15.02 Slopes of tangents and secants
Slopes of tangents
We can find the rate of change of a linear function by finding the slope. How can we find the rate of change of a non-linear function? Consider these scenarios:
- The vertical speed of a projectile
- The reproduction rate of a colony of bacteria
- The revenue growth of a business over a year
- The flow in and out of a body of water
Exploration
We looked at the horizontal velocity of a projectile earlier. Can we use the same technique to find this rate of change? Suppose that the vertical height of the projectile ($y$y, in meters) is given by $y=40t-5t^2$y=40t−5t2. Let’s plot this relationship first:
Notice that the function is non-linear. Does it have a constant rate of change?
We can test this by calculating the rate of change between various points. Let’s choose $\left(1,35\right)$(1,35), $\left(3,75\right)$(3,75), $\left(5,75\right)$(5,75), and $\left(7,35\right)$(7,35):
Calculating the rate of change using the previous technique gives us:
| $\frac{\text{Change in height from }t=1\text{ to }t=3}{\text{Change in time from }t=1\text{ to }t=3}$Change in height from t=1 to t=3Change in time from t=1 to t=3 | $=$= | $\frac{75-35}{3-1}$75−353−1 |
| $=$= | $\frac{40}{2}$402 | |
| $=$= | $20$20 | |
| $\frac{\text{Change in height from }t=3\text{ to }t=5}{\text{Change in time from }t=3\text{ to }t=5}$Change in height from t=3 to t=5Change in time from t=3 to t=5 | $=$= | $\frac{75-75}{5-3}$75−755−3 |
| $=$= | $\frac{0}{2}$02 | |
| $=$= | $0$0 | |
| $\frac{\text{Change in height from }t=5\text{ to }t=7}{\text{Change in time from }t=5\text{ to }t=7}$Change in height from t=5 to t=7Change in time from t=5 to t=7 | $=$= | $\frac{35-75}{7-5}$35−757−5 |
| $=$= | $\frac{-40}{2}$−402 | |
| $=$= | $-20$−20 |
When we choose three different intervals to find the rate of change, we get three different results. The issue is that since the function is non-linear the rate of change is variable.
Since the rate of change is variable, we are looking for a way to find the rate of change at a particular time. We still want to use the slope of a line, but even if we fix one point on the graph, we cannot guarantee that any second point will give us the correct rate of change. And if we choose the same point both times we will get $\frac{0}{0}$00 which is undefined.
A line is a tangent (or tangent line) to a curve at a specific point if it touches and travels in the same direction as the curve at that point. Since it travels in the same direction it has the same rate of change as the function at that point. So the slope of the tangent will give us the rate of change of the function.
In this topic we will only use tangent to refer to lines. However, in other contexts any kind of curve can meet any other kind of curve at a point and travel in the same direction at that point. These two curves are both tangents to each other.
Exploration
Returning to the previous example, tangents to the curve at three points have been drawn in:
The tangent to the curve at $t=1$t=1 is $y=30t-5$y=30t−5. The slope of this tangent is $30$30 and so the rate of change of the height with respect to time after $1$1 second is $30$30 m/s. We can use the other two tangents to find the rate of change at $t=4$t=4 and $t=5$t=5:
| Time | Equation of tangent | Rate of change |
|---|---|---|
| $t=1$t=1 | $y=30t+5$y=30t+5 | $30$30 m/s |
| $t=4$t=4 | $y=80$y=80 | $0$0 m/s |
| $t=5$t=5 | $y=-10t+125$y=−10t+125 | $-10$−10 m/s |
The rate of change of a non-linear function does vary with respect to the independent variable. However we can find the rate of change for any particular value of the independent variable if we know the tangent at this value.
It’s worth considering the situations where we cannot draw a tangent on a graph. The first requirement is that the point is on the domain of the function. If the function is undefined at a horizontal value then the tangent will also be undefined.
Secondly, the function must not suddenly change direction or position at the point. Since the function has two or more directions at these points, they do not have tangents. Examples of these points are given in the graph below:
A line is a tangent (or tangent line) to a curve at a specific point if it touches and travels in the same direction as the curve at that point.
The rate of change of a non-linear function at a specific point is the slope of the tangent to the function at that point.
Practice questions
Question 1
Consider the function $f\left(x\right)=2x-4$f(x)=2x−4 shown here.
If we were to draw a tangent to the function for any $x$x value, what is the slope of that tangent?
Question 2
What is the slope of the tangent at the given point?
Question 3
What is the slope of the tangent at the given point?
Slope of secants
The lines joining pairs of points on the following graph are called secants.
In this chapter, we will be concerned with the slope or slope of secant lines. In the diagram, the green secants have positive slopes and the blue secants have negative slopes. We will see how to be more precise about the slopes and how the secants relate to the slope of the curve itself at a given point.
