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Math lesson plan · 12th grade (Year 13 in England) · 45 minutes

Find a curve from its gradient and a point

Calculus

CCSS HSF.BF.B.4England: A-level Pure Mathematics

Objective and success criteria

Students will: Find a curve from its gradient and a point.

  • I can answer questions on today's skill on my own, without help.
  • I can explain each step and say why it works.
  • I can check my answer and spot a mistake.

Before and after

Needs first: Integrate powers of x (indefinite integrals)

Leads to: the next skill in this area.

Materials: Graphing tool or graph paper, Scientific calculators, Mini whiteboards and pens, Printed student sheet (below).

Warm-up · 5 min

  • Quick retrieval from earlier learning (Integrate powers of x (indefinite integrals)). Students answer on mini whiteboards.
  • Show answers together and note who needs support in the main task.
  1. Find ∫ −18x⁵ dx. Leave out the + C. → -3*x^6Prompt: Add 1 to the power, then divide by the new power.
  2. Find ∫ (−3x² + 8x) dx. Leave out the + C. → -x^3+4x^2Prompt: Integrate each term separately.
  3. Jamal needs ∫ 6/x⁴ dx. Which answer is correct? (6/x⁵ + C / −24/x⁵ + C / −2/x³ + C / 6·ln x + C) → −2/x³ + CPrompt: Rewrite as 6x^(−4).

Hook · 5 min

  • Show a distance-time graph of a journey and ask how fast it was going at one exact moment.
  • Share the objective: Find a curve from its gradient and a point. Read the success criteria together.

I do: model it · 10 min

  • Work through the example out loud, saying why you do each step, not just what you do.
  • Example: dy/dx = −4x − 5 and the curve passes through (3, 4). Find y in terms of x.
  • Step 1: y = −2x² − 5x + C
  • Step 2: 4 = −33 + C → C = 37
  • Step 3: y = −2x² − 5x + 37
  • Answer: -2x^2-5x+37
  • Pause after each step and ask a student to say what comes next.

We do: guided practice · 10 min

  • Do these together. Students try each step on whiteboards before you reveal it.
  • Use the hints as prompts rather than giving the answer.
  1. dy/dx = 6x − 4 and the curve passes through (3, 0). Find y in terms of x. → 3x^2-4x-15Prompt: Integrate to get y, including + C.
  2. A cyclist's velocity is v = 5 + 2t m/s. Starting from s = 0, how far has she travelled after 4 seconds? → 36Prompt: Displacement is the integral of velocity.

You do: independent practice · 10 min

  • Students work on the student sheet. Circulate and check the first two answers of every student early.
  • Students who finish: write a question of their own for a partner, with the answer.

The 8 questions are on the student sheet below, with answers in the key.

Exit ticket · 5 min

  • Two questions on their own. Sort answers into got it / nearly / not yet to plan the next lesson.
  1. dy/dx = 4x + 2 and the curve passes through (−1, −4). Find y in terms of x. → 2x^2+2x-4Prompt: Integrate to get y, including + C.
  2. A cyclist's velocity is v = 3 + 2t m/s. Starting from s = 0, how far has she travelled after 1 seconds? → 4Prompt: Displacement is the integral of velocity.

Watch out for

  • Forgetting the constant of integration.
  • Differentiating a product by multiplying the derivatives.

Support and stretch

Support

  • Review first: Integrate powers of x (indefinite integrals).
  • Use concrete materials: graphing tool or graph paper.
  • Give the first step of each question, or a partly worked example to finish.
  • Pair with a confident partner for the guided practice.

Stretch

  • Explain a common mistake in writing and how to fix it.
  • Write a word problem that uses today's skill, then solve a partner's.

Homework and more practice

Free online practice with hints and printable worksheets for this skill: eurekami.com/math/grade-12.

Free lesson plan from eurekami.com/lesson-plan-builder

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Find a curve from its gradient and a point

Name

Date

Score/ 8

  1. Question 1

    A cyclist's velocity is v = 3 + 8t m/s. Starting from s = 0, how far has she travelled after 3 seconds?

    Answer

  2. Question 2

    dy/dx = 2x + 3 and the curve passes through (2, 7). Find y in terms of x.

    Answer

  3. Question 3

    A cyclist's velocity is v = 3 + 8t m/s. Starting from s = 0, how far has she travelled after 6 seconds?

    Answer

  4. Question 4

    dy/dx = 4x + 2 and the curve passes through (2, −5). Find y in terms of x.

    Answer

  5. Question 5

    A cyclist's velocity is v = 2 + 6t m/s. Starting from s = 0, how far has she travelled after 4 seconds?

    Answer

  6. Question 6

    dy/dx = −6x + 3 and the curve passes through (0, −7). Find y in terms of x.

    Answer

  7. Question 7

    A cyclist's velocity is v = 1 + 2t m/s. Starting from s = 0, how far has she travelled after 2 seconds?

    Answer

  8. Question 8

    dy/dx = 2x and the curve passes through (0, −2). Find y in terms of x.

    Answer

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Answer key: Find a curve from its gradient and a point

  1. 1. 45

  2. 2. x^2+3x-3

  3. 3. 162

  4. 4. 2x^2+2x-17

  5. 5. 56

  6. 6. -3x^2+3x-7

  7. 7. 6

  8. 8. x^2-2