Math 365 - Sample Examinations for Practice

We present samples of a midterm and final examinations for you to practice. Pleasebe aware that the actual examinations for this course may be vastly different to thesesamples. Doing well in a practice exam is no guaranty of a passing mark whendoing the exams.

Use these samples to prepare for the course examinations. We recommend you

• to set a quite place to do the exams and time yourself. Pretend you are in anexamination room.

• to start with the questions you know best. Clearly identify the question youare answering.

• to take note of the time you spend doing the exam. If you go over the 3 hoursrecommended do not stop, keep working and take note of the time youneeded to finish.

• to take note of the number of times you need to consult your textbook orStudy Guide.

• to pay attention on how easily you find what you are looking for.

Once you finish the exam, grade yourself and be critical, that is, put yourself in theplace of the marker. Grade yourself with the solutions and marking schemeprovided. Ask yourself

• is the exam clear and legible?

• are the calculations correct?

• are there any missing steps?

• what can I do to improve my grades?

• what can I do to be more efficient with my time?

Review weak concepts and any material that is not clear yet.

Math 365 Sample of Midterm Exam

Time: 3 hoursPassing grade: 50%Total points: 70

1. Define the following4 points

a) The dot product between two vectors in an 3-dimensional space

b) The norm of a vector in a 3 dimensional space

2. a) Give the parametric equations of the line passing through the points (-4,-6,1)6 pointsand (-2,9,-3).

b) Give the linear equation of a plane that passes through (0,2,1) and isperpendicular to the line in part a).

3. a) Explain why the planes x+ y + z = 1 and x− y + z = 1 are neither parallel4 pointsnor perpendicular.

b) Give the angle between them.

4. Let u =< −1, 3, 6 >, v =< 0,−3, 5 > and w =< −2,−3, 1 >. Evaluate the6 pointsfollowing

a) u · (v ×w)

b) projvw

c) u× (v ×w)

5. Identify the surface given by the equation x2 + y2 + z2 + 2z = 0.3 points

6. a) Identify each one of the surfaces given by the equations 9x2 + y2 + 9z2 = 816 pointsand y = x2 for (z > 0).

b) Find a vector equation for the curve of intersection of the surface in part a) interms of the parameter x = t

7. a) Find the tangent line to the curve r(t) = e−2ti+ cos tj+ 3 sin tk at the point10 points(1, 10)

b) Find the point of intersection of the line in part a) with the yz-plane

c) Give the curvature of the curve in part a) at the same point (1, 1, 0).

8. Find the vectors T,N and B of the curve r(t) = (cos t, sin t, 1) at the point6 pointsr(π/4)

9. Find the length of the curve r(t) = tan−1 ti+ln(1 + t2)

2j on the interval [0, 1].6 points

10. What force is required so that a particle of mass m has the position function4 pointsr(t) = t3i+ t2j+ t3k?

11. The position function of a particle is given by r(t) = t2i+ 5tj+ (t2 − 16t)k.6 pointsWhen is the speed a minimum?

12. Describe the largest region in the xy-plane that corresponds to the domain of the5 pointsfunction

f(x, y) =

√y − x21− x2

.

13. Explain why the limit2 points

lim(x,y)→(0,0)

2x2y

x4 + y2

does not exist.

14. If f(x, y, z) = exy ln(y2 + z) find the second partial derivative∂2f

∂y∂z.2 points

Math 365 Sample of Midterm Exam

Time: 3 hoursPassing grade: 50%Total points: 70

Solutions and Marking

1. Define the following4 points

a) The dot product between two vectors in an 3-dimensional spaceSolution For two vectors u =< u1, u2, u3 > and v =< v1, v2, v3 > the dotproduct is

u · v = u1v1 + u2v2 + u3v3.

(2 pts)

b) The norm of a vector in a 3 dimensional spaceSolution For the vector u =< u1, u2, u3 > its norm is defined as

‖u‖ =√u21 + u22 + u23.

(2 pts)

Solution

2. a) Give the parametric equations of the line passing through the points (-4,-6,1)6 pointsand (-2,9,-3).

SolutionThe vector in the direction of the line isv = (−2, 9,−3)− (−4,−6, 1) = (2, 15,−4) (1 pt)

The parametric equations of the line are

x = −2 + 2t, y = 9 + 15t, z = −3− 4t. (2 pts)

b) Give the linear equation of a plane that passes through (0,2,1) and isperpendicular to the line in part a).

