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The University of Sydney School of Mathematics and Statistics

Solutions to Tutorial 7 (Week 8) MATH2061: Vector Calculus

Semester 1

1. Let u = 3i − j − k, v = −i + 3j + k and w = 2i − 3k. Find (u · v)w, u(v · w), (u × v) · w, u · (v × w), (u × v) × w, u × (v × w). Solution:

u·v = 3×(−1)+(−1) ×3+(−1) ×1 = −7;

(u·v)w = −14j +21k.

v · w = (−1) × 2 + 3 × 0 + 1 × (−3) = −5; u(v · w) = −15i + 5j + 5k.    i j k   u × v =  3 −1 −1 = 2i − 2j + 8k; (u × v) · w = −20. −1 3 1    i j k   v × w = −1 3 1  = −9i − j − 6k; u · (v × w) = −20.  2 0 −3    i j k   (u × v) × w = 2 −2 8  = 6i + 22j + 4k. 2 0 −3     i j k    u × (v × w) =  3 −1 −1  = 5i + 27j − 12k. −9 −1 −6

Note that

(u × v) · w = u · (v × w)

but

(u × v) × w 6= u × (v × w).

2. Find the equations of the straight lines that satisfy each of the following sets of conditions. Give answers in vector form, parametric scalar form and cartesian form. (a) The line ℓ passes through the points P (−2, 1, 3) and Q (1, −4, 2). Solution: The vector

−→ v = P Q = 3i − 5j − k is parallel to ℓ and the position vector of the point P on ℓ is r0 = −2i + j + 3k. Hence the vector form of the equation of ℓ is r(t) = (−2i + j + 3k) + (3i − 5j − k)t,

(−∞ < t < ∞).

The parametric equations for ℓ are x = −2 + 3t,

y = 1 − 5t,

z = 3 − t.

and eliminating t we get the cartesian equations for ℓ : y−1 z−3 x+2 . = = −1 3 −5 c 2019 The University of Sydney Copyright 

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(b) The line ℓ passes through the point P (1, −2, 4) and is parallel to the line ℓ′ given by r = (−i + j + 6k) + (4i − j + k)t.

Solution: The vector u = 4i − j + k. is parallel to ℓ′ and hence to ℓ. Hence the vector equation for ℓ is r(t) = (i − 2j + 4k) + (4i − j + k)t,

(−∞ < t < ∞).

Parametric equations: x = 1 + 4t, y = −2 − t, z = 4 + t. z−4 x−1 y+2 = = Cartesian form: . 4 −1 1 (c) The line ℓ passes through the point P (0, 1, 3) and is perpendicular to the plane 4x − 3y + z = 7. Solution: The normal to the plane is u = 4i − 3j + k. Since the given line ℓ is parallel to u, the vector equation for ℓ is r(t) = (j + 3k) + (4i − 3j + k)t,

(−∞ < t < ∞).

Parametric equations: x = 4t, y = 1 − 3t, z = 3 + t. z−3 x y−1 . = = Cartesian form: 1 4 −3 3.

(a) Find the length of the curve r(t) =

√

2 t i + et j + e−t k; t : 0 → 1.

Solution: The length is Z 1q √ L= ( 2)2 + (et )2 + (−e−t )2 dt Z0 1 √ = 2 + e2t + e−2t dt 0 Z 1q (et + e−t )2 dt = 0 Z 1 = (et + e−t ) dt 0  t 1 = e − e−t 0 =e − e−1 .

(b) Let C be the curve defined by y = f (x); a ≤ x ≤ b. Find a parametric representation of C and the length of C in integral form (assuming f ′ is continuous). Solution: We can parametrise C by x = t, y = f (t); t : a → b. The length is then L=

Z bq

12 + (f ′ (t))2 dt =

Z bp a

a

2

1 + f ′ (t)2 dt.

4. Describe geometrically the vector fields determined by each of the following vector functions: (a) F = −2 i + 3 j;

(b) F = x i + y j.

Solution: (a)

This is a constant vector field which always point to the same direction (−2, 3).

(b) The vector at each point (x, y) has magnitude exactly equals to the distance from 0 and is pointing radially outward from 0 (i.e. the direction from 0 to its position (x, y)).

(a)

(b)

5. In Z each of the following, give a parametric representation of C, and evaluate φ ds. C

(a) φ(x, y, z) = x2 y + 2yz, and C is the straight line segment from (0, 1, 2) to (1, 0, 4). Solution: The straight line segment in R3 from (0, 1, 2) to (1, 0, 4) has parametric representation r(t) = j + 2k + t(i − j + 2k)

= ti + (1 − t)j + (2 + 2t)k;

t : 0 → 1.

