homework 1 – solutionfaculty.uaeu.ac.ae/jaelee/class/1120/09sp_hws.pdf · calculus ii for...

United Arab Emirates UniversityCollege of Sciences

Department of Mathematical Sciences

HOMEWORK 1 – SOLUTIONSection 10.1 Vectors in PlaneSection 10.2 Vectors in Space

Calculus II for EngineeringMATH 1120 SECTION 53 CRN 304483:30 – 5:20 on Sunday & TuesdayDue Date: Tuesday, February 24, 2009

ID No: Solution

Name: Solution

Score: Solution

Calculus II for Engineering 1ST HOMEWORK – SOLUTION Spring, 2009

1. Compute −3a+2b for a= ⟨ 3,2 ⟩ and b= ⟨ 3,−13 ⟩.

Answer.

−3a+2b=−3⟨ 3,2 ⟩+2⟨ 3,−13 ⟩ = ⟨ −9,−6 ⟩+⟨ 6,−26 ⟩

= ⟨ −9+6,−6−26 ⟩ = ⟨ −3,−32 ⟩ . ä

2. Determine whether the vectors a= ⟨ 1,−2 ⟩ and b= ⟨ 2,1 ⟩ are parallel.

Answer. We recall that a and b are parallel if and only if there is a scalar s such thata= sb. If a= sb, then

⟨ 1,−2 ⟩ = s ⟨ 2,1 ⟩ = ⟨ 2s, s ⟩ , 1= 2s and −2= s.

There is no such a scalar s satisfying both 2s = 1 and s =−2. Therefore, given two vectorscannot be parallel. ä

3. Find the vector with initial point A(2,3) and terminal point B(5,4).

Answer.−−→AB = ⟨ 5−2,4−3 ⟩ = ⟨ 3,1 ⟩ . ä

4. Find a vector with the magnitude 4 in the same direction as the vector v = 2i−j.

Answer. Let u be such a vector, i.e., the one with the magnitude 4 in the same direction asthe vector v = 2i−j. Since u is parallel to v, there should be a positive scalar s (negativescalar gives the opposite direction) such that u = sv. Since u has the magnitude 4, wehave

4= ‖u‖ = ‖sv‖ = |s|‖v‖ = |s|‖2i−j‖ = |s|p

5,

i.e., |s| = 4p5= 4

p5

5, i.e., s = 4

p5

5.

Therefore, we deduce such a vector u= 4p

55

(2i−j) . ä5. The thrust of an airplane’s engines produces a speed of 600 mph in still air. The wind

velocity is given by ⟨ −30,60 ⟩. In what direction should the airplane head to fly due west?

Answer. Let v = ⟨ x, y ⟩ and w = ⟨ −30,60 ⟩ be the velocities of the airplane and the wind,respectively. Since we want the airplane to move due west, the sum of two vectors v andw should satisfy v+w = ⟨ c,0 ⟩, where the vector ⟨ c,0 ⟩ points due west with the negativeconstant c. The equation implies

⟨ c,0 ⟩ = v+w = ⟨ x, y ⟩+⟨ −30,60 ⟩ = ⟨ x−30, y+60 ⟩ ,i.e., c = x−30, and y=−60.

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The airplane produces a speed of 600 mph and so we get ‖v‖ = 600, which implies

600= ‖v‖ = ⟨ x, y ⟩ =√

x2+ y2, 6002 = x2+ y2 = x2+ (−60)2,

x =±√

6002−602 =±180p

11.

Because of the condition that x−30= c < 0, we choose x =−180p

11, hence,

v = ⟨ x, y ⟩ = ⟨ −180p

11,−60 ⟩ =−60⟨ 3p

11,1 ⟩ .

This points down and left at an angle of tan−1(

1

3p

11

)≈ 5.73917◦ south of west. ä

6. If vector a has magnitude ‖a‖ = 3 and vector b has magnitude ‖b‖ = 4, what is the largestpossible magnitude for the vector a+b? What is the smallest possible magnitude for thevector a+b? What will be the magnitude of a+b if a and b are perpendicular?

Answer. The largest magnitude of a+b is 7 (if the vectors point in the same direction).The smallest magnitude is 1 (if the vectors point in the opposite directions). If the vectorsare perpendicular, then a+b can be viewed as the hypotenuse of a right triangle withsides a and b, so it has length

p32+42 = 5. ä

7. For the vector 4i−2j +4k, (1) find two unit vectors parallel to the given vector and (2)write the given vector as the product of its magnitude and a unit vector.

Answer. Let v = 4i−2j+4k= 2(2i−j+2k). Then it has the magnitude

‖v‖ = ‖2(2i−j+2k)‖ = 2‖ (2i−j+2k)‖ = 2√

22+ (−1)2+22 = 2p

9= 6.

(1) The unit vector in the same direction as v can be found by

v

‖v‖ = 2(2i−j+2k)6

= 13

(2i−j+2k) .

The unit vector in the opposite direction as v is

− v

‖v‖ =−13

(2i−j+2k) .

Thus, two unit vectors parallel to v are ±13

(2i−j+2k).

(2) Using the results above, we have

v = ‖v‖ v

‖v‖ = 6(23i− 1

3j+ 2

3k

). ä

8. Identify the plane y = 4 as parallel to the xy–plane, xz–plane or yz–plane and sketch agraph.

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Answer. The plane y= 4 is parallel to the xz–plane and passes through (0,4,0). ä9. Find the displacement vectors

−−→PQ and

−−→QR and determine whether the points P(2,3,1),

Q(0,4,2) and R(4,1,4) are colinear (on the same line).

Answer. The displacement vectors are

−−→PQ = ⟨ 0−2,4−3,2−1 ⟩ = ⟨ −2,1,1 ⟩ ,

−−→QR = ⟨ 4−0,1−4,4−2 ⟩ = ⟨ 4,−3,2 ⟩ .

There does not exist any scalar s such that−−→PQ = s

−−→QR, because no scalar s satisfies simul-

taneously−2= 4s, 1=−3s, 1= 2s.

It implies that two vectors−−→PQ and

−−→QR are not parallel. That is, the points P, Q and R

are not colinear. ä10. Use vectors to determine whether the points (2,1,0), (5,−1,2), (0,3,3) and (3,1,5) form a

square.

Answer. We recall that if a square has the side length s, then it has the diagonal of lengthsp

2 by the Pythagorean Theorem.

Let P = (2,1,0), Q = (5,−1,2), R = (0,3,3) and S = (3,1,5). There are six pairs of vectors(−−→PQ,

−−→PR,

−−→PS,

−−→QR,

−−→QS,

−−→RS) and four of them will correspond to sides. Further

−−→PQ,

−−→PR

and−−→PS should form two sides and one diagonal of the square, because they have the same

initial point. When we compute the lengths, we get

‖−−→PQ‖ = ‖⟨ 3,−2,2 ⟩‖ =p

17, ‖−−→PR‖ = ‖⟨ −2,2,3 ⟩‖ =p

17,

‖−−→PS‖ = ‖⟨ 1,0,5 ⟩‖ =p

26.

It implies−−→PS should be the diagonal. However, since

p26 6=

p17

p2, i.e., ‖−−→PS‖ 6= ‖−−→PQ‖

p2,

so−−→PS cannot be the diagonal of a square. That is, those given points cannot form a

square. ä11. In the accompanying figure, two ropes are attached to a 300–pound crate. Rope A exerts

a force of ⟨ 10,−130,200 ⟩ pounds on the crate, and rope B exerts a force of ⟨ −20,180,160 ⟩pounds on the crate.

(11.1) If no further ropes are added, find the net force on the crate and the direction it willmove.

Answer. Let the force due to rope A be a = ⟨ 10,−130,200 ⟩, the force due to rope Bbe b = ⟨ −20,180,160 ⟩, and write the force due to gravity as w = ⟨ 0,0,−300 ⟩. Thenthe net force is

a+b+w = ⟨ 10,−130,200 ⟩+⟨ −20,180,160 ⟩+⟨ 0,0,−300 ⟩ = ⟨ −10,50,60 ⟩ . ä(11.2) If a third rope C is added to balance the crate, what force must this rope exert on the

crate?

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Answer. In order to compensate, rope C must exert a force of ⟨ 10,−50,−60 ⟩ or 78.74(= ‖⟨ 10,−50,−60 ⟩‖) pounds in direction ⟨ 1,−5,−6 ⟩. ä

(11.3) We want to move the crate up and to the right with a constant force of ⟨ 0,30,20 ⟩pounds. If a third rope C is added to accomplish this, what force must the rope exerton the crate?

Answer. Let the force due to rope C be c. We want the net force to be

a+b+c+w = ⟨ 0,30,20 ⟩ , i.e., c+⟨ −10,50,60 ⟩ = ⟨ 0,30,20 ⟩ ,

where a, b and w defined in (1) above are used. That is,

c= ⟨ 0,30,20 ⟩−⟨ −10,50,60 ⟩ = ⟨ 10,−20,−40 ⟩ .

So rope C must exert a force of ⟨ 10,−20,−40 ⟩ or 45.8 (= ‖⟨ 10,−20,−40 ⟩‖) poundsin direction ⟨ 1,−2,−4 ⟩. ä

12. If v ∈ V2 lies in the first quadrant of the xy–plane and makes the angle θ = π/3 with thepositive x–axis and the magnitude 4, then find v in the component form.

Answer. The unit vector making the angle θ = π/3 with the positive x–axis is given by⟨ cos(π/3),sin(π/3) ⟩. (Why?) Hence, the desired vector v ∈V2 is obtained by

v = 4⟨ cos(π/3),sin(π/3) ⟩ = 4⟨ 12

,p

32

⟩ = ⟨ 2,2p

3 ⟩ . ä

13. (Think!) Draw the vectors a = ⟨ 3,2 ⟩, b = ⟨ 2,−1 ⟩ and c = ⟨ 7,1 ⟩. By using the sketchinvolved with those vectors, show that there exist scalars s and t such that c = sa+ tb.Can you prove the existence of such s and t algebraically? Justify your answer.

Answer. The equation c = sa+ tb implies ⟨ 7,1 ⟩ = s ⟨ 3,2 ⟩+ t ⟨ 2,−1 ⟩, i.e., 7 = 3s+2t and

1= 2s− t. Solving the equations for s and t, we get s = 97

and t = 117

. ä

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14. Find the correct figure of the sum a+b with a=−2i−j and b=−3i+2j, where i and jare standard basis vectors of V2.