SolutionThe vector normal to the plane is n =< 2, 15,−4 > (1 pt)

The equation of the plane is

2x+ 15(y − 2)− 4(z − 1) = 0 or 2x+ 15y − 4z = 26 (2 pts)

3. a) Explain why the planes x+ y + z = 1 and x− y + z = 1 are neither parallel4 pointsnor perpendicular.

SolutionThe normals of the planes are n = (1, 1, 1) and u = (1,−1, 1) (1 pt)

The normals are not orthogonal since n · u = 1− 1 + 1 = 1 6= 0 is nonzero

and they are not parallel since n× u = 2i− 2k is nonzero. (1 pt)

b) Give the angle between them.

SolutionThe angle between the normal vectors is

cos θ =n · u‖n‖‖u‖

=1

3(1 pt)

thus the angle is θ ≈ 70.5◦ (1 pt)

4. Let u =< −1, 3, 6 >, v =< 0,−3, 5 > and w =< −2,−3, 1 >. Evaluate the6 pointsfollowing

a) u · (v ×w)

b) projvw

c) u× (v ×w)

Solution (2 points each)

a) v ×w = 12i− 10j− 6k thus

u · (v ×w) =< −1, 3, 6 > · < 12,−10,−6 >= −78

b) v =< 0,−3, 5 > ·w =< −2,−3, 1 >= 14 and ‖v‖2 = 34, thus

projvw =14

34< 0,−3, 5 >=< 0,−21

17,35

17>

c) (u ·w)v − (u · v)w = −1 < 0,−3, 5 > −21 < −2,−3, 1 >=<42, 66,−26 >

5. Identify the surface given by the equation x2 + y2 + z2 + 2z = 0 Solution3 pointsCompleting the perfect squares we get x2 + y2 + (z + 1)2 = 1 (1 pt)

Sphere with center (0, 0,−1) and radius 1. (2 pts)

6. a) Identify each one of the surfaces given by the equations 9x2 + y2 + 9z2 = 816 pointsand y = x2 for (z > 0).

SolutionThe equation 9x2 + y2 + 9z2 = 81 is the upper part (or above the xy-plane) ofan ellipsoid with center at the origin. The equation y = x2 is the upper part (orabove the xy-plane) of an open cylinder in the direction of the z-axis, and traceon the xy-plane the parabola y = x2. (3 pts)

b) Find a vector equation for the curve of intersection of the surface in part a) interms of the parameter x = t

SolutionSince y = x2 the intersection of the two curves gives 9y + y2 + 9z2 = 81solving for z we have

z =

√9− y − y2

9

(1 pt)

and the curve of intersection is

r(t) = ti+ t2j+

√9− t2 − t4

9k (2 pts)

7. a) Find the tangent line to the curve r(t) = e−2ti+ cos tj+ 3 sin tk at the point10 points(1, 10)

SolutionWe have r(0) = (1, 1, 0). Thus the vector in the direction of the line isr′(t) = −2e−2ti− sin tj+ 3 cos tk and r′(0) = −2i+ 3k (2 pts)

The parametric equations of the line are

x = 1− 2t, y = 1, z = 3t (2 pts)

b) Find the point of intersection of the line in part a) with the yz-plane.

SolutionThe intersection with the yz-plane is when x = 0, thus t = 1/2 (1 pt)

and the point of intersection is x = 1− 2(1/2), y = 1, z = 3(1/2) that is(0, 1, 3/2) (1 pt)

c) Give the curvature of the curve in part a) at the same point (1, 1, 0).

SolutionWe have to find κ(0), thus

r′′(t) = 4e−2ti− cos tj− 2 sin tk and r′′(0) = 4i− j (2 pts)

r′(0)× r′′(0) = −3i− 12j+ 2k ‖r′(0)× r′′(0)‖ =√157

(1 pt)

κ(0) =‖r′(0)× r′′(0)‖‖r′(0)‖3

=1

13

√157

13(1 pt)

8. Find the vectors T,N and B of the curve r(t) = (cos t, sin t, 1) at the point6 pointsr(π/4).

Solutionr′(t) = − sin ti+ cos tj and ‖r′(t)‖ = 1.

Hence T(t) = − sin ti+ cos tj and

T(π/4) =1√2(−1, 1, 0). (2 pts)

T′(t) = − cos ti− sin tj; and ‖T′(t)‖ = 1. Thus

N(t) = − cos ti− sin tj and

N(π/4) = − 1√2(1, 1, 0). (2 pts)

Therefore

B(π/4) = −1/2(−1, 1, 0)× (1, 1, 0) = −1/2(0, 0,−2) = (0, 0, 1) = k.

