Austin Cholley Zhu MAT 267 ONLINE A Fall 2021
Assignment Section 10.7 due 08/29/2021 at 11:59pm MST
Problem 1. (1 point)
Find the domain of the vector functions, r(t), listed below.
You may use ”-INF” for −∞and use ”INF” for ∞as necessary,
and use ”U” for a union symbol if a union of intervals is needed.
a) r(t) = Dln(4t),√t+16,1
√7−tE
b) r(t) = √t−8,sin(3t),t2
c) r(t) = e−8t,t
√t2−9,t1/3
Solution:
SOLUTION
(a) The first component is defined when 4t>0⇒t>0. The sec-
ond component is defined when t+16 ≥0⇒t≥ −16. The third
component is defined when 7 −t>0⇒t<7. Thus the domain
is the interval (0,7)
(b) The first component is defined when t−8≥0⇒t≥8. The
second and third component are defined for all real t. Thus the
domain is the interval [8,∞)
(c) The first and third component are defined for all real t. The
first component is defined when t2−9>0⇒t<−3,t>3. Thus
the domain is the interval (−∞,−3)∪(3,∞)
Answer(s) submitted:
•(0,7)
•[8,inf)
•(-inf,-3)U(3,inf)
(correct)
Correct Answers:
•(0,7)
•[8,infinity)
•(-infinity,-3) U (3,infinity)
Problem 2. (1 point)
Let r(t) = (√t+3)i+t2−4
t−2j+sin(2πt)k.
Then
lim
t→1r(t)=i+j+k.
Solution:
SOLUTION:
lim
t→1
√t+3=√4,
lim
t→1
t2−4
t−2=lim
t→1
(t−2)(t+2)
t−2=lim
t→1(t+2) = 3,
lim
t→1sin(2πt) = sin(2π) = 0.
Thus lim
t→1r(t) = √4i+3j+0k
Answer(s) submitted:
•sqrt(4)
•3
•0
(correct)
Correct Answers:
•2
•3
•-2.44929359829471E-16
1
Problem 3. (1 point)
Find the limit:
lim
t→0e−1t−1
t,t4
t5−t4,−5
4+t
h, , i
Solution:
SOLUTION:
lim
t→0
e−1t−1
t[l’Hopital] =lim
t→0−1e−1t
1=−1e0=−1
lim
t→0
t4
t5−t4=lim
t→0
1
t−1=−1
lim
t→0−5
4+t=−5
4.
Thus lim
t→0e−1t−1
t,t4
t5−t4,−5
4+t=−1,−1,−5
4
Answer(s) submitted:
•-1
•-1
•-(5/4)
(correct)
Correct Answers:
•-1
•-1
•-1.25
Problem 4. (1 point)
The curve c(t) = hcost,sint,tilies on which of the following sur-
faces.
Enter Tor Fdepending on whether the statement is true or false.
(You must enter Tor F– True and False will not work.)
1. a circular cylinder
2. a sphere
3. an ellipsoid
4. a plane
Solution:
SOLUTION:
Since x2+y2=cos2t+sin2t=1, the curve lies on a circular cylin-
der.
Answer(s) submitted:
•t
•f
•f
•f
(correct)
Correct Answers:
•T
•F
•F
•F
2
Problem 5. (1 point)
Match the parametric equations with the graphs labeled A - F. As
always, you may click on the thumbnail image to produce a larger
image in a new window (sometimes exactly on top of the old one).
1. x=cos4t,y=t,z=sin4t
2. x=cost,y=sint,z=sin5t
3. x=cost,y=sint,z=lnt
4. x=sin3tcost,y=sin3tsint,z=t
5. x=t,y=1/(1+t2),z=t2
6. x=t2−2,y=t3,z=t4+1
A B C D E F
Solution:
SOLUTION:
1. x2+z2=cos2(4t)+ sin2(4t) = 1, so the curve lies on the cylin-
der with axis the y-axis. A point (x,y,z)on the curve lies directly
above or below (x,0,z), which moves around the unit circle in the
xz-plane with period π/2. At the same time, the z-value increases
at tincreases. So the graph is F.
