Hello, samtheman17!

A man can row at 4 km/h, and run at 3 km/h.

He needs to get from a point $\displaystyle A$, on the south bank of a stretch of still water,

to point $\displaystyle B$ on the north bank of the water.

The direct distance from $\displaystyle A$ to $\displaystyle B$ is 5 km, and the water is 4 km wide.

He starts rowing with an angle θ between North and the direction in which he rows.

Find an expression for the time he will take to get from $\displaystyle A$ to $\displaystyle B$, in terms of θ. Code:

3
C * - - - o B
| *
4 | * 5
| *
A o

The width of the water is: $\displaystyle AC = 4$ km.

The distance from $\displaystyle A$ to $\displaystyle B$ is: $\displaystyle AB = 5$ km.

Pythagorus tells us that: $\displaystyle CB = 3$ km.

He will start at $\displaystyle A$, row to $\displaystyle P$, then run to $\displaystyle B.$

Code:

x P 3-x
C * - - - * - - - - * B
| /
| /
| / y
4 | /
| /
|θ/
|/
A *

Let $\displaystyle \theta \,=\, \angle CAP$

Let $\displaystyle x \,=\, CP$, then: $\displaystyle 3-x \:=\:PB.$

Let $\displaystyle y \,=\,AP$

In right triangle $\displaystyle ACP\!:\;\;y^2 \:=\:x^2 + 4^2 \quad\Rightarrow\quad y \:=\:\sqrt{x^2+16}$

He rows $\displaystyle \sqrt{x^2+16}$ km at 4 km/hr.

. . This will take him: .$\displaystyle \frac{\sqrt{x^2+16}}{4}$ hours.

Then he runs $\displaystyle 3-x$ km at 3 km/hr.

.This will take him: .$\displaystyle \frac{3-x}{3}\:=\:1-\tfrac{1}{3}x$ hours.

Hence, his total time is: .$\displaystyle T \;=\;\tfrac{1}{4}\sqrt{x^2+16} + 1 -\tfrac{1}{3}x$ .[1]

We want this answer in terms of θ.

In right triangle $\displaystyle ACP\!:\;\tan\theta \:=\:\frac{x}{4} \quad\Rightarrow\quad x \:=\:4\tan\theta$ .[2]

Note that: .$\displaystyle x^2+16 \;=\;(4\tan\theta)^2 + 16 \;=\;16\tan^2\!\theta + 16 \;=\;16(\tan^2\!\theta+1) \;=\;16\sec^2\!\theta $

. . Hence: .$\displaystyle \sqrt{x^2+16} \;=\;4\sec\theta$ .[3]

Subnstitute [2] and [3] into [1]: .$\displaystyle T \;=\;\tfrac{1}{4}(4\sec\theta) + 1 - \tfrac{1}{3}(4\tan\theta)$

Therefore: .$\displaystyle T \;=\;\sec\theta + 1 - \tfrac{4}{3}\tan\theta$