
Suppose we want to find the tangent to a curve. Just how can we go about finding one?
Here is one way:
As $Q$ approaches $P$, the secant line approximates the tangent line better and better. The limiting position of the secant line as $Q$ approaches $P$ is the tangent to the curve at $P$. If the curve is given by $y=f(x)$ and $P$ has the coordinates $(x_0,y_0)$, then the slope of the tangent line at $P$ is $f'(x_0)$, the derivative of f evaluated at $x_0$. Let's find the equation of the line tangent to the parabola at $(2,3)$.
The slope of the tangent is just $f'(x)$ evaluated at x. \begin{eqnarray*} f(x) &=& x^21 \\ f'(x) &=& 2x \\ f'(2) &=& 4. \end{eqnarray*} Now, the equation of the line can be written in pointslope form like this: \begin{eqnarray*} yy_0 &=& m(xx_0)\\ yy_0 &=& f'(x_0)(xx_0)\\ y3 &=& 4(x2) \end{eqnarray*} since the line passes through the point $(2,3)$ and has slope $4$. In slopeintercept form, the equation of the tangent line becomes $$ y=4x5. $$
What happens when $x=0$ for this function? What about as $x$ gets large? Now that we can find the tangent to a curve at a point, of what use is this?
Do you notice that as you zoom in on $P$ the curve looks more and more linear and is approximated better and better by the tangent line? Let's get more specific: Near $x_0$, we saw that $y=f(x)$ can be approximated by the tangent line $yy_0=f'(x_0)(xx_0)$. Writing this as $y=y_0+f'(x_0)(xx_0)$ and noting that $y=f(x_0)$, we find that
(Notice that the righthand side is just the 2term Taylor Expansion of $f(x)$.) If we know that value of $f$ at $x_0$, this gives us a way to approximate the value of $f$ at $x$ near $x_0$. We do this by starting at $(x_0,f(x_0))$ and moving along the tangent line to approximate the value of the function at $x$. Look at $f(x) = \arctan{x}$.
Let's use the tangent approximation $f(x) \approx f(x_0)+f'(x_0)(xx_0)$ to approximate $f(1.04)$:
How well does this approximate $\arctan(1.04)$?
Key Concepts
