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

4k

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.

Since D(ln(2 + t3)) = (− 3√

2 ,∞ ), D(3 + t3) = (−∞,∞) and D(− (t3+2)2

4) = (−∞,∞), one gets

D(r) = (− 3√

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)

dr

dt=

{3t2

t3 + 2, 3t2,−3t5

2− 3t2

}.

(b) The magnitude of this vector is

||drdt|| =

√(3t2

2 + t3

)2

+ (3t2)2 +

(3t2 +

3t5

2

)2

= 3t2

√(1

2 + t3

)2

+ 1 +

(1 +

t3

2

)2

= 3t2

√(1

2 + t3

)2

+ 21

2 + t32 + t3

2+

(2 + t3

2

)2

= 3t2

√(1

2 + t3+

2 + t3

2

)2

= 3t2(

1

2 + t3+

2 + t3

2

),

because t3 > −2.(c) The unit tangent vector is

T =drdt

||drdt||

=

{2

t6 + 4t3 + 6,

2 (t3 + 2)

t6 + 4t3 + 6,−t6 − 4t3 − 4

t6 + 4t3 + 6

}.

1

3. Find the vector equation of the line tangent to the graph of r(t) at the point P0(0, 2,−14)

on the curve.Solution: The point P0(0, 2,−1

4) on the curve corresponds to t = −1. We find

r0 = r(−1) = 2j − 1

4k , v0 =

dr

dt(−1) = 3i + 3j− 3

2k .

Thus the tangent line equation is

r = r0 + (t+ 1)v0 = 3(t+ 1) i + (3t+ 5) j− 1

4(6t+ 7)k .

Note that the same line is also described by the following equation which is obtained from theone above by the rescaling and shift of the parameter t: t→ t

3− 1

r = r0 + tv0 = t i + (t+ 2) j− 1

4(2t+ 1)k .

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

L =

∫ 1

−1||drdt|| dt =

∫ 1

−13t2(

1

2 + t3+

2 + t3

2

)dt =

∫ 1

−1

(1

2 + v+

2 + v

2

)dv

=

(ln(2 + v) +

(2 + v)2

4

) ∣∣∣1−1

= 2 + ln 3 ,

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 thecurve having r(1) as its reference point.

Solution: The arc length parameter s can be found as follows

s =

∫ t

1

||drdu|| du = =

∫ t

1

3u2(

1

2 + u3+

2 + u3

2

)du =

∫ t3

1

(1

2 + v+

2 + v

2

)dv

=

(ln(2 + v) +

(2 + v)2

4

) ∣∣∣t31

=t6

4+ t3 − 5

4+ log

(1

3

(t3 + 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+ 3y2 − 6y , k = −11,−9,−8 .

Solution

-2.5 -2.0 -1.5 -1.0

0.5

1.0

1.5

2.0

The level curve equation 2x2 + 8x+ 3y2 − 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 = −11the level curve is just the point (−2, 1).

2. Consider the functionz = 2ey−

π2 sinx− 3ex−

π4 cos y

(i) Find

a)∂z

∂x(π

4,π

2) , b)

∂z

∂y(π

4,π

2) , c)

∂2z

∂x∂y(π

4,π

2) , d)

∂2z

∂y∂x(π

4,π

2) .

Solution:

a)∂z

∂x= 2ey−

π2 cosx− 3ex−

π4 cos y ⇒ ∂z

∂x(π

4,π

2) =√

2 .

b)∂z

∂y= 2ey−

π2 sinx+ 3ex−

π4 sin y ⇒ ∂z

∂y(π

4,π

2) = 3 +

√2 .

c)∂2z

∂x∂y=

∂

∂x

∂z

∂y=

∂

∂x

(2ey−

π2 sinx+ 3ex−

π4 sin y

)= 2ey−

π2 cosx+3ex−

π4 sin y ⇒ ∂2z

∂x∂y(π

4,π

2) = 3+

√2 .

d)∂2z

∂y∂x=

∂

∂y

∂z

∂x=

∂

∂y

(2ey−

π2 cosx− 3ex−

π4 cos y

)= 2ey−

π2 cosx+3ex−

π4 sin y ⇒ ∂2z

∂y∂x(π

4,π

2) = 3+

√2 .

3

(ii) Find the slope of the surface z = 2ey−π2 sinx− 3ex−

π4 cos y

a) in the x-direction at the point (π2, π6);

Solution: The slope kx is equal to

kx =∂z

∂x(π

2,π

6) = −3

√3

2eπ4 .

b) in the y-direction at the point (π3, π).

Solution: The slope ky is equal to

ky =∂z

∂y(π

3, π) =

√3e

π2 .

(iii) Show that the function z = 2ey−π2 sinx− 3ex−

π4 cos y satisfies Laplace’s equation

∂2z

∂x2+∂2z

∂y2= 0 .

Solution: To this end we compute the following derivatives

∂2z

∂x2=

∂

∂x

∂z

∂x=

∂

∂x

(2ey−

π2 cosx− 3ex−

π4 cos y

)= −2ey−

π2 sinx− 3ex−

π4 cos y ,

∂2z

∂y2=

∂

∂y

∂z

∂y=

∂

∂y

(2ey−

π2 sinx+ 3ex−

π4 sin y

)= 2ey−

π2 sinx+ 3ex−

π4 cos y .

The sum of these two expressions is obviously 0.