(2 pts)

9. Find the length of the curve r(t) = tan−1 ti+ln(1 + t2)

2j on the interval [0, 1].6 points

Solution

r′(t) =1

1 + t2i+

t

1 + t2j and

‖r′(t)‖2 =1

(1 + t2)2+

t2

(1 + t2)2=

1 + t2

(1 + t2)2=

1

1 + t2(3 pts)

L =

∫ 1

0

dt√1 + t2

= ln(t+√

1 + t2)∣∣∣10= ln(1+

√2)− ln 1 = ln(1+

√2)

(3 pts)

By trigonometric substitution or using a table of integration.

10. What force is required so that a particle of mass m has the position function4 pointsr(t) = t3i+ t2j+ t3k?

Solutionr(t) = (t3, t2, t3) and a(t) = 6ti+ 2j+ 6tk. (2 pts)

By Newton’s second law

F(t) = ma(t) = 6mti+ 2mj+ 6mtk. (2 pts)

11. The position function of a particle is given by r(t) = t2i+ 5tj+ (t2 − 16t)k.6 pointsWhen is the speed a minimum?

Solutionr′(t) = 2ti+ 5j+ (2t− 16)k and the speed is

‖r′(t)‖ =√

4t2 + 25 + (2t− 16)2 =√8t2 − 64t+ 281 (2 pts)

To find the maximum of this functiond

dt‖r′(t)‖ = 16t− 64

2√8t2 − 64t+ 281

= 0 gives t = 4 and

‖r′(t)‖ > 0 for t > 4 and ‖r′(t)‖ < 0 for t < 4 (3 pts)

Thus the speed is a minimum at t = 4. (1 pt)

12. Describe the largest region in the xy-plane that corresponds to the domain of the5 pointsfunction

f(x, y) =

√y − x21− x2

.

Solutiony = x2 is an open up parabola and y ≥ x2 is the region above the parabola and1− x2 6= 0 implies x 6= ±1. (2 pts)

The domain is the upper part of the parabola y = x2 including the points on theparabola (boundary) and excluding the vertical lines x = 1 and x = −1. (3 pts)

A sketch of the region is also acceptable

Excluding the vertical lines

13. Explain why the limit2 points

lim(x,y)→(0,0)

2x2y

x4 + y2

does not exist.

SolutionThe limit taken along the line x-axis is

lim(x,0)→(0,0)

0

x4= 0 (1 pt)

The limit taken along the parabola y = x2 is

lim(x,x2)→(0,0)

2x4

x4 + x4= 1 (1 pt)

The limits are not equal hence the limit does not exist.

14. If f(x, y, z) = exy ln(y2 + z) find the second partial derivative∂2f

∂y∂z.2 points

Solution∂f

∂z=

exy

y2 + z(1 pt)

and

∂2f

∂y∂z=

∂

∂y

(exy

y2 + z

)=xexy(y2 + z)− exy(2y)

(y2 + z)2

=exy(xy2 + xz − 2y)

(y2 + z)2

(1 pt)

Math 365 Sample of Final Exam

Time: 3 hoursPassing grade: 50%Total points: 70

1. Define the following6 points

a) Gradient of a function f(x, y, z)

b) The tangent plane to the surface f(x, y) at the point P .

c) The derivative of a function f(x, y) in the direction of the vector u.

2. a) Find the derivative of the function f(x, y, z) = exyz at the points (1,1,1) in the4 pointsdirection of u = i− 2j+ k.

b) Find the equation of the tangent plane of the surface in part a) at the samepoint (1,1,1)

3. Find the absolute maximum and minimum values of the function10 pointsf(x, y) = 4x+ 6y − x2 − y2 on the regionD = {(x, y)|0 ≤ x ≤ 4, 0 ≤ y ≤ 5}. Give a sketch of the region D.

4. Reverse the order of integration and evaluate the integral6 points ∫ 1

0

∫ 3

3y

ex2

dx dy

Hint. Sketch the region of integration.

5. Use cylindrical coordinates to find the volume under the paraboloid z = x2 + y26 pointsand above the disk x2 + y2 ≤ 9.

6. Find the area of the surface of the part of the paraboloid z = 4− x2 − y2 that8 pointslies above the xy-plane.

Hint. Choose an appropriate coordinate system.

