UNIVERSITY OF DUBLIN
XMA2E011
TRINITY COLLEGE
Faculty of Engineering, Mathematics
and Science
school of mathematics
SF Engineers
SF MSISS
SF MEMS
Trinity Term 2014
Module 2E011,
???, ???
Final Exam
???
9.30 — 11.30
Dr. Sergey Frolov
ATTEMPT QUESTION 1 and FOUR OTHER QUESTIONS
Log tables are available from the invigilators, if required.
Non-programmable calculators are permitted for this examination,—please indicate the make
and model of your calculator on each answer book used.
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XMA2E011
Each question is worth 18 marks.
1. (a) 14 marks. Solve the following initial value problem by the Laplace transform
y 00 + 4 y = −4 u(t − 2π) + 5 δ(t − 4π) ,
y(0) = −1 ,
y 0 (0) = 0 .
(b) 4 marks. Sketch the input function and the solution.
Show the details of your work.
Solution :
(a) 14 marks. We denote Y (s) = L(y), and then using the formulae
L(y 00 ) = s2 Y (s) − sy(0) − y 0 (0) , L(u(t − a)) =
e−as
, L(δ(t − a)) = e−as ,
s
we get the algebraic equation
(s2 + 4)Y (s) = −4
e−2πs
+ 5e−4πs − s .
s
Solving the equation for Y , we get
(5 marks.)
Y (s) = −
s
4e−2πs
5e−4πs
−
+
.
s2 + 4 s (s2 + 4) s2 + 4
Then we represent
(2 marks.)
−
4
1
s
=− + 2
.
+ 4)
s s +4
s(s2
Finally we use the formulae of the inverse Laplace transform
1
s
−1
−1
L
= 1, L
= cos ωt , L−1 e−as F (s) = f (t − a)u(t − a) ,
2
2
s
s +ω
to get
(7 marks.)
5
y(t) = − cos(2t) + u(t − 2π)(cos(2t) − 1) + u(t − 4π) sin(2t)
2



− cos 2t
if 0 < t < 2π


=
(1)
−1
if 2π < t < 4π .



 −1 + 5 sin 2t
if t > 4π
2
(b) 4 marks. The plot of the input function and the solution is shown below.
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2. Consider the function
f (x, y, z) =
p
2xy + 3z 4 − 6 cos(3x − 2y) ,
and the point P = (2, 3, −1) .
(a) 10 marks. Find a unit vector in the direction in which f increases most rapidly at
the point P .
(b) 1 mark. Sketch the projection of the vector onto the xz-plane
(c) 3 marks. Find a unit vector in the direction in which f decreases most rapidly at
the point P .
(d) 1 mark. Sketch the projection of the vector onto the xy-plane
(e) 3 marks. Find the rate of change of f at the point P in these directions.
Show the details of your work.
Solution:
(a) 10 marks. f increases most rapidly in the direction of its gradient, so we compute
(1 mark.)
p
4
f (2, 3, −1) = 2xy + 3z − 6 cos(3x − 2y) = 3 ,
P
(2 marks.)
(2 marks.)
18 sin(3x − 2y) + 2y
fx (2, 3, −1) = p
= 1,
4
2 2xy − 6 cos(3x − 2y) + 3z P
2x − 12 sin(3x − 2y)
2
fy (2, 3, −1) = p
= ,
4
3
2 2xy − 6 cos(3x − 2y) + 3z P
Page 4 of 14
(2 marks.)
XMA2E011
6z 3
fz (2, 3, −1) = p
= −2 .
2xy − 6 cos(3x − 2y) + 3z 4 P
Thus, the gradient and its magnitude are equal to
7
2
∇f (2, 3, −1) = 1 , , −2 , |∇f (2, 3, −1)| = ≈ 2.33333 .
3
3
Therefore, the unit vector in the direction of the gradient is
2
3
1 , , −2 ≈ (0.428571, 0.285714, −0.857143) .
(3 marks.)
u=
7
3
(b) 1 mark. The projection of the vector onto the xz-plane is the vector
uxz =
3
(1 , 0 , −2) ≈ (0.428571 , 0 , −0.857143) .
