example of Vector Differentiation

Vector Differentiation in Physics: Examples and Complete Worked Solutions

This companion entry develops the ideas from Vector Differentiation in Physics: Vector-Valued Functions, Moving Bases, and Kinematics through a sequence of worked mechanics problems. The problems begin with derivatives in a fixed Cartesian basis, move through dot and cross products and changing vector magnitude, and then progress to moving unit vectors, curved motion, polar coordinates, angular momentum, rotating frames, and the material derivative [1, 2, 3].

All exercises are stated first. Complete solutions follow afterward so that the first half can be used as a problem set without exposing the derivations immediately.

Figure 1. Progression of the problem set from ordinary component differentiation to moving bases and frame-dependent derivatives.

Part I: Exercises

Exercise 1: componentwise vector differentiation

Let

A (t) = t2ex + sint ey + e− tez.

(1)

Assume the Cartesian basis is fixed. Find dA∕dt and evaluate it at t = 0.

Exercise 2: derivative from the limit definition

A particle moves in the plane according to

2 3 r(t) = t ex + t ey.

(2)

Use the difference quotient

r(t + Δt ) − r(t) ------Δt--------

(3)

to derive dr∕dt directly from the limit definition. Then find the velocity at t = 1.

Exercise 3: scalar-vector product rule

Let

f (t) = t2

(4)

and

A (t) = cost ex + sin tey.

(5)

Find

d --[f(t)A (t)] dt

(6)

and evaluate the result at t = π∕2.

Exercise 4: derivative of a dot product

Let

2 A (t) = tex + t ey + ez

(7)

and

2 B (t) = tex + ey + tez.

(8)

Find

d-(A ⋅ B ) dt

(9)

in two ways: first by differentiating the scalar dot product directly, and second by using the vector product rule. Evaluate the result at t = 1.

Exercise 5: rate of change of speed

A particle has velocity

v(t) = 3tex + 4ey.

(10)

Find the acceleration, the speed, and dv∕dt. Evaluate dv∕dt at t = 2 and verify that

dv-= v-⋅ a-. dt v

(11)

Exercise 6: differentiating a rotating unit vector

Let

e(t) = cos(ωt)ex + sin(ωt)ey.

(12)

Show that e is a unit vector. Find de∕dt, prove that it is perpendicular to e, and determine its magnitude.

Exercise 7: uniform circular motion

A particle moves on a circle according to

r (t) = R cos(ωt)ex + R sin(ωt)ey.

(13)

Derive the velocity and acceleration. Show that the speed is constant, that the acceleration points toward the center, and that

2 a = v--. R

(14)

For R = 3 m and ω = 2 rad/s, find the speed and acceleration magnitude.

Exercise 8: tangential and normal acceleration on a curved path

A particle follows

t2 r(t) = tex + --ey. 2

(15)

At t = 1, find the velocity, speed, acceleration, tangential acceleration, normal acceleration, and radius of curvature ρ using

dv- v2- a = dtet + ρ en.

(16)

Figure 2. Geometry for Exercise 8. The acceleration is decomposed into components tangent and normal to the trajectory.

Exercise 9: angular momentum and torque

A particle of mass m moves uniformly on a circle of radius a:

r(t) = acos(ωt)ex + a sin(ωt )ey.

(17)

Find its momentum p, angular momentum L = r × p, and dL∕dt. Explain what the result implies about the torque about the origin.

Exercise 10: velocity in plane polar coordinates

A particle is described by

r(t) = 2 + t,

(18)

𝜃(t) = t2.

(19)

Using

v = ˙r er + r𝜃˙e𝜃,

(20)

find the velocity at t = 1 in the polar basis.

Exercise 11: acceleration in plane polar coordinates

For the same motion as Exercise 10, use

˙2 ¨ ˙ a = (¨r − r𝜃 )er + (r𝜃 + 2r˙𝜃 )e 𝜃

(21)

to find the acceleration at t = 1.

Figure 3. Polar basis for Exercises 10 and 11. Both the scalar components and the basis directions vary during the motion.

Exercise 12: an expanding spiral

A particle follows the polar-coordinate motion

r(t) = at,

(22)

𝜃(t) = ωt,

(23)

where a and ω are constants. Derive general expressions for v and a in the polar basis. Identify the physical origin of each acceleration term.

Exercise 13: a vector fixed in a rotating body

A vector has constant components in a body-fixed frame:

A = 2e1.

