Completely lost: airplane flies N @ 200mph, other flies E @ 400mph; when 500 mi apart?

[allegansveritatem]

Here is the problem:

71 Distance between airplanes
An airplane flying north at 200 mi/hr passed over a point on the ground at 2:00 p.m. Another airplane at the same altitude passed over the point at 2:30 p.m., flying east at 400 mi/hr.

(a) If t denotes the time in hours after 2:30 p.m., express the distance d between the airplanes in terms of t.
(b) At what time after 2:30 p.m. were the airplanes 500 miles apart?

I was able to come up with the notion that the Pythagorian Theorem was going to be involved for part (a) but...well, here is what I got for part (a):



and here is the solution from the solutions manual:



I am able to follow this until I get just beyond the 3rd equals sign. Then the train jumps the rails and I am stuck in the mud. How did the author ever go from the term under the second radical sign to the quadratic equation under the third radical sign??? How is that even possible? I fault the solutions manual on this one: What business had the author of the manual to go eliding so egregiously?

PS: I admit that my solution owes much to the solutions manual. I couldn't have even gotten it, wrong though it is, without some initial guidance.

 

[MarkFL]

If we orient our coordinate axes such that "the point" is the origin, and our units are miles, and the time parameter \(t\) is in hours, then at time \(t=0\), the position of the plane flying north would be:

[MATH]A_1=\left\langle 0,100(2t+1) \right\rangle[/MATH]
And the position of the plane flying east would be:

[MATH]A_2=\left\langle 400t,0 \right\rangle[/MATH]
I am assuming you are good with that. And so yes, the distance between the planes (as a function of time), using the distance formula, is:

[MATH]d(t)=\sqrt{(400t-0)^2+(0-100(2t+1)^2)}=\sqrt{(400t)^2+(100(2t+1))^2}[/MATH]
Now, if we factor \(100^2\) from the radicand, we may write:

[MATH]d(t)=100\sqrt{(4t)^2+(2t+1)^2}[/MATH]
Expand the radicand:

[MATH]d(t)=100\sqrt{16t^2+4t^2+4t+1}[/MATH]
Combine like terms:

[MATH]d(t)=100\sqrt{20t^2+4t+1}[/MATH]
Does that make sense?

 

[allegansveritatem]

If we orient our coordinate axes such that "the point" is the origin, and our units are miles, and the time parameter \(t\) is in hours, then at time \(t=0\), the position of the plane flying north would be:

[MATH]A_1=\left\langle 0,100(2t+1) \right\rangle[/MATH]
And the position of the plane flying east would be:

[MATH]A_2=\left\langle 400t,0 \right\rangle[/MATH]
I am assuming you are good with that. And so yes, the distance between the planes (as a function of time), using the distance formula, is:

[MATH]d(t)=\sqrt{(400t-0)^2+(0-100(2t+1)^2)}=\sqrt{(400t)^2+(100(2t+1))^2}[/MATH]
Now, if we factor \(100^2\) from the radicand, we may write:

[MATH]d(t)=100\sqrt{(4t)^2+(2t+1)^2}[/MATH]
Expand the radicand:

[MATH]d(t)=100\sqrt{16t^2+4t^2+4t+1}[/MATH]
Combine like terms:

[MATH]d(t)=100\sqrt{20t^2+4t+1}[/MATH]
Does that make sense?

yes, I see what you did...but I didn't derive the terms the way you did. I mean, I didn't think it out in terms of coordinate points, at least not precisely. I had a picture of what was happening. Actually there was some kind of graphic in the book but it did not seem to be referring to a grid. My first formula, before I looked at the solutions manual was: (200t)^2 +(400t)^2 = X^2. I did not take into account the fact that the one plane had a 100 mile head start. But after I got that straightened out I see that I was OK until I got to simplifying things under the radical after the 100^2 had been removed. Yes, now I see how the quadratic equation was derived. I will have to redo this tomorrow to make sure I know what is happening but I think I will be OK. Thanks so much for pointing this out. I think my biggest mistake was not properly squaring the (2t +1).

 

[Denis]

Well, you ain't agettin' the pilot's job........