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Year 12 · Vector calculus · 24 min

Differentiating and integrating components

Differentiate and integrate vector functions component by component, preserving vector constants.

01 / LEARN

The idea, step by step

Work component by component

The derivative of r=(x,y,z) is (x′,y′,z′). Apply ordinary scalar rules, including chain-rule factors, to each corresponding component.

Integrating introduces a vector constant (C1,C2,C3). Each component has an independent constant; one scalar constant cannot generally serve all coordinates.

Check a recovered position

Integrate velocity to obtain a position family, then substitute the initial time to determine each constant. Differentiate the result and check the initial vector separately.

Definite vector integration gives displacement: ∫abv⁢(t)d⁢t=r⁢(b)−r⁢(a). It does not give total distance along a curved path.

A common trap

The integration constant is identical in all coordinates.

Instead: Each coordinate has its own constant.

02 / WORKED EXAMPLES

See the thinking

Example 1

Worked example

Differentiate (t3,e2⁢t,sin3⁢t).

  1. Use the power rule on the first coordinate.
  2. The exponential and sine coordinates need chain factors 2 and 3.
  3. Retain component order.
(3t2,2e2⁢t,3cos3⁢t).
Example 2

Worked example

Given v=(2⁢t,3,4t3) and r⁢(0)=(1,−2,5), find position.

  1. Integrate to (t2+C1,3⁢t+C2,t4+C3).
  2. At zero the constants form (1, -2, 5).
  3. Differentiate to check the velocity.
(t2+1,3⁢t−2,t4+5).

Check your understanding

Differentiate (cos2⁢t,t2).

Show answer

(−2sin2⁢t,2⁢t)

Apply the chain rule to the cosine coordinate.

03 / WORKSHEET

Your turn

6 questions · 12 marks · Work at your own pace.

Read the questions online
  1. Differentiate (t4,lnt) for t>0.
    2 marks
  2. Find the second derivative of (t3,2t2,cost).
    2 marks
  3. Integrate (3t2,2⁢t,1) with initial position (0, 1, -1) at zero.
    2 marks
  4. Evaluate ∫02(t,2⁢t,3)d⁢t.
    2 marks
  5. Given v=(et,cost) and r⁢(0)=(2,0), find position.
    2 marks
  6. Evaluate ∫0π⁢/2(cost,sint)d⁢t.
    2 marks
Read the worked solutions

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

  1. Differentiate (t4,lnt) for t>0.
    2 marks

    Use the power rule.

    The logarithmic coordinate requires a positive domain.

    (4t3,1⁢/⁢t)
  2. Find the second derivative of (t3,2t2,cost).
    2 marks

    The first derivative is (3t2,4⁢t,−sint).

    Differentiate again.

    (6⁢t,4,−cost)
  3. Integrate (3t2,2⁢t,1) with initial position (0, 1, -1) at zero.
    2 marks

    Integrate each coordinate.

    Use one constant for each initial coordinate.

    (t3,t2+1,t−1)
  4. Evaluate ∫02(t,2⁢t,3)d⁢t.
    2 marks

    Antiderivative (t2/2,t2,3⁢t).

    Evaluate upper minus lower.

    (2,4,6)
  5. Given v=(et,cost) and r⁢(0)=(2,0), find position.
    2 marks

    Integrate to (et+C1,sint+C2).

    Initial values give C1=1,C2=0.

    (et+1,sint)
  6. Evaluate ∫0π⁢/2(cost,sint)d⁢t.
    2 marks

    Use antiderivative (sint,−cost).

    Subtract values at both endpoints.

    (1,1)
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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 Specialist Mathematics topic 6.3–6.5: polynomial and trigonometric parameterisations, related coordinate rates, motion, tangent directions and arc length. This is school Vector calculus, not vector-field calculus.

Official curriculum source ↗
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