7. a) Give the image of the rectangle with the corners (0,0), (3,0), (0,2), (3,2) under10 pointsthe transformation x = 2u+ 3v, y = u− v.

b) Give the Jacobian of this transformation.

8. Evaluate the line integral8 points ∫C

2x+ 9z ds

where C is the curve x = t, y = t2, z = t3, for 0 ≤ t ≤ 1.

9. Give the parametric representation of the lower half of the ellipsoid4 points2x2 + 4y2 + z2 = 1.

10. Evaluate the surface integral∫ ∫

Syz ds where S is the part of the plane8 points

x+ y + z = 1 that lies in the first octant.

Math 365 Sample of Final Exam

Time: 3 hoursPassing grade: 50%Total points: 70

Solutions and Marking

1. Define the following6 points

a) Gradient of a function f(x, y, z)SolutionIt is the vector∇f defined as

∇f(x, y, z) =< fx(x, y, z), fy(x, y, z), fz(x, y, z) >=∂f

∂xi+

∂f

∂yj+

∂f

∂zk

(2 pts)

b) The tangent plane to the surface f(x, y) at the point P .SolutionTheorem 13.7.2 of the textbook. It is the plane with normal ∇f(P ) andequation z = f(a, b) + fx(x, y)(x− a) + fy(x, y)(y − b) where P = (a, b)(2 pts)

c) The derivative of a function f(x, y) in the direction of the vector u.SolutionIt is the dot product of∇f and u that is fx(x, y)u1 + f(x, y)u2 whereu =< u1, u2 > (2 pts)

2. a) Find the derivative of the function f(x, y, z) = exyz at the points (1,1,1) in the4 pointsdirection of u = i− 2j+ k.

Solutiona)∇f = exyz(yzi+ xzj+ xyk) and ∇f(1, 1, 1) = e < 1, 1, 1 >) and

∇f(1, 1, 1)· < 1,−2, 1 >= e(1− 2 + 1) = 0 (3 pts)

b) Find the equation of the tangent plane of the surface in part a) at the samepoint (1,1,1).

SolutionThe equation of the plane is (x− 1)i+ (y − 1)j+ (z − 1)k = 0 or

x+ y + z = 3 (1 pt)

3. Find the absolute maximum and minimum values of the function10 pointsf(x, y) = 4x+ 6y − x2 − y2 on the regionD = {(x, y)|0 ≤ x ≤ 4, 0 ≤ y ≤ 5}. Give a sketch of the region D.

Solutionfx(x, y) = 4− 2x, fy(x, y) = 6− 2y, critical point is (2,3) and

f(2, 3) = 13 (2 pts)

Line: y = 0; 0 ≤ x ≤ 4; f(x, 0) = 4x− x2 (a CD parabola) maximum at x = 2and minimum at x = 0 and f(2, 0) = 4 and f(0, 0) = 0 = f(4, 0) (2 pts)

Line: x = 0; 0 ≤ y ≤ 5, f(0, y) = 6y − y2 (a CD parabola) maximum at y = 3and minimum at y = 0 and f(0, 3) = 9 and f(0, 0) = 0 (2 pts)

Line: x = 4; 0 ≤ y ≤ 5; f(4, y) = 6y − y2 (a CD parabola) maximum at y = 3and minimum at y = 0 and f(4, 3) = 9 and f(4, 0) = 0 (2 pts)

Line: y = 5; 0 ≤ x ≤ 4; f(x, 5) = −x2 − 4x+ 5 (a CD parabola) maximum atx = 2 and minimum at x = 0 and f(2, 5) = 9 and f(0, 5) = 5 = f(4, 5) (2 pts)

Absolute maximum value 13 = f(2, 3) and absolute minimum value0 = f(0, 0).

4. Reverse the order of integration and evaluate the integral6 points ∫ 1

0

∫ 3

3y

ex2

dx dy

Hint. Sketch the region of integration.

Solution

1 3 On this region 3/0;30 xyx ≤≤≤≤

(2 pts)

Hence

∫ 3

0

∫ x/3

0

ex2

dy dx =1

3

∫ 3

0

xex2

dx =1

6

∫ 3

0

2xex2

dx =ex

2

6

∣∣∣30=

1

6(e9−1)

(4 pts)

5. Use cylindrical coordinates to find the volume under the paraboloid z = x2 + y26 pointsand above the disk x2 + y2 ≤ 9.

