NM9 - Numerical methods in context

NM9 - Numerical methods in context

A question about motion, a pump or a target time may hide a gradient, an area or a root. NM9 brings those numerical methods together: choose the mathematical quantity first, calculate with the available information, then interpret the answer with the right units and accuracy. The context tells you which calculation matters.

Rates from nearby readings

A cart's position is recorded at intervals, but you want its velocity between the readings. Can a table tell you about that instant? It can give an estimate, provided you calculate a rate rather than mistake the recorded position for the answer.

You already know

A central chord estimates a tangent gradient using readings either side of the point. The kinematic relationships v=ds/dtv=\mathrm{d}s/\mathrm{d}t and a=dv/dta=\mathrm{d}v/\mathrm{d}t are on the formulae sheet. The numerical chord calculation itself is not on the formulae sheet: learn it.

Read the vertical quantity and its units before selecting the calculation. Position divided by time is not generally instantaneous velocity: you need change in position over the corresponding change in time. For velocity readings, the same gradient calculation estimates acceleration instead.

Worked example 1: velocity from positions

Question. A cart's displacement ss is recorded in metres. Use the table to estimate its velocity at t=4t=4 seconds.

tt, seconds3.53.5444.54.5
ss, metres10.210.213.013.016.416.4

Plan. Use the displacement readings equally spaced either side of 44 to estimate the displacement-time gradient.

v(4)≈s(4.5)−s(3.5)4.5−3.5central chord=16.4−10.21change in metres over seconds=6.2 m s−1interpret the gradient\begin{aligned} v(4)&\approx\frac{s(4.5)-s(3.5)}{4.5-3.5} &&\text{central chord}\\ &=\frac{16.4-10.2}{1} &&\text{change in metres over seconds}\\ &=6.2\,\mathrm{m\,s^{-1}} &&\text{interpret the gradient} \end{aligned}

Why not use 13/413/4? The displacement 1313 describes position relative to an origin. It does not supply the small displacement change around the requested instant.

Answer. The estimated velocity at 44 seconds is 6.2 m s−16.2\,\mathrm{m\,s^{-1}}.

Check. The one-sided estimates are (13−10.2)/0.5=5.6(13-10.2)/0.5=5.6 and (16.4−13)/0.5=6.8 m s−1(16.4-13)/0.5=6.8\,\mathrm{m\,s^{-1}}. The central value lies between them.

The next example changes only what the instrument measures. Identical numbers can have a different physical meaning.

Worked example 2: acceleration from velocities

Question. An instrument records velocity vv in m s−1\mathrm{m\,s^{-1}}, giving 10.210.2, 13.013.0 and 16.416.4 at 3.53.5, 44 and 4.54.5 seconds respectively. Estimate the acceleration at 44 seconds.

Plan. Estimate the velocity-time gradient with the central pair of readings.

a(4)≈v(4.5)−v(3.5)4.5−3.5acceleration is velocity gradient=16.4−10.21change in velocity over time=6.2 m s−2divide velocity units by seconds\begin{aligned} a(4)&\approx\frac{v(4.5)-v(3.5)}{4.5-3.5} &&\text{acceleration is velocity gradient}\\ &=\frac{16.4-10.2}{1} &&\text{change in velocity over time}\\ &=6.2\,\mathrm{m\,s^{-2}} &&\text{divide velocity units by seconds} \end{aligned}

Why does the unit change? The numerator is now measured in m s−1\mathrm{m\,s^{-1}}, so dividing by seconds gives m s−2\mathrm{m\,s^{-2}}. The same arithmetic no longer describes velocity.

Answer. The estimated acceleration is 6.2 m s−26.2\,\mathrm{m\,s^{-2}}.

Check. The velocity is rising across the interval, consistent with positive acceleration. The recorded velocity at 44 seconds is 13 m s−113\,\mathrm{m\,s^{-1}}, a separate quantity.

This is a useful mathematical fact: changing the labels on the axes can preserve every number in a calculation while changing what the answer means. Units are part of the reasoning.

Watch out

Wrong: “At t=4t=4, acceleration is 1313 because that is the table entry.” Right: 1313 is the velocity reading; acceleration is estimated from its rate of change, here 6.2 m s−26.2\,\mathrm{m\,s^{-2}}.

Use the same idea with closer readings, and keep the physical quantity in the answer.

Totals from rate readings

Area under a rate-time graph estimates an accumulated quantity: the context decides whether that quantity is displacement, distance or something else.

For example, m s−1\mathrm{m\,s^{-1}} multiplied by seconds gives metres; litres s−1\mathrm{litres\,s^{-1}} multiplied by seconds gives litres. This units check often reveals the correct method before any arithmetic. Use the trapezium rule from the formulae sheet when the rate is given at equal time intervals. The kinematic relationship s=∫v dts=\int v\,\mathrm{d}t is also on the formulae sheet; interpreting a general rate-time area as a total is not separately printed there.

For velocity, signed area gives a change in displacement. If velocity is non-negative throughout, it also gives distance travelled. If the object reverses direction, use the signs and split at the turning times before finding total distance. A table alone does not prove that no reversal occurred between readings: use the question's information about the motion.

Worked example 3: distance from a speed table

Question. A cart moves forwards throughout the first 66 seconds. Its speeds are given below. Use the trapezium rule to estimate the distance travelled.

tt, seconds00224466
Speed, m s−1\mathrm{m\,s^{-1}}00335544

Plan. Estimate the speed-time area with three strips of width 22 seconds.

h=6−03=2time widthD≈22[(0+4)+2(3+5)]trapezium rule=20 mrate times time\begin{aligned} h&=\frac{6-0}{3}=2 &&\text{time width}\\ D&\approx\frac{2}{2}\left[(0+4)+2(3+5)\right] &&\text{trapezium rule}\\ &=20\,\mathrm{m} &&\text{rate times time} \end{aligned}

Why area rather than gradient? We want the distance accumulated over the interval, not the rate at which the speed changes at an instant.

Answer. The estimated distance travelled is 20 m20\,\mathrm{m}.

Check. This gives mean speed 20/6≈3.33 m s−120/6\approx3.33\,\mathrm{m\,s^{-1}}, within the observed speed range 00 to 5 m s−15\,\mathrm{m\,s^{-1}}. That is a scale check, not proof of exactness.

The next example changes only the measured rate: a pump replaces the cart, so area now gives a volume.

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