Answer. A simple computation shows

a+b=−2i−j−3i+2j =−5i+j = ⟨ −5,1 ⟩ .

That is, a+b should point in the direction ⟨ −5,1 ⟩. It is easy to see that the red–coloredvector in (2) is the vector ⟨ −5,1 ⟩. Hence, the answer is (2). ä

-5 -4 -3 -2 -1 1 2 3 4 5

-5

-4

-3

-2

-1

1

2

3

4

5

H 1 L

-5 -4 -3 -2 -1 1 2 3 4 5

-5

-4

-3

-2

-1

1

2

3

4

5

H 2 L

-5 -4 -3 -2 -1 1 2 3 4 5

-5

-4

-3

-2

-1

1

2

3

4

5

H 3 L

-5 -4 -3 -2 -1 1 2 3 4 5

-5

-4

-3

-2

-1

1

2

3

4

5

H 4 L

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HOMEWORK 2 – SOLUTIONSection 10.3 The Dot Product

Calculus II for EngineeringMATH 1120 SECTION 53 CRN 304483:30 – 5:20 on Sunday & TuesdayDue Date: Tuesday, March 3, 2009

ID No: Solution

Name: Solution

Score: Solution

Calculus II for Engineering 2ND HOMEWORK – SOLUTION Spring, 2009

1. Compute a ·b.

(1.1) a= ⟨ 3,2,0 ⟩ and b= ⟨ −2,4,3 ⟩.Answer.

a ·b= ⟨ 3,2,0 ⟩ · ⟨ −2,4,3 ⟩ = 3(−2)+2(4)+0(3)= 2. ä(1.2) a= 2i−k and b= 4j−k.

Answer.a ·b= (2i−k) · (4j−k)= 2(0)+0(4)+ (−1)(−1)= 1. ä

2. Compute the angle between the vectors.

(2.1) a= ⟨ 2,0,−2 ⟩ and b= ⟨ 0,−2,4 ⟩.Answer.

cosθ = a ·b‖a‖‖b‖ = ⟨ 2,0,−2 ⟩ · ⟨ 0,−2,4 ⟩

‖⟨ 2,0,−2 ⟩‖‖⟨ 0,−2,4 ⟩‖ =− 2p10

,

θ = cos−1(− 2p

10

)≈ 2.25552 ä

(2.2) a= 3i+j−4k and b=−2i+2j+k.

Answer.

cosθ = a ·b‖a‖‖b‖ = (3i+j−4k) · (−2i+2j+k)

‖3i+j−4‖‖−2i+2j+k‖ = −8

3p

26,

θ = cos−1(− 8

3p

26

)≈ 2.12114 ä

3. Determine whether the vectors are orthogonal.

(3.1) a= ⟨ 4,−1,1 ⟩ and b= ⟨ 2,4,4 ⟩.Answer.

a ·b= ⟨ 4,−1,1 ⟩ · ⟨ 2,4,4 ⟩ = 8 6= 0.

So a and b are not orthogonal. ä(3.2) a= 6i+2j and b=−i+3j.

Answer.a ·b= (6i+2j) · (−i+3j)= 0.

So a and b are orthogonal. ä4. Find a vector perpendicular to the given vector.

(4.1) ⟨ 4,−1,1 ⟩.

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Answer. Let v = ⟨ a,b, c ⟩ be a vector perpendicular to u= ⟨ 4,−1,1 ⟩. Then we shouldhave v ·u= 0, i.e.,

0= v ·u= ⟨ a,b, c ⟩ · ⟨ 4,−1,1 ⟩ = 4a−b+ c, i.e., 4a−b+ c = 0.

There are so many numbers, a, b, and c satisfying the equation. We choose justone, a = 1, b = 4 and c = 0. That is, v = ⟨ 1,4,0 ⟩ is one vector perpendicular to u =⟨ 4,−1,1 ⟩. ä

(4.2) 6i+2j−k.

Answer. Let v = ⟨ a,b, c ⟩ = ai+ bj+ ck be a vector perpendicular to w = 6i+2j−k.Then by the same argument as above, we deduce

0= v ·w = (ai+bj+ ck) · (6i+2j−k)= 6a+2b− c, i.e., 6a+2b− c = 0.

There are so many numbers, a, b, and c satisfying the equation. We choose just one,a = 1, b =−1 and c = 4. That is, v = ⟨ 1,−1,4 ⟩ = i−j+4k is one vector perpendicularto w = 6i+2j−k. ä

5. Find Compba and Projba.

(5.1) a= 3i+j and b= 4i−3j.

Answer.

Compba= a ·b‖b‖ = (3i+j) · (4i−3j)

‖4i−3j‖ = 95

Projba= (Compba

) b

‖b‖ = 95

4i−3j

‖4i−3j‖ = 925

(4i−3j) ä

(5.2) a= ⟨ 3,2,0 ⟩ and b= ⟨ −2,2,1 ⟩.Answer.

Compba= a ·b‖b‖ = ⟨ 3,2,0 ⟩ · ⟨ −2,2,1 ⟩

‖⟨ −2,2,1 ⟩‖ =−23

Projba= (Compba

) b

‖b‖ =−23

⟨ −2,2,1 ⟩‖⟨ −2,2,1 ⟩‖ =−2

9⟨ −2,2,1 ⟩ ä

6. A constant force of ⟨ 60,−30 ⟩ pounds moves an object in a straight line from the point(0,0) to the point (10,−10). Compute the work done.

Answer. The displacement vector is d = ⟨ 10−0,−10−0 ⟩ = ⟨ 10,−10 ⟩. The force is F =⟨ 60,−30 ⟩. By the formula, the work W done is obtained by

W =F ·d= ⟨ 10,−10 ⟩ · ⟨ 60,−30 ⟩ = 10(60)+ (−10)(−30)= 900. ä

7. Label each statement as true or false. If it is true, briefly explain why; if it is false, give acounterexample.

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(7.1) If a ·b=a ·c, then b= c. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . FAnswer. a= ⟨ 1,0,0 ⟩, b= ⟨ 0,1,0 ⟩ and c= ⟨ 0,0,1 ⟩ satisfies a ·b= 0=a ·c. But, obvi-ously, b 6= c. ä

(7.2) If b= c, then a ·b=a ·c. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . TAnswer. If b= c, then b−c= 0 and so

a · (b−c)=a ·0= 0, i.e., a ·b−a ·c= 0, i.e., a ·b=a ·c. ä

(7.3) a ·a= ‖a‖2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . TAnswer. The formula a ·b = ‖a‖‖b‖cosθ, where θ is the angle between a and b, im-plies

a ·a= ‖a‖‖a‖cos0= ‖a‖2, i.e., a ·a= ‖a‖2.

One may compute the dot product with a= ⟨ a,b, c ⟩ and prove the equality. ä(7.4) If ‖a‖ > ‖b‖, then a ·c> b ·c. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . F

Answer. We observe a= ⟨ 2,0 ⟩ and b= ⟨ 0,1 ⟩ satisfy the inequality ‖a‖ = 2> 1= ‖b‖.However, with c= ⟨ 0,3 ⟩, we get a ·c= 0< 3= b ·c. ä

(7.5) If ‖a‖ = ‖b‖, then a= b. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . FAnswer. We observe i = ⟨ 1,0,0 ⟩, j = ⟨ 0,1,0 ⟩ and k = ⟨ 0,0,1 ⟩ have ‖i‖ = 1 = ‖j‖ =‖k‖. However, obviously, i 6= j 6=k. ä

8. By the Cauchy–Schwartz Inequality, |a ·b| ≤ ‖a‖‖b‖. What relationship must exist be-tween a and b to have the equality |a ·b| = ‖a‖‖b‖?

Answer. (1) a = 0 or b = 0. (2) The cosine of the angle between the vectors is ±1. Thishappens exactly when the vectors point in the same or opposite directions. In other words,when a= sb for some scalar s. ä

9. By the Triangle Inequality, ‖a+b‖ ≤ ‖a‖+‖b‖. What relationship must exist betweena and b to have the equality ‖a+b‖ = ‖a‖+‖b‖?

Answer. (1) a= 0 or b= 0. (2) Vectors a and b must be parallel with the same direction sothat a= sb for some positive scalar s. ä

10. The orthogonal projection of vector a along vector b is defined as Orthba=a−Projba.Sketch a picture showing vectors a, b, Projba and Orthba, and explain what is orthogonalabout Orthba.

Answer. Orthba is the component of a that is orthogonal to b:

b ·Orthba= b · (a−Projba)= b ·a−b ·Projba

= b ·a−b · (a ·b)b‖b‖2 = b ·a− (a ·b)(b ·b)

‖b‖2 = b ·a−a ·b= 0,

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where b ·b= ‖b‖2 is used above. ä11. A car makes a turn on a banked road. If the road is banked at 15◦, show that a vector

parallel to the road is ⟨ cos15◦,sin15◦ ⟩. If the car has weight 2500 pounds, find the com-ponent of the weight vector along the road vector. This component of weight provides aforce that helps the car turn.

Answer. The vector b= ⟨ cos15◦,sin15◦ ⟩ represents the direction of the banked road. Theweight of the car is w = ⟨ 0,−2500 ⟩. The component of the weight in the direction of thebank is

Compbw = w ·b‖b‖ =−2500sin15◦ ≈−647.0 lbs

toward the inside of the curve. ä

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HOMEWORK 3 – SOLUTIONSection 10.4 The Cross Product


ID No: Solution

Name: Solution

Score: Solution

Calculus II for Engineering 3RD HOMEWORK – SOLUTION Spring, 2009

1. Compute the determinant: ∣∣∣∣∣∣∣

0 2 −11 −1 21 1 2

∣∣∣∣∣∣∣

Answer. ∣∣∣∣∣∣∣

0 2 −11 −1 21 1 2

∣∣∣∣∣∣∣= 0(−2−2)−2(2−2)− (1+1)=−2. ä

2. Compute the cross product a×b.

(2.1) a= ⟨ 2,−2,0 ⟩ and b= ⟨ 3,0,1 ⟩.Answer.

a×b= ⟨ 2,−2,0 ⟩×⟨ 3,0,1 ⟩ =

∣∣∣∣∣∣∣

i j k2 −2 03 0 1

∣∣∣∣∣∣∣=−2i−2j+6k=−2⟨ 1,1,−3 ⟩ ä

(2.2) a=−2i+j−3k and b= 2j−k.