(24)

The frame rotates relative to an inertial frame with constant angular velocity

⃗ω = 3ez.

(25)

At the instant when e₁ = e_x, find the inertial derivative of A.

Exercise 14: applying the transport theorem

A rotating frame has angular velocity

⃗ω = 2ez.

(26)

A vector expressed in that rotating frame is

2 A (t) = te1 + t e2.

(27)

At t = 1, suppose e₁ = e_x and e₂ = e_y. Use the transport theorem

(dA ) ( dA ) ---- = ---- + ⃗ω × A dt I dt R

(28)

to find the inertial derivative.

Exercise 15: acceleration in a rotating frame

A frame rotates with constant angular velocity

⃗ω = 2ez

(29)

about a fixed origin. At one instant a particle has, relative to the rotating frame,

r = ex,

(30)

vR = 3ey,

(31)

a = 0. R

(32)

Find the inertial acceleration using

a = a + 2 ⃗ω × v + ⃗ω × (⃗ω × r). I R R

(33)

Identify the Coriolis and rotational-position terms separately.

Figure 4. Instantaneous geometry for Exercise 15. The rotating observer measures a relative velocity while the inertial derivative also includes basis rotation.

Exercise 16: material derivative of a vector field

Let the velocity field of a continuum be

v (x,y) = αxex − αyey,

(34)

with α = 0.5 s⁻¹. Let another vector field be

A (x,y,t) = xte + y2e . x y

(35)

Use the material derivative

dA-- ∂A-- dt = ∂t + (v ⋅ ∇ )A

(36)

to find the rate of change experienced by a material particle at x = 1, y = 2, t = 1.

Part II: Complete Worked Solutions

Solution 1: componentwise vector differentiation

Because the Cartesian basis is fixed, differentiate each component:

dA--= 2tex + costey − e− tez. dt

(37)

At t = 0,

dA-- dt = 0ex + 1ey − 1ez.

(38)

Thus

dA-| = e − e . dt t=0 y z

(39)

The key assumption is that the basis vectors themselves have zero derivative.

Solution 2: derivative from the limit definition

First calculate

r(t + Δt ) = (t + Δt )2ex + (t + Δt )3ey.

(40)

Subtract r(t):

r(t + Δt) − r(t) = (2tΔt + Δt2 )ex + (3t2Δt + 3tΔt2 + Δt3)ey.

(41)

Divide by Δt:

r(t +-Δt-) −-r(t) 2 2 Δt = (2t + Δt )ex + (3t + 3tΔt + Δt )ey.

(42)

Taking the limit Δt → 0 gives

dr-= 2te + 3t2e . dt x y

(43)

At t = 1,

v(1) = 2e + 3e . x y

(44)

Solution 3: scalar-vector product rule

Use

d -- (fA ) = ˙f A + fA˙. dt

(45)

Here

f˙= 2t

(46)

and

˙A = − sin tex + cos tey.

(47)

Therefore

-d 2 dt (f A ) = 2t(cos tex + sin tey) + t (− sin tex + cost ey).

(48)

At t = π∕2,

cos(π∕2 ) = 0,

(49)

sin(π∕2) = 1.

(50)

Hence

-d π2- dt(f A )|t=π∕2 = − 4 ex + πey.

(51)

Solution 4: derivative of a dot product

First calculate the dot product itself:

A ⋅ B = t3 + t2 + t.

(52)

Therefore

d-(A ⋅ B ) = 3t2 + 2t + 1. dt

(53)

At t = 1 this equals

6.

(54)

Now use the product rule. The derivatives are

A˙ = ex + 2tey,

(55)

B˙ = 2tex + ez.

(56)

At t = 1,

A = B = ex + ey + ez,

(57)

˙A = ex + 2ey,

(58)

˙B = 2e + e . x z

(59)

Thus

˙A ⋅ B = 3,

(60)

A ⋅B˙ = 3,

(61)

and

d-(A ⋅ B ) = 3 + 3 = 6. dt

(62)

Both methods agree.

Solution 5: rate of change of speed

The acceleration is

a = dv-= 3ex. dt

(63)

The speed is

v = |v | = √9t2-+-16.

(64)

Differentiate:

dv- ----9t---- dt = √ --2-----. 9t + 16

(65)

At t = 2,

dv- = √18--≈ 2.496 mm ∕s2. dt 52

(66)

Now

v ⋅ a = (3t)(3) = 9t,

(67)

v ⋅-a-= √--9t----, v 9t2 + 16

(68)

which matches the direct derivative of the speed.