Substituting x = t, y = 1 − t, p √ and z = 2 + 2t, we have 2 2 2 ds = 1 + (−1) + 2 dt = 6 dt and Z Z 1 √ 2 (x y + 2yz) ds = (t2 (1 − t) + 2(1 − t)(2 + 2t)) 6 dt

0 √ √ √ Z 1 1 11 6 . (4 − 3t2 − t3 ) dt = 6(4 − 1 − ) = = 6 4 4 0

C

(b) φ(x, y, z) = ex+y+z , (0, 5, 5).

and C is the straight line segment from (2, 3, 1) to

Solution: The straight line segment in R3 from (2, 3, 1) to (0, 5, 5) has parametric representation r(t) = 2i + 3j + k + t(−2i + 2j + 4k) = (2 − 2t)i + (3 + 2t)j + (1 + 4t)k; 3

t : 0 → 1.

Substituting x = 2 − 2t, y = 3 + 2t, and z = 1 + 4t we have p √ ds = (−2)2 + 22 + 42 dt = 2 6 dt and Z Z 1 √ x+y +z e ds = e((2−2t)+(3+2t)+(1+4t)) 6 dt C 0  6+4t  1 √ √ Z 1 √ e 6 10  =2 6 (e − e6 ). e6+4t dt = 2 6  =  4 2 0 0

6.

(a) Evaluate

Z

C

F · dr, where F = z i + x j + y k and

r(t) = t i + t2 j + 3 k; t : 0 → 1. Solution:

Z

C

F · dr =

Z

=

Z

z dx + x dy + y dz C 1

(3 + t.2t + 0) dt = 0

11 . 3

(b) Describe the curve given by r(t) in part (a). Solution: The curve given by r(t) is part of the parabola y = x2 , in the plane z = 3, from (0, 0, 3) to (1, 1, 3).

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7. Calculate the work done by the force field F = 2 i−3xy j in moving from A : (0, 0, 0) to B : (2, 4, 0) (a) along the straight line AB ; Solution: The work done by F along a path C is Z Z 2 dx − 3xy dy + 0 dz. F · dr = C

C

The straight line joining AB has equation r = t(2i + 4j) or x = 2t, y = 4t, z = 0; t : 0 → 1. Z 1 Z F · dr = (4 dt − 96t2 dt) = −28. So 0

C

(b) along the straight lines from A to D : (2, 0, 0) and then from D to B ; Solution: Z

C

Z

F · dr =

AD

F · dr +

Z

DB

F · dr.

On AD, y = 0, dy = 0 so Z On DB, x = 2, dx = 0 so Z

DB

Therefore

Z

AD

F · dr =

F · dr =

C

Z

Z

2

2 dx = 4

0

4

0

−6y dy = −48.

F · dr = 4 − 48 = −44.

(c) along the piece of the parabola y = x2 , z = 0 from A to B . Solution: On C, y = x2 , z = 0, so let x = t, t : 0 → 2. Then y = t2 ,

dy = 2t dt,

F · dr =

Z

Hence, Z

C

0

2

dz = 0.

(2 − 6t4 ) dt = −172/5.

8. A particle moves along a curve whose parametric equations are x(t) = sin t,

y(t) = te−t ,

where t is the time.

z(t) = − cos t,

(a) Determine its velocity vector and acceleration vector. Solution: The position vector is r = sin t i + te−t j − cos t k. The velocity vector is dr = cos t i + (1 − t)e−t j + sin t k, dt and the acceleration vector is dv a= = − sin t i + (t − 2)e−t j + cos t k. dt v=

5

(b) Find the magnitudes of the velocity and acceleration at t = 0. Solution: At t = 0, v(0) = i + j and a(0) = −2j + k. √ √ Then magnitude of velocity (i.e., speed) at t = 0 is 1 + 1 = 2, √ √ and magnitude of acceleration at t = 0 is 4 + 1 = 5. 9. Calculate the work done by the force field F = x3 i − z j + 2xy k along the curve C, where C is the straight line segment from (0, 1, −1) to (−2, 1, 2). Solution: The equation of C is: r(t) = j − k + t(−2i + 3k) = −2ti + j + (−1 + 3t)k;

t : 0 → 1.

So x = −2t, y = 1, z = −1 + 3t, dx/dt = −2, dy/dt = 0 and dz/dt = 3. Z Z x3 dx − z dy + 2xy dz Work done = F · dr = C C Z 1 −8t3 × (−2 dt) + 0 + 2 × (−2t) × 1 × 3 dt = 0 Z 1  3  16t − 12t dt = 0  4 1 = 4t − 6t2 0 = −2.

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