Module 2E02 (Frolov), Multivariable Calculus
1
Tutorial Sheet 1
Consider the vector function (with values in R3 )
r(t) = ln(2 + t3 ) i + (3 + t3 ) j −
(t3 + 2)2
k
4
1. Find the domain D(r) of the vector function r(t).
Solution: The domain D(r) of r(t) is the intersection of domains of its component functions.
√
3
2
) = (−∞, ∞), one gets
Since D(ln(2 + t3 )) = (− 3 2 , ∞ ), D(3 + t3 ) = (−∞, ∞) and D(− (t +2)
4
√
3
D(r) = (− 2 , ∞ )
√
that is the vector function r(t) is defined for t > − 3 2.
2. Find
(a) the derivative dr/dt,
(b) the norm ||dr/dt||
(c) the unit tangent vector T for all values of t in D(r).
Simplify the expressions obtained.
Hint: use the formula a2 + 2ab + b2 = (a + b)2 .
Solution:
(a)
3t2
3t5
dr
2
2
= 3
, 3t , −
− 3t .
dt
t +2
2
(b) The magnitude of this vector is
s
s
2
2
2
2
2
5
dr
3t
1
t3
3t
2
2
2
2
|| || =
+ (3t ) + 3t +
= 3t
+1+ 1+
dt
2 + t3
2
2 + t3
2
s
2
2
1
1 2 + t3
2 + t3
2
= 3t
+2
+
2 + t3
2 + t3 2
2
s
2
1
2 + t3
1
2 + t3
2
2
= 3t
+
= 3t
+
,
2 + t3
2
2 + t3
2
because t3 > −2.
(c) The unit tangent vector is
dr
2
2 (t3 + 2) −t6 − 4t3 − 4
dt
T = dr = 6
,
,
.
t + 4t3 + 6 t6 + 4t3 + 6 t6 + 4t3 + 6
|| dt ||
1
3. Find the vector equation of the line tangent to the graph of r(t) at the point P0 (0, 2, − 41 )
on the curve.
Solution: The point P0 (0, 2, − 14 ) on the curve corresponds to t = −1. We find
r0 = r(−1) = 2j −
1
k,
4
v0 =
dr
3
(−1) = 3i + 3j − k .
dt
2
Thus the tangent line equation is
1
r = r0 + (t + 1) v0 = 3(t + 1) i + (3t + 5) j − (6t + 7) k .
4
Note that the same line is also described by the following equation which is obtained from the
one above by the rescaling and shift of the parameter t: t → 3t − 1
1
r = r0 + t v0 = t i + (t + 2) j − (2t + 1) k .
4
4. Find the arc length of the graph of r(t) if −1 ≤ t ≤ 1.
Solution: The arc length of the graph of r(t) is given by the definite integral
Z 1
Z 1
Z 1 dr
1
2 + t3
2+v
1
2
|| || dt =
3t
L=
+
+
dt =
dv
dt
2 + t3
2
2+v
2
−1
−1
−1
(2 + v)2 1
=
ln(2 + v) +
= 2 + ln 3 ,
4
−1
where we have made the substitution v = t3 .
5. Find a positive change of parameter from t to s where s is an arc length parameter of the
curve having r(1) as its reference point.
Solution: The arc length parameter s can be found as follows
Z
s=
1
t
Z t3 1
2 + u3
1
2+v
3u
+
du =
+
dv
2 + u3
2
2+v
2
1
1
(2 + v)2 t3
t6
5
1 3
3
=
ln(2 + v) +
t +2
,
= + t − + log
4
4
4
3
1
dr
|| || du = =
du
Z
t
2
where we have made the substitution v = u3 .
2
2
Tutorial Sheet 2
1. Sketch the level curve z = k for the specified values of k
z = 2x2 + 8x + 3y 2 − 6y ,
k = −11, −9, −8 .
Solution
2.0
1.5
1.0
0.5
-2.5
-2.0
-1.5
-1.0
The level curve equation 2x2 + 8x + 3y 2 − 6y = k can be written as
2(x + 2)2 + 3(y − 1)2 = k + 11 ,
and, therefore, the level curves are ellipses with the centre located at (−2, 1). For k = −11
the level curve is just the point (−2, 1).
2. Consider the function
π
π
z = 2ey− 2 sin x − 3ex− 4 cos y
(i) Find
a)
∂z π π
( , ),
∂x 4 2
b)
∂z π π
( , ),
∂y 4 2
c)
∂ 2z π π
( , ),
∂x∂y 4 2
d)
∂ 2z π π
( , ).
∂y∂x 4 2
Solution:
a)
b)
c)
∂ 2z
∂ ∂z
∂
=
=
∂x∂y
∂x ∂y
∂x
d)
∂ 2z
∂ ∂z
∂
=
=
∂y∂x
∂y ∂x
∂y
π
π
∂z
= 2ey− 2 cos x − 3ex− 4 cos y
∂x
⇒
√
∂z π π
( , ) = 2.
∂x 4 2
√
π
π
∂z
∂z π π
= 2ey− 2 sin x + 3ex− 4 sin y ⇒
( , ) = 3 + 2.
∂y
∂y 4 2
√
π
π
π
π
∂ 2z π π
2ey− 2 sin x + 3ex− 4 sin y = 2ey− 2 cos x+3ex− 4 sin y ⇒
( , ) = 3+ 2 .
∂x∂y 4 2
√
π
π
π
π
∂ 2z π π
2ey− 2 cos x − 3ex− 4 cos y = 2ey− 2 cos x+3ex− 4 sin y ⇒
( , ) = 3+ 2 .
∂y∂x 4 2
3
π
π
(ii) Find the slope of the surface z = 2ey− 2 sin x − 3ex− 4 cos y
a) in the x-direction at the point ( π2 , π6 );
Solution: The slope kx is equal to
√
3 3 π
∂z π π
( , )=−
e4 .
kx =
∂x 2 6
2
b) in the y-direction at the point ( π3 , π).
Solution: The slope ky is equal to
ky =
√ π
∂z π
( , π) = 3e 2 .
∂y 3
π
π
(iii) Show that the function z = 2ey− 2 sin x − 3ex− 4 cos y satisfies Laplace’s equation
∂ 2z ∂ 2z
+
= 0.
∂x2 ∂y 2
Solution: To this end we compute the following derivatives
π
π
∂ 2z
∂ ∂z
∂
y− π2
x− π4
=
=
2e
cos
x
−
3e
cos
y
= −2ey− 2 sin x − 3ex− 4 cos y ,
2
∂x
∂x ∂x
∂x
π
π
π
π
∂ ∂z
∂
∂ 2z
=
=
2ey− 2 sin x + 3ex− 4 sin y = 2ey− 2 sin x + 3ex− 4 cos y .
