Differentiating and integrating components | Year 12

01 / LEARN

Understand the method.

THE CORE RULE

\begin{array}{c} r^{'} = (\begin{array}{cccccccccc} x^{'} \\ y^{'} \\ z^{'} \end{array}) \\ \int (\begin{array}{cccccccccc} x \\ y \\ z \end{array}) d t = (\begin{array}{cccccccccc} X \\ Y \\ Z \end{array}) + C \end{array}

Here $X^{'} = x$ , $Y^{'} = y$ and $Z^{'} = z$ . Keep the same component order and use a vector constant $C = (C_{1}, C_{2}, C_{3})$ for an indefinite integral.

Differentiate one coordinate at a time

A vector function gives a vector for each value of its parameter. Differentiate each coordinate using the usual scalar rules, and keep the coordinates in their original order.

A composite function still needs its chain-rule factor. A constant coordinate has derivative zero; it does not make the entire vector derivative zero.

\frac{d}{d t} (\begin{array}{cccccccccc} t^{2} \\ sin 3 t \\ 5 \end{array}) = (\begin{array}{cccccccccc} 2 t \\ 3 cos 3 t \\ 0 \end{array})

Give every coordinate its own constant

Integrate each coordinate separately. The vector constant $C = (C_{1}, C_{2}, C_{3})$ allows each coordinate to start at a different value.

Substitute the initial vector at the given time to find the constants. Then check both the derivative and the initial value: these are separate conditions.

\int (\begin{array}{cccccccccc} 2 t \\ 3 \\ 4 t^{3} \end{array}) d t = (\begin{array}{cccccccccc} t^{2} + C_{1} \\ 3 t + C_{2} \\ t^{4} + C_{3} \end{array})

Read a definite integral as displacement

Integrating velocity over a time interval gives the change in position. Evaluate the antiderivative at the upper time and subtract its value at the lower time, component by component.

Displacement joins the start to the finish. It does not measure the distance travelled along a curved or retraced path; that requires integrating speed.

\int_{a}^{b} v (t) d t = r (b) - r (a)

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Keep the chain factors

For $f (t) = (t^{3}, e^{2 t}, sin 3 t)$ , find $f^{'} (t)$ and $f^{'} (0)$ .

Worked solution

1
Differentiate the power component.
$\frac{d}{d t} (t^{3}) = 3 t^{2}$
Apply the ordinary power rule to the first component.
2
Differentiate the composite exponential and sine.
$\begin{aligned} \frac{d}{d t} (e^{2 t}) & = 2 e^{2 t} \\ \frac{d}{d t} (sin 3 t) & = 3 cos 3 t \end{aligned}$
The inner derivatives are $2$ and $3$ . Include them before evaluating at a particular parameter.
3
Assemble the derivative and substitute zero.
$\begin{aligned} f^{'} (t) & = (\begin{array}{cccccccccc} 3 t^{2} \\ 2 e^{2 t} \\ 3 cos 3 t \end{array}) \\ f^{'} (0) & = (0, 2, 3) \end{aligned}$
Use $e^{0} = 1$ and $cos 0 = 1$ , keeping the original coordinate order.

Answer

$f^{'} (t) = (3 t^{2}, 2 e^{2 t}, 3 cos 3 t)$ and $f^{'} (0) = (0, 2, 3)$ .

EXAMPLE 02

Recover position from an initial vector

A particle has velocity $v (t) = (2 t, 3, 4 t^{3})$ m/s and initial position $r (0) = (1, - 2, 5)$ m. Find $r (t)$ , with $t$ in seconds.

Worked solution

1
Integrate each velocity component.
$r (t) = (\begin{array}{cccccccccc} t^{2} + C_{1} \\ 3 t + C_{2} \\ t^{4} + C_{3} \end{array})$
The antiderivatives form a family of position functions. Independent constants allow different starting coordinates.
2
Apply all three initial coordinates.
$\begin{aligned} r (0) & = (C_{1}, C_{2}, C_{3}) \\ = (1, - 2, 5) \end{aligned}$
Equality of vectors means equality of each corresponding component.
3
Write the position and check it.
$\begin{aligned} r (t) & = (\begin{array}{cccccccccc} t^{2} + 1 \\ 3 t - 2 \\ t^{4} + 5 \end{array}) \\ r^{'} (t) & = (\begin{array}{cccccccccc} 2 t \\ 3 \\ 4 t^{3} \end{array}) \\ r (0) & = (1, - 2, 5) \end{aligned}$
The derivative recovers the given velocity and the initial vector is correct.

Answer

$r (t) = (t^{2} + 1, 3 t - 2, t^{4} + 5)$ metres.

EXAMPLE 03

Locate a robot and its displacement

A robot starts at $(1, - 2)$ m and moves with $v (t) = (2 t, 3)$ m/s for $0 \leq t \leq 2$ seconds. Find its path, final position and displacement. Does the displacement magnitude measure its curved route?

The curved robot path r(t)=(t squared plus 1,3t minus 2), from (1,-2) to (5,4), with increasing-time direction and a dashed straight displacement vector (4,6).

