Gradients of curves and optimisation — Rates of Change

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Gradients of curves and optimisation — the key ideas

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Step 1 · The tangent

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The gradient of a curve at a point is the gradient of the tangent to the curve at that point — the straight line that just touches the curve there and has the same direction. To estimate it from a graph, draw the tangent with a ruler and calculate rise ÷ run.

Step 2 · Estimating with a small chord

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Without a graph, take a very short chord starting at the point: e.g. for $y = x^2$ at $x = 2$, use $x = 2$ and $x = 2.01$: $\dfrac{4.0401 - 4}{0.01} = 4.01 \approx 4$. For $y = x^2$ the gradient at $x$ follows the pattern $2x$.

Step 3 · Maximum and minimum points

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At the top of a curve (a maximum) or the bottom (a minimum) the tangent is horizontal, so the gradient is 0. To the left of a maximum the gradient is positive and to the right it is negative.

Step 4 · Optimisation

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Optimisation means finding the greatest or least value of something: draw up a table or graph of the quantity and find where it peaks. For example, a rectangle with perimeter 20 m and width $x$ has area $A = x(10 - x)$; the greatest area is where the gradient is 0.

Worked example

A rectangle has perimeter 20 m and width $x$ m. Find the greatest possible area.

Length $= 10 - x$, so $A = x(10 - x)$. Table: $x = 1$: 9; 2: 16; 3: 21; 4: 24; 5: 25; 6: 24; 7: 21.

The greatest area is 25 m², when $x = 5$ (a square). At this point the gradient of the graph is 0.

Test yourself

Past-paper style question

A rectangle has a perimeter of 20 m. Its width is $x$ m and its area is $A$ m², where $A = x(10 - x)$.

(a) Complete the table of $A$ for $x = 1, 2, \dots, 9$.
(b) Find the greatest possible area and the value of $x$ that gives it.
(c) What is the gradient of the graph of $A$ against $x$ at this point? [5 marks]

Show the answer

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(a) $A$: 9, 16, 21, 24, 25, 24, 21, 16, 9. (2 marks)

(b) Greatest area = 25 m² when $x = $ 5. (2 marks)

(c) The gradient is 0 (the tangent at a maximum is horizontal). (1 mark)

Practice

Gradients of curves and optimisation worksheet

Five short questions. Work through them, then reveal the mark scheme to check.

Using the pattern gradient $= 2x$, find the gradient of $y = x^2$ at $x = 4$.
Estimate the gradient of $y = x^2$ at $x = 2$ using the chord from $x = 2$ to $x = 2.01$.
The tangent to a curve at P passes through (1, 2) and (5, 10). Find the gradient of the curve at P.
A rectangle has perimeter 16 m and width $x$ m. Its area is $A = x(8 - x)$. Find the greatest possible area.
What is the gradient of the tangent at the highest point of a curve?

Mark scheme

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$2 \times 4 = $ 8.
$\dfrac{4.0401 - 4}{0.01} = $ 4.01 ≈ 4.
$\dfrac{10 - 2}{5 - 1} = $ 2.
Greatest at $x = 4$: $A = 4 \times 4 = $ 16 m².
0 (the tangent is horizontal).

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