Posted by Jake on Tuesday, December 22, 2009 at 9:20pm.
The positions of A and B are parametric functions of time.
Take the position of ship A at 1:00 pm be (0,0), the
xa(t) = 0
ya(t) = 10t
xb(t) = -15t
yb(t) = 30
Distance between the two ships
D(t) = √((xb(t)-xa(t))²+(yb(t)-ya(t))²)
Substitute xa,xb,ya,yb in the expression for D(t) and differentiate with respect to t.
For the minimum distance, D'(t0) = 0.
Verify that D(t0) is a minimum by confirming that D"(t0)>0, or by comparing values of D'(t0-) and D'(t0+) with D'(t0).
Evaluate D(t0) for the remainder of the question.
Related Questions
Calculus - At 1:00 p.m. ship A is 25 km due north of ship B. If ship A is ...
calculus - At 1:00 p.m. ship A is 25 km due north of ship B. If ship A is ...
calculus - At noon, ship A is 30 nautical miles due west of ship B. Ship A is ...
calculus 1 - At noon, ship A is 30 nautical miles due west of ship B. Ship A is ...
calculus - At noon, ship A is 30 nautical miles due west of ship B. Ship A is ...
Calculus - At 4 PM ship A is 40 miles due South Ship B. Ship A is sailing due ...
Calculus - At noon, ship A is 180 km west of ship B. Ship A is sailing east at ...
Calculus - At noon, ship A is 10 nautical miles due west of ship B. Ship A is ...
calculus - At noon, ship A is 10 nautical miles due west of ship B. Ship A is ...
calculus - At noon, ship A is 30 nautical miles due west of ship B. Ship A is ...
For Further Reading