7
(c) 3 marks. f decreases most rapidly in the direction opposite to its gradient, so the
unit vector is
3
v=−
7
2
1 , , −2 ≈ (−0.428571, −0.285714, 0.857143) .
3
(d) 1 mark. The projection of the vector onto the xy-plane is the vector
3
2
vxy = −
1 , , 0 ≈ (−0.428571, −0.285714, 0) .
7
3
(e) 3 marks. The rate of change of f at P in the direction of u is equal to
|∇f (2, 3, −1)| =
7
≈ 2.33333 ,
3
and the rate of change of f at P in the direction of v is equal to
7
−|∇f (2, 3, −1)| = − ≈ −2.33333 .
3
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XMA2E011
3. Consider the surface
z = f (x, y) = ln
1 2/3 p
3
e
12 sin(x − 2y) + 8y 2 − x3 − 6x2 y + 32
2
.
(a) 7 marks. Find an equation for the tangent plane to the surface at the point
P = (2, 1, z0 ) where z0 = f (2, 1).
(b) 3 marks. Find points of intersection of the tangent plane with the x-, y- and
z-axes.
(c) 2 marks. Sketch the tangent plane, and show the point P = (2, 1, z0 ) on it.
(d) 4 marks. Find parametric equations for the normal line to the surface at the point
P = (2, 1, z0 ).
(e) 2 marks. Sketch the normal line to the surface at the point P = (2, 1, z0 ).
Show the details of your work.
Solution:
(a) 7 marks. We first find
(1 mark.)
12 sin(x − 2y) + 8y 2 − x3 − 6x2 y + 32|P = 8 ,
z0 =
2
,
3
and then simplify
1 2/3 p
3
2
3
2
z = ln
e
12 sin(x − 2y) + 8y − x − 6x y + 32
2
1
2
=
ln(12 sin(x − 2y) + 8y 2 − x3 − 6x2 y + 32) + − ln 2 .
3
3
(2)
Then, we compute the partial derivatives at P
(2 marks.)
∂
−3x2 − 12xy + 12 cos(x − 2y)
z|x=2,y=1 =
= −1 .
∂x
3 (−x3 − 6x2 y + 12 sin(x − 2y) + 8y 2 + 32) P
(2 marks.)
∂ 1
−6x2 − 24 cos(x − 2y) + 16y
4
z|x=2,y=1 =
=− .
3
2
2
∂y 2
3 (−x − 6x y + 12 sin(x − 2y) + 8y + 32) P
3
The tangent plane equation is given by
(2 marks.)
z=
2
4
4
− 1(x − 2) − (y − 1) = 4 − x − y .
3
3
3
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(b) 3 marks. (4, 0, 0), (0, 3, 0) , (0, 0, 4)
(c) 2 marks. The tangent plane is the one through the points in (b).
(d) 4 marks. The normal line to the surface (and the tangent plane) is given by
2
4
r = 2i + j + k + t i + j + k .
3
3
(e) 2 marks. The normal line is perpendicular to the plane.
4. Consider the solid in the first octant bounded by the surface
cos π(1−x)
y e1−x
2
−y
,
z=
1−x
1−x
below by the plane z = 0, and laterally by y = x and y =
√
x.
(a) 3 marks. Sketch the projection of the solid onto the xy-plane.
(b) 10 marks. Use double integration to find the volume of the solid.
(c) 5 marks. Find
e1−x
cos π(1−x)
2
lim
−
.
x→1 1 − x
1−x
Show the details of your work.
Solution:
(a) 3 marks. Below is the projection of the solid onto the xy-plane.
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(b) 10 marks. Therefore, the volume is given by the following double integral
!
ZZ
sin πx
e1−x
2
(5 marks.)
V =
y−
y dA
1−x
1−x
R
#
!
Z 1 " Z √x
sin πx
e1−x
2
=
−
y dy dx
1−x
1−x
0
x
! √x
Z 1
Z
πx
1−x
sin
e
y2 1 1 1−x
πx 2
−
e
− sin
x dx .
