00:01
All right, so in this problem we got a plane that takes off from an airfield at noon, and it's going 80 miles per hour and it's traveling east.
00:10
At two o 'clock, two hours later, there was a plane that was traveling north to the airfield at 110 miles per hour, so it gets there, it arrives at the airfield.
00:23
So that's why i said t is between zero and two, because um, after two hours the plane that was traveling north, the red plane, will land.
00:35
The other plane going east, you know, who knows.
00:40
So wherever these planes are, that green line is their distance, right? so if i called the red distance x, and i cast backwards, i'm going to call the red distance y, because it's going north and south, and i'm going to call the black distance x, we know that the distance, the distance squared, right, pythagorean theorem, is going to equal x squared plus y squared.
01:25
Okay, so when we take the derivative, so two times the distance times the derivative of the distance over time, right, this is what we're looking for, okay, is going to equal 2x dx over dt, this is the 80 miles per hour.
01:52
Now because it's going away from the field, that's going to be positive, and then the derivative of the y is 2y dy over dt, this is our 110 miles per hour, but because it's getting closer to the field, it's, it's going to be negative.
02:11
So we can divide everything by two, we don't need these twos here.
02:18
Okay, so here we go.
02:21
We know, we want to find the green dd over dt.
02:28
Let's put stuff in.
02:31
So we know that this is going to be 80 miles per hour, and we know for the y, the rate is going to be negative 110 miles per hour, and the reason it's negative is because it's getting closer to the airfield.
02:52
The, the plane traveling east is getting farther away, so that's why its rate is, oops, its rate is positive.
03:02
Okay, now we want it at 1 o 'clock or when t equals 1.
03:08
All right, well let's think about this.
03:10
When t equals 1, we know that the plane traveling east is 80 miles away, so we can put that in for x.
03:20
We know that the plane traveling north, the red plane, is 110 miles away, right, because it's going to be at the airfield in an hour, so it has to be 110 miles.
03:35
I'm trying to include the units, so it's going to be that equals.
03:43
Now we're looking for this.
03:47
This is what we want to find, the rate of change in their distance.
03:52
We need a number for d, but we could find it using this little formula right here, because we have numbers for x and for y, so we know that d squared is going to equal 80 squared plus 110 squared, and then we can get our little handy -dandy calculator out and solve for d, and hopefully it comes out as a nice, nice good integer.
04:23
I don't know if it will, but we're going to hope.
04:27
All right, so 110 squared plus 80 squared, does that take the square root? and it's not.
04:38
It's not a good decimal...