2
∂y
∂y ∂y
∂y
The sum of these two expressions is obviously 0.
3. Show that the local linear approximation of the function
f (x, y) = 2xα y β −
xβ
yα
at (1, 1) is
f (x, y) ≈ 1 + (2α − β)(x − 1) + (α + 2β)(y − 1) .
Solution: The local linear approximation of the function at (1, 1) is given by the formula
L(x, y) = f (1, 1) + fx (1, 1)(x − 1) + fy (1, 1)(y − 1) .
We obviously have f (1, 1) = 1, and
fx (x, y) = 2α xα−1 y β − β
xβ−1
yα
⇒
fx (1, 1) = 2α − β ,
fy (x, y) = 2β xα y β−1 + α
xβ
y α+1
⇒
fy (1, 1) = α + 2β ,
which proves the formula.
4
4. Use appropriate forms of the chain rule to find
z = cos
x
sin 2y ;
2
∂z
∂u
and
∂z
∂v
where
x = 3u − 2v , y = u2 − 2v 3 .
Solution: We have
∂z
∂z ∂x ∂z ∂y
1
x
x
=
+
= − sin sin 2y × 3 + 2 cos cos 2y × 2u
∂u
∂x ∂u ∂y ∂u
2
2
2
3
x
x
= − sin sin 2y + 4u cos cos 2y
2
2
2
3
3u − 2v
3u − 2v
2
= − sin
sin 2(u − 2v 3 ) + 4u cos
cos 2(u2 − 2v 3 ) ,
2
2
2
∂z
∂z ∂x ∂z ∂y
1
x
x
=
+
= − sin sin 2y × (−2) + 2 cos cos 2y × (−2v 2 )
∂v
∂x ∂v ∂y ∂v
2
2
2
x
x
= sin sin 2y − 4v 2 cos cos 2y
2
2
3u − 2v
3u − 2v
= sin
sin 2(u2 − 2v 3 ) − 4v 2 cos
cos 2(u2 − 2v 3 ) .
2
2
3
Tutorial Sheet 3
1. Consider the function
p
f (x, y, z) = z 2 + y − x + 2 cos(3x − 2y) ,
and the point P (2, 3, −1) .
(a) Find a unit vector in the direction in which f increases most rapidly at the point P .
(b) Sketch the projection of the vector onto the xz-plane
(c) Find a unit vector in the direction in which f decreases most rapidly at the point
P.
(d) Sketch the projection of the vector onto the yz-plane
(e) Find the rate of change of f at the point P in these directions.
Show the details of your work.
Solution:
(a) f increases most rapidly in the direction of its gradient, so we compute
−6 sin(3x − 2y) − 1
fx (x, y, z) = p
2 2 cos(3x − 2y) − x + y + z 2
4 sin(3x − 2y) + 1
fy (x, y, z) = p
2 2 cos(3x − 2y) − x + y + z 2
z
fz (x, y, z) = p
2 cos(3x − 2y) − x + y + z 2
5
⇒
⇒
⇒
1
fx (2, 3, −1) = − ,
4
fy (2, 3, −1) =
1
,
4
1
fz (2, 3, −1) = − .
2
Thus, the gradient and its magnitude are equal to
√
1 1
6
1
∇f (2, 3, −1) = − , , −
, ||∇f (2, 3, −1)|| =
≈ 0.612372 .
4 4
2
4
Therefore, the unit vector in the direction of the gradient is
1 1
1
4
≈ (−0.408248, 0.408248, −0.816497) .
u = √ − , ,−
4 4
2
6
(b) The projection of the vector u onto the xz-plane is the vector
1
1
4
≈ (−0.408248 , 0 , −0.816497) .
uxz = √ − , 0 , −
4
2
6
It is shown below
z
-0.4-0.3-0.2-0.1
x
-0.2
-0.4
-0.6
-0.8
(c) f decreases most rapidly in the direction opposite to its gradient, so the unit vector is
4
1
1 1
v = −√ − , , −
≈ (0.408248, −0.408248, 0.816497) .
4 4
2
6
(d) The projection of the vector v onto the yz-plane is the vector
1
1
4
vyz = − √ 0 , , −
≈ (0 , −0.408248, 0.816497) .
4
2
6
It is shown below
z
0.8
0.6
0.4
0.2
-0.4-0.3-0.2-0.1
6
y
(e) The rate of change of f at P in the direction of u is equal to
√
6
||∇f (2, 3, −1)|| =
≈ 0.612372 ,
4
and the rate of change of f at P in the direction of v is equal to
√
6
≈ −0.612372 .
−||∇f (2, 3, −1)|| = −
4
2. Let r =
p
x2 + y 2 . Compute
∇f (r) ,
where f (r) = ln(1 + r) .
Solution: We have
∇f (r) =
∂f (r)
∂f (r)
i+
j,
∂x
∂y
∂f (r)
d ln(1 + r) ∂r
1 x
=
=
,
∂x
dr
∂x
1+rr
Thus
∇f (r) =
∂f (r)
d ln(1 + r) ∂r
1 y
=f
=
,
∂y
dr
∂y
1+rr
1 x
1 y
1 r
i+
j=
,
1+rr
1+rr
1+rr
where r = xi + yj.
3. Consider the surface
p
3
2x2 − 3xy 2 + 3 cos(2x + 3y) − 3y 3 + 18
z = f (x, y) = ln
2
(a) Find an equation for the tangent plane to the surface at the point P (3, −2, z0 ) where
z0 = f (3, −2).
(b) Sketch the tangent plane.
(c) Find parametric equations for the normal line to the surface at the point P (3, −2, z0 ).
(d) Sketch the normal line to the surface at the point P (3, −2, z0 ).
Show the details of your work.
Solution:
(a) We first simplify
p
3
2x2 − 3xy 2 + 3 cos(2x + 3y) − 3y 3 + 18
1
z = ln
= ln 2x2 −3xy 2 +3 cos(2x+3y)−3y 3 +18 −ln 2 ,
2
3
7
and compute z0
z0 = z|x=3,y=−2 =
3
1
ln 3 cos(0) + 18 + 24 − 36 + 18 − ln 2 = ln ≈ 0.405465
3
2
Then, we compute the partial derivatives at P (3, −2, z0 )
∂z
−6 sin(2x + 3y) + 4x − 3y 2
=
∂x
3 (2x2 − 3xy 2 + 3 cos(2x + 3y) − 3y 3 + 18)
⇒
∂z
|x=3,y=−2 = 0 .