SolutionIn cylindrical coordinates we have

0 ≤ z ≤ x2 + y2;−3 ≤ x ≤ 3;−3 ≤ y ≤ 3 (2 pts)

thus 0 ≤ r ≤ 3; 0 ≤ θ ≤ 2π; 0 ≤ z ≤ r2 and the integral that gives the volumeis (4 pts)

∫ 3

0

∫ 2π

0

∫ r2

0

r dz dθ dr =

∫ 3

0

∫ 2π

0

r3 dθ dr = 2π

∫ 3

0

r3 dr

=π

2r4∣∣∣30=

81π

2

(4 pts)

6. Find the area of the surface of the part of the paraboloid z = 4− x2 − y2 that8 pointslies above the xy-plane.

Hint. Choose an appropriate coordinate system.

SolutionThe region of integration is −2 ≤ y ≤ 2;−

√4− x2 ≤ y ≤

√4− x2 (2 pts)√

1 + (fx)2 + (fy)2 =√1 + 4x2 + 4y2 =

√1 + 4(x2 + y2) (1 pt)

In polar coordinates f(x, y) = 4− r2; 0 ≤ r ≤ 2; 0 ≤ θ ≤ 2π

∫ 2π

0

∫ 2

0

r√

1 + 4r2 dr dθ =1

2

∫ 2π

0

2r√1 + 4r2 dr dθ

=1

6

∫ 2π

0

(4r2 + 2)3/2∣∣∣20dθ

=1

12(17√17− 1)

∫ 2π

0

dθ

=π

6(17√17− 1)

(5 pts)

7. a) Give the image of the rectangle with the corners (0,0), (3,0), (0,2), (3,2) under10 pointsthe transformation x = 2u+ 3v, y = u− v.

SolutionT (u, v) = (2u+ 3v, u− v)Line: v = 0; 0 ≤ u ≤ 3;T (u, 0) = (2u, u) a line from (0,0) to (6,3)

Line: v = 2; 0 ≤ u ≤ 3;T (u, 2) = (2u+ 6, u− 2) a line from (6,-2) to (12,1)

Line: u = 0; 0 ≤ v ≤ 2;T (0, v) = (2v,−v) a line from (0,0) to (6,-2)

Line: u = 3; 0 ≤ v ≤ 2;T (3, v) = (6 + 2v, 3− v) a line from (6,3) to (12,1)

(6 pts)

b) Give the Jacobian of this transformation.

Solution

(6,3) (12,1) (6,-2)

∣∣∣∣∣∣∣∣2 3

1 −1

∣∣∣∣∣∣∣∣ = −5(4 pts)

8. Evaluate the line integral8 points ∫C

2x+ 9z ds

where C is the curve x = t, y = t2, z = t3, for 0 ≤ t ≤ 1.

SolutionWe have r′(t) = ti+ 2tj+ 3t2k and therefore ||r′|| =

√1 + 4t2 + 9t4 (2 pt)

∫C

2x+ 9z ds =

∫ 1

0

(2t+ 9t3)√

1 + 4t2 + 9t4 dt

=1

4

∫ 1

0

(8t+ 36t3)√

1 + 4t2 + 9t4 dt

=(1 + 4t2 + 9t4)3/2

6

∣∣∣10

=13

6

(6 pt)

9. Give the parametric representation of the lower half of the ellipsoid4 points2x2 + 4y2 + z2 = 1.

Solutionz2 = 1− 2x2 − 4y2 since z < 0 we have z = −

√1− 2x2 − 4y2 with

x = x, y = y (1 pt)

x2

(1/√2)2

+y2

(1/2)2+ z2 = 1 setting

x =u cos v√

2y =

u sin v

2(1 pt)

hence

z = −√1− u2 cos2 v − u2 sin2 v = −

√1− u2

with 0 ≤ u ≤ 1; 0 ≤ v ≤ 2π (2 pts)

10. Evaluate the surface integral∫ ∫

Syz ds where S is the part of the plane8 points

x+ y + z = 1 that lies in the first octant.

SolutionThe region of integration on the xy-plane is

0 ≤ x ≤ 1 and 0 ≤ y ≤ 1− x. (2 pts)

We have√f22 + f2y + f23 =

√3, (1 pt)

then

∫ ∫D

y(1− x− y)√3 dA =

∫ 1

0

∫ 1−x

0

y − xy − y2 dy dx

=√3

∫ 1

0

y

2− xy2

2− y3

3

∣∣∣1−x0

dx

=

√3

6

∫ 1

0

(1− x)3 dx

= −√3

24(1− x)4

∣∣∣10=

√3

24

(5 pts)