# What is the swap curve

## Finding the equation of tangent line to the curve at the given point

EDIT: Let's clarify a couple of things.

The slope of the secant line between $(a, f(a))$ and $(x,f(x)))$ is $$\frac.$$

The slope of the tangent line at $(a, f(a))$ is $$\lim_\frac.$$

To find the equation of a tangent line, one needs to use the point-slope formula, which I've explained below.

Now, in your case, $f(x) = \sqrt$, and we have $a = 1$, $f(a) = 1$. So the slope of the tangent line is $$\lim_\frac<\sqrt - 1>.$$ Now we have to evaluate this limit.

If we try to evaluate this limit by just plugging in $x = 1$, we get $0/0$, which is a problem (dividing by zero is bad), so we need a new strategy.

Idea: When evaluating the limits of fractions, a good trick is to multiply the top and bottom by the "radical conjugate." So:

$$\begin \frac<\sqrt - 1> & = \frac<\sqrt - 1>\frac<\sqrt + 1><\sqrt + 1> \\ & = \frac<(\sqrt - 1)(\sqrt + 1)><(x-1)(\sqrt + 1)> \\ & = \frac<(x-1)(\sqrt + 1)> \\ & = \frac<1><\sqrt + 1>. \end$$

Now we can evaluate $$\lim_ \frac<\sqrt - 1> = \lim_\frac<1><\sqrt + 1>$$ by plugging in $x = 1$ no problem. This will give us the slope of the tangent line. If you want the equation of the tangent line, you need the point-slope formula, explained below.

The point-slope formula says that a line with slope $m$ that passes through $(x_0, y_0)$ has an equation of the form $$y - y_0 = m(x-x_0).$$

In your case, the tangent line passes through $(1,1)$, so you can plug in $x_0 = 1$, $y_0 = 1$. We'll also have the slope, $m$, from the previous section once we evaluate that limit (which I leave to you to do).

http://math.stackexchange.com/questions/63606/finding-the-equation-of-tangent-line-to-the-curve-at-the-given-point what is the swap curve

Source: math.stackexchange.com

Category: Forex