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Derivative of a Vector Function

Differentiate r(t) component by component, then get velocity, speed and the unit tangent
Derivative of r(t) = <t^2, sin(t), e^(3t)>
Unit tangent vector for r(t) = <cos t, sin t, t> at t = pi/2
Derivative of the dot product of <t, t^2, 1> and <1, t, t^3>
Speed of a particle with r(t) = <3t, 4t, 5>

Differentiate Component by Component

A vector-valued function r(t)=⟨f(t), g(t), h(t)⟩\mathbf{r}(t) = \langle f(t),\, g(t),\, h(t)\rangle traces a curve in space as tt varies. Its derivative comes from the same limit as in single-variable calculus,

r′(t)=lim⁡Δt→0r(t+Δt)−r(t)Δt\mathbf{r}'(t) = \lim_{\Delta t \to 0}\frac{\mathbf{r}(t + \Delta t) - \mathbf{r}(t)}{\Delta t}

and because vector subtraction and scalar division act componentwise, this collapses to the rule you actually use:

r′(t)=⟨f′(t), g′(t), h′(t)⟩\mathbf{r}'(t) = \langle f'(t),\, g'(t),\, h'(t)\rangle

It exists only where every component is differentiable — one non-smooth component kills the whole derivative.

What it means geometrically. r′(t)\mathbf{r}'(t) is a vector, not a slope: it points along the curve in the direction of increasing tt and is tangent to it. If r\mathbf{r} is position, then r′\mathbf{r}' is velocity, ∥r′∥\|\mathbf{r}'\| is speed (a scalar), and r′′\mathbf{r}'' is acceleration.

Rules, Speed and the Unit Tangent

With u,v\mathbf{u}, \mathbf{v} differentiable and cc a scalar function:

ddt[u+v]=u′+v′,ddt[c u]=c′u+c u′\frac{d}{dt}[\mathbf{u} + \mathbf{v}] = \mathbf{u}' + \mathbf{v}', \qquad \frac{d}{dt}[c\,\mathbf{u}] = c'\mathbf{u} + c\,\mathbf{u}'

ddt[u⋅v]=u′⋅v+u⋅v′(a scalar)\frac{d}{dt}[\mathbf{u} \cdot \mathbf{v}] = \mathbf{u}' \cdot \mathbf{v} + \mathbf{u} \cdot \mathbf{v}' \quad (\text{a scalar})

ddt[u×v]=u′×v+u×v′(order is fixed)\frac{d}{dt}[\mathbf{u} \times \mathbf{v}] = \mathbf{u}' \times \mathbf{v} + \mathbf{u} \times \mathbf{v}' \quad (\text{order is fixed})

ddt[r(s(t))]=r′(s(t)) s′(t)\frac{d}{dt}[\mathbf{r}(s(t))] = \mathbf{r}'(s(t))\,s'(t)

Speed is the magnitude ∥r′(t)∥=f′2+g′2+h′2\|\mathbf{r}'(t)\| = \sqrt{f'^2 + g'^2 + h'^2}, and arc length on [a,b][a,b] is ∫ab∥r′(t)∥ dt\int_a^b \|\mathbf{r}'(t)\|\,dt.

The unit tangent requires r′(t)≠0\mathbf{r}'(t) \neq \mathbf{0}:

T(t)=r′(t)∥r′(t)∥\mathbf{T}(t) = \frac{\mathbf{r}'(t)}{\|\mathbf{r}'(t)\|}

A curve with r′≠0\mathbf{r}' \neq \mathbf{0} everywhere is called smooth; where r′=0\mathbf{r}' = \mathbf{0} the curve can have a cusp even though every component is differentiable.

