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

`33\; {\rm m}`.

Step-by-step 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[(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 (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)) + (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)) + (1)/(2)\left(f^((2))(t_(0)) \right)\, (t - t_(0))^(2) \\ =\; & f(0) + \left(f^((1))(0)\right)\, (t - 0) + (1)/(2)\left(f^((2))(0) \right)\, (t - 0)^(2) \\ =\; & f(0) + \left(f^((1))(0)\right) \, t + (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 + (1)/(2) * 6 \\ =\; & 33\end{aligned}`.