m273q multivariable calculus afll 2017 review … multivariable calculus review problems for exam 1...

M273Q Multivariable Calculus Fall 2017 Review Problems for Exam 1

The formulas in the box will be provided on the exam.

�(s) =

��dTds

�� (x) =

jf 00(x)j[1 + (f 0(x))2]3=2

�(t) =jjT0(t)jjjjr0(t)jj �(t) =

jjr0(t)� r00(t)jjjjr0(t)jj3

1. True or False? Circle ONE answer for each. Hint: For e�ective study, explain why if `true' and give a

counterexample if `false.'

(a) T or F: If a?b and b?c; then a?c:

Solution: False. Let a = i;b = j; and c =< 1; 0; 1 > : Then a � b = i � j = 0 and b � c =<0; 1; 0 > � < 1; 0; 1 >= 0: However, a � c =< 1; 0; 0 > � < 1; 0; 1 >= 1:

(b) T or F: If a � b = 0; then jja� bjj = jjajjjjbjj:

Solution: True. Assume both a and b are nonzero. (If either is zero the result is obvious). If �is the angle between a and b; then � = �

2 since a and b are orthogonal. jja�bjj = jjajjjjbjj sin � =jjajjjjbjj sin�=2 = jjajjjjbjj:

(c) T or F: For any vectors u;v in R3; jju� vjj = jjv � ujj:

Solution: True. If � is the angle between u and v; then jju�vjj = jjujjjjvjj sin � = jjvjjjjujj sin � =jjv � ujj: (Or, jju� vjj = jj � v � ujj = j � 1jjjv � ujj = jjv � ujj:)

(d) T or F: The vector < 3;�1; 2 > is parallel to the plane 6x� 2y + 4z = 1:

Solution: False. A normal vector to the plane is n =< 6;�2; 4 > : Because < 3;�1; 2 >= 12n;

the vector is parallel to n and hence perpendicular to the plane.

(e) T or F: If u � v = 0; then u = 0 or v = 0:

Solution: False. For example i � j = 0 but i 6= 0 and j 6= 0:

(f) T or F: If u� v = 0; then u = 0 or v = 0:

Solution: False. For example i� i = 0 but i 6= 0:

(g) T or F: If u � v = 0 and u� v = 0; then u = 0 or v = 0:

Solution: True. If both u and v are nonzero, then u � v = 0 implies u and v are orthogonal.But u�v = 0 implies that u and v are parallel. Two nonzero vectors can't be both parallel andorthogonal, so at least one of them must be 0:

(h) T or F: The curve r(t) =0; t2; 4t

�is a parabola.

Solution: True. Parametric equations for the curve are x = 0; y = t2; z = 4t; and since t = z4

we have y = t2 =�z4

�2or y =

z2

16; x = 0: This is an equation of a parabola in the yz�plane.

M273Q Multivariable Calculus Review Problems for Exam 1 - of 9

(i) T or F: If �(t) = 0 for all t; the curve is a straight line.

Solution: True. Notice that �(t) = 0 () jjT0(t)jj = 0 () T0(t) = 0 for all t: But thenT(t) = C; a constant vector, which is true only for a straight line.

(j) T or F: Di�erent parameterizations of the same curve result in identical tangent vectors at a givenpoint on the curve.

Solution: False. For example, r1(t) =< t; t > and r2(t) =< 2t; 2t > both represent the sameplane curve (the line y = x), but the tangent vector r1(t) =< 1; 1 > for all t; while r1(t) =<2; 2 > : In fact, di�erent parametrizations give parallel tangent vectors at a point, but theirmagnitudes may di�er.

