Analyzing related rates problems: equations (Pythagoras) | AP Calculus AB | Khan Academy

- [Instructor] Two cars are driving

towards an intersection from
perpendicular directions.

The first car's velocity is 50 km/h

and the second car's velocity is 90 km/h.

At a certain instant, T sub zero,

the first car is a
distance, X sub T sub zero

or X of T sub zero, of half a kilometer

from the intersection and the second car

is a distance, Y of T sub zero,

of 1.2 kilometers from the intersection.

What is the rate of
change of the distance,

D of T, between the cars at that instant?

So, T sub zero.

Which equation should be
used to solve the problem?

And they give us a
choice of four equations,

right over here, so you
could pause the video

and try to work through it on your own

but I'm about to do it, as well.

So, let's just draw what's going on.

That's always a healthy thing to do.

So, two cars are driving towards

an intersection from
perpendicular directions.

So, let's say that this
is one car right over here

and it is moving in the X direction

towards that intersection,
which is right over there.

And then you have another car

that is moving in the Y direction.

So, let's say it's moving like this.

So, this is the other car.

I should have maybe done a top view.

Well, here we go.

This square represents the car

and it is moving in that direction.

Now, they say at a certain instance.

T sub zero, so let's draw that instant.

So, the first car is a distance,

X of T sub zero, of 0.5 kilometers.

So, this distance, right over here,

let's just call this X of T and let's call

this distance, right over here, Y of T.

Now, how does the distance between the

cars relate to X of T and Y of T?

Well, we could just use
the distance formula,

which is essentially just
the Pythagorean theorem,

to say, well, the
distance between the cars

would be the hypotenuse
of this right triangle.

Remember, they're traveling
from perpendicular directions.

So, that's a right triangle, there.

So, this distance, right over
here, would be X of T squared

plus Y of T squared, and
the square root of that,

and that's just the Pythagorean
theorem, right over here.

This would be D of T, or we could say

that D of T squared is equal to

X of T squared plus Y,
too many parentheses.

Plus Y of T squared.

So, that's the relationship between

D of T, X of T, and Y
of T, and it's useful

for solving this problem because, now,

we can take the derivative of both sides

of this equation with respect to T.

We'd be using various derivative rules,

including the chain
rule, in order to do it,

and then that would give us a relationship

between the rate of change of D of T,

which would be D prime of T,

and the rate of change of X of T, Y of T,

and X of T and Y of T, themselves.

And so, if we look at these
choices, right over here,

we indeed see that D sets up that exact

same relationship that
we just did, ourselves.

That it shows that the distance

squared between the cars is equal to

that X distance from the
intersection squared,

plus the Y distance from
the intersection squared,

and then we can take the
derivative of both sides

to actually figure out this
related, rates question.