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Cambridge IGCSE0606

Calculus

Mathematics Additional 0606 Chapter Notes

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Calculus
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In this chapter

  1. 1Derivatives and the chain rule
  2. 2Products and quotients
  3. 3Gradients, tangents and normals
  4. 4Stationary points and maxima and minima
  5. 5Rates of change and small changes
  6. 6Integration as the reverse of differentiation
  7. 7Definite integrals and areas
  8. 8Kinematics

1. Derivatives and the chain rule

The derived function gives the gradient of a curve at every point. It is written f'(x) or dy/dx. The second derivative is f''(x) or d²y/dx².

The notation comes from small steps. The gradient (delta y)/(deltax) tends to dy/dx as delta x → 0. Only an informal idea of a limit is needed.

For a function of a function, use the chain rule: differentiate the outer function, then multiply by the derivative of the inner function. For trigonometric functions, angles are always in radians.

d²y/dx² = d/dx(dy/dx)

d/dx(xⁿ) = nxⁿ⁻¹

d/dx(sin x) = cos x d/dx(cos x) = -sin x d/dx(tan x) = sec² x

d/dx(e^(x)) = e^(x) d/dx(ln x) = 1/x

dy/dx = dy/du × du/dx

Examiner insight

“Very occasionally, candidates misunderstood the chain rule.” “A significant number of candidates were not awarded the final mark as they gave their answer in degrees rather than radians.”

Worked example 14 marks

It is given that y = tan³ 5x. Find dy/dx in terms of tan 5x.

  1. 1

    Outer function first: dy/dx = 3tan² 5x × d/dx(tan 5x) [1]

  2. 2

    Inner function: d/dx(tan 5x) = 5sec² 5x [1]

  3. 3

    dy/dx = 15tan² 5x sec² 5x [1]

  4. 4

    Use sec² 5x = 1 + tan² 5x: dy/dx = 15tan² 5x (1 + tan² 5x) [1]

2. Products and quotients

For a product y = uv, use the product rule. For a quotient y = u/v, use the quotient rule.

State the rule, then list the four components u, v, u' and v'. In the quotient rule the numerator is vu' - uv' and the denominator is v², so the denominator is squared.

d/dx(uv) = udv/dx + vdu/dx

d/dx(u/v) = (vdu/dx - udv/dx)/v²

Examiner insight

“Some candidates did not use the correct structure of the quotient rule. For example, candidates sometimes reversed the numerator or omitted to square the denominator.”

Worked example 13 marks

A curve has equation y = xe^(2x+1). Find dy/dx.

  1. 1

    Product rule with u = x and v = e^(2x+1) [1]

  2. 2

    d/dx(e^(2x+1)) = 2e^(2x+1) by the chain rule [1]

  3. 3

    dy/dx = x × 2e^(2x+1) + e^(2x+1) × 1 = (2x+1)e^(2x+1) [1]

3. Gradients, tangents and normals

The gradient of a curve at a point is the value of dy/dx there. Substitute the x-coordinate into the derivative. The result is a gradient, not a y-coordinate.

The tangent has that gradient. The normal is perpendicular to the tangent, so its gradient is -1 divided by the tangent's gradient.

Find y from the equation of the curve, then use y - y₁ = m(x - x₁). Leave the answer exact when the question asks for it.

y - y₁ = m(x - x₁)

m_normal = (-1)/m_tangent

Examiner insight

“Some candidates may have improved if they had reread the question as they found the equation of the tangent and not the equation of the normal.” “In weaker responses, candidates found the coordinates of a second point and did not differentiate but instead found the gradient of the chord from (1, 1) to their point and used that as the gradient of the tangent. This was not acceptable.”

Worked example 14 marks

Find, in exact form, the equation of the tangent to the curve y = ln((2x-4)⁵) at the point where x = 3.

  1. 1

    dy/dx = (5(2x-4)⁴ × 2)/((2x-4)⁵) = 10/(2x-4) = 5/(x-2) [1]

  2. 2

    At x = 3: gradient = 5/(3-2) = 5 [1]

  3. 3

    At x = 3: y = ln 2⁵ = ln 32 [1]

  4. 4

    y - ln 32 = 5(x - 3) [1]

4. Stationary points and maxima and minima

At a stationary point dy/dx = 0. Solve for x, then find y from the original equation. Points of inflexion are not included.

Use the second derivative to find the type. A positive value means a minimum. A negative value means a maximum. The first derivative test checks the sign of dy/dx either side of the point.