3. Show that the local linear approximation of the function

f(x, y) = 2xαyβ − xβ

yα

at (1, 1) isf(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) = 2αxα−1yβ − β xβ−1

yα⇒ fx(1, 1) = 2α− β ,

fy(x, y) = 2β xαyβ−1 + αxβ

yα+1⇒ fy(1, 1) = α + 2β ,

which proves the formula.

4

4. Use appropriate forms of the chain rule to find ∂z∂u

and ∂z∂v

where

z = cosx

2sin 2y ; x = 3u− 2v , y = u2 − 2v3 .

Solution: We have

∂z

∂u=∂z

∂x

∂x

∂u+∂z

∂y

∂y

∂u= −1

2sin

x

2sin 2y × 3 + 2 cos

x

2cos 2y × 2u

= −3

2sin

x

2sin 2y + 4u cos

x

2cos 2y

= −3

2sin

3u− 2v

2sin 2(u2 − 2v3) + 4u cos

3u− 2v

2cos 2(u2 − 2v3) ,

∂z

∂v=∂z

∂x

∂x

∂v+∂z

∂y

∂y

∂v= −1

2sin

x

2sin 2y × (−2) + 2 cos

x

2cos 2y × (−2v2)

= sinx

2sin 2y − 4v2 cos

x

2cos 2y

= sin3u− 2v

2sin 2(u2 − 2v3)− 4v2 cos

3u− 2v

2cos 2(u2 − 2v3) .

3 Tutorial Sheet 3

1. Consider the function

f(x, y, z) =√z2 + 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 pointP .

(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

fx(x, y, z) =−6 sin(3x− 2y)− 1

2√

2 cos(3x− 2y)− x+ y + z2⇒ fx(2, 3,−1) = −1

4,

fy(x, y, z) =4 sin(3x− 2y) + 1

2√

2 cos(3x− 2y)− x+ y + z2⇒ fy(2, 3,−1) =

1

4,

fz(x, y, z) =z√

2 cos(3x− 2y)− x+ y + z2⇒ fz(2, 3,−1) = −1

2.

5

Thus, the gradient and its magnitude are equal to

∇f(2, 3,−1) =

(−1

4,1

4,−1

2

), ||∇f(2, 3,−1)|| =

√6

4≈ 0.612372 .

Therefore, the unit vector in the direction of the gradient is

u =4√6

(−1

4,1

4,−1

2

)≈ (−0.408248, 0.408248,−0.816497) .

(b) The projection of the vector u onto the xz-plane is the vector

uxz =4√6

(−1

4, 0 ,−1

2

)≈ (−0.408248 , 0 ,−0.816497) .

It is shown below

-0.4-0.3-0.2-0.1x

-0.8

-0.6

-0.4

-0.2

z

(c) f decreases most rapidly in the direction opposite to its gradient, so the unit vector is

v = − 4√6

(−1

4,1

4,−1

2

)≈ (0.408248,−0.408248, 0.816497) .

(d) The projection of the vector v onto the yz-plane is the vector

vyz = − 4√6

(0 ,

1

4,−1

2

)≈ (0 ,−0.408248, 0.816497) .

It is shown below

-0.4-0.3-0.2-0.1y

0.2

0.4

0.6

0.8

z

6

(e) The rate of change of f at P in the direction of u is equal to

||∇f(2, 3,−1)|| =√

6

4≈ 0.612372 ,

and the rate of change of f at P in the direction of v is equal to

−||∇f(2, 3,−1)|| = −√

6

4≈ −0.612372 .

2. Let r =√x2 + y2. Compute

∇f(r) , where f(r) = ln(1 + r) .

Solution: We have

∇f(r) =∂f(r)

∂xi +

∂f(r)

∂yj ,

∂f(r)

∂x=d ln(1 + r)

dr

∂r

∂x=

1

1 + r

x

r,

∂f(r)

∂y= f

d ln(1 + r)

dr

∂r

∂y=

1

1 + r

y

r,

Thus

∇f(r) =1

1 + r

x

ri +

1

1 + r

y

rj =

1

1 + r

r

r,

where r = xi + yj.

3. Consider the surface

z = f(x, y) = ln3√

2x2 − 3xy2 + 3 cos(2x+ 3y)− 3y3 + 18

2

(a) Find an equation for the tangent plane to the surface at the point P (3,−2, z0) wherez0 = 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

z = ln3√

2x2 − 3xy2 + 3 cos(2x+ 3y)− 3y3 + 18

2=

1

3ln(2x2−3xy2+3 cos(2x+3y)−3y3+18

)−ln 2 ,

7

and compute z0

z0 = z|x=3,y=−2 =1

3ln(3 cos(0) + 18 + 24− 36 + 18

)− ln 2 = ln

3

2≈ 0.405465

Then, we compute the partial derivatives at P (3,−2, z0)

∂z

∂x=

−6 sin(2x+ 3y) + 4x− 3y2

3 (2x2 − 3xy2 + 3 cos(2x+ 3y)− 3y3 + 18)⇒ ∂z

∂x|x=3,y=−2 = 0 .

∂z

∂y=

−6xy − 9 sin(2x+ 3y)− 9y2

3 (2x2 − 3xy2 + 3 cos(2x+ 3y)− 3y3 + 18)⇒ ∂z

∂y|x=3,y=−2 = 0 .

The tangent plane equation is given by

z = z0 + 0(x− 3) + 0(y + 2) = ln3

2.

(b) It is a plane through the point (3,−2, ln 32) parallel to the xy-plane, see the picture.