∂x
∂z
−6xy − 9 sin(2x + 3y) − 9y 2
=
∂y
3 (2x2 − 3xy 2 + 3 cos(2x + 3y) − 3y 3 + 18)
⇒
∂z
|x=3,y=−2 = 0 .
∂y
The tangent plane equation is given by
z = z0 + 0(x − 3) + 0(y + 2) = ln
3
.
2
(b) It is a plane through the point (3, −2, ln 32 ) parallel to the xy-plane, see the picture.
-3.0
4.0
x
3.5
y
-2.5
-2.0
3.0
2.5
2.0
2
z
0.8
0.6
z
0.4
1
4.0
3.5
0.2
0.0
-3.0
3.0
0
x
-2.5
-2.0
y
-1
2.5
-1.5
-1.0
2.0
(c) The normal line to the surface (and the tangent plane) is given by
r = 3i − 2j + t 0i + 0j + k = 3i − 2j + tk .
(d) It is parallel to the z-axis, see the picture above.
4. Consider the function
f (x, y) = x4 − x2 y + y 2 − 3y + 4
Locate all relative maxima, relative minima, and saddle points, if any.
8
-1.5
-1.0
Solution: The graph of the function is shown below
x
1
0
�1
3.0
2.5
2.0
1.5
0.5
1.0
1.0
1.5
2.0
y
2.5
We first find all critical points
fx (x, y) = 4x3 − 2xy = 0 ,
fy (x, y) = −x2 + 2y − 3 = 0 .
From the second equation we find y in terms of x
x2 3
y=
+ ,
2
2
and substituting it to the first equation, we derive the following equation for x
3x3 − 3x = 0 .
There are three solutions to this equation
x = 0 , x = −1 , x = 1 ,
and, therefore, three critical points
3
(x = 0 , y = ) ,
2
(x = −1 , y = 2) ,
(x = 1 , y = 2) .
Computing the values of f at critical points, we get
3
7
f (0, ) = ,
2
4
f (−1, 2) = 1 ,
f (1, 2) = 1 .
To find out if they are maximum, minimum or saddle points we use the second derivative
test. To this end we compute
∂ 2f
∂ 2f
∂ 2f
2
(x,
y)
=
12x
−
2y
,
(x,
y)
=
2
,
(x, y) = −2x ,
∂x2
∂y 2
∂x∂y
9
and
∂ 2f ∂ 2f
−
D(x, y) =
∂x2 ∂y 2
∂2f
∂x2
Computing D and
∂ 2f
∂x∂y
2
= 20x2 − 4y ,
for the three critical points, we get
3
D(0, ) = −6 ,
2
∂ 2f
3
) = −3 ,
(0,
∂x2
2
and therefore (0 , 32 ) is a saddle point.
D(−1, 2) = 12 ,
∂ 2f
(−1, 2) = 8 ,
∂x2
and therefore (−1, 2) is a relative minimum.
D(1, 2) = 12 ,
∂ 2f
(1, 2) = 8 ,
∂x2
and therefore (1, 2) is a relative minimum too.
4
Tutorial Sheet 4
1. Use a double integral to find the volume under the surface
z = 3πex sin y + e−x
and over the rectangle R = {(x, y) : 0 ≤ x ≤ ln 3 , 0 ≤ y ≤ π}.
Solution: The surface is shown below
20
3
10
2
0
0.0
1
0.5
1.0
10
0
The volume is given by the following double integral
ZZ
Z ln 3 Z π
x
−x
x
−x
(3πe sin y + e ) dy dx
V =
(3πe sin y + e ) dA =
0
0
R
Z ln 3
Z ln 3
ln 3
x
−x π
(−3πe cos y + e y) 0 dx =
(6πex + πe−x )dx = π (6ex − e−x )0
=
0
0
1
38
ln 3
− ln 3
0
0
= 6e − e
− (6e − e ) π = 18 − − (6 − 1) π = π
3
3
Thus the volume is equal to
V =
38
π ≈ 39.7935 .
3
2. Consider the solid in the first octant bounded by the surface
e1−x
cos πx
y+
y,
1−x
1−x
√
below by the plane z = 0, and laterally by y = x and y = x.
(a) Sketch the projection of the solid onto the xy-plane.
(b) Use double integration to find the volume of the solid.
(c) Find
e1−x
cos πx y+
y .
lim
x→1 1 − x
1−x
Show the details of your work.
Solution:
(a) Below is the projection of the solid onto the xy-plane, and the solid.
z=
1.0
0.8
0.6
0.4
0.2
0.2
0.4
0.6
0.8
1.0
(b) Therefore, the volume is given by the following double integral
#
Z Z 1−x
Z 1 "Z √x 1−x
e
e
cos πx
cos πx
V =
y+
y dA =
+
y dy dx
1−x
1−x
1−x
1−x
R
0
x
√x
Z 1 1−x
Z
e
cos πx y 2 1 1 1−x
=
+
dx
=
e
+
cos
πx
x dx .
1−x
1 − x 2 x
2 0
0
11
The remaining integral is computed by using integration by parts
Z 1
Z 1
Z 1
1−x
1−x
1−x 1
1−x
e1−x dx = −1 − e1−x |10 = −2 + e ,
d(−x e ) + e
dx = −x e
|0 +
xe
dx =
0
0
0
and
Z
0
1
1
Z 1
1
1
1
1
1
d x sin πx − sin πx dx = x sin πx |0 −
x cos πx dx =
sin πx dx
π
π
π
0
0 π
1
2
= 0 + 2 cos πx |10 = − 2 .
π
π
Z
Thus the volume is equal to
V =
1
e
− 1 − 2 ≈ 0.25782
2
π
(c) We have
e1−x
et cos π(1 − t) et cos πt cos πx lim
y+
y = lim
+
y = lim
−
y
x→1 1 − x
t→0
t→0
1−x
t
t
t
t
2 2
1 + t + t2
1 − π 2t 2
= lim
−
= 1,
t→0
t
t
where we used the Taylor expansions
t2 t3
e ≈ 1 + t + + + ··· ,
2
6
t
t2
t4
cos t = 1 − +
+ ··· .
2
24
2
3. Consider the solid inside the surface r2 + z9 = 19 and outside the surface r = 14 .
Here r2 = x2 + y 2 .
2
(a) What is the surface r2 + z9 = 19 ?
(b) What is the surface r = 41 ?
(c) Sketch the projection of the solid onto the xy-plane, and the solid.
(d) Use double integration and polar coordinates to find the volume of the solid.
Show the details of your work.
