How to Find Changing Distance between Two Moving Objects
In a typical related rates problem, such as when you’re finding a change in the distance between two moving objects, the rate or rates in the given information are constant, unchanging, and you have to figure out a related rate that is changing with time. You have to determine this related rate at one particular point in time.
Here’s an example: One car leaves an intersection traveling north at 50 mph, another is driving west toward the intersection at 40 mph. At one point, the northbound car is threetenths of a mile north of the intersection, and the westbound car is fourtenths of a mile east of it. At this point, how fast is the distance between the cars changing?

Start by creating a diagram.
Calculus — it’s a drive in the country.Before going on with this problem, consider a similar problem that you may run across if you’re using a standard calculus textbook. It involves a ladder leaning against and sliding down a wall. Can you see that the diagram for such a ladder problem would be very similar to this figure except that the yaxis would represent the wall, the xaxis would be the ground, and the diagonal line would be the ladder? These problems are quite similar, but there’s an important difference. The distance between the cars is changing so the diagonal line in the figure is labeled with a variable, s. A ladder, on the other hand, has a fixed length, so the diagonal line in your diagram for the ladder problem would be labeled with a number, not a variable.

List all given rates and the unknown rate.
As Car A travels north, the distance y is growing at 50 miles per hour. That’s a rate, a change in distance per change in time. So,
As Car B travels west, the distance x is shrinking at 40 miles per hour. That’s a negative rate:
You have to figure out how fast s is changing, so,

Write the formula that relates the variables in the problem: x, y, and s.
The Pythagorean theorem, a^{2} + b^{2}^{ }=^{ }c^{2}, will do the trick for this right triangle problem. In this problem, x and y are the legs of the right triangle, and s is the hypotenuse, so x^{2} + y^{2}^{ }=^{ }s^{2}.
The Pythagorean theorem is used a lot in related rates problems. If there’s a right triangle in your problem, it’s quite likely that a^{2} + b^{2}^{ }=^{ }c^{2} is the formula you’ll need.
Because this formula contains the variables x, y, and s which all appear in your list of derivatives in Step 2, you don’t have to tweak this formula.

Differentiate with respect to t.
(Remember, in a related rates problem, all variables are treated like the ys in an implicit differentiation problem.)

Substitute and solve for
“Holy devoid distance lacking length, Batman. How can you solve for
unless you have values for the rest of the unknowns in the equation?” “Take a chill pill, Robin — just use the Pythagorean theorem again.”
You can reject the negative answer because s obviously has a positive length. So s = 0.5.
Now plug everything into your equation:
This negative answer means that the distance, s, is decreasing.
Thus, when car A is 3 blocks north of the intersection and car B is 4 blocks east of it, the distance between them is decreasing at a rate of 2 mph.