The steepness at one point
A chord gives the average rate of change across an interval. Sometimes the question is how steep a curve is at one single point. On the curve , take the point P(2, 1).
Lay a ruler against the curve so that it touches the curve at P without cutting across it there. The line drawn along the ruler is called the tangent to the curve at P. Close to P, the curve and the tangent run in the same direction.
The gradient of the curve at P is defined to be the gradient of the tangent at P. So to estimate how steep the curve is at P, draw the tangent and find its gradient.
The marked point is P(2, 1). The straight line is the tangent: it touches the curve at P, and the curve stays on one side of it.
Read the tangent like any line
The tangent is a straight line, so its gradient is the rise divided by the run between any two points on it. Choose two points on the tangent that are far apart and sit exactly where two gridlines cross, so that their coordinates can be read exactly: (0, −1) and (4, 3).
From (0, −1) to (4, 3), the run is 4 − 0 = 4 and the rise is 3 − (−1) = 4. The gradient is 4 ÷ 4 = 1. So the gradient of the curve at P is about 1.
The two points must be on the tangent, not on the curve. At x = 4 the curve is at y = 4, but the tangent is at y = 3.
The straight line is the tangent at P. Between the two marked points on it, (0, −1) and (4, 3), it rises 4 over a run of 4, so its gradient is 1.
Why the answer is an estimate
A tangent drawn by eye can be tilted slightly, and a different tilt gives a different gradient, perhaps 0.9 or 1.1 instead of 1. So a gradient found from a drawn tangent is an estimate. A careful drawing comes close to the true value.
Three habits make the estimate better. Draw the tangent long, so its direction is clear. Check that the curve stays on one side of it close to P. Read the gradient from two points far apart, because a small triangle turns any error in reading the graph into a large part of the answer.
Chords close in on the tangent
Compare the tangent with chords from P. The chord from P(2, 1) to (4, 4) has gradient (4 − 1) ÷ (4 − 2) = 1.5. The chord from P to (3, 2.25) has gradient 1.25 ÷ 1 = 1.25. The chord from P to (2.5, 1.5625) has gradient 0.5625 ÷ 0.5 = 1.125.
As the second point moves toward P, the chord turns toward the tangent, and its gradient gets closer and closer to 1. Looking ahead: this is how calculus finds the gradient at a point exactly, without drawing.
The white line is the chord from P(2, 1) to (4, 4), with gradient 1.5. It is steeper than the tangent at P, the gold straight line, whose gradient is 1.
Where the curve falls
Where a curve is falling, its tangent falls from left to right too, and the gradient is negative. On a graph of the temperature of a cooling drink against the time in minutes, the tangent has a negative gradient because the temperature is going down, and the size of the gradient is the number of degrees lost per minute at that moment.
The usual mistakes
Giving the rise without dividing by the run. From (0, −1) to (4, 3) the rise is 4, and the gradient is 4 ÷ 4 = 1.
Dividing the coordinates of P. At P(2, 1), 1 ÷ 2 = 0.5 is the gradient of the line from the origin to P, not of the tangent.
Getting the sign wrong. Read left to right: a tangent that climbs has a positive gradient, and a tangent that falls has a negative one.
Worked example: How Fast a Cup of Tea Is Cooling, from a Tangent to Its Cooling Curve
Question The temperature of a cup of tea, y°C, is recorded x minutes after it is poured. The cooling curve passes through (0, 65), (10, 50), (25, 40) and (40, 35). A tangent is drawn to the curve at (10, 50), and it passes through (0, 60) and (30, 30). (a) Estimate the rate at which the tea is cooling 10 minutes after it is poured. (b) Find the average rate of cooling over the first 10 minutes, and explain why it is greater than the answer to part (a).
1.The rate of cooling at 10 minutes is the gradient of the curve at (10, 50). The tangent touches the curve at that point and has the same gradient there, so the gradient is read from the tangent.
The rate of cooling at one moment is the gradient of the curve at that point. It is read from the tangent drawn at (10, 50). 2.Choose two points on the tangent that are far apart, (0, 60) and (30, 30). The gradient is 30 − 6030 − 0 = −3030 = −1.
Take two points on the tangent that are far apart, (0, 60) and (30, 30): the gradient is 30 − 6030 − 0 = −1. 3.(a) The gradient is negative because the temperature is falling. After 10 minutes the tea is cooling at about 1°C per minute.
(a) The gradient is negative because the temperature is falling. After 10 minutes the tea is cooling at about 1°C per minute. 4.The average rate over the first 10 minutes is the gradient of the chord from (0, 65) to (10, 50): 50 − 6510 − 0 = −1510 = −1.5.
The average rate over the first 10 minutes is the gradient of the chord from (0, 65) to (10, 50): 50 − 6510 = −1.5. 5.(b) Over the first 10 minutes the tea cools by 1.5°C per minute on average. The curve is steepest at the start and becomes less steep, so the tea cools fastest when it is hottest, and the chord over the first 10 minutes is steeper than the tangent at x = 10.
(b) The tea cools by 1.5°C per minute on average in the first 10 minutes. The curve is steepest at the start, so the chord is steeper than the tangent at x = 10.
Answer: (a) About 1°C per minute; (b) 1.5°C per minute, which is greater because the curve is steeper at the start than at x = 10
Common mistakes
- Dividing the temperature by the time at the point, 5010 = 5°C per minute. That is the gradient of a line from the origin to the point, which has no meaning here. The rate is a change in temperature divided by a change in time, measured along the tangent.
- Choosing two points on the tangent that are very close together. A small triangle makes any error in reading the graph a large part of the answer. Points that are far apart, such as the ends of the drawn tangent, give a more reliable estimate.