# Thread: Related rates word problem. What steps do you follow to solve this?

1. ## Related rates word problem. What steps do you follow to solve this?

"At t = 0, a single-engine military jet is flying due east at 12 mi/min. At the same altitude and 208 mi directly ahead of it, still at t = 0, is a commercial jet, flying due north at 8 mi/min. When are the two planes closest to each other? What is the minimum distance between them?"

Thanks

2. ## Re: Related rates word problem. What steps do you follow to solve this?

Could you describe/show the method you used?

3. ## Re: Related rates word problem. What steps do you follow to solve this?

Hi Mark from Florida. I drew a diagram first. With this diagram, I set up the planes at their nearest position. The jet is at (x0, 0) and the commercial jet is at (208, y1). I squared the distance formula to avoid radicals. d^2=(x1-x0)^2 + (y1-y0)^2. This gives me d^2=(208-x)^2 + y^2. Differentiation yields 2d dd/dt = 2 (208-x)(dx/dt) + 2y dy/dt. After inputting the knowns (dd/dt is 0, as this is a minimum problem; dx/dt is 12, dy/dt is 8), I simplified and found a formula relating x and y (a ratio). Then I simply made a chart with that ratio per minute. However, I could have just done that all along without the differentiation. What did I miss in the problem solving?

4. ## Re: Related rates word problem. What steps do you follow to solve this?

Here's how I would set this up, which isn't to say it's the best method to say the least!

I would orient my xy-plane such that the positive y-axis is due north and the positive x-axis is due east. The units are in miles. Put the commercial jet at the origin, and the military jet at (-208,0).

To find the parametric equations of motion:

Let t be the parameter representing time measured in minutes.

Military jet:

$\frac{dx}{dt}=12$

$x(t)=12t-208$

Commercial jet:

$\frac{dy}{dt}=8$

$y(t)=8t$

Square of distance D between the jets:

$D^2(t)=x^2(t)+y^2(t)=(12t-208)^2+(8t)^2=208(t^2-24t+208)$

Differentiate with respect to t and equate to zero:

$2D(t)D'(t)=416(t-12)=0$

Critical value at $t=12$

The first derivative test shows the extremum here is a minimum.

Now, to find the minimum distance, we compute:

$D(12)=\sqrt{208(12^2-24(12)+208)}=32\sqrt{13}$

Do these results agree with yours?

5. ## Re: Related rates word problem. What steps do you follow to solve this?

Mark: thanks again. I don't have my text with me to check your answer. This makes sense, although I have to admit I would'nt have thought of this approach (using x(t), y(t)) on my own. Thanks again.