Worked solution

1
Integrate velocity and apply the starting position.
$\begin{aligned} r (t) & = (\begin{array}{cccccccccc} t^{2} + C_{1} \\ 3 t + C_{2} \end{array}) \\ r (0) & = (C_{1}, C_{2}) \\ = (1, - 2) \\ r (t) & = (\begin{array}{cccccccccc} t^{2} + 1 \\ 3 t - 2 \end{array}) \end{aligned}$
The initial position fixes the translated location of the path.
2
Substitute the finishing time.
$\begin{aligned} r (2) & = (\begin{array}{cccccccccc} 2^{2} + 1 \\ 3 (2) - 2 \end{array}) \\ = (5, 4) \end{aligned}$ m.
The graph follows the curved path from $(1, - 2)$ to $(5, 4)$ as time increases.
3
Evaluate the definite velocity integral.
$\begin{aligned} Δ r & = \int_{0}^{2} (2 t, 3) d t \\ = {[(t^{2}, 3 t)]}_{0}^{2} \\ = (4, 6) \end{aligned}$
This agrees with $(5, 4) - (1, - 2)$ . It is the straight start-to-finish vector shown dashed.
4
Interpret the magnitude.
$\begin{aligned} ∣ Δ r ∣ & = \sqrt{4^{2} + 6^{2}} \\ = 2 \sqrt{13} \end{aligned}$ m.
This is the direct separation of the endpoints. The robot travels farther because its route curves between them.

Answer

The path is $r (t) = (t^{2} + 1, 3 t - 2)$ m; the final position is $(5, 4)$ m and displacement is $(4, 6)$ m. Its magnitude $2 \sqrt{13}$ m is endpoint separation, not route length.

03 / YOUR TURN

Try it yourself.

Calculate, explain, then apply.

CHECK 01

Differentiate the components

Find the derivative of $f (t) = (cos 2 t, t^{2}, 7)$ .

Show answer & reasoning

$f^{'} (t) = (- 2 sin 2 t, 2 t, 0)$ .

The cosine needs the chain factor $2$ and a minus sign. The power differentiates to $2 t$ and the constant differentiates to zero.

CHECK 02

Check a proposed position

Given $v (t) = (2 t, 3)$ m/s and $r (0) = (1, - 2)$ m, a student proposes $r (t) = (t^{2} + 1, 3 t + 1)$ m. Its derivative is correct. Is the position correct? Repair it if needed.

Show answer & reasoning

No. Its initial position is $(1, 1)$ m. The correct position is $(t^{2} + 1, 3 t - 2)$ m.

Differentiation removes both constants, so the derivative alone cannot check the starting location. The second coordinate must start at $- 2$ , independently of the first coordinate.

CHECK 03

Find a survey probe’s endpoint

A probe starts at $(2, 0, 1)$ m with velocity $v (t) = (3, 2 t, - 1)$ m/s for $0 \leq t \leq 2$ seconds. Find its displacement and final position.

Show answer & reasoning

Displacement $(6, 4, - 2)$ m; final position $(8, 4, - 1)$ m.

Integrate: $\int_{0}^{2} (3, 2 t, - 1) d t = {[(3 t, t^{2}, - t)]}_{0}^{2} = (6, 4, - 2)$ . Add this to $(2, 0, 1)$ to locate the endpoint. The displacement integral does not include the initial position.

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6 worksheet questions · 12 marks

Open worksheet Worked solutions

Read the questions online

Differentiate $(t^{4}, ln t)$ for $t > 0$ .
2 marks
Find the second derivative of $(t^{3}, 2 t^{2}, cos t)$ .
2 marks
Integrate $(3 t^{2}, 2 t, 1)$ with initial position $(0, 1, - 1)$ at zero.
2 marks
Evaluate $\int_{0}^{2} (t, 2 t, 3) d t$ .
2 marks
Given $v = (e^{t}, cos t)$ and $r (0) = (2, 0)$ , find position.
2 marks
Evaluate $\int_{0}^{\frac{π}{2}} (cos t, sin t) d t$ .
2 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Differentiate $(t^{4}, ln t)$ for $t > 0$ .
2 marks
Use the power rule.
The logarithmic coordinate requires a positive domain.
$(4 t^{3}, \frac{1}{t})$
Find the second derivative of $(t^{3}, 2 t^{2}, cos t)$ .
2 marks
The first derivative is $(3 t^{2}, 4 t, - sin t)$ .
Differentiate again.
$(6 t, 4, - cos t)$
Integrate $(3 t^{2}, 2 t, 1)$ with initial position $(0, 1, - 1)$ at zero.
2 marks
Integrate each coordinate.
Use one constant for each initial coordinate.
$(t^{3}, t^{2} + 1, t - 1)$
Evaluate $\int_{0}^{2} (t, 2 t, 3) d t$ .
2 marks
Antiderivative $(\frac{t^{2}}{2}, t^{2}, 3 t)$ .
Evaluate upper minus lower.
$(2, 4, 6)$
Given $v = (e^{t}, cos t)$ and $r (0) = (2, 0)$ , find position.
2 marks
Integrate to $(e^{t} + C_{1}, sin t + C_{2})$ .
Initial values give $C_{1} = 1, C_{2} = 0$ .
$(e^{t} + 1, sin t)$
Evaluate $\int_{0}^{\frac{π}{2}} (cos t, sin t) d t$ .
2 marks
Use antiderivative $(sin t, - cos t)$ .
Subtract values at both endpoints.
$(1, 1)$

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 Specialist Mathematics school Vector calculus: polynomial and trigonometric parameterisations, related coordinate rates, Bézier paths, componentwise differentiation and integration, motion, tangent directions, circular motion and arc length. Reasoning and proofs use these school topics; vector-field calculus is outside this chapter.

Official curriculum source ↗

In this lesson

Understand the method.

Differentiate one coordinate at a time

Give every coordinate its own constant

Read a definite integral as displacement

See the method in action.

Keep the chain factors

Differentiate the power component.

Differentiate the composite exponential and sine.

Assemble the derivative and substitute zero.

Recover position from an initial vector

Integrate each velocity component.

Apply all three initial coordinates.

Write the position and check it.

Locate a robot and its displacement

Integrate velocity and apply the starting position.

Substitute the finishing time.

Evaluate the definite velocity integral.

Interpret the magnitude.

Try it yourself.

Differentiate the components

Check a proposed position

Find a survey probe’s endpoint

Keep practising