Rates of Change and Tangent Slopes

A question about one instant

A photograph tells us where a moving object is, but it does not directly tell us how fast the object is moving. To calculate a speed from measured positions, we usually compare two moments. The central question of this lesson is how that comparison can tell us a rate at just one moment.

Calculus studies how the value of a function changes when its input changes. Our input will be time, and our output will be distance travelled. The same kind of question can arise when the input is temperature, price, or the amount of water entering a reservoir.

For the falling-body model used here, the distance travelled from the release point is 4.9 times the square of the elapsed time. Time t is measured in seconds and distance s in metres. We use the model for downward travel, so the positive rates below describe motion in that direction.

The table says that the body has travelled 19.6 m by t=2. That is a position measurement relative to the release point. It is not a velocity of 19.6 m/s: the unit and the question are different. To get a velocity, we must compare a change in distance with the time taken for that change.

Average velocity over an interval

Choose a starting time t₁ and a later time t₂. During this interval the distance changes by s(t₂)−s(t₁), while the elapsed time is t₂−t₁. Dividing these changes gives a single rate that would cover the same displacement in the same time.

Approaching two seconds from either side

To estimate the rate at t=2, keep that instant fixed and use shorter intervals ending there. Each new interval removes more of the earlier, slower part of the motion. The averages below increase as the starting time moves towards 2.

We must also check intervals starting at t=2. These include slightly later, faster motion. As their ending times move back towards 2, their averages decrease. The two approaches should identify the same instantaneous rate for this smooth motion.

The pattern has an algebraic explanation. For any different times t₁ and t₂, the difference of their squares has a factor t₂−t₁. Cancelling that factor gives 4.9(t₁+t₂) for the average velocity. When both times are brought close to 2, this expression approaches 4.9(2+2)=19.6.

Another useful version uses an increment h. Compare t=2 with t=2+h, where h may be positive or negative but cannot be zero during the calculation. The difference quotient simplifies to 19.6+4.9h. We can make this as close to 19.6 as we want by choosing a sufficiently small non-zero h.

Seeing the rate as a slope

On a distance–time graph, a horizontal change represents elapsed time and a vertical change represents distance travelled. Therefore the slope of a line joining two points on the curve is their average velocity. Such a line is called a secant.

Fix the point A at t=2 and let the second point B move towards A along the curve. The secant turns towards one limiting line through A: the tangent. Its slope represents the instantaneous velocity. The numerical tables and the geometry express the same limiting process.

The word derivative will be used for this limiting rate of change. We have not yet built the general definition, but we have found its purpose. The next step is to understand limits themselves so that we can justify and calculate these rates for many functions.