- [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.