# Stevie completes a quest by travelling from A to C vi P. The speed along AP is 4 km/hour, and along AB it is 5 km/hour. Solve the following?

## A) Find $D \left(x\right)$ a function for the total distance travelled as a function of $x$ B) Form a function $T \left(x\right)$ for the total journey time. C) What is the minimum time required for Stevie to complete her quest?

Dec 30, 2016

$D \left(x\right) = \sqrt{9 + {x}^{2}} + \left(6 - x\right)$

$T \left(x\right) = \frac{\sqrt{9 + {x}^{2}}}{4} + \frac{6 - x}{5}$

From this we get the minimum time as $1.65$ (hours) which corresponds to a distance of $7$ (km).

#### Explanation:

A) The distance $D \left(x\right)$ from A to C via P as a function of $x$.

By Pythagoras;

$\setminus \setminus \setminus \setminus \setminus A {P}^{2} = A {B}^{2} + B {P}^{2}$
$\therefore A {P}^{2} = {3}^{2} + {x}^{2}$
$\therefore A {P}^{2} = 9 + {x}^{2}$
$\therefore \setminus \setminus A P = \sqrt{9 + {x}^{2}}$ (must be the +ve root)

Then;

$\setminus \setminus \setminus \setminus \setminus D \left(x\right) = A P + P C$
$\therefore D \left(x\right) = \sqrt{9 + {x}^{2}} + \left(6 - x\right)$

B) Find the time, $T \left(x\right)$, that is required to travel from A to C via P.

Using $\text{speed" = "distance"/"time}$:

Along AP the speed is $4$ km/hour, provided $x > 0$ (otherwise we are going along AB at $5$ km/hour) and so:

$\text{ } 4 = \frac{\sqrt{9 + {x}^{2}}}{t} _ 1$
$\therefore {t}_{1} = \frac{\sqrt{9 + {x}^{2}}}{4}$

Along PC (or AB) the speed is $5$ km/hour, and so:

$\text{ } 5 = \frac{6 - x}{t} _ 2$
$\therefore {t}_{2} = \frac{6 - x}{5}$

And so, the total time is given by:

$\setminus \setminus \setminus \setminus \setminus T \left(x\right) = {t}_{1} + {t}_{2}$
$\therefore T \left(x\right) = \frac{\sqrt{9 + {x}^{2}}}{4} + \frac{6 - x}{5}$ for ($x > 0$)

C) Stevie's Quest

In order for Stevie to complete the quest we need to find a critical point of T(x):

Differentiating wrt $x$ we get:

$\text{ } T ' \left(x\right) = \frac{1}{4} \cdot \frac{1}{2} {\left(9 + {x}^{2}\right)}^{- \frac{1}{2}} \cdot 2 x + \frac{1}{5} \left(- 1\right)$
$\therefore T ' \left(x\right) = \frac{x}{4 \sqrt{\left(9 + {x}^{2}\right)}} - \frac{1}{5}$

At a critical point, $T ' \left(x\right) = 0$

$\implies \frac{x}{4 \sqrt{9 + {x}^{2}}} - \frac{1}{5} = 0$
$\therefore 5 x - 4 \sqrt{9 + {x}^{2}} = 0$
$\therefore 5 x = 4 \sqrt{9 + {x}^{2}}$
$\therefore 25 {x}^{2} = 16 \left(9 + {x}^{2}\right)$
$\therefore 25 {x}^{2} = 144 + 16 {x}^{2}$
$\therefore 9 {x}^{2} = 144$
$\therefore {x}^{2} = \frac{144}{9}$
$\therefore {x}^{2} = 16$
$\therefore x = 4$ (must be the +ve root)

When $x = 4$ we have:

$\therefore D \left(4\right) = \sqrt{9 + 16} + \left(6 - 4\right) = 7$
$\therefore T \left(4\right) = \frac{\sqrt{9 + 16}}{4} + \frac{6 - 4}{5} = \frac{33}{20} = 1.65$

We can confirm visually that this corresponds to a minimum by looking at the graph:

graph{sqrt(9 + x^2)/4 + (6-x)/5 [-15, 15, -1, 10]}