Answer.

a×b= ⟨ 2,1,−3 ⟩×⟨ 0,2,−1 ⟩ =

∣∣∣∣∣∣∣

i j k−2 1 −30 2 −1

∣∣∣∣∣∣∣= 5i−2j−4k= ⟨ 5,−2,−4 ⟩ ä

3. Find two unit vectors orthogonal to the two given vectors.

(3.1) a= ⟨ 0,2,1 ⟩ and b= ⟨ 1,0,−1 ⟩.Answer. We recall

Orthogonal Vector

a×b is orthogonal to both a and b. Thus, ± a×b

‖a×b‖ are two unit vectors orthogonal

to both a and b.

a×b= ⟨ 0,2,1 ⟩×⟨ 1,0,−1 ⟩ = ⟨ −2,1,−2 ⟩ , ± a×b

‖a×b‖ =±13⟨ −2,1,−2 ⟩ ä

(3.2) a=−2i+3j−3k and b= 2i−k.

Answer.

a×b= ⟨ −2,3,−3 ⟩×⟨ 2,0,−1 ⟩ = ⟨ −3,−8,−6 ⟩ , ± a×b

‖a×b‖ =± 1p109

⟨ −3,−8,−6 ⟩ ä

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4. Use the cross product to determine the angle between the vectors, assuming that 0 ≤ θ ≤π/2.

(4.1) a= ⟨ 2,2,1 ⟩ and b= ⟨ 0,0,2 ⟩.Answer. We recall

Angle Between Two Vectors

‖a×b‖ = ‖a‖‖b‖sinθ, i.e., sinθ = ‖a×b‖‖a‖‖b‖ , i.e., θ = sin−1

( ‖a×b‖‖a‖‖b‖

)

where 0≤ θ ≤π is the angle between a and b.

a×b= 4⟨ 1,−1,0 ⟩ , ‖a‖ = 3, ‖b‖ = 2, ‖a×b‖ = 4p

2,

θ = sin−1

(4p

26

)≈ 1.23096 rad ≈ 70.5288◦ ä

(4.2) a= i+3j+3k and b= 2i+j.

Answer.

a×b= ⟨ −3,6,−5 ⟩ , ‖a‖ =p

19, ‖b‖ =p

5, ‖a×b‖ =p

70,

θ = sin−1

(p70p95

)≈ 1.03213 rad ≈ 59.1369◦ ä

5. Find the distance from the point Q(1,3,1) to the line through (1,3,−2) and (1,0,−2).

Answer. The distance d from the point Q to the line through the points P and R is ob-tained by the following formula.

Distance from a Point to a Line

d =

∥∥∥−−→PQ×−−→PR

∥∥∥∥∥∥−−→PR

∥∥∥

Letting P = (1,3,−2) and R = (1,0,−2), we have

−−→PQ = 3⟨ 0,0,1 ⟩ ,

−−→PR =−3⟨ 0,1,0 ⟩ ,

−−→PQ×−−→

PR = 9⟨ 1,0,0 ⟩ , d = 9‖⟨ 1,0,0 ⟩‖3‖⟨ 0,1,0 ⟩‖ = 3. ä

6. If you apply a force of magnitude 30 pounds at the end of an 8–inch–long wrench at anangle of π/3 to the wrench, find the magnitude of the torque applied to the bolt.

Answer. We recall

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Torque

Torque τ is defined to be the cross product of the position vector r and force vector F ,i.e.,

τ = r×F , ‖τ‖ = ‖r‖‖F ‖sinθ

where θ is the angle between r and F .

Given information: ‖F ‖ = 30 and θ =π/3 and ‖r‖ = 8inch= 8/12= 2/3feet. Thus we have

‖τ‖ = 23

(30)sinπ

3= 10

p3 ft–lbs ä

7. Label each statement as TRUE or FALSE. If it is true, briefly explain why. If it is false,give a counterexample.

(7.1) If a×b=a×c, then b= c. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . FAnswer. We observe that for a= ⟨ 1,0,0 ⟩, b= ⟨ 1,1,1 ⟩ and c= ⟨ 2,1,1 ⟩,

a×b= ⟨ 0,−1,1 ⟩ =a×c, but b 6= c. ä

(7.2) a×b=−b×a. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . TAnswer. We observe that for a= ⟨ a1,a2,a3 ⟩ and b= ⟨ b1,b2,b3 ⟩,

a×b= ⟨ a2b3−a3b2, a1b3−a3b1, a1b2−a2b1 ⟩ =−b×a. ä

(7.3) a×a= ‖a‖2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . FAnswer. a×a = ⟨ 0,0,0 ⟩ is a vector, while ‖a‖ is a scalar. As a counterexample, fora= i, we get i× i= 0= ⟨ 0,0,0 ⟩ 6= 1= ‖i‖2. ä

(7.4) a · (b×c)= (a ·b)×c. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . FAnswer. (a·b)×c is not possible because a·b is a scalar. A cross product must involvetwo vectors. ä

(7.5) If the force is doubled, the torque doubles. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . TAnswer. Torque is the cross product of direction and force.

r× (2F )= 2(r×F )= 2τ . ä8. Find the area of the parallelogram with two adjacent sides formed by ⟨ −2,1 ⟩ and ⟨ 1,−3 ⟩.

Answer. We recall

Page 3 of 5


Area

The area of the parallelogram with two adjacent sides formed by a and b is the mag-nitude of their cross product:

A = ‖a×b‖ = ‖a‖‖b‖sinθ

where 0≤ θ ≤π is the angle between a and b.

The vectors ⟨ −2,1 ⟩ and ⟨ 1,−3 ⟩ are in the plane (R2). The cross product is defined onlyfor the vectors in the space (R3). But we observe ⟨ −2,1,0 ⟩ and ⟨ 1,−3,0 ⟩ in the space cancorrespond to those vectors in the plane, respectively. So we compute the area with thesevectors:

A = ‖⟨ −2,1,0 ⟩×⟨ 1,−3,0 ⟩‖ = ‖5⟨ 0,0,1 ⟩‖ = 5. ä9. Find the area of the triangle with vertices (0,0,0), (0,−2,1) and (1,−3,0).

Answer. Letting P(0,0,0), Q(0,−2,1) and R(1,−3,0), by the formula, the area of the par-allelogram with two adjacent sides formed by

−−→PQ and

−−→PR is the magnitude of their cross

product:

Aparallelogram =∥∥∥−−→PQ×−−→

PR∥∥∥= ‖⟨ 0,−2,1 ⟩×⟨ 1,−3,0 ⟩‖ = ‖⟨ 3,2,1 ⟩‖ =

p14.

Since the area of the triangle is the half of the area of the parallelogram, hence the desiredarea is

Atriangle =

∥∥∥−−→PQ×−−→PR

∥∥∥2

=p

142

. ä10. Find the volume of the parallelepiped with three adjacent edges formed by ⟨ 0,−1,0 ⟩,

⟨ 0,2,−1 ⟩, and ⟨ 1,0,2 ⟩.

Answer. We recall

Volume

The volume of the parallelepiped determined by the vectors a, b, and c is the magni-tude of their scalar triple product:

V = |c · (a×b)|Letting a = ⟨ 0,−1,0 ⟩, b = ⟨ 0,2,−1 ⟩ and c = ⟨ 1,0,2 ⟩, the formula implies the volumeV = 1. ä

11. Use geometry to identify the cross product. (Do not compute!)

(11.1) j× (j×k).

Answer. Geometry implies

Page 4 of 5


Cross Product on Standard Basis Vectors

i×j =k j×k= i k× i= j

j× i=−k k×j =−i i×k=−j

Hence, we deducej× (j×k)= j× i=−k ä

(11.2) (j× i)×k.

Answer. By the same argument as above, we get

(j× i)×k=−k×k= 0 ä12. Use the parallelepiped volume formula to determine whether the vectors ⟨ 1,1,2 ⟩ and

⟨ 0,−1,0 ⟩ and ⟨ 3,2,4 ⟩ are coplanar.

Answer. Letting a = ⟨ 1,1,2 ⟩, b = ⟨ 0,−1,0 ⟩ and c = ⟨ 3,2,4 ⟩, the formula above impliesthe volume V = 2. Since the volume of the parallelepiped is not zero, it means those threevectors do not lie on the same plane. That is, they are not coplanar. ä

Page 5 of 5



HOMEWORK 4 – SOLUTIONSection 10.5 Lines and Planes in Space


ID No: Solution

Name: Solution

Score: Solution

Calculus II for Engineering 4TH HOMEWORK – SOLUTION Spring, 2009

1. Find (a) parametric equations and (b) symmetric equations of the line.

(1.1) The line through (3,−2,4) and parallel to ⟨ 3,2,−1 ⟩Answer.

(a) x = 3+3t, y=−2+2t, z = 4− t (b)x−3

3= y+2

2= z−4

−1ä

(1.2) The line through (−1,0,0) and parallel to the linex+1−2

= y3= z−2

Answer. The direction comes from the given line, i.e., ⟨ −2,3,1 ⟩.

(a) x =−1−2t, y= 0+3t, z = 0+ t (b)x+1−2

= y3= z

1ä

(1.3) The line through (−3,1,0) and perpendicular to both ⟨ 0,−3,1 ⟩ and ⟨ 4,2,−1 ⟩Answer. ⟨ 0,−3,1 ⟩×⟨ 4,2,−1 ⟩ = ⟨ 1,4,12 ⟩ is in the direction perpendicular to bothvectors.

(a) x =−3+ t, y= 1+4t, z = 0+12t (b)x+3

1= y−1

4= z

12ä

(1.4) The line through (0,−2,1) and normal to the plane y+3z = 4

Answer. ⟨ 0,1,3 ⟩ is normal to the plane.

(a) x = 0+0t = 0, y=−2+ t, z = 1+3t (b) x = 0,y+2

1= z−1

3ä

2. State whether the lines are parallel or perpendicular and find the angle between the lines.

L : x = 4−2t, y= 3t, z =−1+2tM : x = 4+ s, y=−2s, z =−1+3s

Answer. The vectors parallel to lines L and M are respectively vL = ⟨ −2,3,2 ⟩ and vM =⟨ 1,−2,3 ⟩. Since there is no constant c satisfying vL = ⟨ −2,3,2 ⟩ = c ⟨ 1,−2,3 ⟩ = cvM, sothe lines L and M are not parallel.

