How do you take the derivative of a vector function?

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

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

$\mathbf{r}'(t) = \lim_{\Delta t \to 0}\frac{\mathbf{r}(t + \Delta t) - \mathbf{r}(t)}{\Delta t}$

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

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

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

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

Rules, Speed and the Unit Tangent

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

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

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

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

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

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

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

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

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

Common Mistakes to Avoid

Reversing a cross product. $\mathbf{u}' \times \mathbf{v} + \mathbf{u} \times \mathbf{v}'$ — swapping either factor flips a sign, because $\mathbf{a} \times \mathbf{b} = -\,\mathbf{b} \times \mathbf{a}$ .
Confusing $\|\mathbf{r}\|'$ with $\|\mathbf{r}'\|$ . They are different: for $\mathbf{r} = \langle\cos t, \sin t\rangle$ the first is $0$ and the second is $1$ .
Treating $\mathbf{r}'$ as a slope. It is a vector; the "slope" of a space curve is not defined.
Forgetting the chain rule inside a component. $\frac{d}{dt}e^{3t} = 3e^{3t}$ , not $e^{3t}$ .
Normalising before differentiating. $\mathbf{T}'(t)$ is not $\mathbf{r}''/\|\mathbf{r}''\|$ ; differentiate $\mathbf{r}$ first, then normalise.
Dividing by zero at a cusp. Check $\mathbf{r}'(t) \neq \mathbf{0}$ before forming $\mathbf{T}(t)$ .

Examples

Step 1: Differentiate each component separately

Step 2: First:

\dfrac{d}{dt}t^2 = 2t

Step 3: Second:

\dfrac{d}{dt}\sin t = \cos t

Step 4: Third (chain rule):

\dfrac{d}{dt}e^{3t} = 3e^{3t}

Answer:

\mathbf{r}'(t) = \langle 2t,\; \cos t,\; 3e^{3t}\rangle

Step 1:

\mathbf{r}'(t) = \langle -\sin t,\; \cos t,\; 1\rangle

Step 2:

\|\mathbf{r}'(t)\| = \sqrt{\sin^2 t + \cos^2 t + 1} = \sqrt{1 + 1} = \sqrt{2}

— constant speed

Step 3: At

t = \pi/2

\mathbf{r}'(\pi/2) = \langle -1,\; 0,\; 1\rangle

Step 4:

\mathbf{T} = \dfrac{1}{\sqrt{2}}\langle -1,\; 0,\; 1\rangle

Answer: Speed

= \sqrt{2}

;

\mathbf{T}(\pi/2) = \left\langle -\tfrac{1}{\sqrt{2}},\; 0,\; \tfrac{1}{\sqrt{2}}\right\rangle

Step 1:

\mathbf{u}' = \langle 1,\, 2t,\, 0\rangle

and

\mathbf{v}' = \langle 0,\, 1,\, 3t^2\rangle

Step 2:

\mathbf{u}' \cdot \mathbf{v} = (1)(1) + (2t)(t) + (0)(t^3) = 1 + 2t^2

Step 3:

\mathbf{u} \cdot \mathbf{v}' = (t)(0) + (t^2)(1) + (1)(3t^2) = 4t^2

Step 4: Add:

1 + 2t^2 + 4t^2 = 1 + 6t^2

Step 5: Check directly:

\mathbf{u} \cdot \mathbf{v} = t + t^3 + t^3 = t + 2t^3

, whose derivative is

1 + 6t^2

Answer:

1 + 6t^2

(a scalar)

Frequently Asked Questions

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.

Related Solvers

도함수 계산기

Related Guides

Try AI-Math for Free

Get step-by-step solutions to any math problem. Upload a photo or type your question.

Start Solving

Derivative of a Vector Function

Differentiate r(t) component by component, then get velocity, speed and the unit tangent

Differentiate Component by Component

Rules, Speed and the Unit Tangent

Common Mistakes to Avoid

Examples

Frequently Asked Questions

How do you take the derivative of a vector function?

What does the derivative of a vector function represent?

Is there a product rule for dot and cross products?

Why can the unit tangent vector fail to exist?

Related Solvers

Related Guides

Try AI-Math for Free

Derivative of a Vector Function

Differentiate r(t) component by component, then get velocity, speed and the unit tangent

Differentiate Component by Component

Rules, Speed and the Unit Tangent

Common Mistakes to Avoid

Examples

Problem: FindFind Find\mathbf{r}'(t)forforfor\mathbf{r}(t) = \langle t^2,; \sin t,; e^{3t}\rangle$$

Problem: ForthehelixFor the helix Forthehelix\mathbf{r}(t) = \langle \cos t,; \sin t,; t\rangle,findthespeedandtheunittangentat, find the speed and the unit tangent at ,findthespeedandtheunittangentatt = \pi/2$$

Problem: WithWith With\mathbf{u}(t) = \langle t,, t^2,, 1\rangleandandand\mathbf{v}(t) = \langle 1,, t,, t^3\rangle,find, find ,find\dfrac{d}{dt}[\mathbf{u} \cdot \mathbf{v}]$$

Frequently Asked Questions

How do you take the derivative of a vector function?

How do you take the derivative of a vector function?

What does the derivative of a vector function represent?

What does the derivative of a vector function represent?

Is there a product rule for dot and cross products?

Is there a product rule for dot and cross products?

Why can the unit tangent vector fail to exist?

Why can the unit tangent vector fail to exist?

Related Solvers

Related Guides

Try AI-Math for Free