Applications of Differentiation: Gradients, Tangents and Turning Points
Applications of differentiation is the use of the gradient function, dy/dx, to solve problems about a curve, including finding gradients, tangent and normal equations, and the position and nature of turning points using the second derivative. It is a major AQA Level 2 Further Maths topic, often applied to real-life optimisation problems such as maximising area or profit.
Before you start
Make sure you're comfortable with these topics first:
Method
- Differentiate the equation of the curve to find dy/dx, the gradient function.
- For a tangent, substitute the given x-value into dy/dx for the gradient and into y for the point, then use y - y1 = m(x - x1).
- For a normal, use the negative reciprocal of the tangent gradient, -1/m, in the same point-gradient equation.
- For stationary points, set dy/dx = 0 and solve for x, then substitute each x-value back into y to find the coordinates.
- Find the second derivative d2y/dx2 and evaluate it at each stationary point: a positive value means a minimum, a negative value means a maximum.
- In optimisation problems, write the quantity to be maximised or minimised as a function of a single variable before differentiating.
Worked example
A curve has equation y = x^2 + 2x - 3. Find the equation of the normal to the curve at the point where x = 2, giving your answer in the form x + ky + c = 0.
- Differentiate: dy/dx = 2x + 2.
- Find the y-coordinate at x = 2: y = 4 + 4 - 3 = 5, so the point is (2, 5).
- Find the tangent gradient at x = 2: dy/dx = 2(2) + 2 = 6.
- The normal gradient is the negative reciprocal: -1/6.
- Use the point-gradient form: y - 5 = (-1/6)(x - 2).
- Rearrange: 6y - 30 = -(x - 2), so the final answer is x + 6y - 32 = 0.
Practice questions
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Q1Find dy/dx for y = x^2 - 8x + 3.Show answer
Answer: dy/dx = 2x - 8
Q2Find the gradient of the curve y = x^3 - 5x at the point where x = 2.Show answer
Answer: gradient = 7 (dy/dx = 3x^2 - 5, then substitute x = 2)
Q3Find the equation of the tangent to the curve y = x^2 - 2x + 4 at the point where x = 3.Show answer
Answer: y = 4x - 5
Q4Find the coordinates of the turning point of the curve y = x^2 + 4x - 1, and state whether it is a maximum or a minimum.Show answer
Answer: (-2, -5), a minimum (coefficient of x^2 is positive)
Q5A curve has equation y = x^3 - 12x + 2. Find the coordinates of the stationary points and use the second derivative to determine their nature.Show answer
Answer: (2, -14) is a minimum and (-2, 18) is a maximum
Q6A gardener has 44 metres of trellis to enclose a rectangular vegetable bed, using the wall of a shed as one boundary so trellis is only needed for the other three sides. Letting x metres be the width of the bed (perpendicular to the shed wall), show that the area enclosed is A = 44x - 2x^2, then find the value of x that gives the maximum area and state this maximum area.Show answer
Answer: x = 11 m, maximum area = 242 m^2 (set dA/dx = 44 - 4x = 0)
Exam-style questions
Written in the style of a GCSE Further Maths exam paper, with a full mark scheme.
A curve has equation y = x^2 + 6x - 2. Find dy/dx, and state the gradient of the curve at the point where x = 0.
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A curve has equation y = x^2 - 2x - 8. Find the equation of the normal to the curve at the point where x = 4. Give your answer in the form x + ky + c = 0.
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A curve has equation y = x^3 - 12x^2 + 36x - 5. Find the x-coordinates of the stationary points of the curve, and use the second derivative to determine whether each is a maximum or a minimum point.
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Free printable worksheet
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