By the formula on the dot product, we have

cosθ = vL ·vM

‖vL‖‖vM‖ = −2−6+6p17

p14

=− 2p238

, θ = cos−1(− 2p

238

)≈ 1.7 rad,

which is not π/2. Thus, the lines L and M are not perpendicular and the angle betweenthe lines L and M is about 1.7 rad. ä

3. Determine whether the lines are parallel, skew or intersect.

L : x = 3+ t, y= 3+3t, z = 4− t

Page 1 of 5


M : x = 2− s, y= 1−2s, z = 6+2s

Answer. The vectors parallel to lines L and M are respectively vL = ⟨ 1,3,−1 ⟩ and vM =⟨ −1,−2,2 ⟩. Since there is no constant c satisfying vL = ⟨ 1,3,−1 ⟩ = c ⟨ −1,−2,2 ⟩ = cvM,so the lines L and M are not parallel. To determine whether or not the lines intersect, weset the x–, y– and z–values equal:

3+ t = 2− s, 3+3t = 1−2s, 4− t = 6+2s,simpl y, (i) t+ s =−1, (ii) 3t+2s =−2, (iii) t+2s =−2.

From (i) and (ii), we get t = 0 and s =−1. Putting t = 0 and s =−1 into (iii), the equation(iii) holds. This implies that the lines L and M intersect when t = 0 and s =−1:

x = 3, y= 3, z = 4.

Hence, the lines intersect at the point (3,3,4). ä4. Find an equation of the given plane.

(4.1) The plane containing the point (−2,1,0) with normal vector ⟨ −3,0,2 ⟩Answer.

−3(x+2)+0(y−1)+2(z−0)= 0, simpl y, 3x−2z+6= 0. ä(4.2) The plane containing the points (−2,2,0), (−2,3,2) and (1,2,2)

Answer. Letting P(−2,2,0), Q(−2,3,2) and R(1,2,2), we deduce−−→PQ = ⟨ 0,1,2 ⟩ and−−→

PR = ⟨ 3,0,2 ⟩ and−−→PQ ×−−→

PR = ⟨ 2,6,−3 ⟩ is normal to the plane. Hence, using thepoint P(−2,2,0) (one can use Q or R), the equation of the plane is obtained as

2(x+2)+6(y−2)−3(z−0)= 0, simpl y, 2x+6y−3z−8= 0. ä(4.3) The plane containing the point (3,−2,1) and parallel to the plane x+3y−4z = 2

Answer. The vector normal to the plane is ⟨ 1,3,−4 ⟩. Hence, the equation of theplane is

(x−3)+3(y+2)−4(z−1)= 0, simpl y, x+3y−4z+7= 0. ä(4.4) The plane containing the point (3,0,−1) and perpendicular to the planes x+2y− z = 2

and 2x− z = 1

Answer. Let v1 and v2 be the vectors normal to the planes: v1 = ⟨ 1,2,−1 ⟩ and v2 =⟨ 2,0,−1 ⟩. Normal vector must be perpendicular to the normal vectors of both planes.So, the plane that we are looking for has the normal vector v1 ×v2 = ⟨ −2,−1,−4 ⟩.Hence, the equation of the plane is

−2(x−3)− (y−0)−4(z+1)= 0, simpl y, 2x+ y+4z = 2. ä

Page 2 of 5


5. Sketch the given plane.

(5.1) 2x− y+4z = 4

Answer. The equation represents a plane which has the normal vector ⟨ 2,−1,4 ⟩ andpasses through the point P(2,0,0), Q(0,−4,0) and R(0,0,1). We connect those threepoints and sketch the plane roughly. ä

P

Q

R

Normal Vector

-4

-2

0

2

4X-axis

-4

-2

0

2

4Y-axis

-4

-2

0

2

4

Z-axis

(5.2) x+ y= 1

Answer. The equation represents a plane such that(i) it is parallel to the z–axis and perpendicular to the xy–plane,(ii) it has the normal vector ⟨ 1,1,0 ⟩,(iii) it passes through the point P(1,0,0), Q(0,1,0). ä

P

QNormal Vector

-4

-2

0

2

4X-axis

-4-2

02

4Y-axis

-4

-2

0

2

4

Z-axis

6. Find the intersection of the planes 3x+ y− z = 2 and 2x−3y+ z =−1.

Answer. Solve the equations for z and equate:

3x+ y−2= z =−2x−3y−1, i.e., 3x+ y−2=−2x−3y−1,

Page 3 of 5


i.e., 5x+4y= 1, i.e., y= −5x+14

.

Putting it into the first equation, we get

2= 3x+ −5x+14

− z, i.e., z = 7x−74

.

Using x = t as a parameter, we deduce the line:

x = t, y= 14− 5

4t, z =−7

4+ 7

4t.

If one use y= s as a parameter, one can get the line:

x = 15+ 2

5s, y= s, z =−7

5+ 11

5s.

One can use z = u as a parameter and get the equation of the line. ä7. Find the distance between the given objects.

(7.1) The point (1,3,0) and the plane 3x+ y−5z = 2

Answer. The distance formula implies

d = |3(1)+1(3)−5(0)−2|√32+12+ (−5)2

= 4p35

. ä

(7.2) The planes x+3y−2z = 3 and x+3y−2z = 1

Answer. We choose a point P(1,0,0) on the plane x+3y−2z = 1. Applying the distanceformula to the point P and the other plane, we get

d = |1(1)+3(0)−2(0)−3|√32+12+ (−2)2

= 2p14

. ä

8. Find an equation of the plane containing the lines

L : x = 4+ t, y= 2, z = 3+2tM : x = 2+2s, y= 2s, z =−1+4s

Answer. The vectors parallel to the lines are respectively vL = ⟨ 1,0,2 ⟩ and vM = ⟨ 2,2,4 ⟩.The normal vector to the plane is vL ×vM = ⟨ −4,0,2 ⟩. We choose one point (4,2,3) fromL. (One can choose any point from either line.) Hence, the equation of the plane is

−4(x−4)+0(y−2)+2(z−3)= 0, simpl y, 2x− z = 5. ä

9. State whether the statement is true or false (not always true).

(9.1) Two planes either are parallel or intersect. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . T

Page 4 of 5


Answer. Two planes can be even both parallel and intersect if the planes coincide.ä

(9.2) The intersection of two planes is a line. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . FAnswer. It can be a plane if the planes coincide, or can be empty. ä

(9.3) The intersection of three planes is a point. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . FAnswer. It can be a point, a line, or a plane, or can be empty. ä

(9.4) Lines that lie in parallel planes are always skew. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . TAnswer. It is true, unless the parallel planes coincide. ä

(9.5) The set of all lines perpendicular to a given line forms a plane. . . . . . . . . . . . . . . . . . . FAnswer. It is false. However, it is true if we take all lines perpendicular to a givenline through a given point. ä

(9.6) There is one line perpendicular to a given plane. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . FAnswer. It is false. There is one line perpendicular to a given plane through eachpoint of the plane. ä

(9.7) The set of all points equidistant from two given points forms a plane. . . . . . . . . . . . T10. Determine whether the given lines are the same:

L : x = 1+4t, y= 2−2t, z = 2+6tM : x = 9−2s, y=−2+ s, z = 8−3s

Answer. The vectors parallel to the lines are vL = ⟨ 4,−2,6 ⟩ and vM = ⟨ −2,1,−3 ⟩, respec-tively. Since vL = −2vM, the lines L and M are parallel. The point (9,−2,8) lies on theline M when s = 0. We solve for t in the x coordinate of the line L to see that 1+4t = 9implies t = 2. It means when t = 2 and s = 0, the lines L and M have the same x coordinate9. What about the y and z coordinates when t = 2 and s = 0? At this time t = 2 and s = 0,the line L has the y and z coordinates, y= 2−2(2)=−2 and z = 2+6(2)= 14, respectively.However, the line M has the y and z coordinates, y=−2 and z = 8. That is,

t =−2 & s = 0 =⇒ L : (x, y, z)= (9,−2,14), M : (x, y, z)= (9,−2,8).

It implies that these lines L and M are not the same. ä

Page 5 of 5



HOMEWORK 5Chapter 10 Vectors and the Geometry of Space


ID No: Solution

Name: Solution

Score: Solution


Section 10.1 Vectors in the Plane

1. For a given vector v = ⟨ 4,−3 ⟩, (1) find a unit vector in the same direction as the vector vand (2) write the vector v in polar form. (Solve only (1).)

Answer. (1) The unit vector u in the same direction as the vector v is obtained by

u= v

‖v‖ = ⟨ 4,−3 ⟩5

.

(2) Let 0≤ θ <π be the angle between the positive x–axis and the given vector v. Then weobserve

4= ‖v‖cosθ = 5cosθ, −3= ‖v‖sinθ = 5sinθ.

Therefore, the vector v can be expressed in polar form,

v = ⟨ 5cosθ,5sinθ ⟩ = 5⟨ cosθ,sinθ ⟩ ,

where θ satisfies tanθ =−34

. ä

2. Find a vector with the magnitude 3 in the same direction as the vector v = 3i+4j.

Answer. The unit vector u in the same direction as the vector v is obtained by

u= v

‖v‖ = 3i+4j

5.

Therefore, a vector with the magnitude 3 in the same direction as the vector v = 3i+4jshould be

3u= 3(3i+4j)5

. ä

3. The thrust of an airplane’s engines produces a speed of 400 mph in still air. The windvelocity is given by ⟨ −20,30 ⟩. In what direction should the airplane head to fly duenorth?

Answer. Let v = ⟨ x, y ⟩ and w = ⟨ −20,30 ⟩ be the velocities of the airplane and the wind,respectively. Since we want the airplane to move due north, the sum of two vectors v andw should satisfy v+w = ⟨ 0, c ⟩, where the vector ⟨ 0, c ⟩ points due north with the positiveconstant c. The equation implies

⟨ 0, c ⟩ = v+w = ⟨ x, y ⟩+⟨ −20,30 ⟩ = ⟨ x−20, y+30 ⟩ ,i.e., 0= x−20, and c = y+30, i.e., x = 20, and y= c−30.

The airplane produces a speed of 400 mph and so we get ‖v‖ = 400, which implies

400= ‖v‖ = ⟨ x, y ⟩ =√

x2+ y2, 4002 = x2+ y2 = 202+ (c−30)2,

Page 1 of 9


c−30=±√

4002−202, c = 30±20p

399.