Solution:
2
(a) The surface r2 + z9 = 19 is an ellipsoid centred at the origin. Its equation can be written
in rectangular coordinates as
32 x2 + 32 y 2 + z 2 = 1 ,
which shows that it is an prolate spheroid (like a rugby ball).
(b) The surface r = 41 is a cylinder of radius 14 also centred at the origin.
(c) The projection, R, of the solid onto the xy-plane is the annulus with the inner radius 14 ,
and the outer radius 31 .
Below is the projection of the solid onto the xy-plane, and the solid.
12
0.3
0.2
0.1
0.0
-0.1
-0.2
-0.3
-0.3
-0.2
-0.1
0.0
0.1
0.2
0.3
(d) Since the solid is symmetric about the z = 0 plane, the volume of the solid is given by
#
ZZ √
Z 2π "Z 1/3 √
Z 1/3 √
1
2
2
V =2
1 − 9r dA = 2
1 − 9r rdr dθ = 4π
1 − 9r2 d(r2 )
2
R
0
1/4
1/4
1/3
3/2
11
4π 7
7π √
2 3/2 = −4π
7.
(1 − 9r ) =
=
39
27 16
432
1/4
Thus, the volume of the solid is equal to
V =
5
7π √
7 ≈ 0.134683
432
Tutorial Sheet 5
1. Consider the portion of the surface (y − 2)2 + (z + 1)2 = 12 that is above the rectangle
√
√
3
≤ x ≤ 3 , −1 ≤ y ≤ 5}.
R = {(x, y) : −
2
(a) What is the surface?
(b) Sketch the projection of the portion onto the xy-plane.
(c) Use double integration to find the area of the portion.
Show the details of your work.
Solution:
√
a) It is a circular cylinder of radius 12. Its centre line goes through the point (0, 2, −1)
and is parallel to the x-axis.
b) The portion of the cylinder and its projection are shown below
c) The area is given by the formula
s
s
2 2
2 2
ZZ
Z √3 Z 5
∂z
∂z
∂z
∂z
S=
1+
+
dA = √
1+
+
dy dx ,
∂x
∂y
∂x
∂y
R
− 3/2 −1
where
z = −1 +
p
12 − (y − 2)2
13
5
4
3
2
1
-0.5
0.5
1.0
1.5
-1
Computing the derivatives we get
s
√
2 2 s
∂z
∂z
(y − 2)2
12
=p
.
1+
+
= 1+
2
∂x
∂y
12 − (y − 2)
12 − (y − 2)2
Thus, we get
√
Z
S=
3
√
− 3/2
Z
√
5
−1
Z
12
5
Z
1
p
dy dx = 9
12 − (y − 2)2
p
dy = 9
12 − (y − 2)2
−1
Z 3
1
p
= 18
dy .
12 − y 2
0
3
−3
1
p
dy
12 − y 2
To compute the integral we do the substitution
y=
√
√
12 sin t = 2 3 sin t ,
dy =
√
12 cos t dt ,
p
12 − y 2 =
√
12 cos t ,
and get
π/3
Z
S = 18
dt = 6 π .
0
√
2. Consider the solid bounded by the surface z = 4y and the planes
x + y = 2 , x = 0, and z = 0.
(a) Sketch the projection of the portion onto the xy-plane.
(b) Use a triple integral to find the volume of the solid.
Solution:
(a) The solid, G, and its projection, R, onto the xy-plane are shown below
(b) Thus, the volume is equal to
√
ZZZ
V
ZZ Z
=
4y
dV =
Z
= 2
0
G
2
Z
2
Z
dz dA = 2
R
0
0
20
2−x
√
y dy dx
2
42
32 √
(2 − x)3/2 dx = −
(2 − x)5/2 =
2.
3
35
15
0
14
0≤t≤
π
3
2.0
1.5
2
1.0
1
0.0
0.5
0.5
1.0
0
1.5
2.0
0.0
0.5
1.0
2.0
1.5
0.5
1.0
1.5
2.0
3. Consider the lamina with density δ(x, y) = 4π − 5y bounded by
, y = 0 , x = 0, and x = 3.
y = sin πx
3
(a) Sketch the lamina .
(b) Find the mass and centre of gravity of the lamina.
Solution:
(a) The lamina, R, is shown below
1.0
0.8
0.6
0.4
0.2
0.5
1.0
1.5
2.0
2.5
3.0
(b) Its mass is equal to
ZZ
Z
3
Z
sin
πx
3
Z
3
πx 5 2 πx
− sin
) dx
3
2
3
R
0
0
0
Z 2
πx 5 5
2πx
πx 5
15
2πx 3
=
(4π sin
− + cos
) dx = (−12 cos
− x+
sin
)|
3
4 4
3
3
4
4π
3 0
0
15
81
= 12 −
+ 12 =
≈ 20.25 .
4
4
M =
δ(x, y) dA =
(4π − 5y) dy dx =
(4π sin
By the symmetry of the lamina the x-coordinate of its centre of gravity is equal to
xcg =
15
3
.
2
The y-coordinate is given by
ycg
ZZ
Z 2 Z sin πx
Z 3
3
πx 5 3 πx
4
1
4
y(4π − 5y) dy dx =
(2π sin2
− sin
) dx
=
y δ(x, y) dA =
M
81 0 0
81 0
3
3
3
R
Z 3
Z
2πx
πx
4
4 53 3
πx
(1 − cos
=
π
) dx +
(1 − cos2
) d(cos
)
81 0
3
81 3 π 0
3
3
3
4π
4π
20
πx 1
20
2
4π
80
3 πx =
+
(cos
− cos
) =
+
(−2 + ) =
−
.
27 81π
3
3
3 0
27 81π
3
27 243π
Thus, the coordinates of the centre of gravity of the lamina are
xcg = 1 ,
6
80
4π
−
≈ 0.360628 .
27 243π
ycg =
Tutorial Sheet 6
1.
(a) Express rectangular coordinates in terms of spherical coordinates.
Draw the corresponding picture.
(b) Consider the solid G bounded by the surfaces x2 + y 2 + z 2 = 1 and x2 + y 2 + z 2 = 9 and
below by the surface z = 0.
1. What is the surface x2 + y 2 + z 2 = 1?
2. What is the surface x2 + y 2 + z 2 = 9?
3. What is the surface z = 0?
4. Sketch the part of the boundary of the solid G which belongs to the surface z = 0.
5. Use triple integral and spherical coordinates to compute the volume V of the solid G.
6. Use triple integral and spherical coordinates to find the mass M of the solid G if its
density is
√
2
2
2
e− x +y +z
δ(x, y, z) = 2
.
x + y2 + z2
Show the details of your work.
