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Computing Vector Derivatives from Component Formulas

Differentiate vector position formulas component by component, then use the results to find velocity and acceleration at particular times.

Advanced 9 min read

What You'll Learn

  • Differentiate two- and three-component position functions one component at a time
  • Apply polynomial, trigonometric, exponential, and product-rule derivatives
  • Find velocity by differentiating position and acceleration by differentiating velocity
  • Evaluate vector derivatives at specified times using exact values
  • Check derivative calculations by differentiating the results again
  • Recognize and correct a missing chain-rule factor

From Component Formulas to Vector Derivatives

A vector-valued position function can look complicated because it contains several coordinate formulas at once. But its derivative is computed in a systematic way: differentiate each coordinate function with respect to the same parameter, then collect the resulting derivatives into a vector.

This procedure gives the velocity vector from a position vector. Differentiating once more gives the acceleration vector. The main work is therefore careful scalar differentiation, including the chain rule for trigonometric or exponential functions and the product rule when a component is a product.

Key rule: If \(\mathbf{r}(t)=\langle x(t),y(t)\rangle\), then \(\mathbf{r}'(t)=\langle x'(t),y'(t)\rangle\). If \(\mathbf{r}(t)=\langle x(t),y(t),z(t)\rangle\), then \(\mathbf{r}'(t)=\langle x'(t),y'(t),z'(t)\rangle\). Each component is differentiated with respect to \(t\).

Why Differentiation Works Component by Component

The derivative of a vector function is defined using a difference quotient, just as for a scalar function. For a position vector \(\mathbf{r}(t)\), that quotient is \(\frac{\mathbf{r}(t+h)-\mathbf{r}(t)}{h}\). In two dimensions, subtracting vectors and dividing by the scalar \(h\) operate coordinate by coordinate:

$$ \frac{\mathbf{r}(t+h)-\mathbf{r}(t)}{h} = \left\langle \frac{x(t+h)-x(t)}{h}, \frac{y(t+h)-y(t)}{h} \right\rangle. $$

If the two scalar difference quotients approach limits as \(h\) approaches zero, the vector difference quotient approaches the vector formed from those limits. The same reasoning applies to three components. Thus, whenever the component functions are differentiable, their derivatives give the derivative of the vector function.

Component derivative rule: For \(\mathbf{r}(t)=\langle x_1(t),x_2(t),\ldots,x_n(t)\rangle\), if each component is differentiable at \(t\), then \(\mathbf{r}'(t)=\langle x_1'(t),x_2'(t),\ldots,x_n'(t)\rangle\).

For motion, \(\mathbf{r}(t)\) denotes position, \(\mathbf{v}(t)=\mathbf{r}'(t)\) denotes velocity, and \(\mathbf{a}(t)=\mathbf{v}'(t)=\mathbf{r}''(t)\) denotes acceleration. These derivatives are with respect to time \(t\). If position is measured in units of length and time in seconds, velocity has units of length per second and acceleration has units of length per second squared.

Scalar Rules Used in Each Component

Differentiate each coordinate using the ordinary derivative rules for scalar functions. For example, the power rule gives \(\frac{d}{dt}(t^n)=nt^{n-1}\). The chain rule gives \(\frac{d}{dt}\sin(g(t))=\cos(g(t))g'(t)\), \(\frac{d}{dt}\cos(g(t))=-\sin(g(t))g'(t)\), and \(\frac{d}{dt}e^{g(t)}=e^{g(t)}g'(t)\). If a component is a product, use the product rule \((fg)'=f'g+fg'\).

These are scalar rules applied separately to the component formulas; they do not change the component-by-component structure of vector differentiation. Keep track of the inner derivative in every composition, and differentiate again when acceleration is requested.

Worked Examples

Worked Example: Polynomial, Trigonometric, and Exponential Components

Let \(\mathbf{r}(t)=\langle t^3-2t,\sin(2t),e^t\rangle\). Find the velocity and acceleration vectors, then evaluate both at \(t=0\).