Since c > 0, so we take only c = 30+20p

399. (For your information, 30−20p

399≈−369.5<0.) Hence, y= c−30 implies y= 20

p399 and finally

v = ⟨ x, y ⟩ = ⟨ 20,20p

399 ⟩ = 20⟨ 1,p

399 ⟩ .

This points up and right at an angle of tan−1(p

399)≈ 1.52078 radian north of east. ä

4. For vectors a = ⟨ 2,3 ⟩ and b = ⟨ 1,4 ⟩, compare ‖a+b‖ and ‖a‖+‖b‖. Repeat this com-parison for two other choices of a and b. Use the sketch given below to explain why‖a+b‖ ≤ ‖a‖+‖b‖ for any vectors a and b.

Figure 10.6 Adding positionvectors

Answer.

a+b= ⟨ 3,7 ⟩ , ‖a+b‖ =p

58≈ 7.61577< 7.72866≈p

13+p

17= ‖a‖+‖b‖ .

We can find the triangle 4OBC formed by O, B and C in the given figure above.

(i) Let−−→OB = b. Then the length of one side OB in the triangle is ‖b‖.

(ii) Let−−→OA =a. Then the length of another side BC is ‖a‖, because

−−→OA =−−→

BC.

(iii) By the sum of two vectors, we get−−→OC =−−→

OA+−−→OB = a+b. So the length of the other

side OC is ‖a+b‖.

Since the sum of lengths of two sides of a triangle must be bigger than the length of theremaining side, we should get

‖a+b‖ < ‖a‖+‖b‖ .

In addition, when a= sb with a positive scalar s, we have the equality. (Check this out byyourself!) So in general, the following inequality holds:

‖a+b‖ ≤ ‖a‖+‖b‖ ,

which is called the Triangular Inequality. ä

Page 2 of 9


Section 10.2 Vectors in Space

5. Compute a+b, a−3b and ‖4a+2b‖ for a= i−4j−2k and b= i−3j+4k.

Answer.

a+b= 2i−7j+2k, a−3b=−2i+5j−14k,

4a+2b= 6i−22j, ‖4a+2b‖ = 2p

130. ä

6. Find the displacement vectors−−→PQ and

−−→QR and determine whether the points P(2,3,1),

Q(4,2,2) and R(8,0,4) are colinear (on the same line).

Answer.

−−→PQ = ⟨ 4−2,2−3,2−1 ⟩ = ⟨ 2,−1,1 ⟩ ,

−−→PR = ⟨ 8−2,0−3,4−1 ⟩ = ⟨ 6,−3,3 ⟩ .

We observe that for s = 1/3, we have the equation

−−→PQ = ⟨ 2,−1,1 ⟩ = s ⟨ 6,−3,3 ⟩ = s

−−→PR,

i.e., there is a scalar such that−−→PQ = s

−−→PR. It implies that vectors,

−−→PQ and

−−→PR, are parallel

and thus the points are colinear. ä

Section 10.3 The Dot Product

7. Compute the angle between the vectors a= i+3j−2k and b= 2i−3k.

Answer.

cosθ = a ·b‖a‖‖b‖ = (i+3j−2k) · (2i−3k)

‖i+3j−2k‖‖2i+−3k‖ = 8p182

,

θ = cos−1(

8p182

)≈ 0.936017. ä

8. Find Compba and Projba for a= ⟨ 2,0,−2 ⟩ and b= ⟨ 0,−3,4 ⟩.

Answer.

Compba= a ·b‖b‖ = ⟨ 2,0,−2 ⟩ · ⟨ 0,−3,4 ⟩

‖⟨ 0,−3,4 ⟩‖ =−85

,

Page 3 of 9


Projba= (a ·b)b‖b‖2 = (

Compba) b

‖b‖ =−85

⟨ 0,−3,4 ⟩‖⟨ 0,−3,4 ⟩‖ =− 8

25⟨ 0,−3,4 ⟩ . ä

9. Label each statement as True or False. If it is true, briefly explain why; if it is false, givea counterexample.

(9.1) If a ·b=a ·c, then b= c. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . FAnswer. a= ⟨ 1,0,0 ⟩, b= ⟨ 0,1,0 ⟩ and c= ⟨ 0,0,1 ⟩ satisfies a ·b= 0=a ·c. But, obvi-ously, b 6= c. ä

(9.2) If b= c, then a ·b=a ·c. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . TAnswer. If b= c, then b−c= 0 and so

a · (b−c)=a ·0= 0, i.e., a ·b−a ·c= 0, i.e., a ·b=a ·c. ä

(9.3) a ·a= ‖a‖2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . TAnswer. The formula a ·b = ‖a‖‖b‖cosθ, where θ is the angle between a and b, im-plies

a ·a= ‖a‖‖a‖cos0= ‖a‖2, i.e., a ·a= ‖a‖2.

One may compute the dot product with a= ⟨ a,b, c ⟩ and prove the equality. ä(9.4) If ‖a‖ > ‖b‖, then a ·c> b ·c. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . F

Answer. We observe a = 2i and b = j satisfy the inequality ‖a‖ = 2 > 1 = ‖b‖. How-ever, with c= 3j, we get a ·c= 0< 3= b ·c. ä

(9.5) If ‖a‖ = ‖b‖, then a= b. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . FAnswer. We observe i = ⟨ 1,0,0 ⟩, j = ⟨ 0,1,0 ⟩ and k = ⟨ 0,0,1 ⟩ have ‖i‖ = 1 = ‖j‖ =‖k‖. However, obviously, i 6= j 6=k. ä

Section 10.4 The Cross Product

10. Use the cross product to determine the angle between the vectors a= 3i+k and b= 4j+k,assuming that 0≤ θ ≤π/2.

Answer.

a×b= ⟨ −4,−3,12 ⟩ , ‖a×b‖ =p

13, ‖a‖ =p

10, ‖b‖ =p

17,

θ = sin−1

( p13p

170

)≈ 1.49402. ä

Page 4 of 9


11. Use the parallelepiped volume formula to determine whether the vectors ⟨ 2,3,1 ⟩, ⟨ 1,0,2 ⟩,and ⟨ 0,3,−3 ⟩ are coplanar.

Answer. The cross product of a = ⟨ 2,3,1 ⟩ and b = ⟨ 1,0,2 ⟩ is a×b= 3⟨ 2,−1,−1 ⟩. Withc= ⟨ 0,3,−3 ⟩, the volume formula implies

V = |c · (a×b)| = |3⟨ 0,3,−3 ⟩ · ⟨ 2,−1,−1 ⟩| = 0.

Since the parallelepiped has the volume 0, hence, those three vectors are coplanar. ä

12. Show that ‖a×b‖2 = ‖a‖2‖b‖2− (a ·b)2.

Answer. With the angle θ between a and b, we deduce

‖a×b‖2 = (‖a‖‖b‖sinθ)2 = ‖a‖2‖b‖2 sin2θ

= ‖a‖2‖b‖2 (1−cos2θ

)= ‖a‖2‖b‖2−‖a‖2‖b‖2 cos2θ

= ‖a‖2‖b‖2− (‖a‖‖b‖cosθ)2 = ‖a‖2‖b‖2− (a ·b)2 . ä

Section 10.5 Lines and Planes in Space

13. Find (a) parametric equations and (b) symmetric equations of the line through (1,4,1)and parallel to the line x = 2−3t, y= 4 and z = 6+ t.

Answer. The direction comes from the given line, i.e., ⟨ −3,0,1 ⟩.

(a) x = 1−3t, y= 4+0t = 4, z = 1+ t (b)x−1−3

= z−11

, and y= 4. ä

14. State whether the lines are parallel or perpendicular and find the angle between the lines.

L : x = 4−2t, y= 3t, z =−1+2tM : x = 4+ s, y=−2s, z =−1+3s

Answer. The vectors parallel to lines L and M are respectively vL = ⟨ −2,3,2 ⟩ and vM =⟨ 1,−2,3 ⟩. Since there is no scalar c satisfying vL = ⟨ −2,3,2 ⟩ = c ⟨ 1,−2,3 ⟩ = cvM, so thelines L and M are not parallel.

By the formula on the dot product, we have

cosθ = vL ·vM

‖vL‖‖vM‖ = −2−6+6p17

p14

=− 2p238

, θ = cos−1(− 2p

238

)≈ 1.7 rad,

Page 5 of 9


which is not π/2. Thus, the lines L and M are not perpendicular and the angle betweenthe lines L and M is about 1.7 rad. ä

15. Find an equation of the plane containing the point (−2,1,0) with normal vector ⟨ −3,0,2 ⟩.

Answer.

−3(x+2)+0(y−1)+2(z−0)= 0, simpl y, 3x−2z+6= 0. ä

16. Sketch the plane 2x− y+4z = 4.

Answer. (1) On the xy–plane (,i.e., when z = 0), the plane becomes a line 2x− y= 4 passingthrough (x, y)= (0,−4) and (x, y)= (2,0).

(2) On the yz–plane (,i.e., when x = 0), the plane becomes a line −y+4z = 4 passing through(y, z)= (0,1) and (y, z)= (−4,0).

(3) On the xz–plane (,i.e., when y= 0), the plane becomes a line 2x+4z = 4 passing through(x, z)= (0,1) and (x, z)= (2,0).

We draw each line on each plane and combine all lines in the three–dimensional space.Confer the figure below. ä

Plane 2x-y+4z=4

H0,-4,0L

H2,0,0L

H0,0,1L

0

1

2

X

-4

-3

-2

-1

0

Y

-1

0

1

Z

Plane 2x− y+4z = 4

Miscellany

17. (1) Show that a vector (cosθ)i+ (sinθ)j with some angle θ in the plane is a unit vector.

(2) Show that if a vector u in the plane is a unit vector, then it can be expressed in theform of (cosθ)i+ (sinθ)j with some angle θ.

Page 6 of 9


By (1) and (2), we can say that a vector u in the plane is a unit vector if and only ifu= (cosθ)i+ (sinθ)j with some angle θ.

Proof. (1) It is straightforward to see

‖(cosθ)i+ (sinθ)j‖ =√

cos2θ+sin2θ =p

1= 1.

Thus, (cosθ)i+ (sinθ)j is a unit vector.

(2) Let u= xi+ yj be a unit vector in the plane. Then we observe

1= ‖u‖ = ‖xi+ yj‖ =√

x2+ y2, i.e., x2+ y2 = 1.

