These eleven questions cover the whole module: tangent and normal equations, rates of change, related rates and interpreting signs. They are original and ordered from easier to harder. All are written for practice with the methods taught in tangents, normals and rates.
Write each full solution on paper before you open the answer. Compare your method as well as your final value, and note any slip in your mistake log.
Tangents and normals
Question 1. Find the equation of the tangent to y = x² − 2x + 3 at the point where x = 3.
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y = 9 − 6 + 3 = 6, so the point is (3, 6). dy/dx = 2x − 2, so m = 4.
y − 6 = 4(x − 3), so y = 4x − 6. Check: 4(3) − 6 = 6. ✓
Question 2. Find the equation of the normal to the same curve at x = 3, in the form ax + by + c = 0.
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The tangent gradient is 4, so the normal gradient is −1/4.
y − 6 = −(1/4)(x − 3). Multiply by 4: 4y − 24 = −x + 3, so x + 4y − 27 = 0. Check: 3 + 24 − 27 = 0. ✓
Question 3. For y = x³ − 3x², find the tangent and the normal at the point where x = 1.
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y = 1 − 3 = −2, so the point is (1, −2). dy/dx = 3x² − 6x = 3 − 6 = −3.
Tangent: y + 2 = −3(x − 1), so y = −3x + 1. Check: −3 + 1 = −2. ✓
Normal gradient: 1/3. y + 2 = (1/3)(x − 1), so 3y + 6 = x − 1 and x − 3y − 7 = 0. Check: 1 + 6 − 7 = 0. ✓
Question 4. Find the equation of the tangent to y = x² − 4x + 7 that is parallel to the line y = 2x + 1.
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Parallel lines have the same gradient, so dy/dx = 2. Then 2x − 4 = 2, giving x = 3.
y = 9 − 12 + 7 = 4, so the point is (3, 4). y − 4 = 2(x − 3), so y = 2x − 2. Check: 2(3) − 2 = 4. ✓
Question 5. The curve y = 6/x passes through (2, 3). Find the tangent at this point and the coordinates where the tangent meets the axes.
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y = 6x⁻¹, so dy/dx = −6/x². At x = 2 the gradient is −6/4 = −3/2.
y − 3 = −(3/2)(x − 2), so 2y − 6 = −3x + 6 and 3x + 2y − 12 = 0. Check: 6 + 6 − 12 = 0. ✓
When y = 0: 3x = 12, so x = 4. When x = 0: 2y = 12, so y = 6. The tangent meets the axes at (4, 0) and (0, 6).
Rates of change
Question 6. A ball is thrown so that its height is h = 20t − 5t² metres after t seconds. Find dh/dt when t = 3 and interpret it. Find also the greatest height.
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dh/dt = 20 − 10t. At t = 3: 20 − 30 = −10. The ball is falling at 10 m per second when t = 3.
The rate is zero when 20 − 10t = 0, so t = 2. Then h = 40 − 20 = 20 m, the greatest height, because the rate changes from positive to negative.
Question 7. The cost of making x items is C = 0.5x² + 20x + 100 (in RM). Find the rate of change of C with respect to x when x = 30, and state what it means.
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dC/dx = x + 20. At x = 30: 50.
When 30 items are made, the cost is increasing at RM50 per extra item.
Question 8. The temperature of a cooling liquid is T = 100 − 12t + 0.5t² °C after t minutes, for 0 ≤ t ≤ 12. Find dT/dt when t = 4, interpret it, and find the lowest temperature.
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dT/dt = −12 + t. At t = 4: −8. The temperature is decreasing at 8 °C per minute.
The rate is zero when t = 12. T = 100 − 144 + 72 = 28 °C. This is the lowest temperature in the stated range, because the rate is negative for every t below 12.
Related rates
Question 9. The side of a cube is increasing at 0.2 cm per second. Find the rate at which its surface area S = 6x² is increasing when x = 10.
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dS/dx = 12x. dS/dt = 12x × 0.2 = 2.4x. At x = 10: 24 cm² per second.
Question 10. The area of a circular ripple increases at 10 cm² per second. Find the rate of increase of its radius when r = 4, giving your answer to 3 significant figures.
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A = πr², so dA/dr = 2πr = 8π at r = 4. From dA/dt = dA/dr × dr/dt: 10 = 8π × dr/dt.
dr/dt = 10/(8π) = 5/(4π) ≈ 0.398 cm per second.
Question 11. A balloon is inflated so that its volume V = (4/3)πr³ increases at 100 cm³ per second. Find dr/dt when r = 5, in exact form and to 3 significant figures.
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dV/dr = 4πr² = 100π at r = 5. Then 100 = 100π × dr/dt, so dr/dt = 1/π ≈ 0.318 cm per second.
If you got these wrong
| What went wrong | Go back to |
|---|---|
| Used the y-value as the gradient, or forgot to find the point | Find a tangent equation at a given point |
| Flipped the gradient but not the sign, or the reverse | Form a normal using the correct reciprocal sign |
| Gave an average rate, or forgot units | Use a derivative as a rate of change |
| Stopped at dA/dr and forgot dr/dt | Link two changing quantities through a related-rate model |
| Wrote the number without saying increasing or decreasing | Explain a rate sign in context |
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