Differentiate the first component using the power rule:

$$ \frac{d}{dt}(t^3-2t)=3t^2-2. $$

For the second component, apply the chain rule with inner function \(2t\), whose derivative is \(2\):

$$ \frac{d}{dt}\sin(2t)=\cos(2t)\cdot 2=2\cos(2t). $$

The derivative of \(e^t\) is \(e^t\). Therefore,

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

Differentiate each velocity component to obtain acceleration:

$$ \mathbf{a}(t)=\mathbf{v}'(t) =\langle 6t,\;-4\sin(2t),\;e^t\rangle. $$

At \(t=0\), the velocity components are \(3(0)^2-2=-2\), \(2\cos(0)=2\), and \(e^0=1\). The acceleration components are \(6(0)=0\), \(-4\sin(0)=0\), and \(e^0=1\). Thus

$$ \mathbf{v}(0)=\langle -2,2,1\rangle, \qquad \mathbf{a}(0)=\langle 0,0,1\rangle. $$

Check by differentiating the velocity formula: the derivatives of \(3t^2-2\), \(2\cos(2t)\), and \(e^t\) are \(6t\), \(-4\sin(2t)\), and \(e^t\), respectively, which reproduces the stated acceleration. Substitution at \(t=0\) gives the vectors above.

Worked Example: A Two-Component Position Function

Let \(\mathbf{r}(t)=\langle t^2+\cos t,\;3\sin t-t\rangle\). Find \(\mathbf{v}(t)\) and \(\mathbf{a}(t)\), and evaluate them at \(t=\frac{\pi}{3}\).

Differentiate the first component term by term: the derivative of \(t^2\) is \(2t\), and the derivative of \(\cos t\) is \(-\sin t\). For the second component, the derivative of \(3\sin t\) is \(3\cos t\), and the derivative of \(-t\) is \(-1\). Therefore,

$$ \mathbf{v}(t)=\langle 2t-\sin t,\;3\cos t-1\rangle. $$

Differentiate again. The derivative of \(2t-\sin t\) is \(2-\cos t\), and the derivative of \(3\cos t-1\) is \(-3\sin t\). Thus

$$ \mathbf{a}(t)=\langle 2-\cos t,\;-3\sin t\rangle. $$

At \(t=\frac{\pi}{3}\), use \(\sin(\frac{\pi}{3})=\frac{\sqrt{3}}{2}\) and \(\cos(\frac{\pi}{3})=\frac12\). Substituting into velocity gives

$$ \mathbf{v}\left(\frac{\pi}{3}\right) = \left\langle 2\left(\frac{\pi}{3}\right)-\frac{\sqrt{3}}{2}, 3\left(\frac12\right)-1 \right\rangle = \left\langle \frac{2\pi}{3}-\frac{\sqrt{3}}{2}, \frac12 \right\rangle. $$

Substituting into acceleration gives

$$ \mathbf{a}\left(\frac{\pi}{3}\right) = \left\langle 2-\frac12, -3\left(\frac{\sqrt{3}}{2}\right) \right\rangle = \left\langle \frac32, -\frac{3\sqrt{3}}{2} \right\rangle. $$

Check: differentiating the velocity components \(2t-\sin t\) and \(3\cos t-1\) gives \(2-\cos t\) and \(-3\sin t\). At \(t=\frac{\pi}{3}\), these are \(2-\frac12=\frac32\) and \(-3\frac{\sqrt{3}}{2}=-\frac{3\sqrt{3}}{2}\), matching the acceleration evaluation.

Worked Example: A Product and an Exponential Component

Let \(\mathbf{r}(t)=\langle e^{2t},\;te^t,\;\cos(3t)\rangle\). Find velocity and acceleration at \(t=0\).