When we draw the vector u= xi+ yj in the plane, we can measure the angle between thevector u and the x–axis. Let θ be that angle. Then from the triangle, we get

cosθ = x√x2+ y2

, and sinθ = y√x2+ y2

.

Using x2+ y2 = 1, those two equations become

cosθ = x, and sinθ = y.

Therefore, the unit vector u= xi+ yj becomes

u= xi+ yj = (cosθ)i+ (sinθ)j. ä

18. Show that the diagonals of a rhombus∗ are perpendicular to each other.

!" #$%&'(&)*+ $ ,- ,- .#" " (/ 001 1 1 -'!" #2&'(&)*+ - 2&'

!

"

#

a rhombus∗a rhombus or rhomb is an equilateral parallelogram. The rhombus is often called a diamond with the same length of

sides, after the diamonds suit in playing cards.

Page 7 of 9


Proof. Let v and w be the vectors−−→AB and

−−→AD, respectively. That is, v =−−→

AB and w =−−→AD.

Then (1) since a rhombus is a parallelogram, we get v =−−→AB =−−→

DC and w =−−→AD =−−→

BC. (2)Since a rhombus is equilateral, we get ‖v‖ = ‖w‖.

Now, we consider two diagonal vectors−−→AC and

−−→DB. (One can choose other diagonal vec-

tors.) Recalling the sum of two vectors, it is not difficult to see

−−→AD+−−→

AB =−−→AC, i.e., v+w =−−→

AC, and−−→AD−−−→

AB =−−→DB, i.e., v−w =−−→

DB.

To prove that two diagonal vectors−−→AC = v+w and

−−→DB = v−w are perpendicular to each

other, we compute the dot product:

−−→AC ·−−→DB = (v+w) · (v−w)= v ·v−v ·w+v ·w−w ·w = v ·v−w ·w = ‖v‖2−‖w‖2 = 0,

by the property of a rhombus (2) above. Therefore two diagonal vectors−−→AC and

−−→DB are

perpendicular to each other. ä

19. Find the parametric equations of the line L satisfying all the following conditions at thesame time:

(1) The line L lies on the plane z = 3.

(2) Let M be the shadow of the line L onto the xy–plane. Then the shadow line M makesan angle of α with i= ⟨ 1,0 ⟩ and an angle of

(π2−α

)with j = ⟨ 0,1 ⟩, assuming 0≤α< π

2.

Answer. The desired line L can be expressed as follows:

L = { ⟨ x, y,3 ⟩ | ⟨ x, y ⟩ is in the shadow line M. }

M ={⟨ x, y ⟩ | ⟨ x, y ⟩ makes the angle α with ⟨ 1,0 ⟩ and the angle

(π2−α

)with ⟨ 0,1 ⟩.

}

When using the dot product, we get

x = ⟨ x, y ⟩ · ⟨ 1,0 ⟩ = ‖⟨ x, y ⟩‖‖⟨ 1,0 ⟩‖cosα=√

x2+ y2 cosα, i.e., x =√

x2+ y2 cosα,

y= ⟨ x, y ⟩ · ⟨ 0,1 ⟩ = ‖⟨ x, y ⟩‖‖⟨ 0,1 ⟩‖cos(π2−α

)=

√x2+ y2 sinα, i.e., y=

√x2+ y2 sinα,

where cos(π2−α

)= sinα is used. Solving two equations, we get

xcosα

= ysinα

=√

x2+ y2, i.e., y= sinαcosα

x = (tanα) x, or x = cosαsinα

y= (cotα) y.

• Letting x = t (parameter), we deduce y= sinαcosα

t = (tanα) t and so, the parametric equa-tions of the shadow line M are given by

x = t, y= (tanα) t,

Page 8 of 9


and the parametric equations of the desired line L are given by

x = t, y= (tanα) t, z = 3.

• Letting y = s (parameter), we deduce x = cosαsinα

s = (cotα) s and so, the parametric equa-tions of the shadow line M are given by

x = (cotα) s, y= s,

and the parametric equations of the desired line L are given by

x = (cotα) s, y= s, z = 3.

A much simpler answer will be as follows: a line making the angle α with the positive x–axis has the slope tanα. So generally, the line M should be parallel to the line y= (tanα) x.It implies that letting x = t (parameter), we get y= (tanα) t and hence the equations of theline L are

x = t, y= (tanα) t, z = 3. ä

Page 9 of 9



HOMEWORK 6 – SOLUTIONSection 11.2 The Calculus of Vector–Valued Functions


ID No: Solution

Name: Solution

Score: Solution


1. Find the derivative of the given vector–valued function.

(1.1) r(t)= ⟨ t−3t+1

, te2t, t3 ⟩

Answer.

r′(t)= ⟨(

t−3t+1

)′,

(te2t)′ , (

t3)′ ⟩ = ⟨ 4(t+1)2

, (1+2t) e2t, 3t2 ⟩ . ä

(1.2) r(t)= ⟨ cos(5t), tan t, 6sin t ⟩Answer.

r′(t)= ⟨ (cos(5t))′ , (tan t)′ , (6sin t)′ ⟩ = ⟨ −5sin(5t), sec2 t, 6cos t ⟩ . ä

(1.3) r(t)= ⟨√

t2+1, cos t, e−3t ⟩

Answer.

r′(t)= ⟨(√

t2+1)′

, (cos t)′ ,(e−3t)′ ⟩ = ⟨ tp

t2+1, −sin t, −3e−3t ⟩ . ä

Page 1 of 4


2. Evaluate the given indefinite or definite integral.

(2.1)∫

⟨ 3t2 ,

4t⟩ dt

Answer. ∫⟨ 3

t2 ,4t⟩ dt = ⟨

∫3t2 dt,

∫4t

dt ⟩ = ⟨ −3t, 4ln t ⟩+c,

where c is an arbitrary constant vector. ä(2.2)

∫⟨ e−3t, sin(5t), t3/2 ⟩ dt

Answer.∫

⟨ e−3t, sin(5t), t3/2 ⟩ dt = ⟨∫

e−3t dt,∫

sin(5t)dt,∫

t3/2 dt ⟩

= ⟨ − e−3t

3, −cos(5t)

5,

25

t5/2 ⟩+c,


∫⟨ e−3t, t2 cos t3, tcos t ⟩ dt

Answer. Using u = t3 to integrate the second component of the vector and integrationby parts where u = t and dv = cos t dt to integrate the third component of the vectorgives

∫⟨ e−3t, t2 cos t3, tcos t ⟩ dt = ⟨

∫e−3t dt,

∫t2 cos t3 dt,

∫tcos t dt ⟩

= ⟨ − e−3t

3,

sin t3

3, tsin t+cos t ⟩+c,


∫ 4

1⟨p

t, 5 ⟩ dt

Answer. ∫ 4

1⟨p

t, 5 ⟩ dt = ⟨∫ 4

1

pt dt,

∫ 4

15dt ⟩ = ⟨ 14

3, 15 ⟩ . ä

(2.5)∫ 4

0⟨ 2te4t, t2−1,

4tt2+1

⟩ dt

Answer. Using integration by parts where u = t and dv = e4t dt to integrate the first com-ponent and using u = t2+1 to integrate the third component of the vector gives

∫ 4

0⟨ 2te4t, t2−1,

4tt2+1

⟩ dt = ⟨∫ 4

02te4t dt,

∫ 4

0(t2−1)dt,

∫ 4

0

4tt2+1

dt ⟩

= ⟨ 158

e16+ 18

,523

, 2ln17 ⟩ . ä

Page 2 of 4


3. Find t such that r(t) and r′(t) are perpendicular.

(3.1) r(t)= ⟨ 2cos t, sin t ⟩.Answer. It is easy to get r′(t)= ⟨ −2sin t, cos t ⟩.

r(t) ·r′(t)= ⟨ 2cos t, sin t ⟩ · ⟨ −2sin t, cos t ⟩ =−3sin tcos t = 0

⇐⇒ t = 0, ±π2

, ±π, ±3π2

, . . . .

Therefore, r(t) and r′(t) are perpendicular when t = nπ2

for any integer n. ä

(3.2) r(t)= ⟨ t2, t, t2−5 ⟩Answer. It is easy to get r′(t)= ⟨ 2t, 1, 2t ⟩.

r(t) ·r′(t)= ⟨ t2, t, t2−5 ⟩ · ⟨ 2t, 1, 2t ⟩ = 4t3−9t = 0 ⇐⇒ t = 0, ±32

.

Therefore, r(t) and r′(t) are perpendicular when t = 0, t =±32

. ä

4. Find all values of t such that r′(t) is parallel to the xy–plane.

(4.1) r(t)= ⟨ t2, t, sin(t2) ⟩Answer. r′(t) is parallel to the xy–plane when the third component is 0.

ddt

(sin(t2)

)= 2tcos(t2)= 0 ⇐⇒ t = 0, ±√

nπ2

for any odd integer n. ä

(4.2) r(t)= ⟨p

t+1, cos t, t4−8t3 ⟩Answer. r′(t) is parallel to the xy–plane when the third component is 0.

ddt

(t4−8t3)= 4t3−24t2 = 4t2 (t−6)= 0 ⇐⇒ t = 0, 6. ä

Page 3 of 4


5. Label as true or false and explain why.

(5.1) If u(t)= r(t)‖r(t)‖ and u(t) ·u′(t)= 0, then r(t) ·r′(t)= 0. . . . . . . . . . . . . . . . . . . . . . . . . . . . . F

Answer. False. For any function r(t), u(t) is a unit vector for all t, and ‖u(t)‖ = 1 isconstant. Recall the theorem

Relation between Magnitude and Orthogonality

‖w(t)‖ = constant ⇐⇒ w(t) and w′(t) are orthogonal, for all t⇐⇒ w(t) ·w′(t)= 0 for all t

It implies u(t) ·u′(t)= 0 for any function r(t), but there are clearly functions r(t) with‖r(t)‖ 6= constant, so for these functions r(t) ·r′(t) will not be equal 0. ä

(5.2) If r(t0) ·r′(t0)= 0 for some t0, then ‖v(t)‖ is constant. . . . . . . . . . . . . . . . . . . . . . . . . . . . . FAnswer. False. By the Theorem in the solution to the previous problem, ‖r(t)‖ isconstant if and only if r(t) and r′(t) are perpendicular for all t. It is not enough thatthey are perpendicular for some particular value of t = t0. ä

Page 4 of 4

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United Arab Emirates UniversityCollege of Science


HOMEWORK 9 – SOLUTION

Section 12.6 The Gradient and Directional DerivativesSection 12.7 Extrema of Functions of Several Variables

Calculus II for EngineeringMATH 1120 SECTION 53 CRN 304483:30 – 5:20 on Sunday & TuesdayDue Date: Tuesday, April 28, 2009

ID No: Solution

Name: Solution

Score: 20/20


Section 12.6 The Gradient and Directional Derivatives1. Find the gradient of the given function.

