3 Second derivatives

An approach which has been found to work well for second derivatives involves applying the notion of a central difference three times. We begin with

$f^{″}\left(a\right)\approx \frac{f^{\prime }\left(a+\frac{1}{2}h\right)-f^{\prime }\left(a-\frac{1}{2}h\right)}{h}.$

Next we approximate the two derivatives in the numerator of this expression using central differences as follows:

$f^{\prime }\left(a+\frac{1}{2}h\right)\approx \frac{f\left(a+h\right)-f\left(a\right)}{h}\text{and}{f}^{\prime }\left(a-\frac{1}{2}h\right)\approx \frac{f\left(a\right)-f\left(a-h\right)}{h}.$

Combining these three results gives

Example 21

The distance $x$ of a runner from a fixed point is measured (in metres) at intervals of half a second. The data obtained are

$\begin{array}{cccccc}\hfill t\hfill & \hfill 0.0\hfill & \hfill 0.5\hfill & \hfill 1.0\hfill & \hfill 1.5\hfill & \hfill 2.0\hfill \\ ̲& ̲& ̲& ̲& ̲& ̲\\ \hfill x\hfill & \hfill 0.00\hfill & \hfill 3.65\hfill & \hfill 6.80\hfill & \hfill 9.90\hfill & \hfill 12.15\hfill \end{array}$

Use a central difference to approximate the runner’s acceleration at $t=1.5$ s.

Solution

Our aim here is to approximate $x^{″}\left(t\right)$ .

Using data with $t=1.5$ s at its centre we obtain

from which we see that the runner is slowing down.

Exercises

1. Let $f\left(x\right)=cosh\left(x\right)$ and $a=2$ . Let $h=0.01$ and approximate $f^{\prime }\left(a\right)$ using forward, backward and central differences. Work to 8 decimal places and compare your answers with the exact result, which is $sinh\left(2\right)$ .
2. The distance $x$ , measured in metres, of a downhill skier from a fixed point is measured at intervals of 0.25 s. The data gathered are

   $\begin{array}{cccccccc}\hfill t\hfill & \hfill 0\hfill & \hfill 0.25\hfill & \hfill 0.5\hfill & \hfill 0.75\hfill & \hfill 1\hfill & \hfill 1.25\hfill & \hfill 1.5\hfill \\ ̲& ̲& ̲& ̲& ̲& ̲& ̲& ̲\\ \hfill x\hfill & \hfill 0\hfill & \hfill 4.3\hfill & \hfill 10.2\hfill & \hfill 17.2\hfill & \hfill 26.2\hfill & \hfill 33.1\hfill & \hfill 39.1\hfill \end{array}$

   Use a central difference to approximate the skier’s velocity and acceleration at the times

   $t=$ 0.25 s, 0.75 s and 1.25 s. Give your answers to 1 decimal place.

Answer