1. ## Related Rates Question

Two cars travel on perpendicular roads towards the intersection of the roads. The first car is 100 miles from the intersection and is traveling at 55 mph. The second car starts at the same time 250 miles from the intersection and is going 60 mph. How fast is the distance between them changing 1.5 hours later?

I know that I have to take the derivative of the pythagorean theorem and plug in the values for x, dx/dt, y, and dy/dt. I also know that I have to solve for dz/dt. The only thing I don't understand is how I have to incorporate time into the equation.

Thanks.

-Chris

2. Hello, Chris!

A good sketch helps, of course . . .

Two cars travel on perpendicular roads towards the intersection of the roads.
The first car is 100 miles from the intersection and is traveling at 55 mph.
The second car starts at the same time 250 miles from the intersection and is going 60 mph.
How fast is the distance between them changing 1.5 hours later?
Code:
P o
|
55t |
|
A o
|  *
|     *   x
100-55t|        *
|           *
|              *
C o - - - - - - - - o - - - - - o Q
250-60t       B     60t

The intersection is at $\displaystyle C.$

The first car starts at $\displaystyle P$, where $\displaystyle PC = 100$,
. . and drives south at 55 mph.
In $\displaystyle t$ hours, it moves $\displaystyle 55t$ miles to $\displaystyle A.$
. . $\displaystyle AC \:=\:100-5t$

The second car starts at $\displaystyle Q$, where $\displaystyle QC = 250$,
. . and drives west at 60 mph.
Inb $\displaystyle t$ hours, it moves $\displaystyle 60t$ miles to $\displaystyle B.$
. . $\displaystyle BC \:=\:250-60t$

The distance between them, $\displaystyle x \:=\:AB$,
. . is the hypotenuse of right triangle $\displaystyle ABC.$

Therefore: .$\displaystyle x \;=\;\sqrt{(100-55t)^2 + (250-60t)^2}$

Got it?

3. Yes.

Thanks I ton. I completely understand it now.

-Chris