MA 1505 Mathematics ITutorial 9 Solutions

1. We use the criteria of conservative field:

∂

∂y(2xy) = 2x =

∂

∂x(x2 + 2yz),

∂

∂z(2xy) = 0 =

∂

∂x(y2),

∂

∂z(x2 + 2yz) = 2y =

∂

∂y(y2).

So F is conservative, and hence there exists a function f such that ∇f = F.

To find f , first we know that, by i-component of F, we have fx(x, y, z) = 2xy so that

f(x, y, z) = x2y + g(y, z) (∗).

Differentiate (∗) w.r.t. y, we have

fy(x, y, z) = x2 + gy(y, z).

Now by j-component of F, we have fy(x, y, z) = x2 + 2yz, so

gy(y, z) = 2yz ⇒ g(y, z) = y2z + h(z).

Hence (∗) becomesf(x, y, z) = x2y + y2z + h(z)−−(∗∗)

Differentiate (∗∗) w.r.t. z, we have fz(x, y, z) = y2 + h′(z).Now by k-component of F, fz(x, y, z) = y2 so

h′(z) = 0⇒ h(z) = K.

So f(x, y, z) = x2y + y2z +K, where K is a constant.

2. r(t) = ti + 2tj + 3tk, where 0 ≤ t ≤ 1. Then

r′(t) = i + 2j + 3k, ||r′(t)|| =√

14, g(x(t), y(t), z(t)) = t2 − (2t)(3t) + (3t)2 = 4t2.

Therefore ∫Cg(x, y, z) ds =

∫ 1

0(4t2)(

√14) dt =

43

√14.

3. F(r(t)) = t5i + 2t2j− t2k. r′(t) = 1i + 2tj + 3t2k.∫C

F • dr =∫ 1

0(t5i + 2t2j− t2k) • (1i + 2tj + 3t2k) dt

=∫ 1

0(t5 − 3t4 + 4t3) dt =

1730.

MA1505 Tutorial 9 Solutions

4. To parametrize line segment from (a, b, c) to (d, e, f), we can use

r(t) = [ai + bj + ck] + t[(di + ej + fk)− (ai + bj + ck)], 0 ≤ t ≤ 1.

So we havethe line segment C1 joining (0, 0, 0) to (1, 0, 2)

r(t) = ti + 0j + 2tk, 0 ≤ t ≤ 1

the line segment C2 joining (1, 0, 2) to (3, 4, 1)

r(t) = (2t+ 1)i + 4tj + (−t+ 2)k, 0 ≤ t ≤ 1

Then∫C1

2xy dx+ (x2 + z) dy + y dz = 0,

∫C2

2xy dx+ (x2 + z) dy + y dz

=∫ 1

02(2t+ 1)(4t)(2dt) + ((2t+ 1)2 + (−t+ 2))(4dt) + (4t)(−dt)

=∫ 1

0(48t2 + 24t+ 12)dt = 40.

So∫C

2xy dx+ (x2 + z) dy + y dz = 0 + 40 = 40.

5. Let C be the base of the fence which has vector equation

r(t) = 10 cos ti + 10 sin tj, 0 ≤ t ≤ 2π.

So ‖r′(t)‖ = 10.

The area of the fence is then given by∫Ch(x, y)ds =

∫ 2π

0

[4 + 0.01[(10 cos t)2 − (10 sin t)2]

]· 10 dt

=∫ 2π

040 + 10 cos 2t dt

= [40t+ 5 sin 2t]2π0= 80π

Both sides of the fence will give a total area of 160π = 503 m2.

So the amount of paint used is about 5 litre.

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MA1505 Tutorial 9 Solutions

6. We can parametrize Ca by x = t, y = a sin t, 0 ≤ t ≤ π.

Then

I (a) =∫ π0

(1 + a3 sin3 t

)dt+ (2t+ a sin t) a cos tdt

= π − a3∫ π0

(1− cos2 t

)d (cos t) + 2a

∫ π0 td (sin t) + 1

2a∫ π0 sin 2tdt

= π − a3[cos t− 1

3 cos3 t]π0

+ 2a [t sin t]π0 − 2a∫ π0 sin tdt− 1

4a2 [cos 2t]π0

= π + 43a

3 − 4a.

Therefore, I ′ (a) = 4a2 − 4 = 0 implies a = 1 in the domain a > 0.

Since I ′′ (a) = 8a > 0 in the domain a > 0,

we conclude that the minimum value of I (a) in the domain a > 0 is I (1) = π − 83 .

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