2. x2+y2=cos2t+sin2t=1, so the curve lies on the cylinder
with axis the z-axis. A point (x,y,z)on the curve lies directly
above or below (x,y,0), which moves around the unit circle in the
xy-plane with period 2π. At the same time, the z-value oscillates
with a period of π. So the curve repeats itself and the graph is A.
3. x2+y2=cos2t+sin2t=1, so the curve lies on the cylinder
with axis the z-axis. A point (x,y,z)on the curve lies directly
above or below (x,y,0), which moves around the unit circle in the
xy-plane with period 2π. At the same time, the z-value increases
at tincreases. So the graph is D.
4. xand yare bounded and periodic, while zincreases, so the
graph is E.
5. At any point on the curve we have z=x2, so the curve lies on
the parabolyc cylinder parallel to the y-axis. Notice that 0 <y≤1
and z≥0. Also the curve passes through (0,1,0)when t=0 and
y→0,z→∞as t→ ±∞, so the graph is C.
6. x→∞and z→∞as t→ ±∞. Also, y→ ±∞as t→ ±∞. So
the graph is B.
Answer(s) submitted:
•f
•a
•d
•e
•c
•b
(correct)
Correct Answers:
•F
•A
•D
•E
•C
•B
Problem 6. (1 point)
Find a vector function that represents the curve of intersection of
the paraboloid z=7x2+3y2and the cylinder y=4x2. Use the
variable t for the parameter.
r(t) = ht,,i
Solution:
SOLUTION
Let x=t, then the equation of the cylinder give y=4t2. Substi-
tuting x=tand y=4t2into the equation of the paraboloid, yields
z=7t2+3(4t2)2=7t2+48t4. Thus a vector function that rep-
resents the curve of intersection of the two surface is given by
r(t) = ht,4t2,7t2+48t4i
Answer(s) submitted:
•4tˆ(2)
•7tˆ(2)+(3*4*4)*tˆ(4)
(correct)
Correct Answers:
•4*t*t
•7*t*t + (3 * 4 * 4)*t**4
3
Problem 7. (1 point)
Consider the paraboloid z=x2+y2. The plane 5x−4y+z−7=0
cuts the paraboloid, its intersection being a curve.
Find ”the natural” parametrization of this curve.
Hint: The curve which is cut lies above a circle in the xy-plane
which you should parametrize as a function of the variable t so
that the circle is traversed counterclockwise exactly once as t goes
from 0 to 2*pi, and the paramterization starts at the point on the
circle with largest x coordinate. Using that as your starting point,
give the parametrization of the curve on the surface.
c(t) = (x(t),y(t),z(t)), where
x(t) =
y(t) =
z(t) =
Solution:
SOLUTION:
The projection of the curve in the xy−plane has equation 5x−
4y+x2+y2−7=0. Completing the squares gives
x2+5x+25
4+y2−4y+4=25
4+4+7
⇒(x+5
2)2+ (y−2)2=69
4
In the xy−plane, this is a circle of radius q69
4and center (−5
2,2)
and therefore it can be prametrized by x=−5
2+q69
4cost,y=
2+q69
4sint.
Substituting into the equation of the paraboloid, yields
z= (−5
2+q69
4cost)2+ (2+q69
4sint)2.
Thus the ”natural” parametrization of this curve is
x(t) = −5
2+q69
4cost
y(t) = 2+q69
4sint
z(t) = −5
2+q69
4cost2
+2+q69
4sint2
.
