Finding the rate at which the distance from an observer to a plane is changing

[Kamhogo]

" An airplane flies over an observer with a velocity of 400km/h and at an altitude of
500m. If the plane flies horizontally in a straight line, find the rate at which the
distance x from the observer to the plane is changing 0.600 min after the plane passes
over the observer."


Attempt at solution:

1) 400 km/h = 400 000 m/h = 6666.667 m/min = velocity

2) Horizontal position of plane in the sky is given by velocity times time,
==> 66666,67 m/min (t)

3) The altitude, the horizontal position of the plane, and x ( the distance from
the observer ' s point of of view) form a right triangle, with x as the hypotenuse.
Hence,

x^(2)= 500^(2) + ( 6666.667t)^(2)

Distance from observer to the plane is given by x (t) = (500^(2)+(6666.667t)^(2))^(1/2)

4)Rate of change of x with respect to time is dx/dt

dx/dt = ( 1/2)(250000 + (6666.667)^(2)(t^2))^(-1/2)(0+ ((66666.667)^(2))t)

= [2 (66666.667^2)t]/[2500000 + (66666.667^2)(t^2)]^(1/2)

5) dx/dt for t = 0.600 min is equal to 53 333 338,67/4031.129073 = 12 230, 37236 m/min.

Voilà.



This is an even numbered problem from my textbook so the solution is not provided. I'd like
to know if someone can confirm what I found or help me towards the right solution please.

 

[stapel]

"An airplane flies over an observer with a velocity of 400km/h and at an altitude of 500m. If the plane flies horizontally in a straight line, find the rate at which the distance x from the observer to the plane is changing 0.600 min after the plane passes over the observer."

I haven't checked all the numbers, but your general methodology looks sound. However, you can simplify things, starting by noting that they're probably looking for an answer in terms of kilometers per hour. And this is good! It means you'll be working with smaller numbers!

I'd start by doing a different conversion:

. . . . .0.6 min = (6/10 min)*(1 hr / 60 min) = (6*1 / 10*6*10) (hr*min / min) = 1/100 hr = 0.01 hr

That's a lot nicer number! (And you want to stick with nice "exact" forms until the very end. Try not to introduce rounding errors by using decimal approximations too soon.)

The height is fixed at h = 0.5 km. The horizontal distance (from the point directly over the observer) is 400t, where t is in "hours". This gives you your Pythagorean-Theorem equation:

. . . . .\(\displaystyle (x(t))^2\, =\, 0.5^2\, +\, (400t)^2\)

Here's a trick: You don't have to do the square root at this point. You can stick with the equation just as it is, and differentiate:

. . . . .\(\displaystyle (2\, x(t))\, \left(\dfrac{dx}{dt}\right)\, =\, 0\, +\, 2\, (400t)\, (400)\)

. . . . .\(\displaystyle (2\, x(t))\, \left(\dfrac{dx}{dt}\right)\, =\, 320 000t\)

At t = 0.01 hr, the value of x(t) is:

. . . . .\(\displaystyle x(0.01)\, =\, \sqrt{\strut 0.5^2\, +\, 4^2\, }\, =\, \sqrt{\strut 16.25\,}\, =\, \sqrt{\strut \dfrac{65}{4}\,}\, =\, \dfrac{\sqrt{\strut 65\,}}{2}\)

Then:

. . . . .\(\displaystyle (2\, x(0.01))\, \left(\dfrac{dx}{dt}\right)\, =\, 320\, 000(0.01)\)

. . . . .\(\displaystyle 2\, \left(\dfrac{\sqrt{\strut 65\,}}{2}\right)\, \left(\dfrac{dx}{dt}\right)\, =\, 3\, 200\)

. . . . .\(\displaystyle \sqrt{\strut 65\,}\, \left(\dfrac{dx}{dt}\right)\, =\, 3\, 200\)

. . . . .\(\displaystyle \dfrac{dx}{dt}\, =\, \dfrac{3\,200}{\sqrt{\strut 65\,}}\, \approx\, 396.911\,151\, \mbox{km/hr}\)

How does this compare with your result? (Hint: I think you're off by a factor of 2 somewhere.)

 

[Kamhogo]

I changed the conversion ( put everything in km and h) and found 396.9111507 km/h.
I also tried not taking the square root of x ( actually easier and nicer) and found the same answer ?. I was way of with my first result ( 733.82 km/h ). Thanks a lot!