MA005: Calculus I

Our goal is to find a way of exactly determining the slope of the line which is tangent to a function (to the graph of the function) at a point in a way which does not require us to have the graph of the function.

Let's start with the problem of finding the slope of the line $L$ (Fig. 1) which is tangent to $f(x)=x^{2}$ at the point $(2,4).$ We could estimate the slope of $\mathrm{L}$ from the graph, but we won't. Instead, we can see that the line through $(2,4)$ and $(3,9)$ on the graph of $\mathrm{f}$ is an approximation of the slope of the tangent line, and we can calculate that slope exactly: $\mathrm{m}=\Delta \mathrm{y} / \Delta \mathrm{x}=(9-4) /(3-2)=5$. But $\mathrm{m}=5$ is only an estimate of the slope of the tangent line and not a very good estimate. It's too big. We can get a better estimate by picking a second point on the graph of f which is closer to $(2,4)-$ the point $(2,4)$ is fixed and it must be one of the points we use. From Fig. 2, we can see that the slope of the line through the points $(2,4)$ and $(2.5,6.25)$ is a better approximation of the slope of the tangent line at $(2,4): \mathrm{m}=\Delta \mathrm{y} / \Delta \mathrm{x}=(6.25-4) /(2.5-2)=2.25 /.5$ $=4.5$, a better estimate, but still an approximation. We can continue picking points closer and closer to $(2,4)$ on the graph of $\mathrm{f}$, and then calculating the slopes of the lines through each of these points and the point $(2,4)$:

The only thing special about the x-values we picked is that they are numbers which are close, and very close, to $\mathrm{x}=2$. Someone else might have picked other nearby values for $\mathrm{x}.$ As the points we pick get closer and closer to the point $(2,4)$ on the graph of $\mathrm{y}=\mathrm{x}^{2}$, the slopes of the lines through the points and $(2,4)$ are better approximations of the slope of the tangent line, and these slopes are getting closer and closer to $4.$

Practice 1: What is the slope of the line through $(2,4)$ and $(x, y)$ for $y=x_{m}^{2}$ and $x=1.994 ? x=2.0003$ ?

We can bypass much of the calculating by not picking the points one at a time: let's look at a general point near $(2,4)$. Define $\mathrm{x}=2+\mathrm{h}$ so $\mathrm{h}$ is the increment from 2 to $\mathrm{x}$ (Fig. 3). If $\mathrm{h}$ is small, then $\mathrm{x}=2+\mathrm{h}$ is close to 2 and the point $(2+\mathrm{h}, \mathrm{f}(2+\mathrm{h}))=\left(2+\mathrm{h},(2+\mathrm{h})^{2}\right)$ is close to $(2,4)$. The slope $\mathrm{m}$ of the line through the points $(2,4)$ and $\left(2+\mathrm{h},(2+\mathrm{h})^{2}\right)$ is a good approximation of the slope of the tangent line at the point $(2,4)$:

$\begin{aligned} m &=\frac{\Delta y}{\Delta x}=\frac{(2+h)^{2}-4}{(2+h)-2} \\ &=\frac{\left\{4+4 h+h^{2}\right\}-4}{h}=\frac{4 h+h^{2}}{h}=\frac{h(4+h)}{h}=4+h. \end{aligned}$

If $h$ is very small, then $m=4+h$ is a very good approximation to the slope of the tangent line, and $m=4+h$ is very close to the value $4.$ The value $m=4+h$ is called the slope of the secant line through the two points $(2,4)$ and $\left(2+\mathrm{h},(2+\mathrm{h})^{2}\right).$ The limiting value 4 of $\mathrm{m}=4+\mathrm{h}$ as $\mathrm{h}$ gets smaller and smaller is called the slope of the tangent line to the graph of $\mathrm{f}$ at $(2,4)$.

Example 1: Find the slope of the line tangent to $f(x)=x^{2}$ at the point $(1,1)$ by evaluating the slope of the secant line through $(1,1)$ and $(1+\mathrm{h}, \mathrm{f}(1+\mathrm{h}))$ and then determining what happens as $\mathrm{h}$ gets very small (Fig. 4).

Solution: The slope of the secant line through the points $(1,1)$ and $(1+\mathrm{h}, \mathrm{f}(1+\mathrm{h}))$ is

$\begin{aligned} m=& \frac{f(1+h)-1}{(1+h)-1}=\frac{(1+h)^{2}-1}{h}=\frac{\left\{1+2 h+h^{2}\right\}-1}{h} \\&=\frac{2 h+h^{2}}{h}=2+h .\end{aligned}$

As $h$ gets very small, the value of $m$ approaches the value 2, the slope of tangent line at the point $(1,1)$.

Practice 2: Find the slope of the line tangent to the graph of $\mathrm{y}=\mathrm{f}(\mathrm{x})=\mathrm{x}^{2}$ at the point $(-1,1)$ by finding the slope of the secant line, $\mathrm{m}_{\mathrm{sec}}$, through the points $(-1,1)$ and $(-1+\mathrm{h}, \mathrm{f}(-1+\mathrm{h}))$ and then determining what happens to $\mathrm{m}_{\mathrm{sec}}$ as $\mathrm{h}$ gets very small.