-3.0

-2.5

-2.0

-1.5

-1.0

y

2.0

2.5

3.0

3.5

4.0

x

0.0

0.2

0.4

0.6

0.8

z

-3.0

-2.5

-2.0

-1.5

-1.0

y

2.0

2.5

3.0

3.5

4.0

x

-1

0

1

2

z

(c) The normal line to the surface (and the tangent plane) is given by

r = 3i− 2j + ln3

2k + t

(0i + 0j + k

)= 3i− 2j + (t+ ln

3

2)k .

Notice that after shifting t→ t− ln 32, it becomes

r = 3i− 2j + tk .

(d) It is parallel to the z-axis, see the picture above.

8

4. Consider the functionf(x, y) = x4 − x2y + y2 − 3y + 4

Locate all relative maxima, relative minima, and saddle points, if any.

Solution: The graph of the function is shown below

�1

0

1 x

0.5

1.0

1.5

2.0

2.5

y

1.0

1.5

2.0

2.5

3.0

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

y =x2

2+

3

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

(x = 0 , y =3

2) , (x = −1 , y = 2) , (x = 1 , y = 2) .

Computing the values of f at critical points, we get

f(0,3

2) =

7

4, f(−1, 2) = 1 , f(1, 2) = 1 .

9

To find out if they are maximum, minimum or saddle points we use the second derivativetest. To this end we compute

∂2f

∂x2(x, y) = 12x2 − 2y ,

∂2f

∂y2(x, y) = 2 ,

∂2f

∂x∂y(x, y) = −2x ,

and

D(x, y) =∂2f

∂x2∂2f

∂y2−(∂2f

∂x∂y

)2

= 20x2 − 4y ,

Computing D and ∂2f∂x2

for the three critical points, we get

D(0,3

2) = −6 ,

∂2f

∂x2(0,

3

2) = −3 ,

and therefore (0 , 32) is a saddle point.

D(−1, 2) = 12 ,∂2f

∂x2(−1, 2) = 8 ,

and therefore (−1, 2) is a relative minimum.

D(1, 2) = 12 ,∂2f

∂x2(1, 2) = 8 ,

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 belowThe volume is given by the following double integral

V =

∫∫R

(3πex sin y + e−x) dA =

∫ ln 3

0

[∫ π

0

(3πex sin y + e−x) dy

]dx

=

∫ ln 3

0

(−3πex cos y + e−xy)∣∣π0dx =

∫ ln 3

0

(6πex + πe−x)dx = π (6ex − e−x)∣∣ln 3

0

=(6eln 3 − e− ln 3 − (6e0 − e0)

)π =

(18− 1

3− (6− 1)

)π =

38

3π

Thus the volume is equal to

V =38

3π ≈ 39.7935 .

10

0.0

0.5

1.0 0

1

2

3

0

10

20

2. Consider the solid in the first octant bounded by the surface

z =e1−x

1− xy +

cos πx

1− xy,

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

limx→1

( e1−x1− x

y +cosπx

1− xy).

Show the details of your work.Solution:(a) Below is the projection of the solid onto the xy-plane, and the solid.

0.2 0.4 0.6 0.8 1.0

0.2

0.4

0.6

0.8

1.0

11

(b) Therefore, the volume is given by the following double integral

V =

∫∫R

(e1−x

1− xy +

cosπx

1− xy

)dA =

∫ 1

0

[∫ √xx

(e1−x

1− x+

cosπx

1− x

)y dy

]dx

=

∫ 1

0

(e1−x

1− x+

cos πx

1− x

)y2

2

∣∣∣∣√x

x

dx =1

2

∫ 1

0

(e1−x + cos πx

)x dx .

The remaining integral is computed by using integration by parts∫ 1

0

x e1−x dx =

∫ 1

0

(d(−x e1−x) + e1−x dx

)= −x e1−x |10 +

∫ 1

0

e1−x dx = −1− e1−x |10 = −2 + e ,

and ∫ 1

0

x cos πx dx =

∫ 1

0

(d( 1

πx sin πx

)− 1

πsin πx dx

)=

1

πx sin πx |10 −

∫ 1

0

1

πsin πx dx

= 0 +1

π2cos πx |10 = − 2

π2.

Thus the volume is equal to

V =e

2− 1− 1

π2≈ 0.25782

(c) We have

limx→1

( e1−x1− x

y +cosπx

1− xy)

= limt→0

(ett

+cos π(1− t)

t

)y = lim

t→0

(ett− cosπt

t

)y

= limt→0

(1 + t+ t2

2

t−

1− π2t2

2

t

)= 1 ,

where we used the Taylor expansions

et ≈ 1 + t+t2

2+t3

6+ · · · , cos t = 1− t2

2+t4

24+ · · · .

3. Consider the solid inside the surface r2 + z2

9= 1

9and outside the surface r = 1

4.

Here r2 = x2 + y2.(a) What is the surface r2 + z2

9= 1

9?

(b) What is the surface r = 14?

(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:(a) The surface r2 + z2

9= 1

9is an ellipsoid centred at the origin. Its equation can be written

in rectangular coordinates as32x2 + 32y2 + z2 = 1 ,

12

which shows that it is an prolate spheroid (like a rugby ball).(b) The surface r = 1

4is a cylinder of radius 1

4also 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 13.

Below is the projection of the solid onto the xy-plane, and the solid.