In the following diagram, we have chosen a point $x=a$x=a and drawn secants with one end at this point. The idea is that as the other end of a secant moves closer to the point on the curve where $x=a$x=a, the secant will, in the limit, coincide with the tangent to the curve at $x=a$x=a. The tangent has been drawn in blue in the diagram.
The slope of the tangent at $x=a$x=a is what we mean by the slope of the curve at $x=a$x=a.
To be precise about the slope of a secant, we need information about both endpoints where the secant touches the curve. In the diagram below, the secant illustrated has endpoints at coordinates $(a,f(a))$(a,f(a)) and $(b,f(b))$(b,f(b)).
The slope of the secant is the ratio of the distances $f(b)-f(a)$f(b)−f(a) and $b-a$b−a. We form the fraction $\frac{f(b)-f(a)}{b-a}$f(b)−f(a)b−a. This is often called the rise over the run.
If we were interested in the slope of the tangent at $x=a$x=a, we might imagine letting $b$b move closer to $a$a so that eventually the secant becomes the tangent at $a$a. When $b$b moves closer to $a$a, both quantities $b-a$b−a and $f(b)-f(a)$f(b)−f(a) become smaller. Although they both approach zero and the ratio $\frac{0}{0}$00 conveys no information, the fraction $\frac{f(b)-f(a)}{b-a}$f(b)−f(a)b−a settles down to a fixed value as the distances become very small.
The limiting value of this ratio as $b\rightarrow a$b→a is taken to be the slope of the tangent at $a$a.
Worked examples
Question 4
We wish to find the slope of the tangent to the curve given by $y=x^2+3x-1$y=x2+3x−1 at the point $x=2$x=2.
We can think of secant lines touching the curve at $x=2$x=2. At this point, $y=2^2+3\times2-1=9$y=22+3×2−1=9. We could choose the other endpoint of the secant line to be at $x=3,y=17$x=3,y=17. Then, the slope of this secant is $\frac{17-9}{3-2}=8$17−93−2=8.
This is a first estimate of the slope of the tangent at $x=2$x=2. For a better estimate, we might put the other endpoint of the secant line at $x=2.5,y=12.75$x=2.5,y=12.75. Then, the slope is $\frac{12.75-9}{2.5-2}=7.5$12.75−92.5−2=7.5.
For an even better estimate, we could bring the endpoint much closer to $x=2$x=2. We might try $x=2.01$x=2.01. For this value of $x$x, the function value is $9.0701$9.0701 and so, the slope of this secant is $\frac{9.0701-9}{2.01-2}=7.01$9.0701−92.01−2=7.01.
We might try an endpoint on the other side of $x=2$x=2, say $x=1.99$x=1.99. Then, the function value is $8.9301$8.9301 and the slope of this secant is $\frac{8.9301-9}{1.99-2}=6.99$8.9301−91.99−2=6.99.
The slope of the tangent at $x=2$x=2 appears to be somewhere between $6.99$6.99 and $7.01$7.01. and we could narrow down the value further by making the distance between the endpoints of the secant even smaller.
We could investigate the idea of making the endpoints of a secant very close together to approximate the slope of a tangent by letting the endpoints be at $x=a$x=a and $x=a+\delta$x=a+δ, where $\delta$δ is a small positive or negative number which we can allow to become as close to zero as we like.
Question 5
Consider the function given by $f(x)=x^2-x+4$f(x)=x2−x+4. We wish to know the slope of the tangent at $x=3$x=3.
We have $f(3)=10$f(3)=10. So, we construct secants with one endpoint at $(3,10)$(3,10). For the other endpoint, we choose a point a small distance away from $x=3$x=3. We can write it as $x=3+\delta$x=3+δ.
Now, $f(3+\delta)=(3+\delta)^2-(3+\delta)+4$f(3+δ)=(3+δ)2−(3+δ)+4 which distributes and simplifies to $10+5\delta+\delta^2$10+5δ+δ2.
The slope of this secant must be $\frac{f(3+\delta)-f(3)}{3+\delta-3}=\frac{10+5\delta+\delta^2-10}{\delta}.$f(3+δ)−f(3)3+δ−3=10+5δ+δ2−10δ.
That is, the slope of the secant is $\frac{5\delta+\delta^2}{\delta}$5δ+δ2δ. Now, $\delta$δ is a common factor in the numerator of this fraction and therefore it can be canceled with the $\delta$δ in the denominator provided $\delta$δ is not zero, which it is not.
So, the slope of a secant with one endpoint at $x=3$x=3 is $5+\delta$5+δ, however closely $\delta$δ approaches zero. We must conclude that the slope of the tangent is the limiting value as $\delta\rightarrow0$δ→0. So, the tangent at $x=3$x=3 has slope $5$5.