Answer(s) submitted:
•
•
•
(incorrect)
Correct Answers:
•- 5/2 + 4.15331193145904*cos(t)
•- -4/2 + 4.15331193145904*sin(t)
•(- 5/2 + 4.15331193145904*cos(t))**2 + (- -4/2 + 4.15331193145904*sin(t))**2
Problem 8. (1 point)
Find the derivative of the vector function
r(t) = ln(5−t2)i+√17 +tj−3e−3tk
r0(t) = h, , i
Solution:
SOLUTION:
r0(t) = d
dt [ln(5−t2)] i+d
dt [√17 +t]j+d
dt [−3e−3t]k
=−2t
5−t2i+1
2√17+tj−3(−3)e−3tk
=−2t
5−t2,1
2√17 +t,9e−3t
Answer(s) submitted:
•-((2t)/(5-tˆ(2)))
•(1/(2sqrt(17+t)))
•9eˆ(-3t)
(correct)
Correct Answers:
•-2*t/(5 - t*t)
•1/(2*sqrt(17 + t))
•-3 * -3 * exp(-3 * t)
4
Problem 9. (1 point)
For the given position vectors r(t)compute the unit tangent vector
T(t)for the given value of t.
A) Let r(t) = hcos4t,sin4ti.
Then T(π
4)h,i
B) Let r(t) = ht2,t3i.
Then T(1) = h,i
C) Let r(t) = e4ti+e−tj+tk.
Then T(1)=i+j+
k.
Solution:
SOLUTION:
(A) r0(t) = h−4sin(4t),4cos(4t)i
⇒r0π
4=h−4sin(1π),4 cos(1π)i
⇒
r0π
4
=q42sin2(1π) + 42cos2(1π) = q42sin2(1π) + cos2(1π)=
4
Thus
Tπ
4=r0(π
4)
|r0(π
4)|
=1
4h−4sin(1π),4 cos(1π)i
=1
4h−4sin(1π),4 cos(1π)i
=h−sin(1π),cos(1π)i
(B) r0(t) = h2t,3t2i ⇒ r0(1) = h2,3i
Thus
T(1) = r0(1)
|r0(1)|
=1
√(2)2+(3)2h2,3i
=1
1√13 h2,3i
=D2
√13 ,3
√13 E
(C) r0(t) = 4e4ti−1e−1tj+k⇒r0(1) = 4e4i−1e−1j+k
Thus
T(1) = r0(1)
|r0(1)|
=1
√16e8+1e−2+1(4e4i−1e−1j+k)
=4e4
√16e8+1e−2+1i+−1e−1
√16e8+1e−2+1j+1
√16e8+1e−2+1k
Answer(s) submitted:
•0
•-1
•0
•-1
•0
•0
•0
(score 0.285714285714286)
Correct Answers:
•-1.22464679914735E-16
•-1
•0.554700196225229
•0.832050294337844
•0.999988098257553
•-0.00168446670144392
•0.00457885522517935
Problem 10. (1 point)
Find parametric equations for the tangent line at the point
cos5
6π,sin5
6π,5
6πon the curve x=cost,y=sint,z=t
x(t)=
y(t)=
z(t)=
(Your line should be parametrized so that it passes through the
given point at t=0).
Solution:
SOLUTION
The point cos5
6π,sin5
6π,5
6π=−√3
2,1
2,5π
6corre-
sponds to t=5π
6, so the tangent vector there is r05π
6=
h−sin 5π
6,cos5π
6,1i=h−1
2,−√3
2,1i.
Thus the tangent line goes through the point −√3
2,1
2,5π
6and is
parallel to the vector h−1
2,−√3
2,1i.
Parametric equations are x=−√3
2−1
2t,y=1
2−√3
2t,z=5π
6+t
Answer(s) submitted:
•cos((5/6)pi )+t(-sin((5/6)pi ))
•sin((5/6)pi )+t(cos((5/6)pi ))
•(5/6)pi +t
(correct)
Correct Answers:
•cos(5*3.14159265358979/6)+(-sin(5*3.14159265358979/6))*t
•sin(5*3.14159265358979/6)+(cos(5*3.14159265358979/6))*t
•5*3.14159265358979/6 +t
5
Problem 11. (1 point)
Find the parametric equations for the tangent line to the curve
x=t5−1,y=t4+1,z=t5
at the point (0,2,1). Use the variable tfor your parameter.
x=,
y=,
z=
Solution:
SOLUTION:
The vector equation for the curve is r(t) = ht5−1,t4+1,t5iso
r0(t) = h5t4,4t3,5t4i. The point (0,2,1)corresponds to t=1,
so the tangent vector there is r0(1) = h5,4,5i. Thus, the tangent
line goes through the point (0,2,1)and is parallel to the vector
h5,4,5i.