-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

V = 2

∫∫R

√1− 9r2 dA = 2

∫ 2π

0

[∫ 1/3

1/4

√1− 9r2 rdr

]dθ = 4π

∫ 1/3

1/4

√1− 9r2

1

2d(r2)

= −4π1

3

1

9(1− 9r2)3/2

∣∣∣∣1/31/4

=4π

27

(7

16

)3/2

=7π

432

√7 .

Thus, the volume of the solid is equal to

V =7π

432

√7 ≈ 0.134683

5 Tutorial Sheet 5

1. Consider the portion of the surface (y − 2)2 + (z + 1)2 = 12 that is above the rectangle

R = {(x, y) : −√

3

2≤ x ≤

√3 , −1 ≤ y ≤ 5}.

(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

13

-0.5 0.5 1.0 1.5

-1

1

2

3

4

5

c) The area is given by the formula

S =

∫∫R

√1 +

(∂z

∂x

)2

+

(∂z

∂y

)2

dA =

∫ √3−√3/2

∫ 5

−1

√1 +

(∂z

∂x

)2

+

(∂z

∂y

)2

dy dx ,

wherez = −1 +

√12− (y − 2)2

Computing the derivatives we get√1 +

(∂z

∂x

)2

+

(∂z

∂y

)2

=

√1 +

(y − 2)2

12− (y − 2)2=

√12√

12− (y − 2)2.

Thus, we get

S =

∫ √3−√3/2

∫ 5

−1

√12√

12− (y − 2)2dy dx = 9

∫ 5

−1

1√12− (y − 2)2

dy = 9

∫ 3

−3

1√12− y2

dy

= 18

∫ 3

0

1√12− y2

dy .

To compute the integral we do the substitution

y =√

12 sin t = 2√

3 sin t , dy =√

12 cos t dt ,√

12− y2 =√

12 cos t , 0 ≤ t ≤ π

3

and get

S = 18

∫ π/3

0

dt = 6π .

2. Consider the solid bounded by the surface z =√

4y and the planesx+ 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:

14

0.0

0.5

1.0

1.5

2.00.0

0.51.0

1.52.0

0

1

2

0.5 1.0 1.5 2.0

0.5

1.0

1.5

2.0

(a) The solid, G, and its projection, R, onto the xy-plane are shown below(b) Thus, the volume is equal to

V =

∫∫∫G

dV =

∫∫R

∫ √4y0

dz dA = 2

∫ 2

0

∫ 2−x

0

√y dy dx

= 2

∫ 2

0

2

3(2− x)3/2 dx = −4

3

2

5(2− x)5/2

∣∣∣∣20

=32

15

√2 .

3. Consider the lamina with density δ(x, y) = 4π − 5y bounded byy = sin πx

3, y = 0 , x = 0, and x = 3.

(a) Sketch the lamina .(b) Find the mass and centre of gravity of the lamina.Solution:(a) The lamina, R, is shown below

0.5 1.0 1.5 2.0 2.5 3.0

0.2

0.4

0.6

0.8

1.0

15

(b) Its mass is equal to

M =

∫∫R

δ(x, y) dA =

∫ 3

0

∫ sin πx3

0

(4π − 5y) dy dx =

∫ 3

0

(4π sinπx

3− 5

2sin2 πx

3) dx

=

∫ 2

0

(4π sinπx

3− 5

4+

5

4cos

2πx

3) dx = (−12 cos

πx

3− 5

4x+

15

4πsin

2πx

3)|30

= 12− 15

4+ 12 =

81

4≈ 20.25 .

By the symmetry of the lamina the x-coordinate of its centre of gravity is equal to

xcg =3

2.

The y-coordinate is given by

ycg =1

M

∫∫R

y δ(x, y) dA =4

81

∫ 2

0

∫ sin πx3

0

y(4π − 5y) dy dx =4

81

∫ 3

0

(2π sin2 πx

3− 5

3sin3 πx

3) dx

=4

81π

∫ 3

0

(1− cos2πx

3) dx+

4

81

5

3

3

π

∫ 3

0

(1− cos2πx

3) d(cos

πx

3)

=4π

27+

20

81π(cos

πx

3− 1

3cos3

πx

3)

∣∣∣∣30

=4π

27+

20

81π(−2 +

2

3) =

4π

27− 80

243π.

Thus, the coordinates of the centre of gravity of the lamina are

xcg = 1 , ycg =4π

27− 80

243π≈ 0.360628 .

6 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 + y2 + z2 = 1 and x2 + y2 + z2 = 9 and

below by the surface z = 0.

1. What is the surface x2 + y2 + z2 = 1?

2. What is the surface x2 + y2 + z2 = 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 itsdensity is

δ(x, y, z) =e−√x2+y2+z2

x2 + y2 + z2.

16

Show the details of your work.Solution :(a) We have

x = r cos θ sinφ , y = r sin θ sinφ , z = r cosφ , 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, andθ 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.(b) 4. The plot is shown below

-3 -2 -1 0 1 2 3

-3

-2

-1

0

1

2

3

(b) 5. We use the spherical coordinates to get

V =

∫∫∫G

dV =

∫ 2π

0

(∫ π/2

0

[∫ 3

1

r2dr

]sinφ dφ

)dθ =

2π

3(33 − 13) =

52π

3≈ 54.4543

(b) 6. We use the spherical coordinates to get

M =

∫∫∫G

δ(x, y, z)dV =

∫ 2π

0

(∫ π/2

0

[∫ 3

1

e−r

r2r2dr

]sinφ dφ

)dθ

= 2π

∫ 3

1

e−rdr = 2π(e−1 − e−3) ≈ 1.99863 (6.1)

17

2.

