4.1, 4.2 - The gradient function of a curve

4.1, 4.2 - The gradient function of a curve

A curve can be steep at one point and almost flat at another, so one gradient cannot describe the whole curve. You will interpret its gradient function, use supplied gradient functions to work out local gradients and explain what those gradients mean as rates of change.

The gradient at one point

At (1,1)(1,1) on y=x2y=x^2, should the curve's gradient be 22 or 44? Both numbers come from sensible-looking lines through that point. The choice depends on whether you want the gradient at one point or an average between two points.

You already know

A straight line has gradient m=change in ychange in xm=\frac{\text{change in }y}{\text{change in }x}. Its gradient is constant. A positive gradient slopes upwards as you move right; a negative gradient slopes downwards; a horizontal line has gradient zero.

The blue curve below has a tangent of gradient 22 at PP. The other line passes through two points on the curve, so it gives the average gradient between them. It does not give the gradient at PP.

[DIAGRAM: asset_name: Tangent and chord at the same point; asset_slug: l2fm-tangent-and-chord; caption: The tangent gives gradient 2 at P while the chord gives average gradient 4 from P to Q; description: Coordinate axes show y equals x squared, P at 1 comma 1 and Q at 3 comma 9. A sage tangent through P has equation y equals 2x minus 1 and gradient 2. A clay chord through P and Q has equation y equals 4x minus 3 and gradient 4. Clear separate labels distinguish local tangent gradient at P from average gradient between P and Q]
The tangent gives gradient 2 at P while the chord gives average gradient 4 from P to Q

A tangent follows the curve's direction at its point of contact. If you inspect a smooth curve very close to that point, its direction approaches the direction of the tangent. A chord uses a second, distinct point, so it measures an average over an interval instead.

Gradient of a curve

The gradient of a curve at a point is the gradient of the tangent to the curve at that point. The gradient function, written dydx\frac{\mathrm{d}y}{\mathrm{d}x}, gives that gradient for each allowed value of xx.

Read dydx\frac{\mathrm{d}y}{\mathrm{d}x} as “dee y by dee x”. Here it names a new function; it is not y/xy/x. A tangent is defined by its local direction, so it need not meet the curve only once overall, and it may cross the curve.

Key idea

The gradient at one point comes from the tangent there. A line joining two distinct points gives an average gradient between them.

Use that distinction to settle the opening choice.

Using a supplied gradient function

A curve equation gives heights; its gradient function gives slopes. Substitute into the expression that answers the question you have actually been asked.

For y=x2−3x+1y=x^2-3x+1, suppose you are given dydx=2x−3\frac{\mathrm{d}y}{\mathrm{d}x}=2x-3. The first expression gives a point's yy-coordinate. The second gives the tangent gradient there. You will learn to obtain gradient functions in the next lesson; for now they are supplied.

Gradient at a specified value

If the gradient function is dydx=g(x)\frac{\mathrm{d}y}{\mathrm{d}x}=g(x), then the tangent gradient at x=ax=a is

m=g(a)m=g(a)

The point must lie in the curve's domain and the gradient function must be defined there.
Not on the formulae sheet: learn it.

First compare the height and the gradient at the same input.

Worked example 1: gradient versus coordinate

Question. A curve has equation y=x2−3x+1y=x^2-3x+1 and gradient function dydx=2x−3\frac{\mathrm{d}y}{\mathrm{d}x}=2x-3. Work out its gradient at x=2x=2.

Plan. Substitute 22 into the gradient function, rather than the curve equation.

m=2(2)−3use the gradient function=1evaluate\begin{aligned} m&=2(2)-3 &&\text{use the gradient function}\\ &=1 &&\text{evaluate} \end{aligned}

Why this step? The question asks for a slope. Substituting into y=x2−3x+1y=x^2-3x+1 would instead give the height −1-1.

Answer. The gradient is 11.

Check. The tangent rises 11 unit for each 11-unit movement right. This is a positive slope, even though the point is below the horizontal axis.

Change only the sign of the input, from 22 to −2-2. Keep the negative value in brackets during substitution.

Continue this lesson.

You’re previewing NovaLearn without an account. Create a free account to read the rest.

Create free account