Could you describe/show the method you used?
"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?"
I already have the answer, verified by the text, but had to do some unkosher gymnastics to arrive at the answer.
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?
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.
Square of distance D between the jets:
Differentiate with respect to t and equate to zero:
Critical value at
The first derivative test shows the extremum here is a minimum.
Now, to find the minimum distance, we compute:
Do these results agree with yours?
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.