=
dx =
1−x
1−x
2
2 0
2
0
x
The remaining integral is computed by using integration by parts
Z 1
Z 1
1−x
(2 marks.)
xe
dx =
d(−x e1−x ) + e1−x dx
0
0
Z 1
1−x 1
e1−x dx = −1 − e1−x |10 = −2 + e ,
= −x e
|0 +
0
and
Z
(3 marks.)
0
1
πx
x sin
dx =
2
Z
1
2
πx 2
πx
d − x cos
+ cos
dx
π
2
π
2
0
Z 1
2
πx 1
2
πx
= − x cos
|0 +
cos
dx
π
2
2
0 π
4
πx 1
4
= 0 + 2 sin
|0 = 2 .
π
2
π
Thus the volume is equal to
V =
2
e
− 1 − 2 ≈ 0.156499
2
π
(c) 5 marks. We have
e1−x
et sin π(1−t) et cos πt sin πx
2
2
2
lim
−
= lim
−
= lim
−
x→1 1 − x
t→0
t→0
1−x
t
t
t
t
2 2
1 + t + t2
1 − π 4t 2
= lim
−
= 1,
t→0
t
t
where we used that
et ≈ 1 + t +
t2 t3
+ + ··· ,
2
6
cos t = 1 −
t2
t4
+
+ ··· .
2
24
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XMA2E011
5. (a) Let the path C between the points (−π/2 , π) and (3π/2 , −2π/3) be a curve
formed from two line segments C1 and C2 , where C1 is joining (−π/2 , π) and
(3π/2 , π), and C2 is joining (3π/2 , π) and (3π/2 , −2π/3).
i. 2 marks. Plot the path C, and show its orientation on the plot.
ii. 5 marks. Parameterise C1 and C2 , and evaluate
Z
3y
(−6xy + 3π 3 sin 3x) dx − (3x2 + 2π 3 cos ) dy .
2
C
(b) 2 marks. Show that for any integration path C the integral above depends only
on the initial and terminal points (−π/2 , π) and (3π/2 , −2π/3) of the path C.
(c) 7 marks. Find the potential function φ(x, y).
(d) 2 marks. Use the Fundamental Theorem of Line Integrals to find the value of the
integral above for an integration path between the initial point (−π/2 , π) and the
terminal point (3π/2 , −2π/3).
Show the details of your work.
Solution:
(a) i. 2 marks. Below is the path C.
3
2
1
1
-1
2
3
4
5
-1
-2
(a) ii. 5 marks. The line segments C1 is parametrized as x = t, y = π, −π/2 ≤ t ≤ 3π/2
and one gets
Z
Z 3π/2
3y
3
2
3
(−6xy + 3π sin 3x) dx − (3x + 2π cos ) dy =
(−6πt + 3π 3 sin 3t) dt
2
C1
−π/2
27
3
9π
π
3π/2
= (−3πt2 − π 3 cos 3t)|−π/2 = − π 3 + π 3 − π 3 cos
+ π 3 cos = −6π 3 ≈ −186.038 .
4
4
2
2
(3)
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XMA2E011
The line segments C2 is parametrized as x = 3π/2, y = t, π ≥ t ≥ −2π/3 and one gets
Z −2π/3
3y
27
3t
−( π 2 + 2π 3 cos ) dt
(−6xy + 3π sin 3x) dx − (3x + 2π cos ) dy =
2
4
2
π
C2
4
27
4
3π
45
4
119 3
9
= π3 − π3 =
π ≈ 307.479 .
= π 3 + π 3 sin π + π 3 + π 3 sin
2
3
4
3
2
4
3
12
(4)
Z
3
The sum gives
47 3
π
12
2
3
≈ 121.441.
(b) 2 marks. We have
f (x, y) = −6xy + 3π 3 sin 3x ,
g(x, y) = −(3x2 + 2π 3 cos
3y
).
2
Thus
∂y f (x, y) = −6x ,
∂x g(x, y) = −6x ,
and therefore the integral is independent of the path.
(c) 7 marks. To compute the integral we find the potential function φ(x, y)
∂φ
= −6xy + 3π 3 sin 3x
∂x
(3 marks.)
⇒
φ(x, y) = −3x2 y − π 3 cos 3x + C(y) .
To find C(y) we use that
(3 marks.)