2. Which of the following are vectors?

(a)p

Vector Scalar Nonsense [(a � b)c]� a

(b) Vector Scalarp

Nonsense c� [(a � b)� c]

(c)p

Vector Scalar Nonsense c� [(a � b)c](d) Vector

pScalar Nonsense (a� b) � c

3. Which of the following are meaningful?

(a)p

Meaningful Nonsense jjwjj(u� v)

(b) Meaningfulp

Nonsense (u � v)�w

(c)p

Meaningful Nonsense u � (v �w)

4. Find the values of x such that the vectors < 3; 2; x > and < 2x; 4; x > are orthogonal.

Solution: For the two vectors to be orthogonal, we need < 3; 2; x > � < 2x; 4; x >= 0: That is,3(2x) + 2(4) + x(x) = 0; or x = �2 or x = �4:

5. Find the decomposition a = akb + a?b of a =< 1; 1; 1 > along b =< 2;�1;�3 > :

Solution: akb =a � bb � bb =

��27;1

7;3

7

�: Finally, a?b = a � akb =< 1; 1; 1 > �

��27;1

7;3

7

�=�

9

7;6

7;4

7

�:

6. Let a =

�1p2;1p2

�;b =

�1p2;�1p2

�; and u =< 3; 0 > :

(a) Show that a and b are orthogonal unit vectors.

Solution: a � b = 0 and jjajj =p12+12p

2= 1; and jjbjj =

p12+(�1)2p

2= 1:

(b) Find the decomposition of u along a:

Solution: ujja = (u � a)a =3p2

�1p2;1p2

�=

�3

2;3

2

�: Thus, u?a =< 3; 0 > � 32 ; 32� =

32 ;� 3

2

�:

(c) Find the decomposition of u along b:


Solution: Similarly, ujjb = (u � b)b =

�< 3; 0 > �

�1p2;� 1p

2

��1p2;� 1p

2

�=

�3

2;�3

2

�:

Thus, u?b =< 3; 0 > � 32 ; 32� = 32 ;� 3

2

�= u?a:

7. (a) Find an equation of the sphere that passes through the point (6, -2, 3) and has center (-1,2,1).

Solution: Use the distance formula to �nd the distance between (6, -2, 3) and (-1,2,1). Then,the equation for the circle is (x+ 1)2 + (y � 2)2 + (z � 1)2 = 69:

(b) Find the curve in which this sphere intersects the yz-plane.

Solution: The intersection of this sphere with the yz�plane is the set of points on the spherewhose x�coordinate is 0. Putting x=0 in to the equation, (y � 2)2 + (z � 1)2 = 68; whichrepresents a circle in the yz�plance with center (0; 2; 1) and radius

p68:

8. For each of the following quantities (cos �; sin �; x; y; z; and w) in the picture below, �ll in the blank withthe number of the expression, taken from the list to the right, to which it is equal.

x

y

z

w

a

b

θ

cos � = 5 1.a � bjjajj

sin �= 4 2.a � bjjbjj

x = 2 3.jja� bjjjjbjj

y= 3 4.jja� bjjjjajjjjbjj

z= 7 5.ja � bjjjajjjjbjj

w= 6 6. jjb� ajj

7.(b� a) � bjjbjj

9. Find an equation for the line through (4;�1; 2) and (1; 1; 5):

Solution: The line has direction v =< �3; 2; 3 > : Letting P0 = (4;�1; 2); parametric equations are

x = 4� 3t; y = �1 + 2t; z = 2 + 3t:


10. Find an equation for the line through (�2; 2; 4) and perpendicular to the plane 2x� y + 5z = 12:

Solution: A direction vector for the line is a normal vector for the plane, n =< 2;�1; 5 >; andparametric equations for the line are

x = �2 + 2t; y = 2� t; z = 4 + 5t:

11. Find an equation of the plane through (2; 1; 0) and parallel to x+ 4y � 3z = 1:

Solution: Since the two planes are parallel, they will have the same normal vectors. Then we cantake n =< 1; 4;�3 > and an equation of the plane is 1(x� 2) + 4(y � 1)� 3(z � 0) = 0:

12. Find an equation of the plane that passes through the point (�1;�3; 2) and contains the line x(t) =�1� 2t; y(t) = 4t; z(t) = 2 + t:

