the distance z between the girls at time t seconds (t > 60) is
z^2 = (10t)^2 + (8(t-60))^2
2z dz/dt = 2(10t) + 64*2(t-60) = 148t-7680 = 4(37t-1920)
so now just figure z(120) and use that to find dz/dt
480√29 dz/dt = 10080
dz/dt = 21/√29 = 3.9 ft/s
A girl starts at a point A and runs east at the rate of 10 ft/sec. One
minute later, another girl starts at A and runs north at a rate of 8 ft/sec.
At what rate is the distance between them changing 1 minute after the
second girl starts ?
2 answers
At a time of t seconds,
let the distance covered by the east-bound girl be x ft
let the distance covered by the north-bounder by y ft
let the distance between them be d ft
given: dx/dt = 10 ft/s, dy/dt = 8 ft/s
find dd/dt when t = 60
d^2 = x^2 + y^2
2d dd/dt = 2x dx/dt + 2y dy/dt
dd/dt = (x(10) + y(8) )/d
when t = 60, x = 600, y = 480, d = 120√41
dd/dt = ( 600(10) + 480(8) )/(120√41) = ...... ft/s
let the distance covered by the east-bound girl be x ft
let the distance covered by the north-bounder by y ft
let the distance between them be d ft
given: dx/dt = 10 ft/s, dy/dt = 8 ft/s
find dd/dt when t = 60
d^2 = x^2 + y^2
2d dd/dt = 2x dx/dt + 2y dy/dt
dd/dt = (x(10) + y(8) )/d
when t = 60, x = 600, y = 480, d = 120√41
dd/dt = ( 600(10) + 480(8) )/(120√41) = ...... ft/s