Solution 6: differentiating a rotating unit vector

The magnitude satisfies

e ⋅ e = cos2(ωt ) + sin2(ωt) = 1.

(69)

Thus e is a unit vector. Differentiate:

de --- = − ω sin(ωt )ex + ω cos(ωt)ey. dt

(70)

The dot product with e is

de e ⋅---= − ω cos(ωt)sin(ωt) + ω sin(ωt )cos(ωt) = 0. dt

(71)

Therefore the derivative is perpendicular to the unit vector. Its magnitude is

| | ||de|| = ω. |dt|

(72)

The vector changes direction at angular rate ω while preserving unit length.

Solution 7: uniform circular motion

Differentiate the position:

v = − R ω sin (ωt)ex + Rω cos(ωt)ey.

(73)

Its magnitude is

v = R ω.

(74)

This is constant. Differentiate again:

2 2 a = − R ω cos(ωt)ex − R ω sin(ωt)ey.

(75)

Since

r = R cos(ωt)ex + R sin (ωt)ey,

(76)

we have

a = − ω2r.

(77)

The acceleration therefore points inward toward the origin. Its magnitude is

a = R ω2.

(78)

Because v = Rω,

v2- R2-ω2- 2 R = R = R ω = a.

(79)

For R = 3 m and ω = 2 rad/s,

v = 6 mm ∕s,

(80)

2 a = 12 mm ∕s .

(81)

Solution 8: tangential and normal acceleration on a curved path

Differentiate the position:

v = ex + tey.

(82)

At t = 1,

v = ex + ey.

(83)

The speed is

v = √1-+-t2,

(84)

so at t = 1,

v = √2.-

(85)

The acceleration is

a = ey.

(86)

The tangential acceleration is the rate of change of speed:

a = dv-= √--t----. t dt 1 + t2

(87)

At t = 1,

a = √1-. t 2

(88)

Since the total acceleration magnitude is 1,

∘ ------ ∘ ------- 1 1 an = a2 − a2t = 1 − --= √--. 2 2

(89)

Finally,

2 an = v-. ρ

(90)

Therefore

2 -- ρ = v--= --2√---= 2√ 2. an 1∕ 2

(91)

Thus at t = 1,

-1-- at = an = √2-,

(92)

√ -- ρ = 2 2.

(93)

Solution 9: angular momentum and torque

The velocity is

v = − aω sin (ωt)ex + aω cos(ωt)ey.

(94)

Thus

p = mv.

(95)

The angular momentum is

L = r × p.

(96)

Only the z component survives. Its magnitude is

( ) Lz = ma2 ω cos2(ωt ) + sin2(ωt) .

(97)

Therefore

2 L = ma ωez.

(98)

This is constant, so

dL ---= 0. dt

(99)

Since

⃗τ = dL-, dt

(100)

the net torque about the origin is zero. The inward force is radial, so its moment arm is parallel to the force direction and r × F = 0.

Solution 10: velocity in plane polar coordinates

We have

r = 2 + t,

(101)

˙r = 1.

(102)

Also

𝜃 = t2,

(103)

𝜃˙= 2t.

(104)

At t = 1,

r = 3,

(105)

˙r = 1,

(106)

˙𝜃 = 2.

(107)

Thus

v = ˙rer + r˙𝜃e𝜃 = er + 6e𝜃.

(108)

Solution 11: acceleration in plane polar coordinates

The needed derivatives are

r˙= 1, ¨r = 0,

(109)

˙𝜃 = 2t, 𝜃¨= 2.

(110)

At t = 1, r = 3 and 𝜃 = 2. The radial component is

ar = ¨r − r˙𝜃2 = 0 − 3(2)2 = − 12.

(111)

The transverse component is

a𝜃 = r¨𝜃 + 2˙r˙𝜃 = 3(2) + 2(1)(2) = 10.

(112)

Therefore

a = − 12er + 10e𝜃.

(113)

The −12e_r term contains the centripetal contribution generated by rotation of the radial basis direction.

Solution 12: an expanding spiral

For

r = at,

(114)

we have

r˙= a, ¨r = 0.

(115)

For

𝜃 = ωt,

(116)

we have

𝜃˙= ω, ¨𝜃 = 0.

(117)

The velocity is

v = aer + atωe𝜃.

