## Two Jets Separating at 872 mi/hr ... when?

Limits, differentiation, related rates, integration, trig integrals, etc.

### Two Jets Separating at 872 mi/hr ... when?

A jet flying due north at 640 mi/hr passes over this town at noon. 15 minutes later another jet flying due west passes over the same town. The planes are flying at the same altitude. At what time will they be separating at 872 mi/hr?

(edited) The answer is 1:15 p.m.

I tried. I think I must be missing something very simple.

Drew the figure, a right triangle with the right angle in the lower right hand corner. Distance y (vertical leg) represents distance covered by the 640 mi/hr jet and the distance x (horizontal leg) represents distance covered by the 600 mi/hr jet, and distance r (hypotenuse) the distance between the two jets.

$x^2\,+y^2\,=\,r^2$ ... courtesy of Pythagoras

$x\,\frac{dx}{dt}\,+\,y\frac{dy}{dt}\,=\,r\frac{dr}{dt}$ ... differentiated

I believe my knowns are
$\frac{dx}{dt}\,=600,\,\frac{dy}{dt}\,=640,\,and\,\frac{dr}{dt}\,=\,872$

I was thinking that I need to find $t$ in hours, and that I should be able to set it up like this:

The x-distance would be
$600(t\, -\, 0.25)$

and the y-distance would be
$640t$

making the r-distance
$\sqrt{(600(t\, -\, 0.25))^2\,+\,(640t)^2}$

Plugging everything into the 1st derivative equation above,
$600(t\, -\, 0.25)\,*600\,+\,640t\,*640\,=\,872\sqrt{(600(t\, -\, 0.25))^2\,+\,(640t)^2}$

(edited) This should have worked out so that $t\,=\,\frac{5}{4}$. Then we add that, (in the form of 1:15) to 12:00 noon and get 1:15 p.m.

Alas, things got gnarly when I tried to solve the above equation. What am I missing? Is there a way on this one to "work smarter, not harder?"
Last edited by jaybird0827 on Thu Jun 04, 2009 8:30 pm, edited 4 times in total.

jaybird0827

Posts: 24
Joined: Tue May 26, 2009 6:31 pm
Location: NC

jaybird0827 wrote:I think I must be missing something very simple....

I'm sure not seeing anything simple...

stapel_eliz

Posts: 1797
Joined: Mon Dec 08, 2008 4:22 pm

### Re: Two Jets Separating at 872 mi/hr ... when?

Eliz,

Thanks for taking a look. I almost gave up on this, but I just couldn't. I decided to experiment by creating a similar problem in which the rates were known, the time was known, and the requirement was to find the rate at which the two jets were separating at a given time. I contrived it to work out. Then I tried to solve the same problem except changing the unknown to "when will the two jets be separating at such and such a rate?" It was just as impossible as the above!

As a result, my hypothesis is that the method attempted, as described in the prior post, is not valid because it involves making time a variable in a context in which quantities are changing with respect to time. IOW, we can't mix $t$ with $\frac{dx}{dt}$. At least, I don't know of a way to do that.

I went back to the original problem and experimented. Since 640 and 600 are multiples of 4 and the difference in the times is 1/4 hour, I figured, let's see where the two jets are every 15 minutes once the second jet passes over the town.

Click on link for spreadsheet: http://spreadsheets.google.com/pub?key=r9P9eSgXFuOShFXnCzElu8w&hl=en. One thing I noticed was how the changes in rate of separation decreased over the succession of 15-minute intervals. I thought that was interesting.

I had noticed with other problems of this type, where, e.g. the unknown is the rate of one of the vehicles, a rate of separation or approach, or a distance, it turns out to be contrived to involve a convenient Pythagorean triple (extended to rationals with terminating decimals). When I was working the above out and came to the 1:00 p.m. entry, I realized immediately that the x-distance = 600 and y-distance = 800 were a Pythagorean triple in the making and that the problem would have to be so contrived to make the 872 work out.

So, getting back to
$r^2\,=\,x^2\,+y^2$ and $r\,\frac{dr}{dt}\,=\,x\,\frac{dx}{dt}\,+\,y\,\frac{dy}{dt}$

$1000^2\,=\,600^2\,+\,800^2$, so $r\,=\,1000$, and

$1000\,\frac{dr}{dt}\,=\,600\times600\,+\,800\times640$

$1000\,\frac{dr}{dt}\,=\,360\,000\,+\,512\,000$

$\frac{dr}{dt}\,=\,872\,mi/hr\,\,at\,\,1:15\,p.m.$

Not exactly a "plug it in and crank it out" solution. If you (or anyone out there) know of a more methodical approach for a related rate problem with this type of requirement and conditions, I'd be very interested in learning. I hope that these posts are contributing something worthwhile.

jaybird0827

Posts: 24
Joined: Tue May 26, 2009 6:31 pm
Location: NC