∂φ
dC(y)
3y
= −3x2 +
= −(3x2 + 2π 3 cos )
∂y
dy
2
4 3
3y
⇒ C(y) = − π sin
+C.
3
2
⇒
dC(y)
3y
= −2π 3 cos
dy
2
(5)
Thus, we get
(1 mark.)
4
3y
φ(x, y) = −3x2 y − π 3 cos 3x − π 3 sin
+C.
3
2
(d) 2 marks. By using the formula, we obtain
Z
(3π/2 ,−2π/3)
(−6xy + 3π 3 sin 3x) dx − (3x2 + 2π 3 cos
(−π/2 ,π)
3y
47
(3π/2 ,−2π/3)
) dy = φ(x, y)|(−π/2 ,π)
= π3 ,
2
12
which agrees with the result in (a)ii.
6. (a) 3 marks. Express rectangular coordinates in terms of cylindrical coordinates
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XMA2E011
(b) Consider the solid G bounded by the surfaces x2 + y 2 = 1 and x2 + y 2 = 4, above
by the surface z = 5 − x2 − y 2 , and below by the surface z = 0.
i. 1 mark. What are the surfaces x2 + y 2 = 1 and x2 + y 2 = 4?
ii. 1 mark. What is the surface z = 5 − x2 − y 2 ?
iii. 1 mark. What is the surface z = 0? Sketch the part of the boundary of the
solid G which belongs to the surface z = 0.
iv. 5 marks. Use triple integral to compute the volume V of the solid G.
v. 7 marks. Use triple integral to find the mass M of the solid G if its density is
δ(x, y, z) = e5−x
2 −y 2 −z
.
Show the details of your work.
Solution :
(a) 3 marks. We have
x = r cos θ ,
y = r sin θ ,
z=z
(b) i. 1 mark. Cylinder of radius 1, and cylinder of radius 2.
(b) ii. 1 mark. Paraboloid.
(b) iii. 1 mark. The xy-plane. The plot is shown below
(b) iv. 5 marks. We use the cylindrical coordinates to get
# !
ZZZ
Z 2π Z 2 "Z 5−r2
Z
V =
dV =
rdz dr dθ = 2π
G
0
1
0
1
2
1
5
r(5 − r2 )dr = 2π( r2 − r4 )|21
2
4
5 1
15
= 2π(10 − 4 − + ) = π ≈ 23.5619 .
2 4
2
(6)
(b) v. 7 marks. We use the cylindrical coordinates to get
# !
ZZZ
Z 2π Z 2 "Z 5−r2
2
M =
δ(x, y, z)dV =
e5−r −z rdz dr dθ
G
Z
= 2π
1
2
0
2
1
0
2
(e5−r − 1)rdr = π(−e5−r − r2 )|21 = π(e4 − e − 3) ≈ 153.561 .
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XMA2E011
Useful Formulae
1. Let r(t) be a vector function with values in R3 :
(a) Its derivative is
dr
dt
=
df
dt
,
dg
dt
,
dh
dt
r(t) = f (t) i + g(t) j + h(t) k .
.
(b) The magnitude of this vector is
|| dr
||
dt
(c) The unit tangent vector is T =
dr
dt
|| dr
||
dt
=
q
df 2
dt
+
dg 2
dt
+
dh 2
dt
.
.
(d) The vector equation of the line tangent to the graph of r(t) at the point P =
(x0 , y0 , z0 ) corresponding to t = t0 on the curve is R(t) = r0 + (t − t0 ) v0 , where
r0 = r(t0 ) and v0 =
dr
(t ) .
dt 0
R2
|| dt .
|| dr
dt
R t dr
(f) The arc length parameter s having r(t0 ) as its reference point is s = t0 || du
|| du .
(e) The arc length of the graph of r(t) between t1 and t2 is L =
1
2. Let σ be a surface in R3 : z = f (x, y)
(a) The slope kx of the surface in the x-direction at the point (x0 , y0 ) is kx =
∂z
(x0 , y0 ) .
∂x
(b) The slope ky of the surface in the y-direction at the point (x0 , y0 ) is ky =
∂z
(x0 , y0 ) .
