Gradients of tangents
We can find the rate of change of a linear function by finding the gradient. 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 metres) 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 gradient 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 gradient 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 gradient 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 gradient 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 gradient of that tangent?
Question 2
What is the gradient of the tangent at the given point?
Question 3
What is the gradient of the tangent at the given point?
Outcomes
M7-10
Apply differentiation and anti-differentiation techniques to polynomials
91262
Apply calculus methods in solving problems