Question

A car is moving with speed 30 m/s and acceleration 6 m/s2 at a given instant. Using a second-degree Taylor polynomial, estimate how far the car moves in the next second.

Answer 1

Answer:

$33\; {\rm m}$.

Explanation:

Let for positive integer $n$, let $f^{(n)}(x)$ denote the $n$th derivative of $f(x)$ at $x$.

The $n$th order Taylor polynomial expansion of $f(x)$ at $x_{0}$ could be written in the form:

$\begin{aligned}f(x_{0}) + \sum\limits_{j = 1}^{n} \left[\frac{1}{j!}\, \left(f^{(j)}(x_{0})\right)\, (x - x_{0})^{j}\right]\end{aligned}$.

For example, when $n = 2$, the polynomial would include three terms:

• $f(x_{0})$,
• $\begin{aligned}\left(f^{(1)}(x_{0})\right)\, (x - x_{0})\end{aligned}$ (for $j = 1$,) and
• $\displaystyle \frac{1}{2}\left(f^{(2)}(x) \right)\, (x - x_{0})^{2}$ (for $j = 2$.)

$\begin{aligned}f(x_{0}) + \left(f^{(1)}(x_{0})\right)\, (x - x_{0}) + \frac{1}{2}\left(f^{(2)}(x) \right)\, (x - x_{0})^{2}\end{aligned}$.

Let $f(t)$ denote the distance that this vehicle travelled at time $t$.

• The first derivative $f^{(1)}(t)$ would denote the speed of the vehicle at time $t$.
• The second derivative $f^{(2)}(t)$ would denote the acceleration of the vehicle at time $t$.

At $t_{0} = 0$, the distance that the vehicle travelled would be $0$. The question states that:

• The speed of the vehicle at $t_{0} = 0$ was $f^{(1)}(t_{0}) = 30$.
• The acceleration of the vehicle at $t_{0} = 0$ was $f^{(2)}(t_{0}) = 6$.

The second-degree Taylor polynomial expansion of $f(t)$ at $t_{0} = 0$ would be:

$\begin{aligned}& f(t_{0}) + \left(f^{(1)}(t_{0})\right)\, (t - t_{0}) + \frac{1}{2}\left(f^{(2)}(t_{0}) \right)\, (t - t_{0})^{2} \\ =\; & f(0) + \left(f^{(1)}(0)\right)\, (t - 0) + \frac{1}{2}\left(f^{(2)}(0) \right)\, (t - 0)^{2} \\ =\; & f(0) + \left(f^{(1)}(0)\right) \, t + \frac{1}{2}\left(f^{(2)}(0) \right) \, t \end{aligned}$.

Substituting in $f(0) = 0$, $f^{(1)}(0) = 30$, and $f^{(2)}(0) = 6$, the polynomial estimate of $f(1)$ at $t_{0} = 0$ would be:

$\begin{aligned} & 0 + 30 + \frac{1}{2} \times 6 \\ =\; & 33\end{aligned}$.