Full justification is expected: state the sign and the conclusion. In a practical problem, write one quantity in terms of one variable, differentiate, set it to zero, and answer the question asked.

dy/dx = 0

d²y/dx² > 0 ⇒ minimum

d²y/dx² < 0 ⇒ maximum

Examiner insight

“Some candidates did not show that the value of v gave a minimum point.” “As has been common in this type of question in the past, many candidates have not answered the question completely by omitting in this case to find the stationary value of S.”

Worked example 15 marks

A ship is sailing with speed v km h⁻¹. The sailing cost per hour, textdollarC, is given by C = v² + 3000/v + 100. Find the speed that makes C a minimum. Justify that this value of C is a minimum.

  1. 1

    dC/dv = 2v - 3000/v² [2]

  2. 2

    Set dC/dv = 0: 2v³ = 3000, so v³ = 1500 [1]

  3. 3

    v = √[3]1500 = 11.4 km h⁻¹ [1]

  4. 4

    d²C/dv² = 2 + 6000/v³ > 0 for v > 0, so it is a minimum [1]

5. Rates of change and small changes

For connected rates of change, link the rates with the chain rule. Write down which rate you are given and which you want. The units help: dV/dt is volume per unit time.

For a small change, a small increase h in x changes y by about dy/dx times h. Evaluate dy/dx at the starting value of x.

A decreasing quantity has a negative rate.

dy/dt = dy/dx × dx/dt

delta y ≈ dy/dx × delta x

Examiner insight

“A few candidates omitted to use x = 4 and gave their answer as a product of h and an expression in x.” “The most common error made by candidates was not appreciating that y was decreasing in value.”

Worked example 14 marks

The variables x and y are related by y = 3√x + 1/(√x). Use calculus to find the approximate change in y when x increases from 4 to 4 + h, where h is small.

  1. 1

    y = 3x^(1/2) + x^(-1/2), so dy/dx = 3/2x^(-1/2) - 1/2x^(-3/2) [2]

  2. 2

    Small change: delta y ≈ dy/dx × h, with x = 4 [1]

  3. 3

    dy/dx = 3/2×1/2 - 1/2×1/8 = 11/16, so delta y ≈ 11/16h [1]

6. Integration as the reverse of differentiation

Integration is the reverse process of differentiation. Check an answer by differentiating it.

An indefinite integral needs an arbitrary constant, + c. Without it the answer is incomplete.

For a linear function ax + b inside the function, divide by a. This covers (ax+b)ⁿ for any rational n, sin(ax+b), cos(ax+b), sec²(ax+b) and e^(ax+b). The case n = -1 gives a logarithm.

int xⁿ dx = (xⁿ⁺¹)/(n+1) + c (n ≠ -1)

int 1/x dx = ln x + c int 1/(ax+b) dx = 1/aln(ax+b) + c

int (ax+b)ⁿ dx = ((ax+b)ⁿ⁺¹)/(a(n+1)) + c (n ≠ -1)

int sin(ax+b) dx = -1/acos(ax+b) + c int cos(ax+b) dx = 1/asin(ax+b) + c

int sec²(ax+b) dx = 1/atan(ax+b) + c int e^(ax+b) dx = 1/ae^(ax+b) + c

Examiner insight

“A few candidates offered a correct expression but omitted the constant of integration. This was not condoned for full credit.”

Worked example 13 marks

Find int sec²x/3 dx.

  1. 1

    The answer has the form ktanx/3 [1]

  2. 2

    Divide by the coefficient 1/3: k = 3 [1]

  3. 3

    3tanx/3 + c [1]

7. Definite integrals and areas

A definite integral has limits. Integrate first, then substitute the upper limit and subtract the value at the lower limit. No constant is needed.

For the area between a curve and the x-axis, the limits are often the x-intercepts. For the area between a line and a curve, first find where they meet. The area is the integral of the upper function minus the lower function. Some areas are a sum of two areas.

Show the integration and the substitution of the limits. When the question asks for an exact answer, do not use decimals.

int_a^(b) f(x) dx = F(b) - F(a)

area between curves = int_a^(b) (y_upper - y_lower)dx

Examiner insight

“Many candidates were unable to identify a suitable strategy to find the limits they needed.” “All too often, candidates resorted to the use of their calculators when evaluating the square brackets using the limits.”

Worked example 16 marks

A curve has equation y = -x² + 3ax - 2a² where a is a positive constant. The area of the region enclosed by the curve and the x-axis is 9/16. Find the value of a.

