+0

# [Precalc]Optimization Problem

0
152
6

An air traffic controller sees two aircraft flying at the same altitude on his screen. One, a piper cub, is headed due west at 150 miles per hour. The other, a Lear jet, is 15 miles due north of the Piper and is headed due south at 400 miles per hour. How close will the two aircraft come to each other?

My target function is
$$d=\sqrt{(150t)^2+(15-400t)^2}$$

not sure how to approach the problem from there

Oct 11, 2018
edited by Guest  Oct 11, 2018

#1
+2

Where is 'd' a minumum?   (this is basically pythagorean theorom)

d = sqrt( 22500t^2 + 225 +160000t^2 - 12000t)      This is a quadratic equation.....

(1)  d^2 = ( 182500t^2 -12000t + 225)                             find the minumum....take the derivative and look for slope = 0

365000t - 12000 = 0

t = .032876 hr

Substitute this value in to the 'd' equation

d = sqrt   (  150 (.032876))^2    +  (15 - 400(.032876))^2  )  = 5.2668 miles

Or take equation (1) and graph it to find the minimum point of the parabola......

Oct 11, 2018
#2
+97586
+2

That is a good start and I got the same.

Now just like guest said you take the first derivative and set it equal to 0.

When you get the answer I suppose you should check that it give a the time when the distance is minimum but just looking at the problem this will obviouly be the case.

I am also going to change your d into an x,

$$x=\sqrt{(150t)^2+(15-400t)^2}\\ x=[(150t)^2+(15-400t)^2]^{0.5}\\ \frac{dx}{dt}=0.5[(150t)^2+(15-400t)^2]^{-0.5}[300t*150+2(15-400t)*-400]\\ \frac{dx}{dt}=\frac{0.5[300t*150+2(15-400t)*-400]}{ [(150t)^2+(15-400t)^2]^{0.5} }\\ \frac{dx}{dt}=\frac{0.5[45000t+800(400t-15)]}{ [(150t)^2+(15-400t)^2]^{0.5} }\\ \frac{dx}{dt}=0\;\;when\\ 0.5[45000t+800(400t-15)]=0\\ 450t+8(400t-15)=0\\ 3650t-120=0\\ t=\frac{12}{365}\;\;hours\\ When \;\;t=\frac{12}{365}\;\;hours\\ x=[(150*\frac{12}{365})^2+(15-400*\frac{12}{365})^2]^{0.5}\\$$

((150*12/365)^2+(15-400*12/365)^2)^0.5 = 5.266851623825875 = 5.27 miles

There you go. My answer is exactly the same as guests!

Good work guest. Can 2 'great' minds both be wrong ?

Oct 11, 2018
#3
+2

If you've been instructed to do it this way fine, that's what you have to do, but you should be aware that there is a much easier method that avoids the calculus.

With problems like this, usually best is to bring one of the moving objects, boats, planes, cars, whatever, to rest.

In this case, think of a wind of 150mph blowing from the west.  As a result, relative to the ground, the Piper is at rest and the Lear will travel in a direction arctan(150/400) = 20.556 deg east of south.

Now drop a perpendicular from the Piper onto this line. Its length will be 15sin(20.556) = 5.2669 miles.

Tiggsy

Oct 11, 2018
#4
+97586
+1

That is an interesting way to do it Tiggsy, I have not seen it done like that before.  :)

Melody  Oct 11, 2018
#5
+96302
+1

That is pretty cool  !!!

CPhill  Oct 11, 2018
#6
+16454
+1

Another NON-Caculus method:

Minimum pt of the parabola   =   - b / 2a  = 12000/182500 = t = .032876 hr

Then substitute this value of 't' in to the d equation to get the same answer...

Oct 12, 2018