NM4, NM5 - Estimating gradients with chords

NM4, NM5 - Estimating gradients with chords

A table of points may describe a curve without giving you a formula to differentiate. NM4 uses a chord to estimate the gradient of a tangent; NM5 improves the estimate by choosing points more carefully. This gives a practical bridge between a measured change and an instantaneous rate.

A chord as a tangent estimate

How can two positions from a moving object's journey tell you how fast it is moving at one instant? A chord gives the average rate between two readings. If the readings are close enough to the instant we care about, that average can estimate the tangent's gradient there.

You already know

The tangent gradient is dy/dx\mathrm{d}y/\mathrm{d}x at a point. For two distinct points on a straight line, gradient is change in yy divided by change in xx. A displacement-time gradient represents velocity.

A chord joins two points on a curve. It is a straight line, so its gradient can be calculated even when the curve's formula is unknown. The chord and tangent generally have different gradients: an estimate is not an exact derivative merely because the arithmetic is exact.

Chord estimate

tangent gradient near P≈yQ−yPxQ−xP\text{tangent gradient near }P\approx \frac{y_Q-y_P}{x_Q-x_P}

Here PP is the point of interest and QQ is a nearby point on the same curve, with xQ≠xPx_Q\ne x_P. This is a one-sided estimate; nearby points on both sides can give a better estimate.
Not on the formulae sheet: learn it.

Keep the subtractions in the same order. Reversing both is fine; reversing just one changes the sign.

Worked example 1: a forward chord

Question. A curve passes through P(2,8)P(2, 8) and Q(2.4,13.824)Q(2.4, 13.824). Use the chord PQPQ to estimate the tangent gradient at PP.

Plan. Divide the change in height by the corresponding change in horizontal position.

Δy=13.824−8=5.824change from P to QΔx=2.4−2=0.4same directionmPQ=5.8240.4=14.56gradient of the chord\begin{aligned} \Delta y&=13.824-8=5.824 &&\text{change from }P\text{ to }Q\\ \Delta x&=2.4-2=0.4 &&\text{same direction}\\ m_{PQ}&=\frac{5.824}{0.4}=14.56 &&\text{gradient of the chord} \end{aligned}

Why subtract 88 from 13.82413.824? We used the movement from x=2x=2 to x=2.4x=2.4 in the denominator, so the numerator must describe the same movement.

Answer. The tangent gradient at PP is approximately 14.5614.56.

Check. Both changes are positive, so a positive gradient makes sense. The chord rises about 5.85.8 in a horizontal distance of 0.40.4, so a gradient near 1515 is plausible.

The next example changes only the second point: it is closer to PP, allowing the chord to follow the curve more locally.

Worked example 2: a closer forward chord

Question. A second point on the curve is R(2.2,10.648)R(2.2, 10.648). Use PRPR to estimate the tangent gradient at P(2,8)P(2, 8).

Plan. Repeat the chord calculation using the closer point.

mPR=10.648−82.2−2matched differences=2.6480.2evaluate the changes=13.24divide\begin{aligned} m_{PR}&=\frac{10.648-8}{2.2-2} &&\text{matched differences}\\ &=\frac{2.648}{0.2} &&\text{evaluate the changes}\\ &=13.24 &&\text{divide} \end{aligned}

Why is this a more local estimate? It averages the gradient over a shorter piece of curve, so changes farther from PP have less influence.

Answer. The tangent gradient is approximately 13.2413.24, using the closer chord.

Check. The new estimate differs from 14.5614.56; the earlier answer was a chord gradient, not an exact tangent gradient.

Writing the fraction before evaluating it makes both the point choice and the sign visible.

Watch out

Wrong: m=(10.648−8)/(2−2.2)=−13.24m=(10.648-8)/(2-2.2)=-13.24. Only the denominator was reversed. Right: use (10.648−8)/(2.2−2)=13.24(10.648-8)/(2.2-2)=13.24, or reverse both subtractions.

The next question keeps the same calculation but moves the point of interest.

Using points on both sides

When readings are available on both sides of the point, a chord centred on that point usually gives a better estimate than a chord reaching only one way.

For an estimate at x=ax=a, use points at a−ha-h and a+ha+h, where h>0h>0. The horizontal gap is 2h2h, not hh. The chord joins these two curve points; it does not have to pass through the curve point at x=ax=a.

[DIAGRAM: asset_name: Chords approaching a tangent; asset_slug: nm4-closer-chords; caption: A chord centred at x equals 2 approaches the tangent as its endpoints move closer; description: The cubic curve y equals x cubed, its tangent at (2, 8) of gradient 12, a wide chord from (1, 1) to (3, 27) of gradient 13, and a closer chord from (1.8, 5.832) to (2.2, 10.648) of gradient 12.04]
A chord centred at x equals 2 approaches the tangent as its endpoints move closer

The two-sided calculation is often called a central estimate.

Central estimate of a gradient

dydx∣x=a≈f(a+h)−f(a−h)2h\left.\frac{\mathrm{d}y}{\mathrm{d}x}\right|_{x=a} \approx \frac{\mathrm{f}(a+h)-\mathrm{f}(a-h)}{2h}

Use two points equally spaced either side of aa, with small positive hh, on a smooth curve. The denominator is the whole distance between the two points.
Not on the formulae sheet: learn it.

With a table, choose the nearest available points that straddle the requested value. If equally spaced points are available, use them rather than a wider or one-sided chord.

Continue this lesson.

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