(a) Express rectangular coordinates in terms of cylindrical coordinates. Draw the correspond-ing picture.

(b) Consider the solid G bounded by the surfaces x2 + y2 = 1 and x2 + y2 = 4, above by thesurface z = 5− x2 − y2, and below by the surface z = 0.

i. What are the surfaces x2 + y2 = 1 and x2 + y2 = 4?

ii. What is the surface z = 5− x2 − y2?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−x2−y2−z .

Show the details of your work.Solution :(a) We have

x = r cos θ , 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(b) iv. We use the cylindrical coordinates to get

V =

∫∫∫G

dV =

∫ 2π

0

(∫ 2

1

[∫ 5−r2

0

rdz

]dr

)dθ = 2π

∫ 2

1

r(5− r2)dr = 2π(5

2r2 − 1

4r4)|21

= 2π(10− 4− 5

2+

1

4) =

15

2π ≈ 23.5619 .

(6.2)

18

-2 -1 0 1 2

-2

-1

0

1

2

(b) v. We use the cylindrical coordinates to get

M =

∫∫∫G

δ(x, y, z)dV =

∫ 2π

0

(∫ 2

1

[∫ 5−r2

0

e5−r2−zrdz

]dr

)dθ

= 2π

∫ 2

1

(e5−r2 − 1)rdr = π(−e5−r2 − r2)|21 = π(e4 − e− 3) ≈ 153.561 .

7 Tutorial Sheet 7

1. Consider the surface that extends upward from the semicircle y =√

4− x2 in the xy-planeto the surface

z =1

12y2 +

π

4x2y

(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(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 , 0 ≤ t ≤ π .

The area of the surface is given by the following line integral

A =

∫C

z ds =

∫C

(1

12y2 +

π

4x2y) ds .

Thus, we get

A =

∫ π

0

(1

12y2 +

π

4x2y)

√(dx

dt

)2

+

(dy

dt

)2

dt

=

∫ π

0

(1

3sin2 t+ 2π cos2 t sin t)

√(−2 sin t)2 + (2 cos t)2 dt

=1

3

∫ π

0

(1− cos 2t) dt+ 4π

∫ π

0

cos2 t sin t dt =π

3− 4π

3

∫ π

0

cos2 t d cos t =π

3−4π

3cos3 t

∣∣∣∣π0

= 3π .

19

2.

(a) Let the path C between the points (−π/2 , π) and (3π/2 ,−2π/3) be a curve formed fromtwo 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

I1 ≡∫C1

(−6xy + 3π3 sin 3x) dx− (3x2 + 2π3 cos3y

2) dy .

iii. Parameterise C2, and evaluate

I2 ≡∫C2

∫C

(−6xy + 3π3 sin 3x) dx− (3x2 + 2π3 cos3y

2) dy .

iv. Compute the sum I = I1 + I2.

(b) Show that for any integration path C the integral above depends only on the initial andterminal 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 anintegration 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

-1 1 2 3 4 5

-2

-1

1

2

3

(a) ii. The line segments C1 is parametrized as x = t, y = π,−π/2 ≤ t ≤ 3π/2 and one gets

I1 =

∫C1

(−6xy + 3π3 sin 3x) dx− (3x2 + 2π3 cos3y

2) dy =

∫ 3π/2

−π/2(−6πt+ 3π3 sin 3t) dt

= (−3πt2 − π3 cos 3t)|3π/2−π/2 = −27

4π3 +

3

4π3 − π3 cos

9π

2+ π3 cos

π

2= −6π3 ≈ −186.038 .

(7.3)(a) iii. The line segments C2 is parametrized as x = 3π/2, y = t, π ≥ t ≥ −2π/3 and one gets

I2 =

∫C2

(−6xy + 3π3 sin 3x) dx− (3x2 + 2π3 cos3y

2) dy =

∫ −2π/3π

−(27

4π2 + 2π3 cos

3t

2) dt

=9

2π3 +

4

3π3 sin π +

27

4π3 +

4

3π3 sin

3π

2=

45

4π3 − 4

3π3 =

119

12π3 ≈ 307.479 .

(7.4)(a) iv. The sum gives I = 47

12π3 ≈ 121.441.

(b) We have

f(x, y) = −6xy + 3π3 sin 3x , g(x, y) = −(3x2 + 2π3 cos3y

2) .

Thus∂yf(x, y) = −6x , ∂xg(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

∂x= −6xy + 3π3 sin 3x ⇒ φ(x, y) = −3x2y − π3 cos 3x+ C(y) .

To find C(y) we use that

∂φ

∂y= −3x2 +

dC(y)

dy= −(3x2 + 2π3 cos

3y

2) ⇒ dC(y)

dy= −2π3 cos

3y

2

⇒ C(y) = −4

3π3 sin

3y

2+ C .

(7.5)

Thus, we get

φ(x, y) = −3x2y − π3 cos 3x− 4

3π3 sin

3y

2+ C .

21

(d) By using the formula, we obtain∫ (3π/2 ,−2π/3)

(−π/2 ,π)(−6xy + 3π3 sin 3x) dx− (3x2 + 2π3 cos

3y

2) dy = φ(x, y)|(3π/2 ,−2π/3)(−π/2 ,π) =

47

12π3 ,

which agrees with the result in (a)iv.

3. Consider the integral ∮C

(e−2x + 5x2y − 2y2) dx− (3ey − 3x3) dy .

