# Math Help - Related Rates

1. ## Related Rates

A man, who is in a rowboat 2 miles from the nearest point A on a straight shoreline, wishes to reach a house located at point B which is 6 miles further down the coast. He plans to row to point P that is between A and B. P is also x miles from the house. He will then walk the remainder of the distance to the house. Suppose he can row at the rate of 3mph. He can walk at the rate of 5mph. If T is the total time to reach the house, express T as a function of x.
[--------------------------6miles------------------------]
..........................................[----------- x ------------]
A_________________________P______________________B (house)
|......................................../
|2................................./
|m........................./
|i..................../
|l.............../
|e........./
|s...../
|./
.
(rowboat)

2. We know that whatever value we have for $x$, the best course to the shore is a straight line, with no curves. To find this distance, we simply use geometry.

We know that the shortest distance from ship to shore makes a right angle. The distance from the boat to shore such that $x$ is an unknown variable, therefore, is the hypotenuse of a triangle made with the shortest route to shore and the shore itself. We know the former: it's $2$. We also know the latter, which is $6-x$. So, the hypotenuse, then, using the Pythagorean theorem, is $\sqrt{4+(6-x)^2}$. That should be enough for you to make the function $T(x)$.

Just in case it's not, though, here you go: Next we plug in time values. $(1/3)\sqrt{4+(6-x)^2}$ is the time it takes to row to shore, and $x/5$ is the time it takes to walk the remaining distance. Therefore:

$T(x)=(1/3)\sqrt{4+(6-x)^2}+x/5$

EDIT: fixed

3. Originally Posted by hatsoff
We know that whatever value we have for $x$, the best course to the shore is a straight line, with no curves. To find this distance, we simply use geometry.

We know that the shortest distance from ship to shore makes a right angle. The distance from the boat to shore such that $x$ is an unknown variable, therefore, is the hypotenuse of a triangle made with the shortest route to shore and the shore itself. We know the former: it's $2$. We also know the latter, which is $6-x$. So, the hypotenuse, then, using the Pythagorean theorem, is $\sqrt{4+(6-x)^2}$. That should be enough for you to make the function $T(x)$.

Just in case it's not, though, here you go: Next we plug in time values. $3\sqrt{4+(6-x)^2}$ is the time it takes to row to shore, and $5x$ is the time it takes to walk the remaining distance. Therefore:

$T(x)=3\sqrt{4+(6-x)^2}+5x$
so i don't need to use any rates like dx/dt

4. edit

5. Originally Posted by smschaefer
A man, who is in a rowboat 2 miles from the nearest point A on a straight shoreline, wishes to reach a house located at point B which is 6 miles further down the coast. He plans to row to point P that is between A and B. P is also x miles from the house. He will then walk the remainder of the distance to the house. Suppose he can row at the rate of 3mph. He can walk at the rate of 5mph. If T is the total time to reach the house, express T as a function of x.
[--------------------------6miles------------------------]
..........................................[----------- x ------------]
A_________________________P______________________B (house)
|......................................../
|2................................./
|m........................./
|i..................../
|l.............../
|e........./
|s...../
|./
.
(rowboat)
I think the immage should look like this...

Then using the pythagorean theorem we can solve for $d_1$.

$d_1=\sqrt{4+(6-x)^2}$

so since $d=rt$ t is equal to $t=\frac{d}{r}$

So he rows 3mph and walks 5mph we

$t_1=\frac{\sqrt{4+(6-x)^2}}{3}$

and the time he walks is

$t_2=\frac{x}{5}$

So the total time T is

$T=t_1+t_2=\frac{\sqrt{4+(6-x)^2}}{3}+\frac{x}{5}$

6. Originally Posted by TheEmptySet
I think the immage should look like this...

Then using the pythagorean theorem we can solve for $d_1$.

$d_1=\sqrt{4+(6-x)^2}$

so since $d=rt$ t is equal to $t=\frac{d}{r}$

So he rows 3mph and walks 5mph we

$t_1=\frac{\sqrt{4+(6-x)^2}}{3}$

and the time he walks is

$t_2=\frac{x}{5}$

So the total time T is

$T=t_1+t_2=\frac{\sqrt{4+(6-x)^2}}{3}+\frac{x}{5}$
This is correct, yes. I accidentally multiplied by 3 and 5, instead of dividing, as I should have. Sorry for the mixup. I have fixed the above posts.

7. Originally Posted by smschaefer
so i don't need to use any rates like dx/dt
We have to take the derivative (dT/dx, not dx/dt), then plug in zero for the function value (dT/dx).

$t(x)=(-1/3)(6-x)[4+(6-x)^2]^{-1/2}+(1/5)$

$0=(-1/3)(6-x)[4+(6-x)^2]^{-1/2}+(1/5)$

$x=6-\sqrt{6}$

That should do it.