Parametric equations are
x=5t
y=2+4t
z=1+5t
Answer(s) submitted:
•5t
•2+4t
•1+5t
(correct)
Correct Answers:
•5*t
•2+4*t
•1+5*t
Problem 12. (1 point)
Evaluate
Z8
0ti+t2j+t3kdt =i+j+k.
Solution:
SOLUTION:
R8
0ti+t2j+t3kdt =R8
0t dti+R8
0t2dtj+R8
0t3dtk
=ht2
2i8
0i+ht3
3i8
0j+ht4
4i8
0k
=32i+512
3j+1024k
Answer(s) submitted:
•40
•
•
(incorrect)
Correct Answers:
•32
•170.666666666667
•1024
Problem 13. (1 point)
If r(t) = cos(3t)i+sin(3t)j−10tk
compute r0(t)=i+j+k
and Rr(t)dt=i+j+k+C
with Ca constant vector.
Solution:
SOLUTION:
r0(t) = d
dt [cos(3t)]i+d
dt [sin(3t)]j+d
dt [−10t]k
=−3sin(3t)i+3cos(3t)j−10k
Rr(t)dt = (Rcos(3t)dt)i+ (Rsin(3t)dt)j+ (R(−10t)dt)k
=sin(3t)
3i−cos(3t)
3j−10t2
2k+C
Answer(s) submitted:
•-3sin(3t)
•3cos(3t)
•-10
•(1/3)sin(3t)
•-(1/3)cos(3t)
•-5tt
(correct)
Correct Answers:
•- 3*sin(3*t)
•3*cos(3*t)
•2*-5
•(1/3)*sin(3*t)
•(-1/3)*cos(3*t)
•-5*t*t
Problem 14. (1 point)
Find a vector parametrization of the curve x=−3z2in the xz-
plane. Use tas the parameter in your answer.
~r(t) =
Solution:
SOLUTION:
Since the curve is in the xz- plane, it must be y=0. Letting
z=tyields x=−3t2.
Thus a vector parametrization of the curve is
~r(t) = h−3t2,0,ti
Answer(s) submitted:
•-3tˆ(2)i+kt
(correct)
Correct Answers:
•<-3*tˆ2,0,t>
6
Problem 15. (1 point)
Are the following statements true or false?
? 1. The line parametrized by x=7,y=5t,z=6+tis parallel
to the x-axis.
? 2. The parametric curve x= (3t+4)2,y=5(3t+4)2−9, for
0≤t≤3 is a line segment.
? 3. A parametrization of the graph of y=ln(x)for x>0 is
given by x=et,y=tfor −∞<t<∞.
Solution:
SOLUTION
1. The direction vector of the line is h0,5,1i. This vector is not
parallel to the x-axis. Thus the statement is False.
2. Eliminating the parameter tyields the equation y=5x−
9,16 ≤x≤169. Thus the parametric curve is a line segment
and the statement is True.
3. Substituting x=et,y=tinto the equation y=ln(x)yields
t=ln(et) = t. For the given values of t,x>0. Thus the statement
is True.
Answer(s) submitted:
•F
•T
•T
(correct)
Correct Answers:
•F
•T
•T
Problem 16. (1 point)
Find a vector parametric equation ~r(t)for the line through the
points P= (−4,0,2)and Q= (1,2,0)for each of the given con-
ditions on the parameter t.