(118)

The acceleration is

a = − atω2er + 2aωe 𝜃.

(119)

The radial term

− atω2er

(120)

is the inward centripetal-type contribution associated with angular motion at radius r = at. The transverse term

2aωe 𝜃

(121)

arises because the radius changes while the particle also rotates. It is the polar-coordinate analogue of a Coriolis-type coupling term.

Solution 13: a vector fixed in a rotating body

A vector whose components are constant in the rotating frame still changes in the inertial frame because the basis rotates. The transport theorem gives

( ) dA-- = ⃗ω × A dt I

(122)

for a vector fixed in the rotating frame.

At the stated instant,

⃗ω = 3e , z

(123)

A = 2ex.

(124)

Therefore

( ) dA-- dt = (3ez) × (2ex) = 6ey. I

(125)

The vector has zero derivative relative to the body but a nonzero inertial derivative.

Solution 14: applying the transport theorem

First differentiate only the rotating-frame components:

( ) dA-- dt = e1 + 2te2. R

(126)

At t = 1 and at the stated alignment,

( ) dA ---- = ex + 2ey. dt R

(127)

Also

A = ex + ey.

(128)

The basis-rotation contribution is

⃗ω × A = 2ez × (ex + ey).

(129)

Using

ez × ex = ey,

(130)

ez × ey = − ex,

(131)

we obtain

⃗ω × A = − 2ex + 2ey.

(132)

Hence

( ) dA-- dt = (ex + 2ey) + (− 2ex + 2ey ), I

(133)

( dA ) ---- = − ex + 4ey. dt I

(134)

Solution 15: acceleration in a rotating frame

The Coriolis contribution is

2⃗ω × vR = 2(2ez) × (3ey).

(135)

Since

ez × ey = − ex,

(136)

we get

2⃗ω × vR = − 12ex.

(137)

Now calculate the position-dependent rotational term. First,

⃗ω × r = 2ez × ex = 2ey.

(138)

Then

⃗ω × (⃗ω × r) = 2ez × 2ey = − 4ex.

(139)

Since a_R = 0,

aI = − 12ex − 4ex = − 16ex.

(140)

Thus

aI = − 16ex.

(141)

The −12e_x term is the Coriolis contribution and the −4e_x term is the double-cross-product contribution caused by the rotating position vector.

Solution 16: material derivative of a vector field

The vector field is

A = xtex + y2ey.

(142)

Its explicit time derivative is

∂A-- = xe . ∂t x

(143)

At x = 1 this is

∂A--= e . ∂t x

(144)

The velocity field at x = 1, y = 2 is

v = 0.5ex − 1ey.

(145)

For the x component A_x = xt,

∂Ax ∂Ax -∂x- = t, -∂y- = 0.

(146)

At t = 1,

(v ⋅ ∇ )A = (0.5)(1) + (− 1)(0 ) = 0.5. x

(147)

For the y component A_y = y²,

∂Ay ∂Ay ----= 0, ----= 2y. ∂x ∂y

(148)

At y = 2,

(v ⋅ ∇ )Ay = (0.5)(0) + (− 1)(4) = − 4.

(149)

Therefore

(v ⋅ ∇ )A = 0.5ex − 4ey.

(150)

Adding the explicit time derivative,

dA--= e + (0.5e − 4e ), dt x x y

(151)

dA ----= 1.5ex − 4ey. dt

(152)

The result contains both local time variation and variation caused by the particle moving through a spatially nonuniform field.

Summary of recurring methods

The examples illustrate a small set of ideas that repeatedly appear in mechanics:

in a fixed Cartesian basis, differentiate vector components directly;
use ordinary product rules for scalar-vector, dot-product, and cross-product expressions;
a perpendicular derivative changes direction without changing vector magnitude to first order;
curved motion naturally separates into tangential and normal acceleration;
polar-coordinate differentiation requires derivatives of the basis vectors as well as the scalar components;
rotating-frame derivatives differ from inertial derivatives by ω × A;
the material derivative combines explicit time change with change caused by motion through a spatial field.

References

[1] J. R. Taylor, Classical Mechanics, University Science Books, 2005.

[2] J. B. Marion and S. T. Thornton, Classical Dynamics of Particles and Systems, 5th ed., Brooks/Cole, 2004.

[3] H. Goldstein, C. Poole, and J. Safko, Classical Mechanics, 3rd ed., Addison-Wesley, 2002.

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