Related Rates (3rd Time)

[scrum]

Okay these are still confusing me. I got further than I ever have before but my answer comes out wrong so here's the one i was working on. Please tell me which of my steps are wrong.

A plane flying horizontally at an altitude of 1 mile and a speed of of 500 mi/h passes directly over a radar station. Find the rate at which the distance from the plane to the station is increasing when it is 2 mi away from the station.

1. made diagram

the blank is 1, x is 2, and y is the distance that is increasing.

2. I said dx/dt is 500mi/hr because that is the speed of the plane.

3. I said x^2 + 1^2 = y^2

4. I implicitly differentiated that to be 2x(dx/dt)=2y(dy/dt)

5. I plugged 500 in for (dx/dt), 2 in for x, and sqrt(5) in for y.

6. solved for dy/dt and got 1000/sqrt(5)

wrong answer

sorry for the repeated questions i am just way confused by these. [/quote]

 

[stapel]

Is x = 2, or is y? Which one indicates the distance from the station? :wink:

Eliz.

 

[soroban]

Hello, scrum!

You were close . . .

A plane flying horizontally at an altitude of 1 mile
and a speed of of 500 mi/h passes directly over a radar station.
Find the rate at which the distance from the plane to the station is increasing
when it is 2 mi away from the station.

1. made diagram

               x
      * - - - - - - - - *
      |              *
      |           *
    1 |        * y
      |     *
      |  *
      * - - - - - - - - - -

2. I said: \(\displaystyle \,\frac{dx}{dt} \,=\,50\) mph, because that is the speed of the plane. . . . . Yes!

3. I said: \(\displaystyle \:x^2\,+\,1^2\:=\:y^2\) . . . . Good!

4. I implicitly differentiated: \(\displaystyle \:2x\left(\frac{dx}{dt}\right) \:=\:2y\left(\frac{dy}{dt}\right)\) . . . . Right!

5. I plugged 500 in for (dx/dt), 2 in for x, and sqrt(5) in for y . . . . no

It says, "when it is 2 miles away from the station".
. . This means that: \(\displaystyle \:y \:=\:2\) . . . not \(\displaystyle x.\)

 

[Dr. Flim-Flam]

You want to find dy/dt.

dy/dt = (x/y)dx/dt. dx/dt = 500. When y=2, x=sqrt(3).

Hence dy/dt = 250*sqrt(3) m/hr. QED