Assume that the curve C is oriented counterclockwise, and it is the boundary of the region Rbetween 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

0.5 1.0 1.5 2.0

0.5

1.0

1.5

2.0

(b) By Green’s Theorem we have∮C

(e−2x + 5x2y − 2y2) dx− (3ey − 3x3) dy =

∫∫R

(∂(−3ey + 3x3)

∂x− ∂(e−2x + 5x2y − 2y2)

∂y

)dA

=

∫∫R

(4x2 + 4y

)dA =

∫ 2

0

∫ x

x2/2

(4x2 + 4y

)dy dx =

∫ 2

0

(4x2(x− x2

2) + 2(x2 − x4

4)

)dx

= (2

3x3 + x4 − 1

2x5)

∣∣∣∣20

=16

3+ 16− 16 =

16

3.

22

8 Tutorial Sheet 8

1. Consider the lamina that is the portion of the surface x2 + y2− 3z− 1 = 0 inside the surfacex2 + y2 = 9/4. The density of the lamina is

δ(x, y, z) = δ0(9

4+ x2 + y2)

where δ0 is a constant(a) What is the surface x2 + y2 − 3z − 1 = 0?(b) What is the surface x2 + y2 = 9

4?

(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 + y2 − 3z − 1 = 0 is a paraboloid.(b) The surface x2 + y2 = 9

4is 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 0.5 1.0 1.5

-1.5

-1.0

-0.5

0.5

1.0

1.5

The projection of the surface onto the xy-plane is the disk R: x2 + y2 ≤ 94.

(d) The mass of the lamina is equal to the following surface integral

M =

∫∫σ

δ(x, y, z) dS = δ0

∫∫R

(9

4+x2+y2)

√1 +

(∂z

∂x

)2

+

(∂z

∂y

)2

dA =2

3δ0

∫∫R

(9

4+x2+y2)3/2 dA .

The simplest way to compute the double integral over R is to use the polar coordinates

M =2

3δ0

∫∫R

(9

4+ x2 + y2)3/2 dA =

4π

3δ0

∫ 3/2

0

(9

4+ r2)3/2 rdr =

4π

3δ0

1

5(9

4+ r2)5/2

∣∣∣∣3/20

=4π

15δ0

35

25(4√

2− 1) =81π

40δ0 (4√

2− 1) ≈ 29.6256 δ0

2. Consider the surface σ which is the portion of the surface z = x2+y2−13

below the planez = 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

9y j + 4k ;

23

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 + y2 ≤ 94.

(b) Taking into account that through the surface z = f(x, y) oriented by down, the vectorfield F(x, y, z) = M i +N j + P k has

flux =

∫∫σ

F · n dS =

∫∫R

(M∂f

∂x+N

∂f

∂y− P

)dA ,

we get by using the polar coordinates (M = 0 , N = 169y , P = 4)

flux =

∫∫R

(32

27y2 − 4

)dA =

∫ 2π

0

∫ 3/2

0

(32

27r2 sin2 φ− 4) rdr dφ

=

∫ 2π

0

(3 sin2 φ− 9

2)dφ =

3

2π − 9π = −15

2π .

3.(a) Express rectangular coordinates in terms of spherical coordinates. Draw the correspond-

ing 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+ 12xy2 + 4z3) i− (2y2 − y + 4x2) j + (4 + 3z + 5yz − 4y3)k

across the surface σ with outward orientation where σ is given by the equationx2 + y2 + z2 = 9.Show the details of your work.Solution :(a) We have, see the picture below

x = r cos θ sinφ , y = r sin θ sinφ , z = r cosφ , 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, andθ 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)

V =

∫∫∫V

dV =

∫ 2π

0

(∫ π

0

[∫ R

0

r2dr

]sinφ dφ

)dθ =

4

3πR3 .

(c) The surface is a sphere of radius 3. We have

flux =

∫∫σ

F · n dS =

∫∫∫V

divF dV =

∫∫∫V

(−3 + 12y2 − 4y + 1 + 3 + 5y) dV

=

∫∫∫V

(1 + y + 12y2) dV =

∫∫∫V

dV +

∫∫∫V

12y2 dV =4

3π 33 + 4

∫∫∫V

r2 dV

= 36π + 4

∫∫∫V

r2 dV ,

where we first used that that the volume of a ball of radius R is 43πR3, and then that for an

odd function f(−y) = −f(y) ∫∫∫V

f(y) dV = 0

because a ball of radius R is symmetric under the reflection y → −y, and finally that∫∫∫V

3y2 dV =

∫∫∫V

(x2 + y2 + z2) dV =

∫∫∫V

r2 dV

due to the symmetry between x, y, z coordinates for a sphere (rotational symmetry).To compute the last integral we use again spherical coordinates∫∫∫

V

r2 dV =

∫ 2π

0

(∫ π

0

[∫ 3

0

r4dr

]sinφ dφ

)dθ =

4

5π35 =

972

5π .

Thus the flux is

flux =4068

5π .

4. Let C be the triangle in the plane z = 12y 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 onto the xy-plane.(b) Use Stokes’ Theorem to evaluate

∮CF · 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, 0) and (0, 2, 1), (0, 2, 1) and (0, 0, 0), respectively. Parameterize C1, C2 andC3, and evaluate

∮CF · dr.

