The Gradient of a Curve as a Limit
See that a curve has a different gradient at every point, and that the gradient of a chord settles on a single value as the second point slides in. That limiting value is the gradient of the tangent.
What a learner can do afterwards
- Compute chord gradients on y = x² for h = 1, 0.1 and 0.01 at x = 2
- Say what number the chord gradients are approaching and why it is not reached
- Explain why a curve has no single gradient the way a line does
1 · Read
Last stop you found a tangent to a circle without any new machinery, because a circle hands you a radius to be square to. The Ridgeway ramp gives you nothing of the sort. Its profile on the plan is a curve like y = x², and the crew wants to know how steep it is where a board leaves it. There is a harder problem underneath. A straight line has one gradient everywhere, but a curve is gentle low down and steep further up, so asking for the gradient of the curve is asking the wrong question. You have to ask at a point.
You already know how to find the gradient between two points: divide the change in y by the change in x. Do that between two points on a curve and you get the gradient of the chord joining them, which is the average steepness across that stretch. It is not the steepness at either end. But there is a way in. Keep the first point fixed, then slide the second one towards it and watch the chord gradients. As the stretch shrinks, the average has less and less room to differ from the steepness at the fixed point.
Take y = x² and fix the point where x = 2, so y = 4. Now walk the second point in, using h for the gap in x. With h = 1 the second point is (3, 9), so the chord gradient is (9 - 4) / 1 = 5. With h = 0.1 it is (2.1, 4.41), and the gradient is 0.41 / 0.1 = 4.1. With h = 0.01 it is (2.01, 4.0401), and the gradient is 0.0401 / 0.01 = 4.01. Notice the run is h itself, not the new x value.
Read that list: 5, 4.1, 4.01. The numbers close in on 4, and smaller h keeps closing in. What they never do is arrive. Putting h = 0 would land the second point on the first, so rise and run both vanish and 0 divided by 0 names no number. So the answer is not a chord gradient. It is the value they approach, called their limit. The chords swing round as they shrink, settling on one line through the fixed point. That line is the tangent, and its gradient is the limit. That limit is what we mean by the gradient at x = 2, and it is 4.
Two checks are worth making. First, come in from the other side. With h = -0.1 the second point is (1.9, 3.61), and the gradient is -0.39 divided by -0.1, which is 3.9. So the left-hand chords climb to 4 while the right-hand ones fall to 4, and the answer is caught between them. Second, try a different point on the same curve. At x = 5, h = 0.1 gives 10.1 and h = 0.01 gives 10.01, heading for 10. Same curve, different place, different gradient.
A curve has no single gradient, so the question has to be asked at a point. Fix that point, join it to a second point on the curve, and the chord gradient measures the average steepness between them. Slide the second point in and the chord gradients settle on one number without ever reaching it, because h = 0 would give 0 divided by 0. That number is their limit. It is the gradient of the curve at the fixed point, and the gradient of the tangent there.
2 · Watch
Take it off screen
Where it sits
24 questions wait behind this lesson, each with its answer explained. Every answer feeds the sky: stars light as they are learned, and dim when it is time to come back.