Solution :
(a) We have
x = r cos θ sin φ ,
y = r sin θ sin φ ,
z = r cos φ ,
r ≥ 0 , 0 ≤ φ ≤ π , 0 ≤ θ ≤ 2π ,
where r is the radial coordinate, φ is the angle between the radius-vector and the z-axis, and
θ is the angle between the projection of the radius-vector onto the xy-plane and the x-axis.
(b) 1. Sphere of radius 1.
(b) 2. Sphere of radius 3.
(b) 3. The xy-plane.
16
(b) 4. The plot is shown below
3
2
1
0
-1
-2
-3
-3
-2
0
-1
1
2
3
(b) 5. We use the spherical coordinates to get
ZZZ
Z
V =
2π
Z
π/2
Z
dV =
G
0
0
3
!
r2 dr sin φ dφ dθ =
1
2π 3
52π
(3 − 13 ) =
≈ 54.4543
3
3
(b) 6. We use the spherical coordinates to get
ZZZ
Z
M =
2π
Z
π/2
Z
δ(x, y, z)dV =
G
Z
= 2π
0
0
1
3
!
e−r 2
r dr sin φ dφ dθ
r2
3
e−r dr = 2π(e−1 − e−3 ) ≈ 1.99863
(6.1)
1
2.
(a) Express rectangular coordinates in terms of cylindrical coordinates. Draw the corresponding picture.
(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.
17
i. What are the surfaces x2 + y 2 = 1 and x2 + y 2 = 4?
ii. What is the surface z = 5 − x2 − y 2 ?
iii. What is the surface z = 0? Sketch the part of the boundary of the solid G which
belongs to the surface z = 0.
iv. Use triple integral to compute the volume V of the solid G.
v. Use triple integral to find the mass M of the solid G if its density is
δ(x, y, z) = e5−x
Show the details of your work.
Solution :
(a) We have
x = r cos θ ,
2 −y 2 −z
y = r sin θ ,
z=z
(b) i. Cylinder of radius 1, and cylinder of radius 2.
(b) ii. Paraboloid.
(b) iii. The xy-plane. The plot is shown below
2
1
0
-1
-2
-2
-1
0
18
1
.
2
(b) iv. We use the cylindrical coordinates to get
# !
ZZZ
Z
Z "Z
2
2π
V =
2
5−r
dV =
Z
rdz dr dθ = 2π
0
G
= 2π(10 − 4 −
1
1
0
2
5
1
r(5 − r2 )dr = 2π( r2 − r4 )|21
2
4
5 1
15
+ ) = π ≈ 23.5619 .
2 4
2
(6.2)
(b) v. We use the cylindrical coordinates to get
ZZZ
Z
Z "Z
2π
M =
2
δ(x, y, z)dV =
G
Z
= 2π
2
0
1
2
#
5−r2
e5−r
2 −z
!
rdz dr dθ
0
2
(e5−r − 1)rdr = π(−e5−r − r2 )|21 = π(e4 − e − 3) ≈ 153.561 .
1
7
Tutorial Sheet 7
1. Consider the surface that extends upward from the semicircle y =
to the surface
1
π
z = y 2 + x2 y
12
4
(a) Make a rough sketch of the surface.
(b) Use a line integral to find the area of the surface.
Solution:
(a) Below is the surface
√
4 − x2 in the xy-plane
(b) Since the curve is the semicircle of radius 2 centred at the origin, it is convenient to
parametrize it as follows
x = 2 cos t ,
y = 2 sin t ,
19
0 ≤ t ≤ π.
The area of the surface is given by the following line integral
Z
Z
1
π
z ds =
( y 2 + x2 y) ds .
A=
4
C
C 12
Thus, we get
s 2
2
1 2 π 2
dx
dy
A =
( y + x y)
+
dt
12
4
dt
dt
0
Z π
q
1 2
2
( sin t + 2π cos t sin t) (−2 sin t)2 + (2 cos t)2 dt
=
3
0
π
Z π
Z π
Z
1
π 4π π
π 4π
3 2
2
=
(1 − cos 2t) dt + 4π
cos t sin t dt = −
cos t d cos t = − cos t = 3π .
3 0
3
3 0
3
3
0
0
Z
π
2.
(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. Plot the path C, and show its orientation on the plot.
ii. Parameterise C1 , and evaluate
Z
3y
I1 ≡
(−6xy + 3π 3 sin 3x) dx − (3x2 + 2π 3 cos ) dy .
2
C1
iii. Parameterise C2 , and evaluate
Z Z
3y
(−6xy + 3π 3 sin 3x) dx − (3x2 + 2π 3 cos ) dy .
I2 ≡
2
C2 C
iv. Compute the sum I = I1 + I2 .
(b) 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) Find the potential function φ(x, y).
(d) 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. Below is the path C.
20
3
2
1
1
-1
2
3
4
5
-1
-2
(a) ii. 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
(−6πt + 3π 3 sin 3t) dt
I1 =
(−6xy + 3π sin 3x) dx − (3x + 2π cos ) dy =
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
(7.3)
(a) iii. The line segments C2 is parametrized as x = 3π/2, y = t, π ≥ t ≥ −2π/3 and one gets
Z −2π/3
Z
27
3t
3y
3
2
3
−( π 2 + 2π 3 cos ) dt
I2 =
(−6xy + 3π sin 3x) dx − (3x + 2π cos ) dy =
2
4
2
π
C2
9
4
27
4
3π
45
4
119 3
= π 3 + π 3 sin π + π 3 + π 3 sin
= π3 − π3 =
π ≈ 307.479 .
2
3
4
3
2
4
3
12
(7.4)
47 3
(a) iv. The sum gives I = 12
π ≈ 121.441.
(b) 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) To compute the integral we find the potential function φ(x, y)
∂φ
= −6xy + 3π 3 sin 3x
∂x
⇒
φ(x, y) = −3x2 y − π 3 cos 3x + C(y) .
To find C(y) we use that
∂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
Thus, we get
4
3y
φ(x, y) = −3x2 y − π 3 cos 3x − π 3 sin
+C.
3
2
21
(7.5)
(d) By using the formula, we obtain
Z
(3π/2 ,−2π/3)
(−6xy + 3π 3 sin 3x) dx − (3x2 + 2π 3 cos
(−π/2 ,π)
47
3y
(3π/2 ,−2π/3)
= π3 ,
) dy = φ(x, y)|(−π/2 ,π)
2
12
which agrees with the result in (a)iv.
3. Consider the integral
I
(e−2x + 5x2 y − 2y 2 ) dx − (3ey − 3x3 ) dy .