∂y
(c) The equation for the tangent plane to the surface at the point P = (x0 , y0 , z0 ) is
z = z0 + kx (x − x0 ) + ky (y − y0 ) .
(d) Parametric equations for the normal line to the surface at P = (x0 , y0 , z0 ) are
r(t) = r0 + t(−kx i − ky j + k) , r0 = x0 i + y0 j + z0 k .
(e) The volume under the surface and over a region R in the xy-plane is
RR
V = R f (x, y) dA .
(f) The area of the portion of the surface that is above a region R in the xy-plane is
r
2
RR
RR
∂z 2
∂z
S = σ dS = R 1 + ∂x
+ ∂y
dA .
(g) The mass of the lamina with the density δ(x, y, z) that is the portion of the surface
that is above a region R in the xy-planeris
RR
RR
M = σ δ(x, y, z) dS = R δ(x, y, z) 1 +
∂z 2
∂x
+
2
∂z
∂y
dA .
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3. The local linear approximation of the function z = f (x, y) at the point (x0 , y0 ) is
L(x, y) = f (x0 , y0 ) + fx (x0 , y0 )(x − x0 ) + fy (x0 , y0 )(y − y0 ) .
4. Let f (x, y, z) be a function of three variables
(a) The gradient of f is ∇f = (fx , fy , fz ) .
(b) f increases most rapidly in the direction of its gradient, and the rate of change of
f in this direction is equal to ||∇f ||.
(c) If f is smooth then its critical points satisfy fx = fy = fz = 0.
5. Let R be a region in the xy-plane bounded by the curves y = g(x) , y = h(x) , x =
a , x = b, and g ≤ h for a ≤ x ≤ b. Then the double integral over the region is
i
RR
R b hR h(x)
f (x, y) dA = a g(x) f (x, y)dy dx .
R
6. Let R be a region in the xy-plane bounded by the curves (in polar coordinates)
r = r1 (θ) , r = r2 (θ) , θ = α , θ = β and r1 ≤ r2 for α ≤ θ ≤ β. Then the double
integral over the region is
i
RR
RR
R β hR r2 (θ)
f
(r,
θ)
dA
=
f
(r,
θ)
rdr
dθ
=
f
(r,
θ)rdr
dθ .
R
R
α
r1 (θ)
7. Let R be a plain lamina with density δ(x, y).
(a) Its mass is equal to M =
RR
R
δ(x, y) dA .
(b) The x-coordinate of its centre of gravity is equal to xcg =
1
M
RR
(c) The y-coordinate of its centre of gravity is equal to ycg =
1
M
RR
R
x δ(x, y) dA .
R
y δ(x, y) dA .
8. Let G be a simple solid whose projection onto the xy-plane is a region R. G is bounded
by a surface z = g(x, y) from below and by a surface z = h(x, y) from above.
i
RRR
RR hR h(x,y)
(a) The triple integral over the solid is
f
(x,
y,
z)
dV
=
f
(x,
y,
z)
dz
dA .
G
R
g(x,y)
RRR
RR
(b) The volume of the solid is V =
dV = R [h(x, y) − g(x, y)] dA .
G
9. Let G be a solid enclosed between the two surfaces (in spherical coordinates)
r = g(θ , φ) , r = h(θ , φ).
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(a) The triple integral over the solid is
i
RRR
R 2π R π hR h(θ ,φ)
2
f
(r,
θ,
φ)
dV
=
f
(r,
θ,
φ)
r
dr
sin
φ
dφ
dθ .
g(θ ,φ)
0
0
G
RRR
R 2π R π hR h(θ ,φ) 2 i
r
dr
sin
φ
dφ
dθ .
(b) The volume of the solid is V =
dV
=
g(θ ,φ)
0
G
0
RRR
(c) The mass of the solid with the density δ(r, θ, φ) is M =
δ(r, θ, φ) dV .
G
10. Let a region Rxy in the xy-plane be mapped to a region Ruv in the uv-plane under the
change of variables u = u(x, y) , v = v(x, y).
∂u ∂v
(a) The magnitude of the Jacobian of the change is ∂(u,v)
= ∂x ∂y −
∂(x,y) (b) The integral over Rxy is
RR
Rxy
f (x, y) dAxy =
RR
Ruv
∂u ∂v ∂y ∂x .