  1. 1

    Curve meets the x-axis: -(x-a)(x-2a) = 0 [1]

  2. 2

    x = a and x = 2a [1]

  3. 3

    int_a^(2a)(-x² + 3ax - 2a²)dx = 9/16 [1]

  4. 4

    [-x³/3 + 3ax²/2 - 2a²x]_a^(2a) [1]

  5. 5

    (-2a³/3) - (-5a³/6) = a³/6 [1]

  6. 6

    a³/6 = 9/16, so a³ = 27/8 and a = 3/2 [1]

8. Kinematics

For a particle on a straight line, displacement is s, velocity is v = ds/dt and acceleration is a = dv/dt. Going the other way, integrate: v = int a dt and s = int v dt.

Each integration brings a constant. Find it from the condition the question gives, such as the velocity at a given time.

On a displacement–time graph the gradient is the velocity. On a velocity–time graph the gradient is the acceleration. Distance travelled is not the final displacement if the particle turns round. Find the times when v = 0, find the displacement at each, and add the sizes of the separate movements.

v = ds/dt a = dv/dt = d²s/dt²

v = int a dt s = int v dt

Examiner insight

“Many candidates did not use a correct method, with many making a substitution of t = 6 into their response to part (b) and not considering any other displacements.” “Some candidates, having considered the correct times, did not change a displacement of – 4.5 to 4.5 in their final calculation.”

Worked example 14 marks

In this question time t is in seconds and displacement s is in metres. For t geqslant 0, a particle moves on the positive x-axis. At time t the acceleration of the particle is 2t - 9. At time t = 7 the velocity of the particle is 4. Find the times when the velocity of the particle is 0.

  1. 1

    v = int (2t - 9) dt = t² - 9t + c [1]

  2. 2

    4 = 7² - 9× 7 + c, so c = 18 [1]

  3. 3

    t² - 9t + 18 = 0, so (t-3)(t-6) = 0 [1]

  4. 4

    t = 3 and t = 6 [1]

Recap

  • dy/dx is the gradient. Use radians for trigonometric functions.
  • Chain rule: dy/dx = dy/du×du/dx.
  • Product rule: u v' + v u'. Quotient rule: (vu' - uv')/v².
  • Tangent: y - y₁ = m(x - x₁). Normal gradient = (-1)/m.
  • Stationary point: dy/dx = 0. Second derivative positive: minimum. Negative: maximum.
  • Small change: delta y ≈ dy/dx delta x. Connected rates: multiply the chain.
  • Integrate, then add c. Divide by a for (ax+b).
  • Definite integral: F(b) - F(a). Area between curves: upper minus lower.
  • v = ds/dt, a = dv/dt. Distance travelled: split where v = 0.

End-of-chapter exercise

Test yourself on the whole chapter. Work through these before moving on.

  1. Given that f(x) = 2x³ - 4/x, find f'(x) and f''(x).3 marks
  2. Differentiate y = (4x² + 5)^(3/2) with respect to x.3 marks
  3. Find the gradient of the curve y = sin(2x+1) at the point where x = 0.5. Give your answer correct to 3 significant figures.3 marks
  4. Do not use a calculator. Find the gradient of the curve y = (2x+1)/(x-3) at the point where x = 4.4 marks
  5. Do not use a calculator. Find the equation of the normal to the curve y = x³ - 4x + 1 at the point where x = 2. Give your answer in the form x + by + c = 0, where b and c are integers.4 marks
  6. Do not use a calculator. The variables x and y are related by y = x³ - 2/x. (a) Use calculus to find the approximate change in y when x increases from 2 to 2 + h, where h is small. (b) When x = 2, x is decreasing at a rate of 0.4 units per second. Find the corresponding rate of change of y.6 marks
  7. A farmer has 120 m of fencing. She uses all of it to make three sides of a rectangular field. The fourth side is a straight wall. The two sides at right angles to the wall have length x metres. Find the greatest possible area of the field. Justify that it is a maximum.5 marks
  8. Do not use a calculator. Find int((3x-2)⁴ + 8/(2x+5))dx.4 marks
  9. Do not use a calculator. The line y = k - 2x, where k is a constant, meets the curve y = 12/(2x+1) at the point where x = 1/2 and at one other point. Find the exact area of the region enclosed by the line and the curve. Give your answer in the form p - qln 3, where p and q are integers.8 marks
  10. A particle moves in a straight line. At time t seconds after passing through a fixed point O, its velocity is v m s⁻¹, where v = e^(2t) - 6e^(t) + 8. (a) Find the exact times when the particle is at instantaneous rest. (b) Find the acceleration of the particle at the first of these times. (c) Find the exact distance travelled by the particle between t = 0 and t = ln 4.10 marks

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