For the first component, the chain rule gives \(\frac{d}{dt}e^{2t}=2e^{2t}\). For the second, use the product rule with \(f(t)=t\) and \(g(t)=e^t\):

$$ \frac{d}{dt}(te^t) =(1)e^t+t(e^t) =e^t(t+1). $$

For the third component, the chain rule gives \(\frac{d}{dt}\cos(3t)=-3\sin(3t)\). Therefore,

$$ \mathbf{v}(t)=\langle 2e^{2t},\;e^t(t+1),\;-3\sin(3t)\rangle. $$

Differentiate again. The first component gives \(4e^{2t}\). For the second component, apply the product rule to \(e^t(t+1)\):

$$ \frac{d}{dt}\big(e^t(t+1)\big) =e^t(t+1)+e^t =e^t(t+2). $$

The derivative of \(-3\sin(3t)\) is \(-9\cos(3t)\). Hence

$$ \mathbf{a}(t)=\langle 4e^{2t},\;e^t(t+2),\;-9\cos(3t)\rangle. $$

At \(t=0\), use \(e^0=1\), \(\sin(0)=0\), and \(\cos(0)=1\). Then

$$ \mathbf{v}(0)=\langle 2(1),\;1(0+1),\;-3(0)\rangle =\langle 2,1,0\rangle, $$ $$ \mathbf{a}(0)=\langle 4(1),\;1(0+2),\;-9(1)\rangle =\langle 4,2,-9\rangle. $$

Check by differentiating the velocity components: \((2e^{2t})'=4e^{2t}\), \((e^t(t+1))'=e^t(t+2)\), and \((-3\sin(3t))'=-9\cos(3t)\). These agree with the acceleration formula, whose value at zero is \(\langle4,2,-9\rangle\).

A Reliable Calculation Process

When finding velocity and acceleration from component formulas, organize the work so that each derivative and each substitution can be checked independently.

1
Write the components clearly.
Identify every scalar function in the position vector and the variable with respect to which you are differentiating.
2
Differentiate each component.
Use the scalar derivative rules, including the chain rule for composed functions and the product rule for products.
3
Collect the results into velocity.
Preserve the component order used in the original position vector.
4
Differentiate velocity for acceleration.
Differentiate every velocity component once more, again with respect to the same parameter.
5
Evaluate only after finding the formulas.
Substitute the requested time into each component, then simplify exact values when available.

Common Mistake: Losing the Chain-Rule Factor

A frequent error is treating the derivative of \(\sin(2t)\) as though its input were just \(t\). The incorrect calculation is

$$ \frac{d}{dt}\sin(2t)=\cos(2t). $$

This omits the derivative of the inner function \(2t\). Since \(\frac{d}{dt}(2t)=2\), the chain rule requires

$$ \frac{d}{dt}\sin(2t) = \cos(2t)\cdot 2 = 2\cos(2t). $$

The error also affects the acceleration. Differentiating the incorrect velocity component \(\cos(2t)\) would give \(-2\sin(2t)\), while the correct velocity component \(2\cos(2t)\) differentiates to \(-4\sin(2t)\). Including the inner derivative at each stage prevents this missing-factor error.

Why Component Checks Matter

A vector derivative can be incorrect even if most of its entries are right. Each coordinate may involve a different rule: one may use the power rule, another the chain rule, and another the product rule. Checking by differentiating the proposed velocity to recover acceleration is a useful way to detect errors in the second derivative. For a requested time, substituting into each component separately also helps prevent sign or arithmetic mistakes.

The central method remains the same for two or three components: differentiate each scalar formula, preserve the component order, and evaluate the resulting vectors at the requested time. The vector notation organizes the answers; the correctness of each entry comes from the corresponding scalar derivative.

Check Your Understanding

For each question, differentiate component by component and keep the components in their original order.

  1. If \(\mathbf{r}(t)=\langle t^4,\cos(3t)\rangle\), find \(\mathbf{v}(t)\) and \(\mathbf{a}(t)\).
  2. If \(\mathbf{r}(t)=\langle e^t,t^2,\sin t\rangle\), find \(\mathbf{v}(0)\).
  3. For the second component \(te^{2t}\), use the product rule and chain rule to find its first derivative.
  4. Find \(\mathbf{a}(t)\) for \(\mathbf{r}(t)=\langle 2\sin t,e^t\rangle\).
  5. Explain why differentiating \(\sin(5t)\) produces a factor of \(5\).

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