Imagine you're tracking a drone flying in the sky. At any instant, it has a position — say, 30 metres east and 40 metres north of you. That's a vector: r=30i^+40j^. A second later, it's moved. The question kinematics asks is: how fast is that position changing? That rate of change is velocity, and to get it, you differentiate the position vector.
But here's the key difference from school calculus: in school, you differentiated a scalar function like y=x2. Here, you're differentiating a vector function — something that has both magnitude and direction, and both can change with time.
The Intuition First
Think of a vector as an arrow. When time passes, that arrow can do two things:
It can get longer or shorter (magnitude changes).
It can rotate (direction changes).
Velocity is the total rate of change of that arrow. If the drone flies straight away from you, only the length changes. If it flies in a circle around you, only the direction changes. Most real motion does both.
So vector differentiation is just: take the derivative of each component separately, because components are independent scalars.
The Precise Statement
If a position vector is written in Cartesian coordinates as:
r(t)=x(t)i^+y(t)j^+z(t)k^
where i^,j^,k^ are fixed unit vectors (they don't change direction with time), then:
dtdr=dtdxi^+dtdyj^+dtdzk^
That's it. You differentiate each component function x(t),y(t),z(t) exactly as you would in single-variable calculus, and the unit vectors stay put.
dtd(f(t)u^)=dtdfu^(if u^ is constant)
Why This Works
The derivative of a vector is defined the same way as for a scalar — as a limit:
dtdr=limΔt→0Δtr(t+Δt)−r(t)
The numerator is a vector difference. When you write r in components, the difference splits into component differences. The limit then acts on each component separately because the unit vectors are constant. So the definition forces component-wise differentiation.
A Concrete Example
A particle moves such that:
r(t)=(3t2)i^+(5sint)j^+(2e−t)k^
Its velocity is:
v(t)=dtdr=(6t)i^+(5cost)j^+(−2e−t)k^
Notice: the x-component grows linearly, the y-component oscillates, the z-component decays. Each derivative is just the ordinary derivative of that component's function.
The One Trap: Non-Constant Unit Vectors
The rule above assumes i^,j^,k^ are fixed. That's true in Cartesian coordinates. But in polar coordinates, the unit vectors r^ and θ^rotate as the particle moves. Differentiating a vector in polar coordinates requires the product rule because the unit vectors themselves depend on time. …
Momentum is mass times velocity; force is mass times acceleration. At t=2 s, p=8i^+8j^ kg·m/s and F=4i^+8j^ N.
The problem gives velocity as a function of time, and asks for momentum and force at a specific instant. Both are vector quantities, so we treat each component separately.
Momentum is straightforward: p=mv. Since mass is constant, we just plug in the velocity at t=2 s.
Force requires acceleration. Acceleration is the time derivative of velocity: a=dtdv. Because velocity is given as a vector function, we differentiate each component with respect to t, then evaluate at t=2 s. Then F=ma.
Let's work through it.
Find the velocity at t=2 s.
v(t)=2ti^+t2j^
At t=2:
v(2)=2(2)i^+(2)2j^=4i^+4j^ m/s.
Compute momentum at t=2 s.
p=mv=2×(4i^+4j^)=8i^+8j^ kg·m/s.
Watch out
A common mistake is to forget that momentum is a vector — don't just give a magnitude. The answer must have both components.