Show the details of your work.Solution :(a) Below is the triangle, σ, and its projection, R, onto the xy-plane

25

0.0

0.5

1.0

1.5

2.0 0.0

0.5

1.0

1.5

2.0

0.0

0.5

1.0

0.5 1.0 1.5 2.0

0.5

1.0

1.5

2.0

(b) According to Stokes’ Theorem∮C

F · dr =

∫∫σ

(curlF) · n dS =

∫∫R

(curlF) ·(−∂z∂x

i− ∂z

∂yj + k

)dA ,

where n is the normal vector to the surface z = 12y looking up the positive z-axis.

To use Stokes’ Theorem we first compute

curlF =

(∂(3y − 2xz)

∂y− ∂(3x2 − 4y)

∂z

)i +

(∂(4x+ 3z)

∂z− ∂(3y − 2xz)

∂x

)j

+

(∂(3x2 − 4y)

∂x− ∂(4x+ 3z)

∂y

)k

= 3 i + (3 + 2z) j + 6xk ,

and

−∂z∂x

i− ∂z

∂yj + k = −1

2j + k .

Thus we have∮C

F · dr =

∫∫R

(6x− z − 3

2) dA =

∫ 2

0

∫ 2−x

0

(6x− y

2− 3

2) dy dx

=

∫ 2

0

(6x(2− x)− 1

4(2− x)2 − 3

2(2− x)) dx

=

∫ 2

0

(−25x2

4+

29x

2− 4) dx = −25

1223 +

29

422 − 8 =

13

3.

(c) The line segments C1 is parametrized as x = t, y = 0, z = 0, 0 ≤ t ≤ 2 and one gets∫C1

F · dr =

∫ 2

0

4t dt = 8 .

26

The line segments C2 is parametrized as x = 2− t, y = t, z = t/2, 0 ≤ t ≤ 2 and one gets∫C2

F · dr =

∫C1

(4x+ 3z) dx− (4y − 3x2) dy + (3y − 2xz)dz

=

∫ 2

0

(−4(2− t)− 3t

2− 4t+ 3(2− t)2 +

1

2(3t− (2− t)t)) dt =

∫ 2

0

(7t2

2− 13t+ 4) dt

=7

623 − 13

222 + 8 = −26

3.

The line segments C3 is parametrized as x = 0, y = 2− t, z = 1− t2, 0 ≤ t ≤ 2 and one gets∫

C3

F · dr =

∫C1

(4x+ 3z) dx− (4y − 3x2) dy + (3y − 2xz)dz

=

∫ 2

0

(4(2− t)− 3

2(2− t)) dt =

∫ 2

0

(5− 5

2t) dt

= 10− 5

422 = 5 .

Summing up the contributions one gets∮C

F · dr =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′′ + 4y = 2(cos2 t− sin2 t) , y(0) = 1 , y′(0) = 3 .

Solution : (a) We denote Y (s) = L(y). By using the formulae

L(y′′) = s2Y (s)− sy(0)− y′(0) , L(cos2 t− sin2 t) = L(cos 2t) =s

s2 + 4,

we get the algebraic equation

(s2 + 4)Y (s)− s− 3 =2s

s2 + 4.

Solving the equation for Y , we get

Y (s) =3

s2 + 4+

s

s2 + 4+

2s

(s2 + 4)2.

27

Finally we use the formulas of the inverse Laplace transform

L−1(

2

s2 + 4

)= sin 2t , L−1

(s

s2 + 4

)= cos 2t , L−1

(4s

(s2 + 4)2

)= t sin 2t ,

to get

y(t) = (3

2+t

2) sin 2t+ cos 2t .

The term t sin 2t in the solution shows that the system is unstable, and the amplitude of theoscillations increases in time.

(b) The plots of the input function and the solution are below

2 4 6 8 10 12

-6

-4

-2

2

4

6

2.y′′ + y/4 = e−t/2 , y(0) = 3/2 , y′(0) = −1 .

Solution : (a) We denote Y (s) = L(y). By using the formulae

L(y′′) = s2Y (s)− sy(0)− y′(0) , L(e−t/2) =1

s+ 1/2,

we get the algebraic equation

(s2 + 1/4)Y (s)− 3s

2+ 1 =

1

s+ 1/2.

Solving the equation for Y , we get

Y (s) = − 1

s2 + 14

+3s

2s2 + 12

+1

(s+ 12)(s2 + 1

4)

= − 1

s2 + 14

+3s

2s2 + 12

+2

s+ 12

− 2s− 1

2

s2 + 14

=2

s+ 12

− s

2s2 + 12

,

where we used the partial fraction decomposition.Finally we use the formulae of the inverse Laplace transform

L−1(

s

s2 + 1/4

)= cos t/2 , L−1

(1

s+ 1/2

)= e−t/2 ,

to get

y(t) = 2e−2t − 1

2cos

t

2.

28

(b) The plots of the input function and the solution are below

5 10 15 20 25

-0.5

0.5

1.0

1.5

Note that the oscillations stabilize very quickly and become harmonic.

3.

y′′ + 4y =

{2t if 0 < t < 2π0 if t > 2π

, y(0) = 1 , y′(0) = −1 .

Solution : (a) We denote Y (s) = L(y) and represent the function on the right-hand side of theequation as

r(t) = 2t (u(t)− u(t− 2π)) .

By using the formulae

L(y′′) = s2Y (s)−sy(0)−y′(0) , L(f(t−a)u(t−a)) = e−asF (s) ⇒ L(r) =2

s2−2e−2πs

s2−4πe−2πs

s,

we get the algebraic equation

(s2 + 4)Y (s)− s+ 1 =2

s2− 2e−2πs

s2− 4πe−2πs

s.

