The Gradient of a Curve

A gradient that keeps changing

A straight line has a single gradient — it is equally steep everywhere. A curve is different: it is gentle in some places and steep in others, so its gradient changes from point to point. The whole challenge is finding the gradient at one chosen spot.

If straight-line gradients are not fresh, revisit straight-line graphs and gradients first, since the gradient of a curve is defined using a straight line.

Tangent and chord

Two straight lines help us think about a curve:

• A chord joins two points on the curve. Its gradient is the average rate of change between them.
• A tangent touches the curve at one point, matching its direction exactly. Its gradient is the gradient of the curve at that point.

Imagine sliding the second end of a chord closer and closer to the first. The chord pivots until it becomes the tangent. That limiting idea is explained in introduction to calculus: rates of change; here we use the shortcut, differentiation, to get the gradient directly.

Finding the gradient at a point

The method has three steps:

1. Differentiate the curve to get the gradient function dy/dx.
2. Substitute the x-value of your point.
3. Read off the gradient.

Worked example 1: Find the gradient of y = x² at x = 3.

1. Differentiate: dy/dx = 2x.
2. Substitute x = 3: 2 × 3 = 6.
3. The curve has gradient 6 at x = 3 (quite steep, sloping up).

Worked example 2: Find the gradient of y = x² − 4x + 1 at x = 1.

1. dy/dx = 2x − 4.
2. At x = 1: 2(1) − 4 = −2.
3. Gradient −2 — the curve slopes downwards here.

Notice the same curve gives different gradients at different x-values. The table shows this for y = x²:

x	dy/dx = 2x	Steepness/direction
−2	−4	steep, downward
0	0	flat (the bottom)
1	2	gentle, upward
3	6	steep, upward

The equation of a tangent line

A tangent is a straight line, so it has its own equation y = mx + c. To find it you need the gradient at the point and the coordinates of the point.

Worked example 3: Find the equation of the tangent to y = x² at the point where x = 2.

1. Point: at x = 2, y = 2² = 4, so the point is (2, 4).
2. Gradient: dy/dx = 2x, so at x = 2 the gradient is m = 4.
3. Use y = mx + c with the point: 4 = 4(2) + c, so 4 = 8 + c, c = −4.
4. Tangent: y = 4x − 4.

The equation of a normal line

The normal is the line perpendicular to the tangent at the same point. Since perpendicular gradients multiply to −1, the normal's gradient is the negative reciprocal of the tangent's gradient.

Worked example 4: Find the equation of the normal to y = x² at x = 2.

1. From example 3, the tangent gradient is 4 and the point is (2, 4).
2. Normal gradient = negative reciprocal of 4 = −1/4.
3. y = mx + c with the point: 4 = (−1/4)(2) + c = −0.5 + c, so c = 4.5.
4. Normal: y = −¼x + 4.5.

Turning points: where the gradient is zero

At the top of a hill or the bottom of a valley on a curve, the curve is momentarily flat: its gradient is 0. These are called turning points (or stationary points). Find them by solving dy/dx = 0.

Worked example 5: Find the turning point of y = x² − 6x + 5.

1. dy/dx = 2x − 6.
2. Set to 0: 2x − 6 = 0, so x = 3.
3. Find y: y = 3² − 6(3) + 5 = 9 − 18 + 5 = −4.
4. Turning point at (3, −4) — the minimum of this upward parabola.

Worked example 6: Find the turning points of y = x³ − 3x.

1. dy/dx = 3x² − 3.
2. Set to 0: 3x² − 3 = 0, so x² = 1, x = ±1.
3. At x = 1: y = 1 − 3 = −2, point (1, −2). At x = −1: y = −1 + 3 = 2, point (−1, 2).
4. Two turning points: a maximum at (−1, 2) and a minimum at (1, −2).

Why gradient zero marks a peak or trough: Just before a maximum the curve climbs (positive gradient); just after, it falls (negative gradient). The only way to switch from positive to negative smoothly is to pass through zero — so the very top has gradient 0. The same logic, reversed, gives a minimum.

Where the gradient of a curve matters

The gradient of a curve answers "how fast, right now?" On a curved distance–time graph it is the instantaneous speed. On a profit curve, the turning point with gradient 0 is the maximum profit. Engineers find the steepest point of a stress curve; designers minimise material by locating a curve's lowest point. Tangent and normal lines are used in physics for reflection and in computer graphics for lighting and collisions.

Practice activity

1. Find the gradient of y = x² at x = 5.
2. Find the gradient of y = x² + 2x at x = −1.
3. Find the equation of the tangent to y = x² at x = 1.
4. Find the turning point of y = x² − 8x + 3.
5. Find both turning points' x-values for y = x³ − 12x.

Answers:

1. dy/dx = 2x; at x = 5 → 10.
2. dy/dx = 2x + 2; at x = −1 → −2 + 2 = 0 (a turning point).
3. Point (1, 1), gradient 2 → 1 = 2(1) + c → c = −1, so y = 2x − 1.
4. 2x − 8 = 0 → x = 4; y = 16 − 32 + 3 = −13, point (4, −13).
5. 3x² − 12 = 0 → x² = 4 → x = ±2.

Summary

A curve's gradient changes at every point. The gradient at a point equals the gradient of the tangent touching the curve there. Find it by differentiating to get dy/dx, then substituting the x-value. The tangent uses that gradient and the point in y = mx + c; the normal uses the negative reciprocal gradient. Where dy/dx = 0 the curve is flat — these turning points are the maximums and minimums that make differentiation so useful.