C
Assume that the curve C is oriented counterclockwise, and it is the boundary of the region R
between y = x2 /2 and y = x.
(a) Sketch the region R.
(b) Use Green’s Theorem to evaluate the integral.
Show the details of your work.
Solution:
(a) Below is the region R
2.0
1.5
1.0
0.5
0.5
1.0
1.5
2.0
(b) By Green’s Theorem we have
I
ZZ ∂(−3ey + 3x3 ) ∂(e−2x + 5x2 y − 2y 2 )
−2x
2
2
y
3
(e
+ 5x y − 2y ) dx − (3e − 3x ) dy =
−
dA
∂x
∂y
C
R
ZZ
Z 2Z x
Z 2
x4
x2
2
2
2
2
=
4x + 4y dA =
4x + 4y dy dx =
4x (x − ) + 2(x − ) dx
2
4
R
0
x2 /2
0
2
2
1 16
16
= ( x3 + x4 − x5 ) =
+ 16 − 16 =
.
3
2
3
3
0
22
8
Tutorial Sheet 8
1. Consider the lamina that is the portion of the surface x2 + y 2 − 3z − 1 = 0 inside the surface
x2 + y 2 = 9/4. The density of the lamina is
9
δ(x, y, z) = δ0 ( + x2 + y 2 )
4
where δ0 is a constant
(a) What is the surface x2 + y 2 − 3z − 1 = 0?
(b) What is the surface x2 + y 2 = 94 ?
(c) Sketch the projection of the lamina onto the xy-plane.
(d) Find the mass of the lamina.
Show the details of your work.
Solution:
(a) The surface x2 + y 2 − 3z − 1 = 0 is a paraboloid.
(b) The surface x2 + y 2 = 94 is a circular cylinder of radius 3/2.
(c) Below is the lamina σ, and its projection onto the xy-plane
1.5
1.0
0.5
-1.5
-1.0
-0.5
0.5
1.0
1.5
-0.5
-1.0
-1.5
The projection of the surface onto the xy-plane is the disk R: x2 + y 2 ≤ 49 .
(d) The mass of the lamina is equal to the following surface integral
s
2 2
ZZ
ZZ
ZZ
9 2 2
∂z
∂z
2
9
M=
δ(x, y, z) dS = δ0
( +x +y ) 1 +
+
dA = δ0
( +x2 +y 2 )3/2 dA .
∂x
∂y
3
σ
R 4
R 4
The simplest way to compute the double integral over R is to use the polar coordinates
3/2
ZZ
Z 3/2
2
9
4π
9
4π 1 9
2
2 3/2
2 3/2
2 5/2 M =
δ0
( + x + y ) dA =
δ0
( + r ) rdr =
δ0 ( + r ) 3
3
4
3 5 4
R 4
0
0
5
√
√
4π 3
81π
=
δ0 5 (4 2 − 1) =
δ0 (4 2 − 1) ≈ 29.6256 δ0
15 2
40
2. Consider the surface σ which is the portion of the surface z =
z = 5/12, oriented by downward unit normals.
(a) Sketch the projection of the surface onto the xy-plane.
(b) Find the flux of the vector field F across σ.
F(x, y, z) =
16
y j + 4k ;
9
23
x2 +y 2 −1
3
below the plane
Show the details of your work.
Solution:
(a) The surface σ is the same as the lamina in Problem 1. Thus, the projection of the
surface onto the xy-plane is the disk R: x2 + y 2 ≤ 94 .
(b) Taking into account that through the surface z = f (x, y) oriented by down, the vector
field F(x, y, z) = M i + N j + P k has
ZZ
ZZ ∂f
∂f
+N
− P dA ,
flux =
F · n dS =
M
∂x
∂y
σ
R
we get by using the polar coordinates (M = 0 , N = 16
y , P = 4)
9
ZZ Z 2π Z 3/2
32 2
32
( r2 sin2 φ − 4) rdr dφ
flux =
y − 4 dA =
27
27
R
0
0
Z 2π
9
3
15
(3 sin2 φ − )dφ = π − 9π = − π .
=
2
2
2
0
3.
(a) Express rectangular coordinates in terms of spherical coordinates. Draw the corresponding picture.
(b) Use triple integral and spherical coordinates to derive the formula for the volume of a
ball of radius R.
(c) Use the Divergence Theorem to find the flux of the vector field
F(x, y, z) = (−3x + 12xy 2 + 4z 3 ) i − (2y 2 − y + 4x2 ) j + (4 + 3z + 5yz − 4y 3 ) k
across the surface σ with outward orientation where σ is given by the equation
x2 + y 2 + z 2 = 9.
Show the details of your work.
Solution :
(a) We have, see the picture below
x = r cos θ sin φ ,
y = r sin θ sin φ ,
z = r cos φ ,
r ≥ 0 , 0 ≤ φ ≤ π , 0 ≤ θ ≤ 2π ,
where r is the radial coordinate, φ is the angle between the radius-vector and the z-axis, and
θ is the angle between the projection of the radius-vector onto the xy-plane and the x-axis.
24
(b) We use the spherical coordinates to get (you must give more details here)
ZZZ
Z 2π Z π Z R
4
2
V =
dV =
r dr sin φ dφ dθ = πR3 .
3
V
0
0
0
(c) The surface is a sphere of radius 3. We have
ZZ
ZZZ
ZZZ
flux =
F · n dS =
div F dV =
(−3 + 12y 2 − 4y + 1 + 3 + 5y) dV
V
V ZZZ
ZZZ
Z ZσZ
ZZZ
4 3
2
2
(1 + y + 12y ) dV =
dV +
12y dV = π 3 + 4
r2 dV
=
3
V
V
V
V
ZZZ
= 36π + 4
r2 dV ,
V
where we first used that that the volume of a ball of radius R is 34 πR3 , and then that for an
odd function f (−y) = −f (y)
ZZZ
f (y) dV = 0
V
because a ball of radius R is symmetric under the reflection y → −y, and finally that
ZZZ
ZZZ
ZZZ
2
2
2
2
3y dV =
(x + y + z ) dV =
r2 dV
V
V
V
due to the symmetry between x, y, z coordinates for a sphere (rotational symmetry).
To compute the last integral we use again spherical coordinates
Z 2π Z π Z 3
ZZZ
972
4
4
2
π.
r dr sin φ dφ dθ = π35 =
r dV =
5
5
0
0
0
V
Thus the flux is
flux =
4068
π.
5
4. Let C be the triangle in the plane z = 12 y with vertices (2, 0, 0), (0, 2, 1) and (0, 0, 0) with a
counterclockwise orientation looking down the positive z-axis.