Solving the equation for Y , we get

Y (s) = − 1

s2 + 4+

s

s2 + 4+

2

s2(s2 + 4)− 2e−2πs

s2(s2 + 4)− 4πe−2πs

s(s2 + 4).

Then we use4

s2(s2 + 4)=

1

s2− 1

s2 + 4,

4

s(s2 + 4)=

1

s− s

s2 + 4,

and get

Y (s) =1

2s2− 3/2

s2 + 4+

s

s2 + 4− πe−2πs

s− e−2πs

2s2+

e−2πs

2(s2 + 4)+πse−2πs

s2 + 4.

Finally we use the formulae of the inverse Laplace transform

L−1(

2

s2 + 4

)= sin 2t , L−1

(s

s2 + 4

)= cos 2t , L−1

(e−asF (s)

)= f(t− a)u(t− a) ,

29

to get

y(t) =t

2− 3

4sin 2t+ cos 2t if 0 < t < 2π ,

and

y(t) =t

2− 3

4sin 2t+ cos 2t− π − (t− 2π)/2 +

1

4sin 2(t− 2π) + π cos 2(t− 2π) if t > 2π .

Simplifying the expression above, we get

y(t) =

{t2− 3

4sin 2t+ cos 2t if 0 < t < 2π

(1 + π) cos 2t− 12

sin 2t if t > 2π.

(b) The plots of the input function and the solution are shown below.

2 4 6 8 10 12

5

10

10 Tutorial Sheet 10

1. Consider the following initial value problem

y′′ + 9 y = −9u(t− π) + 6 δ(t− 2π) , y(0) = 1 , y′(0) = 0 .

(a) Plot the function

−9u(t− π) + 6(u(t− 2π +

1

2)− u(t− 2π − 1

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

-8

-6

-4

-2

(b) We denote Y (s) = L(y), and then using the formulae

L(y′′) = s2Y (s)− sy(0)− y′(0) , L(u(t− a)) =e−as

s, L(δ(t− a)) = e−as ,

we get the algebraic equation

(s2 + 9)Y (s) = −9e−πs

s+ 6e−2πs + s .

Solving the equation for Y , we get

Y (s) = − 9e−πs

s(s2 + 9)+

6e−2πs

s2 + 9+

s

s2 + 9.

Then we use the formula

− 9

s(s2 + 9)= −1

s+

s

s2 + 9,

and write Y (s) in the form

Y (s) = −e−πs 1

s+ e−πs

s

s2 + 9+ e−2πs

6

s2 + 9+

s

s2 + 9.

Finally we use the formulas of the inverse Laplace transform

L−1(

1

s

)= 1 , L−1

(s

s2 + ω2

)= cosωt , L−1

(ω

s2 + ω2

)= sinωt ,

L−1(e−asF (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

. (10.7)

(c) The plot of the solution is shown below.

31

2 4 6 8 10 12

-3

-2

-1

1

2. Applying convolution,(a) find the solution.(b) sketch the input function and the solution.Show the details of your work.

y′′ + 3y′ + 2y =

{6 if 0 < t < 10 if t > 1

, y(0) = 2 , y′(0) = −2 .

Solution : We denote Y (s) = L(y) and represent the function on the right-hand side of theequation as

r(t) = 6(u(t)− u(t− 1)

).

By using the formulas

L(y′) = sY (s)−y(0) , L(y′′) = s2Y (s)−sy(0)−y′(0) , L(u(t−a)) =e−as

s⇒ L(r) =

6

s−6

e−s

s,

we get the algebraic equation

(s2 + 3s+ 2)Y (s)− 2s− 4 =6

s− 6

e−s

s⇒ Y (s) = 2(s+ 2)Q(s) +R(s)Q(s) ,

where

R(s) =6

s− 6

e−s

s, Q(s) =

1

s2 + 3s+ 2=

1

(s+ 1)(s+ 2)=

1

s+ 1− 1

s+ 2.

By using the convolution theorem we get the integral representation

y(t) = 2q′(t) + 4q(t) +

∫ t

0

q(t− τ)r(τ) dτ ,

where

q(t) = L−1(Q(s)) = e−t−e−2t , q′(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∫ t

0

q(t−τ)r(τ) dτ = 6

∫ t

0

q(t−τ) dτ = 6

∫ t

0

(e−t+τ−e−2t+2τ ) dτ = 3−6e−t+3e−2t if 0 < t < 1 ,

32

∫ t

0

q(t−τ)r(τ) dτ = 6

∫ 1

0

q(t−τ) dτ = 6

∫ 1

0

(e−t+τ−e−2t+2τ ) dτ = 6e−t(e−1)−3e−2t(e2−1) if t > 1 ,

Adding q′(t) + q(t), we get

y(t) =

2e−t if t < 0

3− 4e−t + 3e−2t if 0 < t < 1−3e2−2t + 6e1−t + 3e−2t − 4e−t if t > 1

,

The plot of the input function and the solution is shown below.

-0.5 0.5 1.0 1.5 2.0 2.5 3.0

1

2

3

4

5

6

3. Using Laplace transforms, solve the integral equation. (Show the details of your work.)

y(t) = − sin t+

∫ t

0

y(τ) sin(t− τ) dτ .

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) = − 1

s2 + 1+ Y (s)

1

s2 + 1.

Solving for Y (s), we obtain

Y (s) = − 1

s2,

and thereforey(t) = −t .

33