Solution: To �nd an equation of the plane we must �rst �nd two nonparallel vectors in the plane,then their cross product will be a normal vector to the plane. But since the given line lies in theplane, its direction vector a =< �2; 4; 1 > is one vector parallel to the plane. To �nd another vectorb which lies in the plane pick any point on the line [say (�1; 0; 2); found by setting t = 0] and let b bethe vector connecting this point to the given point in the plane (�1;�3; 2): (But beware; we should�rst check that the given point is not on the given line. If it were on the line, the plane wouldnt beuniquely determined. What would n then be when we set n = a � b?) Thus b =< 0; 3; 0 > andn = a�b =< 0+ 3; 0� 0; 6� 0 >=< 3; 0; 6 > and an equation of the plane is 3(x+1)+ 6(z� 2) = 0or x+ 2z = 3:

13. Find the point at which the line x(t) = 1� t; y = t; z(t) = 1 + t and the plane z = 1� 2x+ y intersect.

Solution: Substitute the line into the equation of the plane. z = 1�2x+y ) 1+t = 1�2(1�t)+t)t = 1) the point of intersection is (0; 1; 2):

14. (a) Find an equation of the plane that passes through the points A(2; 1; 1); B(�1;�1; 10); and C(1; 3;�4:)

Solution: The vector��!AB =< �3;�2; 9 > and

�!AC =< �1; 2;�5 > lie in the plane, so n =��!

AB � �!AC =< �8;�24;�8 > or equivalently, < 1; 3; 1 > is a normal vector to the plane. Thepoint A(2; 1; 1) lies on the plane so an equation for the plane is 1(x�2)+3(y�1)+1(z�1) = 0:

(b) A second plane passes through (2; 0; 4) and has normal vector < 2;�4;�3 > : Find an equation forthe line of intersection of the two planes.

Solution: The point (2,0,4) lies on the second plane, but the point also satis�es the equationof the �rst plane, so the point lies on the line of intersection of the planes. A vector v in thedirection of this intersecting line is perpendicular to the normal vectors of both planes, so takev =< 1; 3; 1 > � < 2;�4;�3 >=< �5; 5;�10 > or just use < 1;�1; 2 > : Parametric equationsfor the line are

x = 2 + t; y = �t; z = 4 + 2t:

15. Provide a clear sketch of the following traces for the quadratic surface y =px2 + z2 + 1 in the given

planes. Label your work appropriately.

x = 0;x = 1; y = 0; y = 2; z = 0:


x

y

x

z

y

z

__-axis

__-axis

__-axis

__-axis

Solution:

16. Match the equations with their graphs. Give reasons for your choices.

(a) II 8x+ 2y + 3z = 0

(b) I z = sinx+ cos y

(c) IV z = sin

��

2 + x2 + y2

�(d) III z = ey

17. Find a vector function that represents the curve of intersection of the cylinder x2+ y2 = 16 and the planex+ z = 5:

Solution: The projection of the curve C of intersection onto the xy�plane is the circle x2 + y2 =16; z = 0: So we can write x = 4 cos t; y = 4 sin t; 0 � t � 2�: From the equation of the plane, we havez = 5 � x = 5 � 4 cos t; so parametric equations for C are x = 4 cos t; y = 4 sin t; z = 5 � 4 cos t; 0 �t � 2�; and the corresponding vector function is r(t) =< 4 cos t; 4 sin t; z = 5� 4 cos t >; 0 � t � 2�:

18. Find an equation for the tangent line to the curve x = 2 sin t; y = 2 sin 2t; and z = 2 sin 3t at the point(1;p3; 2):


Solution: The curve is given by r(t) =< 2 sin t; 2 sin 2t; 2 sin 3t >; so r0(t) =< 2 cos t; 4 cos 2t; 6 cos 3t >: The point (1;

p3; 2) corresponds to t = �=6; so the tangent vector there is r0(�=6) =<

p3; 2; 0 > :

Then the tangent line has direction vector <p3; 2; 0 > and includes the point (1;

p3; 2); so parametric

equations are x = 1 +p3t; y =

p3 + 2t; z = 2:

19. A helix circles the z�axis, going from (2; 0; 0) to (2; 0; 6�) in one turn.

(a) Parameterize this helix.

Solution: r(t) =< 2 cos t; 2 sin t; 3t > : (Note that 1 revolution is 2�; so 2�(3) = 6�:)

(b) Calculate the length of a single turn.

Solution: For 0 � t � 2�; jjr0(t)jj = p4 + 9 =

p13: Thus s =

Z 2�

0

p13dt = 13(2�):

(c) Find the curvature of this helix.