(a) Sketch the triangle and its projection
H onto the xy-plane.
(b) Use Stokes’ Theorem to evaluate C F · dr.
F(x, y, z) = (4x + 3z) i − (4y − 3x2 ) j + (3y − 2xz) k .
(c) Represent C as a union of three line segments C1 , C2 , C3 connecting the points (0, 0, 0)
and (2, 0, 0), (2, 0,H0) and (0, 2, 1), (0, 2, 1) and (0, 0, 0), respectively. Parameterize C1 , C2 and
C3 , and evaluate C F · dr.
Show the details of your work.
Solution :
(a) Below is the triangle, σ, and its projection, R, onto the xy-plane
25
2.0
1.5
1.0
2.0
0.5
1.0
1.5
0.0
0.0
1.0
0.5
0.5
0.5
1.0
1.5
2.0
0.0
0.5
1.0
1.5
(b) According to Stokes’ Theorem
ZZ
ZZ
I
∂z
∂z
(curl F) · − i −
(curl F) · n dS =
F · dr =
j + k dA ,
∂x
∂y
R
σ
C
where n is the normal vector to the surface z = 12 y looking up the positive z-axis.
To use Stokes’ Theorem we first compute
∂(4x + 3z) ∂(3y − 2xz)
∂(3y − 2xz) ∂(3x2 − 4y)
−
i+
−
j
curl F =
∂y
∂z
∂z
∂x
∂(3x2 − 4y) ∂(4x + 3z)
+
−
k
∂x
∂y
= 3 i + (3 + 2z) j + 6x k ,
and
−
Thus we have
I
C
∂z
1
∂z
i−
j + k = − j + k.
∂x
∂y
2
Z 2 Z 2−x
3
y 3
F · dr =
(6x − z − ) dA =
(6x − − ) dy dx
2
2 2
0
0
Z 2R
1
3
=
(6x(2 − x) − (2 − x)2 − (2 − x)) dx
4
2
Z0 2
2
25x
29x
25
29
13
=
(−
+
− 4) dx = − 23 + 22 − 8 =
.
4
2
12
4
3
0
ZZ
(c) The line segments C1 is parametrized as x = t, y = 0, z = 0, 0 ≤ t ≤ 2 and one gets
Z
Z 2
F · dr =
4t dt = 8 .
C1
0
26
2.0
The line segments C2 is parametrized as x = 2 − t, y = t, z = t/2, 0 ≤ t ≤ 2 and one gets
Z
Z
F · dr =
(4x + 3z) dx − (4y − 3x2 ) dy + (3y − 2xz)dz
C2
C
Z 2
Z 21
3t
1
7t2
2
(−4(2 − t) − − 4t + 3(2 − t) + (3t − (2 − t)t)) dt =
(
− 13t + 4) dt
=
2
2
2
0
0
7 3 13 2
26
=
2 − 2 +8=− .
6
2
3
The line segments C3 is parametrized as x = 0, y = 2 − t, z = 1 − 2t , 0 ≤ t ≤ 2 and one gets
Z
Z
F · dr =
(4x + 3z) dx − (4y − 3x2 ) dy + (3y − 2xz)dz
C3
C
Z 21
Z 2
3
5
(4(2 − t) − (2 − t)) dt =
=
(5 − t) dt
2
2
0
0
5
= 10 − 22 = 5 .
4
Summing up the contributions one gets
I
F · dr =
C
13
3
which agrees with the result obtained by using Stokes’ Theorem.
9
Tutorial Sheet 9
(a) Solve the following initial value problems by the Laplace transform.
(b) Sketch the input function and the solution.
Show the details of your work.
1.
y 00 + 4y = 2(cos2 t − sin2 t) ,
y(0) = 1 ,
y 0 (0) = 3 .
Solution : (a) We denote Y (s) = L(y). By using the formulae
L(y 00 ) = s2 Y (s) − sy(0) − y 0 (0) ,
L(cos2 t − sin2 t) = L(cos 2t) =
we get the algebraic equation
(s2 + 4)Y (s) − s − 3 =
2s
.
+4
s2
Solving the equation for Y , we get
Y (s) =
s2
3
s
2s
+ 2
+ 2
.
+ 4 s + 4 (s + 4)2
27
s
,
s2 + 4
Finally we use the formulas of the inverse Laplace transform
2
s
4s
−1
−1
−1
L
= sin 2t , L
= cos 2t , L
= t sin 2t ,
s2 + 4
s2 + 4
(s2 + 4)2
to get
3 t
y(t) = ( + ) sin 2t + cos 2t .
2 2
The term t sin 2t in the solution shows that the system is unstable, and the amplitude of the
oscillations increases in time.
(b) The plots of the input function and the solution are below
6
4
2
2
4
6
8
10
12
-2
-4
-6
2.
y 00 + y/4 = e−t/2 ,
y(0) = 3/2 ,
y 0 (0) = −1 .
Solution : (a) We denote Y (s) = L(y). By using the formulae
L(y 00 ) = s2 Y (s) − sy(0) − y 0 (0) ,
L(e−t/2 ) =
1
,
s + 1/2
we get the algebraic equation
(s2 + 1/4)Y (s) −
3s
1
+1=
.
2
s + 1/2
Solving the equation for Y , we get
3s
1
1
1
1 +
1 +
1
1 = − 2
2
2
2
s +4
2s + 2
(s + 2 )(s + 4 )
s +
2
s
=
,
1 −
2
s+ 2
2s + 12
Y (s) = −
1
4
+
3s
2
2s +
1
2
+
where we used the partial fraction decomposition.
Finally we use the formulae of the inverse Laplace transform
s
1
−1
−1
L
= cos t/2 , L
= e−t/2 ,
s2 + 1/4
s + 1/2
to get
y(t) = 2e−2t −
28
1
t
cos .
2
2
2
s+
1
2
−2
s − 12
s2 + 41
(b) The plots of the input function and the solution are below
1.5
1.0
0.5
5
10
15
20
25
-0.5
Note that the oscillations stabilize very quickly and become harmonic.
3.
00
y + 4y =
2t if 0 < t < 2π
,
0 if t > 2π
y(0) = 1 ,
y 0 (0) = −1 .
Solution : (a) We denote Y (s) = L(y) and represent the function on the right-hand side of the
equation as
r(t) = 2t (u(t) − u(t − 2π)) .
