The Gradient of a Curve as a Limit

Mathematics · Calculus & Analysis · ages 16-17

Name ______________________ Date ____________

On y = x², what is the gradient of the chord joining (4, 16) and (4.1, 16.81)?
Answer: ______________
A curve is steeper in some places than in others. What does that force you to do when you ask for its gradient?
- use the steepest part of the curve
- name the point you are asking about
- average the steepness over the whole curve
- measure it on a chord instead
Setting h = 0 puts the second point exactly on top of the first one.
Circle one: True False
On y = x^2, find the gradient of the chord from x = 2 to x = 3.
Answer: ______________
On y = x³, what is the gradient of the chord joining (2, 8) and (2.1, 9.261)?
- 1.261
- 12.61
- 4.41
- 9.261
On y = x², work out the gradient of the chord from x = 1 to x = 1.1.
Answer: ______________
Amir works out the chord gradient on y = x² from x = 0 to x = 2, gets 2, and says the gradient of the curve is 2. Where is the slip?
- the chord gradient should have been (4 - 0) / 4, which is 1
- a chord gradient can never be a whole number
- that is an average over a stretch, and a curve has a gradient at each point
- he should have used the y values rather than the x values on the bottom
On y = x^2, find the gradient of the chord from x = 2 to x = 2.01.
Answer: ______________
On y = 3x², chords at x = 1 give gradients of 9 when h = 1, 6.3 when h = 0.1 and 6.03 when h = 0.01. What is the gradient of the curve at x = 1?
- 3
- 9
- 6
- 6.03
Kai says that because the chord gradients get closer and closer to 4 but never equal 4, the gradient at the point cannot really be 4. Where is the slip?
- the chord gradients do reach 4, at h = 0
- the gradient at the point is the average of the chord gradients instead
- a gradient that is never reached does not exist at all
- the limit is the value they close in on, and that value is the gradient

Answer key

For grown-ups. Fold this page away before handing over the rest.

The Gradient of a Curve as a Limit W1-mt_LAlhcwVjln-s1

8.1 · The change in y is 16.81 - 16 = 0.81 and the change in x is 0.1, so the gradient is 0.81 / 0.1 = 8.1.
name the point you are asking about · Because the steepness changes along the curve, no single number describes the whole of it. The question only has an answer once a point is named.
True · With h = 0 there is no gap in x left at all, so the two ends of the chord land in the same place and there is nothing left to join.
5 · From x = 2 to x = 3, y grows from 4 to 9, so the chord gradient is (9 - 4)/1 = 5.
12.61 · The rise is 9.261 - 8 = 1.261 and the run is 0.1, so the gradient is 1.261 / 0.1 = 12.61.
2.1 · At x = 1, y = 1. At x = 1.1, y = 1.21. So the gradient is (1.21 - 1) / 0.1 = 0.21 / 0.1 = 2.1.
that is an average over a stretch, and a curve has a gradient at each point · His arithmetic is right: (4 - 0) / (2 - 0) = 2. But 2 is the average steepness from 0 to 2, and the curve is gentler than that at the start and steeper at the end.
4.01 · 2.01^2 = 4.0401, so the chord gradient is (4.0401 - 4)/0.01 = 4.01.
6 · The gradients are closing in on 6: the gaps are 3, then 0.3, then 0.03. The limit is 6.
the limit is the value they close in on, and that value is the gradient · No chord is asked to equal 4. The gradient at a point is defined as the value the chords approach, and approaching 4 from both sides is what pins it down.