Exam November 2014, questions and answers PDF

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MATH1002: Mathematical Methods 2 – Semester 2, 2014

Mid Semester Test Name:

Student Number:

Signature:

I am taking this test in Room:

Tear off this page and put it on your desk with your Student Card, then fill the next page and wait for the signal from the invigilators to start the test.

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MATH1002: Mathematical Methods 2 – Semester 2, 2014

Mid Semester Test Name:

Student Number:

Total Time: 60 minutes. Total Marks: 40 All answers should be fully explained. It is not sufficient to just give simple yes or no solutions without justification.

Tick the tutorial you are attending (this question is for the purpose of handing your copy back to you after it is marked). Mon 9-11 MLR2 Andrew Bassom Mon 11-1 MLR3 Isaac May Mon 11-1 MLR2 Michael McPhail Mon 1-3 MLR3 Gregory Collin Mon 1-3 MLR2 Blake Segler Mon 3-5 MLR2 Blake Segler Tue 9-11 MLR2 Kyle Rosa Tue 11-1 MLR3 David Dragojevich Tue 11-1 MLR2 Zainab Alsheekhhussain Tue 2-4 MLR2 Alice Niemeyer Tue 2-4 MLR3 Alice Devillers Tue 4-6 MLR2 Konstantinos Sakellariou Wed 9-11 MLR2 Lyle Noakes Wed 11-1 MLR2 Ben Luo Wed 12-2 MLR3 Claire Delides Wed 1-3 MLR2 Martin Paesold Wed 2-4 MLR3 Claire Delides Wed 3-5 MLR2 Martin Paesold Thur 11-1 MLR3 Stephen Glasby Thur 1-3 MLR2 Stephen Glasby Thur 3-5 MLR3 Ayham Zaitouny Thur 3-5 MLR2 Phil Schrader Fri 9-11 MLR2 Lexi Goff Fri 11-1 MLR2 Lexi Goff Fri 2-4 MLR3 Ayham Zaitouny Fri 2-4 MLR2 Phil Schrader

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Question 1. Find an antiderivative for x6 ln(x). Solution: We are going to use integration by parts. We take f ′ = x6 and g = ln(x). 7 So that f = x7 and g ′ = 1x . Hence an antiderivative is Z 7 Z x7 ln(x) 1 x 1 x7 ln(x) − F (x) = x6 dx · dx = − 7 7 7 x 7 x7 ln(x) x7 − = 7 49 Alternative method. We take f ′ = ln(x) and g = x6 . So that f = x ln(x) − x and g ′ = 6x5 . Hence Z Z 6 6 I = x ln(x)dx = (x ln(x) − x)x − (x ln(x) − x)6x5 dx Z Z 7 7 6 = x ln(x) − x − 6 x ln(x)dx + 6 x6 dx

So re-arranging I =

x7 ln(x) 7

= x7 ln(x) − x7 + 6 −

x7 − 6I 7

x7 49

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Question 2. Determine if the improper integral Z ∞ 1 dx 2 x +x 1 is convergent or divergent. If it is convergent, what is its value? You might find it useful to recall that ln(ab) = ln(a) + ln(b) and ln(a/b) = ln(a) − ln(b). Solution: The function is undefined at 0 and −1 which are not contained in the domain of integration so this is a simple type II improper integral. This is a rational function so we need to do partial fractions. 1 Cover-the-zero method (or any other method) yields x21+x = x1 − x+1 . Rt 1 R∞ 1 Now 1 x2 +x dx = limt→∞ 1 x2 +x dx. For R t t1 > 1: R t 1 1 dx = 1 x − x+1 dx = [ln |x| − ln |x + 1|]t1 1 x2 +x = ln(t) − ln(t + 1) − ln(1) + ln(2) = ln(2) − ln(t + 1) + ln(t). Rt So limt→∞ 1 x21+x dx = limt→∞ (ln(2) − ln(t + 1) + ln(t)) ) = ln(2) − limt→∞ ln(1 + t1) = ln(2) = ln(2) − limt→∞ ln( t+1 t

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Question 3. A function f (x) has values given as below. x 0 0.5 1 1.5 2 2.5 3 f (x) 1 1.2 1.5 1.3 0.9 0.8 0.6 It is also known that |f ′′ (x)| ≤ 2.4 for x ∈ (0, 3). 3 Recall the error formula for the composite mid-point rule is Nh f ′′ (c). 24 (a) Use the composite mid-point rule with 3 subintervals to approximate the value of

R3 0

f (x) dx.

We have to add the areas of 3 rectangles to get: RSolution: 3 f (x) dx ≈ f (0.5) + f (1.5) + f (2.5) = 1.2 + 1.3 + 0.8 = 3.3. 0 (b) Find an upper bound for the absolute value of the error made in part (a). Deduce the R3 smallest range [M1 , M2 ] in which the actual value of 0 f (x) dx must lie. Solution: Here N = 3, h = 1, so applying the formula: 3 |error| ≤ 2.4 = 0.3 24 Thus the real value for the integral lies inside [3, 3.6]. R3 (c) Suppose we need to estimate 0 f (x) dx with an error of at most 10−3 . What is the least number N of intervals (all of the same length) we need to use to ensure this? Solution: For general N , h = 3/N so 33 N 2.7 2.4 = 2 . 3 N 24N We want |error| < 10−3 , so it is enough to require that which implies N > 51.9. As N is an integer, we choose N = 52. |error| ≤

2.7 N2

< 10−3 . Thus N 2 > 2700,

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Question 4. Let I =

R 4 R √y 0

y/2

x4 dx dy.

