⟨et−7,0,−7t−2

⟩and at t = 7

r′(7) =⟨e7−7,0,−7 ·7−2⟩=

⟨1,0,−1

7

⟩.

The speed is the magnitude of the velocity vector, that is,

v(7) = ‖r′(7)‖=

√12 +02 +

(−1

7

)2

=

√5049≈ 1,01


1.01015


1.01015

30. (1 pt) From Rogawski ET 2e section 13.3, exercise 23.Find a path that traces the circle in the plane y =−3 with radius r = 4 and center (2,−3,0) with constant speed 12.r1(s) = 〈 , , 〉

Solution:We start with the following parametrization of the circle:r(t) = 〈2,−3,0〉+4〈cos t,0,sin t〉= 〈2+4cos t,−3,0+4sin t〉We need to reparametrize the curve by making a substitution on t = φ(s), so that the new parametrization r1(s) = r(φ(s)) satisfies||r′1(s)||= 12 for all s. We find r′1 using the Chain Rule:r′1(s) = d

ds r(φ(s)) = φ′(s)r′(φ(s))Next, we differentiate r(t) and then replace t by φ(s):r′(t) = 〈−4sin t,0,4cos t〉r′(φ(s)) = 〈−4sin(φ(s)),0,4cos(φ(s))〉Now, we get: r′1(s) = φ′(s)〈−4sin(φ(s)),0,4cos(φ(s))〉= 4φ′(s)〈−sin(φ(s)),0,cos(φ(s))〉Hence,||r′1(s)||= 4|φ′(s)|

√(−sin(s))2 +(cos(s))2 = 4|φ′(s)|

To satisfy ||r′1(s)||= 12 for all s, we choose φ′(s) = 3.We may take the antiderivative of φ(s) = 3s and obtain the following parametrization:r1(s) = r(φ(s)) = r(3s) = 〈2+4cos(3s),−3,0+4sin(3s)〉


4cos(3s)+2-34sin(3s)


2+4*cos(3*s)-34*sin(3*s)

12

31. (1 pt) From Rogawski ET 2e section 13.3, exercise 3.Compute the length of the curve r(t) =

⟨2t, ln t, t2

⟩over the interval 1≤ t ≤ 4

L =Solution: The derivative of r(t) is r′(t) =

⟨2, 1

t ,2t⟩. We use the Arc Length Formula to obtain:

L =Z 4

1‖r′(t)‖dt =

Z 4

1

√22 +

(1t

)2

+(2t)2 dt

=Z 4

1

√(2t +

1t

)2

dt =Z 4

1

(2t +

1t

)dt = t2 + ln t

∣∣∣∣41

= (16+ ln4)− (1+ ln1) = 15+ ln4


16.3863


16.3863

32. (1 pt) From Rogawski ET 2e section 13.3, exercise 29.Evaluate s(t) =

R t−∞||r′(u)||du for the Bernoulli spiral r(t) = 〈et cos(3t),et sin(3t)〉.

It is convenient to take −∞ as the lower limit since s(−∞) = 0. Then use s to obtain an arc length parametrization of r(t).r1(s) = 〈 , 〉

Solution:We differentiate r(t) and compute the norm of the derivative vector. This gives:r′(t) = 〈et cos(3t)−3et sin(3t),et sin(3t)+3et cos(3t)〉= et 〈cos(3t)−3sin(3t),sin(3t)+3cos(3t)〉||r′(t)||= et

√(cos(3t)−3sin(3t))2 +(sin(3t)+3cos(3t))2

= et√

cos2(3t)+ sin2(3t)+9(cos2(3t)+ sin2(3t)) =√

10et

We now evaluate the improper integral:s(t) =

R t−∞||r′(u)||du

= lımR→−∞

R tR

√10eudu = lım

R→−∞

√10eu

∣∣tR = lım

R→−∞

√10(et − eR

)=√

10et

An arc length parametrization of r(t) is r1(s) = r(φ(s)) where t = φ(s) is the inverse function of s(t).We find t = φ(s) by solving s =

√10et for t:

t = φ(s) = ln s√10

An arc length parametrization of r(t) is:⟨s

3,16228 cos(

3ln(

s3,16228

)), s

3,16228 sin(

3ln(

s3,16228

))⟩Answer(s) submitted:


s/3.16228*cos(3*ln(s/3.16228))s/3.16228*sin(3*ln(s/3.16228))

13

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