Gradient of a Curve (NSW Specific)

The lines joining pairs of points on the following graph are called secants. 

In this chapter, we will be concerned with the slope or gradient of secant lines. In the diagram, the green secants have positive gradients and the blue secants have negative gradients. We will see how to be more precise about the gradients and how the secants relate to the gradient 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 gradient of the tangent at $x=a$x=a is what we mean by the gradient of the curve at $x=a$x=a.

 

To be precise about the gradient 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 gradient 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 gradient 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 gradient of the tangent at $a$a.

 

Example 1

We wish to find the gradient 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 gradient of this secant is $\frac{17-9}{3-2}=8$17−93−2​=8.

This is a first estimate of the gradient 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 gradient 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 gradient 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 gradient of this secant is $\frac{8.9301-9}{1.99-2}=6.99$8.9301−91.99−2​=6.99.

The gradient 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 gradient 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.

Example 2

Consider the function given by $f(x)=x^2-x+4$f(x)=x2−x+4. We wish to know the gradient 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 expands and simplifies to $10+5\delta+\delta^2$10+5δ+δ2.  

The gradient 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 gradient 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 cancelled with the $\delta$δ in the denominator provided $\delta$δ is not zero, which it is not.

So, the gradient 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 gradient of the tangent is the limiting value as $\delta\rightarrow0$δ→0. So, the tangent at $x=3$x=3 has gradient $5$5.