Q.The position of a particle is given by r=3.0ti^+2.0t2j^+5.0k^ where t is in seconds and the coefficients have the proper units for r to be in metres.
(a) Find v(t) and a(t) of the particle.
(b) Find the magnitude and direction of v(t) at t=1.0s.
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. …
The velocity and acceleration are found by differentiating the position vector component‑wise. At t=1.0s, the velocity has magnitude 5.0m/s and makes an angle of about 53∘ with the x-axis.
The problem gives the position vector as a function of time:
r(t)=3.0ti^+2.0t2j^+5.0k^
All coefficients are in SI units so that r comes out in metres. The z-component is constant — the particle never moves in the z-direction.
Why differentiate?
In kinematics, velocity is the rate of change of position, and acceleration is the rate of change of velocity. Since r(t) is given in Cartesian components, we can differentiate each component separately — the unit vectors i^,j^,k^ are fixed in direction, so they behave like constants.
v(t)=dtdr,a(t)=dtdv
(a) Finding v(t) and a(t)
Differentiate r(t) component by component:
x-component: dtd(3.0t)=3.0
y-component: dtd(2.0t2)=4.0t
z-component: dtd(5.0)=0
So the velocity vector is
v(t)=3.0i^+4.0tj^
Notice the z-component is zero — the motion is confined to the xy-plane.
Differentiate v(t) to get acceleration:
x-component: dtd(3.0)=0
y-component: dtd(4.0t)=4.0
z-component: 0
Hence
a(t)=4.0j^
The acceleration is constant, purely in the +y direction, with magnitude 4.0m/s2.
Tip
Because the x-velocity is constant (3.0m/s) and the y-velocity increases linearly, the particle follows a parabolic path — just like projectile motion, but here the acceleration is in the y-direction only.
Concept: Velocity and Acceleration as Limits, in Vector Form
Method: First-Principles Limit Definition (no differentiation rules quoted)
Rather than applying dtd(tn)=ntn−1 to each component directly, this method constructs v(t) and a(t) from the formal limit definitions v=limΔt→0ΔtΔr and a=limΔt→0ΔtΔv -- showing explicitly where each term in the final answer comes from.