By using the formulae
L(y 00 ) = s2 Y (s)−sy(0)−y 0 (0) , L(f (t−a)u(t−a)) = e−as F (s) ⇒ L(r) =
2 2e−2πs 4πe−2πs
− 2 −
,
s2
s
s
we get the algebraic equation
(s2 + 4)Y (s) − s + 1 =
2e−2πs 4πe−2πs
2
−
−
.
s2
s2
s
Solving the equation for Y , we get
Y (s) = −
Then we use
1
s
2
2e−2πs
4πe−2πs
+
+
−
−
.
s2 + 4 s2 + 4 s2 (s2 + 4) s2 (s2 + 4) s(s2 + 4)
4
1
1
= 2− 2
,
+ 4)
s
s +4
s2 (s2
4
1
s
= − 2
,
+ 4)
s s +4
s(s2
and get
Y (s) =
1
3/2
s
πe−2πs e−2πs
e−2πs
πse−2πs
−
+
−
−
+
+
.
2s2 s2 + 4 s2 + 4
s
2s2
2(s2 + 4)
s2 + 4
Finally we use the formulae of the inverse Laplace transform
2
s
−1
−1
L
= sin 2t , L
= cos 2t , L−1 e−as F (s) = f (t − a)u(t − a) ,
2
2
s +4
s +4
29
to get
y(t) =
t 3
− sin 2t + cos 2t if 0 < t < 2π ,
2 4
and
y(t) =
t 3
1
− sin 2t + cos 2t − π − (t − 2π)/2 + sin 2(t − 2π) + π cos 2(t − 2π) if t > 2π .
2 4
4
Simplifying the expression above, we get
t 3
− 4 sin 2t + cos 2t if 0 < t < 2π
2
.
y(t) =
(1 + π) cos 2t − 21 sin 2t if t > 2π
(b) The plots of the input function and the solution are shown below.
10
5
2
10
4
6
8
10
12
Tutorial Sheet 10
1. Consider the following initial value problem
y 00 + 9 y = −9 u(t − π) + 6 δ(t − 2π) ,
y(0) = 1 ,
y 0 (0) = 0 .
(a) Plot the function
1
1 −9 u(t − π) + 6 u(t − 2π + ) − u(t − 2π − )
2
2
(b) Solve the initial value problem by the Laplace transform.
(c) Plot the solution.
Show the details of your work.
Solution :
(a) The plot of the input function is shown below.
30
5
10
15
-2
-4
-6
-8
(b) We denote Y (s) = L(y), and then using the formulae
e−as
L(y ) = s Y (s) − sy(0) − y (0) , L(u(t − a)) =
, L(δ(t − a)) = e−as ,
s
00
0
2
we get the algebraic equation
(s2 + 9)Y (s) = −9
e−πs
+ 6e−2πs + s .
s
Solving the equation for Y , we get
Y (s) = −
Then we use the formula
−
9e−πs
6e−2πs
s
+
+
.
s(s2 + 9) s2 + 9 s2 + 9
1
s
9
=− + 2
,
+ 9)
s s +9
s(s2
and write Y (s) in the form
Y (s) = −e−πs
1
s
6
s
+ e−πs 2
+ e−2πs 2
+ 2
.
s
s +9
s +9 s +9
Finally we use the formulas of the inverse Laplace transform
1
s
ω
−1
−1
−1
= cos ωt , L
= sin ωt ,
L
= 1, L
s
s2 + ω 2
s2 + ω 2
L−1 e−as F (s) = f (t − a)u(t − a) ,
(10.6)
to get
y(t) = −u(t − π) + cos 3(t − π)u(t − π) + 2 sin 3(t − 2π)u(t − 2π) + cos 3t

 cos 3t if 0 < t < π
−1 if π < t < 2π
=
.

−1 + 2 sin 3t if t > 2π
(c) The plot of the solution is shown below.
31
(10.7)
1
2
4
6
8
10
12
-1
-2
-3
2. Applying convolution,
(a) find the solution.
(b) sketch the input function and the solution.
Show the details of your work.
6 if 0 < t < 1
00
0
y + 3y + 2y =
,
0 if t > 1
y(0) = 2 ,
y 0 (0) = −2 .
Solution : We denote Y (s) = L(y) and represent the function on the right-hand side of the
equation as
r(t) = 6 u(t) − u(t − 1) .
By using the formulas
L(y 0 ) = sY (s)−y(0) , L(y 00 ) = s2 Y (s)−sy(0)−y 0 (0) , L(u(t−a)) =
6
e−s
e−as
⇒ L(r) = −6
,
s
s
s
we get the algebraic equation
(s2 + 3s + 2)Y (s) − 2s − 4 =
6
e−s
−6
s
s
⇒
Y (s) = 2(s + 2)Q(s) + R(s)Q(s) ,
where
R(s) =
e−s
6
−6
,
s
s
Q(s) =
1
1
1
1
=
=
−
.
s2 + 3s + 2
(s + 1)(s + 2)
s+1 s+2
By using the convolution theorem we get the integral representation
Z t
0
y(t) = 2q (t) + 4q(t) +
q(t − τ )r(τ ) dτ ,
0
where
q(t) = L−1 (Q(s)) = e−t −e−2t , q 0 (t) = L−1 (sQ(s)) = −e−t +2e−2t , r(t) = L−1 (R(s)) = 6 u(t)−u(t−1) .
Computing the integral, we get
Z t
Z t
Z t
q(t−τ )r(τ ) dτ = 6
q(t−τ ) dτ = 6
(e−t+τ −e−2t+2τ ) dτ = 3−6e−t +3e−2t
0
0
0
32
if 0 < t < 1 ,
Z
t
1
Z
(e−t+τ −e−2t+2τ ) dτ = 6e−t (e−1)−3e−2t (e2 −1) if t > 1 ,
q(t−τ ) dτ = 6
q(t−τ )r(τ ) dτ = 6
0
0
0
1
Z
0
Adding q (t) + q(t), we get


2e−t
if t < 0
3 − 4e−t + 3e−2t
if 0 < t < 1
y(t) =
,

2−2t
1−t
−2t
−t
−3e
+ 6e + 3e − 4e
if t > 1
The plot of the input function and the solution is shown below.
6
5
4
3
2
1
-0.5
0.5
1.0
1.5
2.0
2.5
3.0
3. Using Laplace transforms, solve the integral equation. (Show the details of your work.)
Z t
y(τ ) sin(t − τ ) dτ .
y(t) = − sin t +
0
Solution : We denote Y (s) = L(y) and represent the equation in the following form
y = − sin t + y ∗ sin(t) .
By the convolution theorem,
Y (s) = −
s2
1
1
+ Y (s) 2
.
+1
s +1
Solving for Y (s), we obtain
Y (s) = −
1
,
s2
and therefore
y(t) = −t .
33