(a) If~r(0) = h−4,0,2iand~r(2) = h1,2,0i, then
~r(t) =
(b) If~r(3) = Pand~r(6) = Q, then
~r(t) =
(c) If the points Pand Qcorrespond to the parameter values
t=0 and t=−4, respectively, then
~r(t) =
Solution:
SOLUTION
The line has direction ~
PQ =h5,2,−2i
(a) Using the point P= (−4,0,2)as a base point, the vector
equation of the line is
~r(t) = (−4,0,2) + t
2h5,2,−2i
(b) The vector equation of the line is
~r(t) = (−4,0,2) + t−3
3h5,2,−2i
(c) The vector equation of the line is
~r(t) = (−4,0,2) + −t
4h5,2,−2i
Answer(s) submitted:
•
•
•
(incorrect)
Correct Answers:
•(-4,0,2)+t/2*<5,2,-2>
•(-4,0,2)+(t-3)/3*<5,2,-2>
•(-4,0,2)+[-(t/4)]*<5,2,-2>
7
Problem 17. (1 point)
The function r(t)traces a circle. Determine the radius, center, and
plane containing the circle
r(t) = 9i+ (7cos(t))j+ (7 sin(t))k
Plane : x=
Circle’s Center : ( , , )
Radius :
Solution:
Solution: We have:
x(t) = 9,y(t) = 7cos(t),z(t) = 7 sin(t)
Hence,
y(t)2+z(t)2=49cos2(t)+49 sin2(t) = 49(cos2(t)+sin2(t)) = 49
This is the equation of a circle in the vertical plane x=9.
The circle is centered at the point (9,0,0)and its radius is √49 =7
Answer(s) submitted:
•9
•9
•0
•0
•7
(correct)
Correct Answers:
•9
•9
•0
•0
•7
Problem 18. (1 point)
Use cos(t)and sin(t), with positive coefficients, to parametrize
the intersection of the surfaces x2+y2=9 and z=2x4.
r(t) = h, , i
Solution:
Solution:
The points on the cylinder x2+y2=9 and on z=2x4can be writ-
ten in the form:
x2+y2=9→(3cost,3 sint,z)
z=2x4→x,y,2x4
The points (x,y,z)on the intersection curve must satisfy the fol-
lowing equations:
x=3cost
y=3sint
z=2x4=2(3cost)4
We obtain the vector parametrization:
r(t) = 3cost,3sin t,2(3cos t)4
Answer(s) submitted:
•3cost
(correct)
Correct Answers:
•3*cos(t); 3*sin(t); 2*[3*cos(t)]ˆ4
8
Problem 19. (1 point)
Find a parametrization, using cos(t)and sin(t), of the following
curve:
The intersection of the plane y=2 with the sphere x2+y2+z2=
29
r(t) = h, , i
Solution:
Solution: Substituting y=2 in the equation of the sphere gives:
x2+ (2)2+z2=29 ⇒x2+z2=25
This circle in the horizontal plane y=2 has the parametrization
x=√25cost,z=√25 sint.Therefore, the points on the intersec-
tion of the plane
y=2 and the sphere x2+y2+z2=29, can be written in the form
(5cost,2,5 sint), yielding the following parametrization:
r(t) = h5cost,2,5sin ti
Answer(s) submitted:
•5cost
(correct)
Correct Answers:
•5*cos(t); 2; 5*sin(t)
Problem 20. (1 point)
Find the solution r(t)of the differential equation with the given
initial condition:
r0(t) = hsin9t,sin6t,3ti,r(0) = h5,6,4i
r(t) = h,,i
Solution:
Solution: We first integrate the vector r0(t)to find the general
solution:
r(t) = Zhsin9t,sin6t,3tidt
=Zsin9tdt,Zsin 6tdt,Z3tdt=−1
9cos9t,−1
6cos6t,3
2t2+c
Substituting the initial condition we obtain:
r(0) = −1
9cos0,−1
6cos0,3
202+c
=h5,6,4i=−1
9,−1
6,0+c
Hence,
c=h5,6,4i−−1
9,−1
6,0=46
9,37
6,4
Hence the solution to the differential equation with the given ini-
tial condition is:
r(t) = −1
9cos9t,−1
6cos6t,3
2t2+46
9,37
6,4
=1
9(46 −cos9t),1
6(37 −cos6t),4+3
2t2
Answer(s) submitted:
•
•
•
(incorrect)
Correct Answers:
•5.11111-[cos(9*t)]/9
•6.16667-[cos(6*t)]/6
•4+1.5*tˆ2
9
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