Common Mistakes to Avoid

  • Reversing a cross product. u′×v+u×v′\mathbf{u}' \times \mathbf{v} + \mathbf{u} \times \mathbf{v}' — swapping either factor flips a sign, because a×b=− b×a\mathbf{a} \times \mathbf{b} = -\,\mathbf{b} \times \mathbf{a}.
  • Confusing ∥r∥′\|\mathbf{r}\|' with ∥r′∥\|\mathbf{r}'\|. They are different: for r=⟨cos⁡t,sin⁡t⟩\mathbf{r} = \langle\cos t, \sin t\rangle the first is 00 and the second is 11.
  • Treating r′\mathbf{r}' as a slope. It is a vector; the "slope" of a space curve is not defined.
  • Forgetting the chain rule inside a component. ddte3t=3e3t\frac{d}{dt}e^{3t} = 3e^{3t}, not e3te^{3t}.
  • Normalising before differentiating. T′(t)\mathbf{T}'(t) is not r′′/∥r′′∥\mathbf{r}''/\|\mathbf{r}''\|; differentiate r\mathbf{r} first, then normalise.
  • Dividing by zero at a cusp. Check r′(t)≠0\mathbf{r}'(t) \neq \mathbf{0} before forming T(t)\mathbf{T}(t).

示例题目

Step 1: Differentiate each component separately
Step 2: First: ddtt2=2t\dfrac{d}{dt}t^2 = 2t
Step 3: Second: ddtsin⁡t=cos⁡t\dfrac{d}{dt}\sin t = \cos t
Step 4: Third (chain rule): ddte3t=3e3t\dfrac{d}{dt}e^{3t} = 3e^{3t}
Answer: r′(t)=⟨2t,  cos⁡t,  3e3t⟩\mathbf{r}'(t) = \langle 2t,\; \cos t,\; 3e^{3t}\rangle

Step 1: r′(t)=⟨−sin⁡t,  cos⁡t,  1⟩\mathbf{r}'(t) = \langle -\sin t,\; \cos t,\; 1\rangle
Step 2: ∥r′(t)∥=sin⁡2t+cos⁡2t+1=1+1=2\|\mathbf{r}'(t)\| = \sqrt{\sin^2 t + \cos^2 t + 1} = \sqrt{1 + 1} = \sqrt{2} — constant speed
Step 3: At t=π/2t = \pi/2: r′(π/2)=⟨−1,  0,  1⟩\mathbf{r}'(\pi/2) = \langle -1,\; 0,\; 1\rangle
Step 4: T=12⟨−1,  0,  1⟩\mathbf{T} = \dfrac{1}{\sqrt{2}}\langle -1,\; 0,\; 1\rangle
Answer: Speed =2= \sqrt{2}; T(π/2)=⟨−12,  0,  12⟩\mathbf{T}(\pi/2) = \left\langle -\tfrac{1}{\sqrt{2}},\; 0,\; \tfrac{1}{\sqrt{2}}\right\rangle

Step 1: u′=⟨1, 2t, 0⟩\mathbf{u}' = \langle 1,\, 2t,\, 0\rangle and v′=⟨0, 1, 3t2⟩\mathbf{v}' = \langle 0,\, 1,\, 3t^2\rangle
Step 2: u′⋅v=(1)(1)+(2t)(t)+(0)(t3)=1+2t2\mathbf{u}' \cdot \mathbf{v} = (1)(1) + (2t)(t) + (0)(t^3) = 1 + 2t^2
Step 3: u⋅v′=(t)(0)+(t2)(1)+(1)(3t2)=4t2\mathbf{u} \cdot \mathbf{v}' = (t)(0) + (t^2)(1) + (1)(3t^2) = 4t^2
Step 4: Add: 1+2t2+4t2=1+6t21 + 2t^2 + 4t^2 = 1 + 6t^2
Step 5: Check directly: u⋅v=t+t3+t3=t+2t3\mathbf{u} \cdot \mathbf{v} = t + t^3 + t^3 = t + 2t^3, whose derivative is 1+6t21 + 6t^2
Answer: 1+6t21 + 6t^2 (a scalar)

常见问题

Differentiate each component with respect to the parameter and reassemble them into a vector. For r(t) = <f, g, h> the derivative is r'(t) = <f', g', h'>. No new technique is needed beyond ordinary single-variable differentiation.

It is the tangent vector to the curve, pointing in the direction of increasing t. If r(t) is position then r'(t) is velocity, its magnitude is the speed, and r''(t) is acceleration.

Yes, and both look like the ordinary product rule. The dot version gives a scalar. The cross version requires you to preserve the order of the factors, since the cross product is anticommutative.

T(t) = r'(t)/||r'(t)|| is undefined wherever r'(t) is the zero vector, because you would divide by zero. Such points are where the curve can have a cusp, even if each component function is perfectly differentiable.

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