Math 8: Calculus in one and several variablesSpring 2018 - Homework 3

Return date: Wednesday 04/18/18

keywords: limits and integrals with Taylor series, vectors, dot product

Instructions: Write your answers neatly and clearly on straight-edged paper, use completesentences and label any diagrams. Please show your work; no credit is given for solutions withoutwork or justification.

exercise 1. (4 points) Use Taylor series to evaluate the limit

a) limx→0sin(x)−x−x3

62x5 .

Solution: We know that

sin(x) =

∞∑n=0

(−1)n · x2n+1

(2n+ 1)!= x− x3

3!+x5

5!− x7

7!...

Hence for x 6= 0 we have

sin(x)− x− x3

6

2x5= − 1

3! · x2+

1

2 · 5!− x2

2 · 7!...

Therefore

limx→0

sin(x)− x2x3

= limx→0− 1

3! · x2+

1

2 · 5!− x2

2 · 7!... = −∞ .

b) limx→2cos(x−2)−1+ (x−2)2

2(x−2)4 .

Solution: We know that

cos(y) =

∞∑n=0

(−1)2n · y2n

(2n)!= 1− y2

2!+y4

4!− y6

6!...

Substituting y = x− 2, we get

cos(x− 2) =∞∑n=0

(−1)2n · (x− 2)2n

(2n)!= 1− (x− 2)2

2!+

(x− 2)4

4!− (x− 2)6

6!...

Hence for x 6= 2 we have

cos(x− 2)− 1 + (x−2)22

(x− 2)4=

1

4!− (x− 2)2

6!+ ...

Therefore

limx→2

cos(x− 2)− 1 + (x−2)22

(x− 2)4= lim

x→2

1

4!− (x− 2)2

6!+ ... =

1

4!=

1

24,

as all powers of x− 2 in the last series vanish at x = 2.

Math 8: Calculus in one and several variablesSpring 2018 - Homework 3

Return date: Wednesday 04/18/18

exercise 2. (3 points) Use Taylor series to evaluate∫e−2xdx as an infinite series.

Solution: Taylor series expansion for ey is

ey =

∞∑n=0

yn

n!= 1 +

y

1!+y2

2!+y3

3!...

Substitute y = −2x, we get

e−2x =

∞∑n=0

(−1)n · 2n · xn

n!= 1− 2x

1!+

22 · x2

2!− 23 · x3

3!...

∫e−2xdx = C+

∞∑n=0

(−1)n · 2n · xn+1

(n+ 1) · n!= C+

∞∑n=0

(−1)n · 2n · xn+1

(n+ 1)!= C+x−2x2

2!+22 · x3

3!−23 · x4

4!...,

where C is a constant.

exercise 3. (8 points) Describe in words the regions of R3 represented by the followingequation(s) or inequalities. Make a simple sketch to clarify your answer.

a) z ≥ −2.Solution: The half-space which consists of the set of points above the plane z = −2depicted below.

b) x2 = 4.Solution: The equation x2 = 4 is equal to x = 2 or x = −2. These are two planes parallelto the yz-plane.

Math 8: Calculus in one and several variablesSpring 2018 - Homework 3

Return date: Wednesday 04/18/18

c) y = z.Solution: In the yz-plane (x = 0) the points satisfying this equation lie on the straightline y = z. This equation is independent of x. Hence we get the shifted line at x = c,where c is a constant. In total we get a plane which is formed by parallel translates of theline in the yz-plane.

d) x2 + y2 = 4 and z = 2.Solution: x2 + y2 = 4 is a cylinder of radius 2 around the z-axis. z = 2 is a plane parallelto the xy-plane. If both equations are satisfied we get the intersection which is a circle ofradius 2.

Math 8: Calculus in one and several variablesSpring 2018 - Homework 3

Return date: Wednesday 04/18/18

exercise 4. (2 points) For the following vectors u and v calculate

u+ v , 2u+ 3v , |u| and |u− v|

a) u = 〈2, 3〉 and v = 〈−2,−1〉.Solution:

u+ v = 〈2 + (−2), 3 + (−1)〉 = 〈0, 2〉.

2u+ 3v = 〈4, 6〉+ 〈−6,−3〉 = 〈−2, 3〉.

|u| =√

22 + 32 =√4 + 9 =

√13.

|u− v| =√42 + 42 =

√32.

b) u = 〈6, 1,−3〉 and v = 〈2,−2, 1〉.Solution:

u+ v = 〈6 + 2, 1 + (−2),−3 + 1〉 = 〈8,−1,−2〉.

2u+ 3v = 〈12, 2,−6〉+ 〈6,−6, 3〉 = 〈18,−4,−3〉.

|u| =√

62 + 12 + (−3)2 =√36 + 1 + 9 =

√46.

|u− v| =√

42 + 32 + (−4)2 =√41.

exercise 5. (3 points) Let α = ∠(u,v) be the angle between the vectors u and v. For thefollowing vectors calculate

|u| , |v| , u

|u|, α = ∠(u,v) and proju(v)

a) u = 〈1, 4〉 and v = 〈2, 3〉.Solution:

|u| =√12 + 42 =

√1 + 16 =

√17.

|v| =√

22 + 32 =√4 + 9 =

√13.

u

|u|= 〈 1√

17,

4√17〉.

cosα =u · v|u||v|

=2 + 12√17 ·√13

=14√

17 ·√13.

Math 8: Calculus in one and several variablesSpring 2018 - Homework 3

Return date: Wednesday 04/18/18

α = cos−1(

14√17 ·√13

)' cos−1(0.942) ' 19.65◦.

proju(v) =u · v|u|· u

|u|=

14√17〈 1√

17,

4√17〉 = 〈14

17,56

17〉.

b) u = 〈−1, 4, 8〉 and v = 〈12, 1, 2〉.Solution:

|u| =√(−1)2 + 42 + 82 =

√1 + 16 + 64 =

√81 = 9.

|v| =√

122 + 12 + 22 =√144 + 1 + 4 =

√149.

u

|u|= 〈−1

9,4

9,8

9〉.

cosα =u · v|u||v|

=−12 + 4 + 16

9 ·√149

=8

9 ·√149

.

α = cos−1(

8

9 ·√149

)' cos−1(0.0728) ' 85.82◦.

proju(v) =u · v|u|· u

|u|=

8

9〈−19,4

9,8

9〉 = 〈−8

81,32

81,64

81〉

c) For the vectors in 5 a) draw a picture of u and v and the projection proju(v).Solution:

Figure 1: The vectors u = 〈1, 4〉 (red) and v = 〈2, 3〉 (blue). The orthogonal component orthu(v)(green). The endpoint of orthu(v) on u marks the tip of the vector proju(v).