A second opinion

[BigGlenntheHeavy]

I solved this problem, but somehow I have a bad taste in my mouth, ergo a second opinion is in order.

At noon ship A was 100 miles due east of ship B. Ship A is sailing west at 12 mph, and ship B is sailing south at 10 mph. At what time will the ships be nearest to each other, and what will this distance be?

Any help will be appreciated, please show all work, as I want to see how you solve this problem.

 

[wjm11]

At noon ship A was 100 miles due east of ship B. Ship A is sailing west at 12 mph, and ship B is sailing south at 10 mph. At what time will the ships be nearest to each other, and what will this distance be?

Hi, Glenn,

I just put ship B at the origin and A at 100 on the x-axis.
I wrote position versus time equations for them.
I put these expressions into the distance formula and found the derivative.
I set the derivative equal to zero and calculated t.

t = 4.918 hours
A is at position 40.9836 mi
B is at position –49.18 mi
And finally,
d = 64.02 mi

Bill

 

[BigGlenntheHeavy]

Good show wjm11, as I went the related rate route, thinking it had to be something profound, when all it was, was a distance problem, Thanks again, wjm11, BGtH.

 

[chrisr]

that's beautiful!

as one goes west faster than the other goes south,
the sum of their perpendicular distances from the south-going one's
initial position is decreasing steadily,
yet even though that sum is a minimum when the ships are lined up
on a line of longitude (one is south of the other),
their distance apart has increased from a minimum,
which was reached prior to being vertically apart.

example; 3+5 = 8 and 3^2 + 5^2 = 34
1+6 = 7 and 6^2 + 1^2 = 37