(1.1) f(x, y) = x3e3y − y4

Answer.

Grad f(x, y) = ∇f(x, y) = 〈 fx(x, y), fy(x, y) 〉 =⟨

3x2e3y, 3x3e3y − 4y3⟩. �

(1.2) f(x, y) = e3y/x − x2y3

Answer.

Grad f(x, y) = ∇f(x, y) = 〈 fx(x, y), fy(x, y) 〉 =⟨−3yx2 e

3y/x − 2xy3, 3xe3y/x − 3x2y2

⟩

=⟨−3yx2 e

3y/x,3xe3y/x

⟩+⟨−2xy3, −3x2y2

⟩

= − 3x2 e

3y/x 〈 y, −x 〉 − xy2 〈 2y, 3x 〉 . �

2. Find the gradient of the given function at the indicated point.

(2.1) f(x, y) = sin(3xy) + y2 at (π, 1)

Answer.

Grad f(x, y) = ∇f(x, y) = 〈 3y cos(3xy), 3x cos(3xy) + 2y 〉 = 3 cos(3xy) 〈 y, x 〉+ 2y 〈 0, 1 〉

Grad f(π, 1) = ∇f(π, 1) = 3 cos(3π) 〈 1, π 〉+ 2(1) 〈 0, 1 〉 = 〈 −3, 2− 3π 〉 . �

(2.2) f(x, y, z) = z2e2x−y − 4xz2 at (1, 2, 2)

Answer.

Grad f(x, y, z) = ∇f(x, y, z) =⟨

2z2e2x−y − 4z2, −z2e2x−y, 2ze2x−y − 8xz⟩

= ze2x−y 〈 2z, −z, 2 〉 − 4z 〈 z, 0, 2x 〉

Grad f(1, 2, 2) = ∇f(1, 2, 2) = 2e2−2 〈 2(2), −2, 2 〉 − 4(2) 〈 2, 0, 2(1) 〉 = −4 〈 2, 1, 3 〉 .�

Page 1 of 11


3. Compute the directional derivative of f at the given point in the direction of the indicated vector.

(3.1) f(x, y) = x3y − 4y2 at (2,−1) with u =⟨

1√2,

1√2

⟩

Answer.

Du f(x, y) = ∇f(x, y) · u = 〈 fx(x, y), fy(x, y) 〉 ·⟨

1√2,

1√2

⟩

=⟨

3x2y, x3 − 8y⟩·⟨

1√2,

1√2

⟩= 3x2y + x3 − 8y√

2

Du f(2,−1) = 3(22)(−1) + 23 − 8(−1)√2

= 4√2

= 2√

2. �

(3.2) f(x, y) = e4x2−y at (1, 4), u in the direction of 〈 −2,−1 〉

Answer. The unit vector u in the direction 〈 −2,−1 〉 is obtained by

u = 〈 −2,−1 〉‖〈 −2,−1 〉‖ = − 1√

5〈 2, 1 〉 .

Using this unit vector u, we get

Du f(x, y) = ∇f(x, y) · u = 〈 fx(x, y), fy(x, y) 〉 · − 1√5〈 2, 1 〉

=⟨

8xe4x2−y,−e4x2−y ⟩ · − 1√5〈 2, 1 〉 = −(16x− 1)e4x2−y

√5

Du f(1, 4) = −(16(1)− 1)e4(12)−4√

5= −3

√5. �

(3.3) f(x, y) = x2 sin(4y) at (−2, π/8), u in the direction from (−2, π/8) to (0, 0)

Answer. The vector in the direction from (−2, π/8) to (0, 0) is 〈 2,−π/8 〉. So the unit vectoru in the direction from (−2, π/8) to (0, 0) is obtained by

u = 〈 2,−π/8 〉‖〈 2,−π/8 〉‖ = 1√

π2 + 256〈 16,−π 〉 .


Du f(x, y) = ∇f(x, y) · u = 〈 fx(x, y), fy(x, y) 〉 · 1√π2 + 256

〈 16,−π 〉

=⟨

2x sin(4y), 4x2 cos(4y)⟩· 1√π2 + 256

〈 16,−π 〉

= 32x sin(4y)− 4πx2 cos(4y)√π2 + 256

Du f(−2, π/8) = 32(−2) sin (4(π/8))− 4π(−2)2 cos (4(π/8))√π2 + 256

= − 64√π2 + 256

. �

Page 2 of 11


(3.4) f(x, y) = y2 + 2ye4x at (0,−2), u in the direction from (0,−2) to (−4, 4)

Answer. The vector in the direction from (0,−2) to (−4, 4) is 〈 −4, 6 〉. So the unit vectoru in the direction from (0,−2) to (−4, 4) is obtained by

u = 〈 −4, 6 〉‖〈 −4, 6 〉‖ = 1√

13〈 −2, 3 〉 .


Du f(x, y) = ∇f(x, y) · u = 〈 fx(x, y), fy(x, y) 〉 · 1√13〈 −2, 3 〉

=⟨

8ye4x, 2y + 2e4x⟩· 1√

13〈 −2, 3 〉 = (6− 16y)e4x + 6y√

13

Du f(0,−2) = (6− 16(−2))e4(0) + 6(−2)√13

= 2√

13. �

(3.5) f(x, y, z) =√x2 + y2 + z2 at (1,−4, 8), u in the direction of 〈 1, 1,−2 〉

Answer. The unit vector u in the direction 〈 1, 1,−2 〉 is obtained by

u = 〈 1, 1,−2 〉‖〈 1, 1,−2 〉‖ = 1√

6〈 1, 1,−2 〉 .


Du f(x, y, z) = ∇f(x, y, z) · u = 〈 fx(x, y, z), fy(x, y, z), fz(x, y, z) 〉 · 1√6〈 1, 1,−2 〉

= 1√x2 + y2 + z2

〈 x, y, z 〉 · 1√6〈 1, 1,−2 〉 = x+ y − 2z√

6(x2 + y2 + z2)

Du f(1,−4, 8) = 1− 4− 2(8)√6(12 + (−4)2 + 82)

= − 199√

6. �

Page 3 of 11


4. Find the directions of maximum and minimum change of f at the given point, and the values ofthe maximum and minimum rates of change.

(4.1) f(x, y) = x2 − y3 at (−1,−2)

Answer.

Grad f(x, y) = ∇f(x, y) =⟨

2x, −3y2⟩, Grad f(−1,−2) = 〈 −2, −12 〉 = −2 〈 1, 6 〉 .

It implies(a) Direction of maximum change of f is −2 〈 1, 6 〉,(b) Direction of minimum change of f is 2 〈 1, 6 〉,(c) Value of maximum rate of change is ‖−2 〈 1, 6 〉‖ = 2

√37,

(d) Value of minimum rate of change is −‖−2 〈 1, 6 〉‖ = −2√

37. �(4.2) f(x, y) = y2e4x at (3,−1)

Answer.

Grad f(x, y) = ∇f(x, y) =⟨

4y2e4x, 2ye4x⟩

= 2ye4x 〈 2y, 1 〉 ,Grad f(3,−1) = −2e12 〈 −2, 1 〉 = −2e12 〈 −2, 1 〉 .

It implies(a) Direction of maximum change of f is −2e12 〈 −2, 1 〉,(b) Direction of minimum change of f is 2e12 〈 −2, 1 〉,(c) Value of maximum rate of change is

∥∥∥−2e12 〈 −2, 1 〉∥∥∥ = 2e12√5,

(d) Value of minimum rate of change is −∥∥∥−2e12 〈 −2, 1 〉

∥∥∥ = −2e12√5. �(4.3) f(x, y) = x cos(3y) at (−2, π)

Answer.

Grad f(x, y) = ∇f(x, y) = 〈 cos(3y), −3x sin(3y) 〉 ,Grad f(−2, π) = 〈 cos(3π), −3(−2) sin(3π) 〉 = 〈 −1, 0 〉 .

It implies(a) Direction of maximum change of f is 〈 −1, 0 〉,(b) Direction of minimum change of f is −〈 −1, 0 〉,(c) Value of maximum rate of change is ‖〈 −1, 0 〉‖ = 1,(d) Value of minimum rate of change is −‖〈 −1, 0 〉‖ = −1. �

(4.4) f(x, y) =√x2 + y2 at (3,−4)

Page 4 of 11


Answer.

Grad f(x, y) = ∇f(x, y) = 1√x2 + y2

〈 x, y 〉 , Grad f(3,−4) = 15 〈 3, −4 〉 .

It implies

(a) Direction of maximum change of f is 15 〈 3, −4 〉,

(b) Direction of minimum change of f is −15 〈 3, −4 〉,

(c) Value of maximum rate of change is∥∥∥∥15 〈 3, −4 〉

∥∥∥∥ = 1,

(d) Value of minimum rate of change is −∥∥∥∥15 〈 3, −4 〉

∥∥∥∥ = −1. �

(4.5) f(x, y, z) =√x2 + y2 + z2 at (1, 2,−2)

Answer.

Grad f(x, y, z) = ∇f(x, y, z) = 1√x2 + y2 + z2

〈 x, y, z 〉 ,

Grad f(1, 2,−2) = 13 〈 1, 2, −2 〉 .

It implies

(a) Direction of maximum change of f is 13 〈 1, 2, −2 〉,

(b) Direction of minimum change of f is −13 〈 1, 2, −2 〉,

(c) Value of maximum rate of change is∥∥∥∥13 〈 1, 2, −2 〉

∥∥∥∥ = 1,

(d) Value of minimum rate of change is −∥∥∥∥13 〈 1, 2, −2 〉

∥∥∥∥ = −1. �

Page 5 of 11


Section 12.7 Extrema of Functions of Several Variables5. Locate all critical points and classify them using the Second Derivative Test.

(5.1) f(x, y) = cos2 x+ y2

Answer. Step 1 Critical Points We compute the partial derivatives and find the critical points:

fx = −2 sin x cosx = 0 =⇒ sin x = 0 or cosx = 0 =⇒ x = 0, ±π2 ,±π,±3π2 ,±2π, . . .

fy = 2y = 0 =⇒ y = 0.

