At noon, ship A is 30 nautical miles due west
of ship B. So if this is ship A and this is ship B
the distance between the two ships would be 30 miles.
Next, ship A is sailing west at 16 knots,
so ship A is sailing in this direction,
and ship B is sailing north at 15 knots,
so ship B is sailing in this direction.
How fast in knots is the distance between
the ships changing at 4 P.M.?
Well, first thing you'll notice is from noon to 4 P.M.
would be four hours and therefore for this problem,
t, time, equals four hours.
Now going back to our diagram,
let's let this length here be equal to x miles,
this length here be equal to y miles,
and the length between the two ships, s miles.
Now let's begin to list the given information.
Because ship A is sailing west at 16 knots,
x is changing at that rate and therefore dx dt
is equal to 16 knots,
and because ship B is sailing north at 15 knots,
dy dt would be 15 knots.
And because we're trying to find how fast
the distance is changing between the two ships,
we're looking for ds dt.
To find ds dt, we need an equation that relates s, x and y
which we can then differentiate to find ds dt.
And because we have a right triangle here,
we can use the Pythagorean theorem to form our equation.
s squared must equal x squared plus y squared.
Before we differentiate this with respect to t though,
let's determine some more information.
Let's find x, y and s when t equals four.
When t equals four, x would be equal to
this distance here which would be thirty miles, or thirty
plus the rate of change times the time.
And because dx dt equals 16 knots,
this length would be 30 plus 16 knots times
the time of four hours which is equal to 94 miles.
y would be equal to the rate of change times four hours
which would be 15 times four or 60 miles.
But let's also find the length of s
when x equals 94 and y equals 60.
So s squared would be equal to
94 squared plus 60 squared.
And we know s has to be positive, so now if we take
the square root of both sides of the equation,
we would have s equals the square root of
94 squared plus 60 squared,
which would be the square root of 12,436,
which would be approximately 111.52 miles.
Now that we have all the information we need
let's find the derivative of this equation here
with respect to t on the next slide.
So we'd have the derivative of
s squared with respect to t must equal the derivative
of x squared plus y squared with respect to t.
So that would be two s times ds dt
equals two x times dx dt
plus two y times dy dt.
Notice we can divide both sides by two
that would give us s times ds dt
equals x times dx dt
plus y times dy dt
and now we'll divide both sides by s
to solves for ds dt.
So we have ds dt
equals x times dx dt
plus y times dy dt
divided by s.
Now that we have this equation here for ds dt
we can perform substitutions for x, y, s
and dx dt and dy dt to find ds dt.
So when t equals four, ds dt is going to be equal to
x which is 94
times dx dt which is 16
plus y which is 60
times dy dt which is 15
divided by s we'll use the exact value here
of the square root of 12,436.
Let's continue on the next slide.
This comes out to 2,404 divided by
the square root of 12,436, and this would be knots.
Let's get a decimal approximation for this
ds dt would be approximately,
we can see here it'd be approximately 21.56 knots.
So we can say at 4 P.M. the distance between the ships
is changing at a rate of approximately 21.56 knots.
I hope you found this explanation helpful.