Calculus Optimization

Ship #1 passes by a sea buoy traveling on a bearing of S 40 E at a speed of 15 miles per hour. Just as the first ship passes the buoy, a second ship that is 10 miles from the buoy is approaching the buoy at 20 miles per hour on a bearing of N 70 E. How fast will the distance between the ships be changing

a) after 20 minutes
b) after 30 minutes

angle between ship1 and ship2 directions: ie 20° from N 70 E and 50° from S 40 E
angle = 70°

distance between the two ships at any time t be equal to x
x² = (15t)² + (10 - 20t)² - 2(15t)(10 - 20t)cos70
x² = 225t² + 100 - 400t + 400t² - 300tcos70 + 600t²cos70
x² = 625t² + 100 - 400t - 300tcos70 + 600t²cos70

a) t = 20 mins
x = (5/3) √{13 - 12sin(π/9) }
x² = 625t² + 100 - 400t - 300tcos70 + 600t²cos70
2x dx/dt = 1250t - 400 - 300cos70 + 1200tcos70
2x dx/dt = 50/3 + 100sin(π/9)
x dx/dt = 25/3 + 50sin(π/9)

dx/dt = [ 25/3 + 50sin(π/9) ] / [ (5/3) √{13 - 12sin(π/9) } ]
dx/dt = 5.1166 mph
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b)
x² = 625t² + 100 - 400t - 300tcos70 + 600t²cos70
x = √( 625t² + 100 - 400t - 300tcos70 + 600t²cos70 )
dx/dt = ½ (( 625t² + 100 - 400t - 300tcos70 + 600t²cos70 )^(-½) * (1250t - 400 - 300cos70 + 1200cos70)

when t = 30 mins
dx/dt = 21.8404 mph
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