Question

An airplane is flying at an altitude of 7 mi and passes directly over a radar antenna. When the plane is 10 mi from the antenna (s=10), the radar detects that the distance s is changing at the rate 300 mph. What is the speed of the airplane at that moment?

Answer 1

@zepdrix

Answer 2

Is this a Doppler effect problem?

Answer 3

no, it's a related rates problem

Answer 4

this look right so far?

Answer 5

lol yea, mine is close

Answer 6

Okay, so I'm going to assume that the radar is actually measuring \(s'\) and what we really want is \(x'\).

Answer 7

well... i know that I want to find dx/dt right?

Answer 8

I'm also going to assume it's not changing altitude \((y'=0)\) and that \(x'\) is constant.

Answer 9

so is 7 constant right?

Answer 10

and I would do the derivative of Pythagorean theorem equation?

Answer 11

We have \[ \begin{array}{rcl}
s^2 &=& y^2 + x^2 \\
2ss' &=& 2yy' + 2xx'
\end{array}\]

Answer 12

This is just the chain rule.

Answer 13

but what you did looks similar to Pythagorean lol

Answer 14

Yeah there is Pythagorean in there.

Answer 15

yea i got the same thing... and then i stopped lol

Answer 16

Well you know \(s=10, s'=300, y=7,y'=0,x=\sqrt{10^2-7^2} \)

Answer 17

What do you get when you plug that in and solve for \(x'\)? Is it correct?

Answer 18

Wait, hold on, I think I might have missed something.

Answer 19

Are you getting the correct answer?

Answer 20

well if y is constant, wouldn't there be no dy/dx * y

Answer 21

i got square root of 51

Answer 22

Yeah, it would essentially be 0.

Answer 23

for x

Answer 24

yeah, but for \(x'\)?

Answer 25

isn't that what we're trying to find?

Answer 26

Yeah

Answer 27

\[(2)(10)(300)=(2)(\sqrt{51})x'\]

Answer 28

so 300 mph isn't ds/dt?

Answer 29

Yes.

Answer 30

i got 420.08 mph :)