Solution: The unit tangent vector is T(t) = 1p13

< �2 sin t; 2 cos t; 3 >; so T0(t) = �2p13

<

cos t; sin t; 0 >. Thus, jjT0(t)jj = 2p13: Since jjr0(t)jj = p

13; � = 213

20. (a) Sketch the curve with vector function r(t) = ht; cos�t; sin�ti ; t � 0:

Solution: The corresponding parametric equations for the curve are x = t; y = cos�t; z = sin�t:Since y2+ z2 = 1; the curve is contained in a circular cylinder with axis the x�axis. Since x = t;the curve is a helix.

(b) Find r0(t) and r00(t):

Solution: Since r(t) = ht; cos�t; sin�ti ; r0(t) = h1;�� sin�t; � cos�ti ; and r00(t) = 0;�pi2 cos�t;��2 sin�t� :

21. Which curve below is traced out by r(t) =

�sin�t; cos�t;

1

4t2�; 0 � t � 2:

Solution: Graph 1. Note that r(0) =< 0; 1; 0 > :

22. Find a point on the curve r(t) =t+ 1; 2t2 � 2; 5

�where the tangent line is parallel to the plane x+2y�

4z = 5:


Solution: The plane has normal vector n =< 1; 2;�4 > : Since r0(t) =< 1; 4t; 0 >; we want <1; 4t; 0 > � < 1; 2;�4 >= 0: That is, 1 + 8t = 0; and r(� 1

8 ) =<78 ;

�6342 ; 5 > :

23. Let r(t) =p

2� t; (et � 1)=t; ln(t+ 1)�:

(a) Find the domain of r:

Solution: The expressionsp2� t; (et� 1)=t; and ln(t+1) are all de�ned when 2� t � 0: Thus,

t � 2; t 6= 0; and t+ 1 > 0) t > �1: Finally, the domain of r is (�1; 0) [ (0; 2]:

(b) Find limt!0

r(t):

Solution: limt!0

r(t) =Dp

2; 1; 0E: Note that in the y� component we use l'Hospital's Rule.

(c) Find r0(t):

Solution: r0(t) =D� 1

2p2�t ;

tet�et+1t2 ; 1

t+1

E:

24. Suppose that an object has velocity v(t) =3p1 + t; 2 sin(2t); 6e3t

�at time t; and position r(t) =<

0; 1; 2 > at time t = 0: Find the position, r(t); of the object at time t:

Solution: r(t) =

Zv(t)dt =

�Z3p1 + tdt;

Z2 sin(2t)dt;

Z6e3tdt

�=D2(1 + t)3=2 + c1;� cos(2t) + c2; 2e

3t + c3

E:

Thus, r(0) =< 2 + c1;�1 + c2; 2 + c3 >=< 0; 1; 2 >) c1 = �2; c2 = 2; c3 = 0: So, r(t) =D2(1 + t)3=2 � 2;� cos(2t) + 2; 2e3t

E:

25. If r(t) =t2; t cos�t; sin�t

�; evaluate

Z 1

0

r(t)dt:

Solution:

Z 1

0

r(t)dt =

�Z 1

0

t2dt;

Z 1

0

t cos�tdt;

Z 1

0

sin�tdt

�=

�1

3;� 2

�2;2

�

�:

26. Find the length of the curve: x = 2 cos(2t); y = 2t3=2; and z = 2 sin(2t); 0 � t � 1:

Solution: s =

Z 1

0

s�dx

dt

�2

+

�dy

dt

�2

+

�dz

dt

�2

dt =

Z 1

0

q16 sin2(2t) + 16 cos2(2t) + 9t dt =

Z 1

0

p16 + 9tdt =

122

27:

27. Reparameterize the curve r(t) =< et; et sin t; et cos t > with respect to arc length measured from the point(1; 0; 1) in the direction of increasing t:

Solution: The parametric value corresponding to the point (1; 0; 1) is t = 0: Since r0(t) =< et; et(cos t+

sin t); et(cos t�sin t) >; jjr0(t)jj = etp1 + (cos t+ sin t)2 + (cos t� sin t)2 =

p3et; and s(t) =

R t0eup3du =p

3(et�1)) t = ln�1 + 1p

3s�: Therefore, r(t(s)) =

D1 + 1p

3s;�1 + 1p

3s�sin ln

�1 + 1p

3s�;�1 + 1p

3s�cos ln

�1 + 1p

3s�E

:


28. Find the tangent line to the curve of intersection of the cylinder x2 + y2 = 25 and the plane x = z at thepoint (3; 4; 3):

Solution: Let x(t) = 5 cos t; y(t) = 5 sin t; and z(t) = 5 cos t; so that r(t) =< 5 cos t; 5 sin t; 5 cos t >and r0(t) =< �5 sin t; 5 cos t;�5 sin t > : When (x; y; z) = (3; 4; 3); x(t) = z(t) = 5 cos t = 3; andy(t) = 5 sin t = 4; so r0(t0) =< �4; 3;�4 > is a direction vector for the line. The tangent line hasparametric equations x(t) = 3 � 4t; y(t) = 4 + 3t; and z(t) = 3 � 4t: ANOTHER SOLUTION: Let

x(t) = t; y =p25� x2 =

p25� t2; and z(t) = x(t) = t: Then r(t) =

Dt; (25� t)1=2; t

E: In this

case, r0(t) =D1; �tp

25�t2 ; 1E: When x = 3; t = 3; and r0(3) =< 1; �34 ; 1 > : So the tangent line has

parametric equations x(t) = 3 + t; y(t) = 4� 34 t; and z(t) = 3 + t:

29. For the curve given by r(t) =13 t

3; t2; 2t�; �nd

(a) the unit tangent vector

Solution: T(t) =r0(t)jjr0(t)jj =

< t2; 2t; 2 >pt4 + t2 + 1

=< t2; 2t; 2 >

(t2 + 2):

(b) the unit normal vector

Solution:

T0(t) =< 4t; 4� 2t2;�4t >

(t2 + 2)2) jjT0(t)jj =

p4(t4 + 4t2 + 4)

(t2 + 2)2=

2

t2 + 2andN(t) =

< 2t; 2� t2;�2t >t2 + 2

(c) the curvature

Solution: �(t) =jjT0(t)jjjjr0(t)jj =

2

(t2 + 2)2

30. A particle moves with position function r(t) =< t ln t; t; e�t >. Find the velocity, speed, and accelerationof the particle.

Solution: v(t) = r0(t) =< 1 + ln t; 1;�e�t > : �(t) = jjv(t)jj =p(1 + ln t)2 + 1 + (�e�t)2 =p

2 + 2 ln t+ (ln t)2 + e�2t: a(t) = v0(t) =<1

t; 0; e�t > :

31. A particle starts at the origin with initial velocity< 1;�1; 3 > and its acceleration is a(t) =< 6t; 12t2;�6t >: Find its position function.

Solution: v(t) =

Za(t)dt =

�Z6t dt;

Z12t2 dt;

Z�6t dt

�=3t2; 4t3;�3t2�+C; but< 1;�1; 3 >=

v(0)+C; soC =< 1;�1; 3 > and v(t) =3t2 + 1; 4t3 � 1;�3t2 + 3

�: r(t) =

Zv(t)dt =

t3 + t; t4 � t; 3t� t3

�=

D: But r(0) = 0; so D = 0; and r(t) =t3 + t; t4 � t; 3t� t3

�:

32. A ying squirrel has position r(t) =

�t+

t2

2; 1� t; 2 + t2

�at time t: Compute the following at time t = 1:

(a) The velocity at time t = 1; v(1) = < 2;�1; 2 >.


Solution: v(t) = r0(t) =< 1 + t;�1; 2t > :

(b) The speed at time t = 1; �(1) = 3 .

Solution: � = jjv(t)jj: jjv(1)jj = p4 + 1 + 4 = 3:

33. Consider the vector valued function r(t) = describing the curve shown below. Put the curvature of r atA;B and C in order from smallest to largest. Draw the osculating circles at those points.

Solution: B, A, C.

m273q multivariable calculus afll 2017 review … multivariable calculus review problems for exam 1...

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