calculus
posted by sal on .
A car leaves an intersection traveling east. Its position t sec later is given by
x = t2 + t
ft. At the same time, another car leaves the same intersection heading north, traveling
y = t2 + 3t ft
in t sec. Find the rate at which the distance between the two cars will be changing 3 sec later. (Round your answer to one decimal place.)

x = t^2 + 1t which is 10 at t=3
y = t^2 + 3t which is 18 at t=3
dx/dt = 2 t + 1 which is 7 at t=3
dy/dt = 2 t + 3 which is 9 at t=3
r^2 = x^2 + y^2 = 434 at t = 3
so at t = 3, r = 20.6
2 r dr/dt = 2 x dx/dt + 2 y dy/dt
20.6 dr/dt = 10(7) + 18(9)
dr/dt = 11.3 ft/s 
apple

x would actually be 12 not 10