−1
dAuv .
f (x(u, v), y(u, v)) ∂(u,v)
∂(x,y) 11. The area of the surface that extends upward from the curve x = x(t) , y = y(t) , a ≤
t ≤ b in the xy-plane to the surface z = f (x, y) is given by the following line integral
q R
Rb
dy 2
dx 2
A = C z ds = a f (x(t), y(t))
+
dt .
dt
dt
12. Consider a line integral
R
C
f (x, y) dx + g(x, y) dy , and let P = (xP , yP ) and Q =
(xQ , yQ ) be the endpoints of the curve C.
(a) The line integral is independent of the path if ∂y f (x, y) = ∂x g(x, y) .
(b) Then there is a potential function φ(x, y) satisfying
∂φ
∂x
= f (x, y) ,
∂φ
∂y
= g(x, y) ,
(c) and the Fundamental Theorem of Line Integrals says that
R
R Q ∂φ
f
(x,
y)
dx
+
g(x,
y)
dy
=
dx + ∂φ
dy = φ(x, y)|Q
P = φ(xQ , yQ ) −
∂y
C
P ∂x
φ(xP , yP ) .
13. Let a closed curve C be oriented counterclockwise, and be the boundary of a simply
connected region R in the xy-plane. By Green’s Theorem we have
H
RR ∂g(x,y) ∂f (x,y) − ∂y
dA
f
(x,
y)
dx
+
g(x,
y)
dy
=
∂x
C
R
14. Let F(x, y, z) = M i + N j + P k be a vector field.
(a) If σ is the surface z = f (x, y), oriented by upward unit normals n, and R is the
projection of σ onto the xy-plane then
RR
RR ∂f
flux = σ F · n dS = R −M ∂f
−
N
+
P
dA .
∂x
∂y
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(b) If σ is the surface z = f (x, y), oriented by downward unit normals n, and R is the
projection of σ onto the xy-plane then
RR
RR ∂f
flux = σ F · n dS = R M ∂f
+
N
−
P
dA .
∂x
∂y
(c) According to the Divergence Theorem the flux of F across a closed surface σ with
outward orientation is
RR
RRR
flux = σ F · n dS =
div F dV ,
V
div F =
∂M
∂x
+
∂N
∂y
+
∂P
∂z
.
(d) If σ is an oriented smooth surface that is bounded by a simple, closed, smooth
boundary curve C with positive orientation then, according to Stokes’ Theorem
∂N ∂M H
RR
∂P
∂M
∂N
∂P
F·dr
=
(curl
F)·n
dS
,
curl
F
=
i+
−
−
j+ ∂x − ∂y k .
∂y
∂z
∂z
∂x
C
σ
15. The Laplace transform of a function f (t) is the function F (s) defined by
R∞
F (s) = L(f (t)) = 0 e−st f (t)dt , f (t) = L−1 (F (s)) .
F unction T ransf orm
F unction T ransf orm
eat
1
s−a
eat tn
n!
(s−a)n+1
eat sin ωt
ω
(s−a)2 +ω 2
eat cos ωt
s−a
(s−a)2 +ω 2
eat sinh ωt
ω
(s−a)2 −ω 2
eat cosh ωt
s−a
(s−a)2 −ω 2
t sin ωt
2ωs
(s2 +ω 2 )2
t cos ωt
s2 −ω 2
(s2 +ω 2 )2
u(t − a)
e−as
s
δ(t − a)
e−as
16. Let F (s) = L(f (t)), then L(f (t − a)u(t − a)) = e−as F (s);
L (eat f (t)) = F (s − a) ; L (tf (t)) = − dFds(s) ; L (f (kt)) = k1 F
s
k
.
17. Let Y (s) = L(y), then L(y 0 ) = sY (s) − y(0) , L(y 00 ) = s2 Y (s) − sy(0) − y 0 (0) .
18. Convolution. Let f (t) ∗ g(t) =
Rt
0
f (τ )g(t − τ ) dτ . Then L(f (t) ∗ g(t)) = F (s)G(s)
c UNIVERSITY OF DUBLIN 2015