(a) Sketch the region of integration. Solution: y 4

2

x

(b) Change the order of integration and hence determine I . Solution: We can describe the region as 0 ≤ x ≤ 2, x2 ≤ y ≤ 2x, so Z 2 Z 2x x4 dy dx I= 0

=

Z

0

2 4

x

[y]2x x2

=



6

x2

dx =

Z

2 0

 7 2

x x − 7 3



 2x5 − x6 dx

= 64/21

0

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Question 5. Let R denote the solid bounded by the paraboloid x2 + y 2 + z = 3 between the planes z = 0 and z = 1. Let (x, y, z) = g(ρ, θ, ξ) denote the transformation into cylindrical coordinates. (a) State the definition of g(ρ, θ, ξ). Solution: (x, y, z) = g(ρ, θ, ξ) = (ρ cos θ, ρ sin θ, ξ ). (b) Describe the region S such that drical coordinates.

RRR

R

f (x, y, z )dxdydz is the triple integral over S in cylin-

Solution: Then the region S is S = {(ρ, θ, ξ) | 0 6 ξ 6 1, 0 6 ρ 6

or

p

3 − ξ, 0 6 θ 6 2π}

√ S = {(ρ, θ, ξ) | 0 6 ρ 6 2, 0 6 ξ 6 1, 0 6 θ 6 2π} √ √ ∪{(ρ, θ, ξ) | 2 6 ρ 6 3, 0 6 ξ 6 3 − ρ2 , 0 6 θ 6 2π}.

RRR (x2 + y 2 )dxdydz in terms of cylindrical coordinates. DO NOT (c) Express the integral R ATTEMPT TO EVALUATE THE INTEGRAL. Solution: Z ZZZ 2 2 (x + y )dxdydz = R

=

Z

2π θ=0 2π θ=0

Z Z

1 ξ=0 1 ξ=0

Z

√ 3−ξ

(ρ2 cos2 θ + ρ2 sin2 θ)ρ dρdξdθ

ρ=0

Z

√ 3−ξ

ρ3 dρdξdθ

ρ=0

Or with other description longer (marks for similar steps): ZZZ Z 2π Z √2 Z 1 Z 2π Z √3 Z x2 + y 2 dxdydz = ρ3 dξdρdθ + √ R

θ=0

ρ=0

ξ=0

θ=0 ρ= 2

3−ρ2

ρ3 dξdρdθ

ξ=0

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Question 6. (a) In R2 , find a parameterisation for the circle of radius 2 centered at the origin, traversed in the clockwise direction, starting and ending at the point (2, 0). y radius = 2

x

Solution: Using polar coordinates, we find a parameterisation is r(t) = (2 cos(2π − t), 2 sin(2π − t)) = (2 cos(t), −2 sin(t)) for t ∈ [0, 2π]. (b) In R3 , let S be the surface of the cylinder of radius 2 about the z-axis. Find a parameterisation for the curve, starting at (2, 0, 0) and finishing at (2, 0, 1), which winds up regularly once around S, as in the picture. z

(2,0,1)

x (2,0,0)

y

Solution: We use cylindrical coordinates and find Equivalently

r(t) = (2 cos(t), −2 sin(t), t/(2π)) for t ∈ [0, 2π]. r(t) = (2 cos(2πt), −2 sin(2πt), t) for t ∈ [0, 1].

(c) Find the length L of the curve described in Part (b). Solution: Say they found r(t) = (2 cos(t), −2 sin(t), t/(2π)) for t ∈ [0, 2π]. Then r˙ (t) q= (−2 sin(t), −2 cos(t), 1/(2π)), q 2 2t+ 1 = r (t) |= 4 sin t + 4 cos 4 + 4π12 . so | ˙ 4π 2 q R 2π q √ Thus L = 0 4 + 4π12 dt = 2π 4 + 4π12 = 16π 2 + 1.

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Question 7. (a) Let S be a surface described by the parameterisation S(u, v ) for (u, v ) ∈ D.

Define the surface integral of a function g(x, y, z) over the surface S, explaining all notation you use. Solution: ZZ

g(x, y, z)dS = S

ZZ

g(S(u, v ))|N(u, v )|dudv, D

where N(u, v) = Su × Sv . (b) Let S denote the surface area of the northern hemisphere of the unit sphere centred at the origin. In terms of spherical coordinates, S is the parametric surface S(u, v) = (cos u sin v, sin u sin v, cos v) with 0 6 v 6 π/2 and 0 6 u 6 2π, and the normalRRvector N(u, v) satisfies |N(u, v)| = sin v. Determine S z 2 dS. Solution: ZZ

2

z dS = S

Z

0

π/2 Z 2π

= 2π

cos2 v sin v dudv

0

Z

0

π/2

π/2  cos3 v cos v sin v dv = 2π − 3 0 2

2π = 3

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