Hence, all critical points are as follows:

(x, y) = (0, 0),(nπ

2 , 0), and (mπ, 0) ,

where n = ±1,±3,±5, . . . and m = ±1,±2,±3, . . . .Step 2 Second Derivatives Test Using the Second Derivatives Test, we classify the criticalpoint.

fxx = −2(cos2 x− sin2 x

)= −2

(1− 2 sin2 x

)= 2(2 sin2 x− 1

), fyy = 2, fxy = 0,

D(x, y) = fxxfyy − f 2xy = 4

(2 sin2 x− 1

).

Putting each critical point into the determinant D(x, y), we get

D(0, 0) = −4 < 0, D (mπ, 0) = −4 < 0,

D(nπ

2 , 0)

= 4 > 0, and fxx

(nπ

2 , 0)

= 2 > 0.

Therefore, we conclude (0, 0) and (mπ, 0) are saddle points and (nπ/2, 0) gives the localminimum. �

(5.2) f(x, y) = 4xy − x4 − y4 + 4


fx = 4y − 4x3 = 4(y − x3) = 0 =⇒ y = x3, (1)fy = 4x− 4y3 = 4(x− y3) = 0 =⇒ x = y3. (2)

Solving the equations (1) and (2) for x and y, we get

y = x3 = (y3)3, i.e., y9 = y, 0 = y9 − y0 = y(y8 − 1) = y(y4 + 1)(y4 − 1) = y(y4 + 1)(y2 + 1)(y2 − 1)

= y(y4 + 1)(y2 + 1)(y − 1)(y + 1) =⇒ y = 0, or y = −1, or y = 1.

Page 6 of 11


By the same argument (or by putting into the equation (1) or (2)), we get x = 0, x = −1 orx = 1. Hence, all critical points are (x, y) = (0, 0), (−1,−1), and (1, 1).Step 2 Second Derivatives Test Using the Second Derivatives Test, we classify the criticalpoint.

fxx = −12x2, fyy = −12y2, fxy = 4,D(x, y) = fxxfyy − f 2

xy = (12xy)2 − 16 = 16 (3xy − 1) (3xy + 1) .


D(0, 0) = −16 < 0, D(1, 1) = 96 > 0, and fxx(1, 1) = −12 < 0,D(−1,−1) = 96 > 0, and fxx(−1,−1) = −12 < 0.

Therefore, we conclude (0, 0) is a saddle point and (1, 1) and (−1,−1) give the local maximum.�

(5.3) f(x, y) = 2x2 + y3 − x2y − 3y


fx = 4x− 2xy = 2x(2− y) = 0 =⇒ x = 0 or y = 2, (3)fy = 3y2 − x2 − 3 = 0. (4)

When x = 0 from the result (3), the equation (4) becomes 3y2 − 3 = 0, i.e., 0 = 3(y2 − 1) =3(y − 1)(y + 1), i.e., y = ±1. So we have the critical points (0,±1).When y = 2 from the result (3), the equation (4) becomes 3(22)−x2−3 = 0, i.e., 0 = x2−9 =(x− 3)(x+ 3), i.e., x = ±3. So we have the critical points (±3, 2).That is, all the critical points are (0,±1) and (±3, 2).Step 2 Second Derivatives Test Using the Second Derivatives Test, we classify the criticalpoint.

fxx = 4− 2y = 2(2− y), fyy = 6y, fxy = −2x,D(x, y) = fxxfyy − f 2

xy = 12y(2− y)− 4x2 = 4[3y(2− y)− x2

].


D(0, 1) = 12 > 0, and fxx(0, 1) = 2 > 0,D(0,−1) = −36 < 0, and D(−3, 2) = −36 < 0, and D(3, 2) = −36 < 0.

Therefore, we conclude (0,−1) and (±3, 2) are saddle points and (0, 1) gives the local mini-mum. �

(5.4) f(x, y) = x sin y

Page 7 of 11



fx = sin y = 0 =⇒ y = 0,±π,±2π, . . .

fy = x cos y = 0 =⇒ x = 0, or y = ±π2 ,±3π2 , . . . .

Hence, the critical points are (x, y) = (0, 0), and (0, nπ), where n is any integer.Step 2 Second Derivatives Test Using the Second Derivatives Test, we classify the criticalpoint.

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D(0, 0) = −1 < 0, D(0, nπ) = −1 < 0

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Page 8 of 11


6. Find the absolute extrema of the function on the region.

(6.1) f(x, y) = x2 + y2 − 4xy, region bounded by y = x, y = −3 and x = 3.

Answer. Let R be the region bounded by the given graphs.Step 1 Critical Points We find the critical points on the region R.

fx = 2x− 4y = 2(x− 2y) = 0 =⇒ x = 2y,fy = 2y − 4x = 2(y − 2x) = 0 =⇒ y = 2x.

Solving two equations for x and y, we get

y = 2x = 2(2y), i.e., y = 4y2, y(4y − 1) = 0, y = 0 or y = 14 .

Putting into the equation x = 2y, we have the critical points (x, y) = (0, 0), (1/2, 1/4),which are on the region R. At (x, y) = (0, 0) and (1/2, 1/4), we have f(0, 0) = 0 andf(1/2, 1/4) = −3/16.Step 2 Local Extrema on Boundary of R The region R has three boundaries: (a) y = x for−3 ≤ x ≤ 3; (b) y = −3 for −3 ≤ x ≤ 3; (c) x = 3 for −3 ≤ y ≤ 3. We find the local extremaof f(x, y) on each boundary.(a) y = x for −3 ≤ x ≤ 3: On this boundary, f(x, y) becomes

g(x) = f(x, x) = x2 + x2 − 4x2 = −2x2, −3 ≤ x ≤ 3,

which has the absolute maximum value g(0) = 0 at x = 0 and the absolute minimumvalue g(−3) = −18 = g(−3) at x = ±3. That is, on this boundary, f(x, y) has the absolutemaximum value 0 at (x, y) = (0, 0) and the absolute minimum value −18 at (x, y) = (−3,−3).(b) y = −3 for −3 ≤ x ≤ 3: On this boundary, f(x, y) becomes

g(x) = f(x,−3) = x2 + 12x+ 9, −3 ≤ x ≤ 3,

which has the absolute maximum value g(3) = 54 at x = 3 and the absolute minimum valueg(−3) = −18 at x = −3. That is, on this boundary, f(x, y) has the absolute maximum value54 at (x, y) = (3,−3) and the absolute minimum value −18 at (−3,−3).(c) x = 3 for −3 ≤ y ≤ 3: On this boundary, f(x, y) becomes

g(x) = f(3, y) = y2 − 12y + 9, −3 ≤ y ≤ 3,

which has the absolute maximum value g(−3) = 54 at y = −3 and the absolute minimumvalue g(3) = −18 at y = 3. That is, on this boundary, f(x, y) has the absolute maximumvalue 54 at (x, y) = (3,−3) and the absolute minimum value −18 at (3, 3).Step 2 Conclusion Collecting all the results above, we deduce finally, f(x, y) has the followingtable:

Page 9 of 11


(x, y) (0, 0) (1/2, 1/4) (3,−3) (3, 3) (−3,−3)f(x, y) 0 −3/16 54 −18 −18

Hence, we conclude that the surface of f(x, y) over the given region has the absolute maximumvalue f(3,−3) = 54 at (x, y) = (3,−3) and the absolute minimum value f(−3,−3) = −18 at(x, y) = (−3,−3) and (3, 3). �

(6.2) f(x, y) = x2 + y2 − 2x− 4y, region bounded by y = x, y = 3 and x = 0.

Answer. Let R be the region bounded by the given graphs.Step 1 Critical Points We find the critical points on the region R.

fx = 2x− 2 = 2(x− 1) = 0 =⇒ x = 1,fy = 2y − 4 = 2(y − 2) = 0 =⇒ y = 2.

So we have the critical point (x, y) = (1, 2), which is on the region R and f(1, 2) = −5.Step 2 Local Extrema on Boundary of R The region R has three boundaries: (a) y = x for0 ≤ x ≤ 3; (b) y = 3 for 0 ≤ x ≤ 3; (c) x = 0 for 0 ≤ y ≤ 3. We find the local extrema off(x, y) on each boundary.(a) y = x for 0 ≤ x ≤ 3: On this boundary, f(x, y) becomes

g(x) = f(x, x) = x2 + x2 − 6x = 2x2 − 6x,

which has the absolute minimum value g(3/2) = −9/2 at x = 3/2 and the absolute maximumvalue g(0) = 0 = g(3) at x = 3 and x = 0. That is, on this boundary, f(x, y) has theabsolute minimum value −9/2 at (x, y) = (3/2, 3/2) and the absolute maximum value 0 at(x, y) = (3, 3) and (x, y) = (0, 0)(b) y = 3 for 0 ≤ x ≤ 3: On this boundary, f(x, y) becomes

g(x) = f(x, 3) = x2 − 2x− 3,

which has the absolute minimum value g(1) = −4 at x = 1 and the absolute maximum valueg(3) = 0 at x = 3. That is, on this boundary, f(x, y) has the absolute maximum value 0 at(x, y) = (3, 3) and the absolute minimum value −4 at (1, 3).(c) x = 0 for 0 ≤ y ≤ 3: On this boundary, f(x, y) becomes

g(x) = f(0, y) = y2 − 4y,

which has the absolute minimum value g(2) = −4 at y = 2 and the absolute maximum valueg(0) = 0 at y = 0. That is, on this boundary, f(x, y) has the absolute maximum value 0 at(x, y) = (0, 0) and the absolute minimum value −4 at (0, 2).Step 2 Conclusion Collecting all the results above, we deduce finally, f(x, y) has the followingtable:

Page 10 of 11


(x, y) (1, 2) (3/2, 3/2) (0, 0) (3, 3) (1, 3) (0, 2)f(x, y) −5 −9/2 0 0 −4 −4

Hence, we conclude that the surface of f(x, y) over the given region has the absolute maximumvalue f(3, 3) = 0 at (x, y) = (3, 3) or (0, 0) and the absolute minimum value f(1, 2) = −5 at(x, y) = (1, 2). �

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Page 11 of 11

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homework 1 – solutionfaculty.uaeu.ac.ae/jaelee/class/1120/09sp_hws.pdf · calculus ii for...

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