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Question

Amy ran west at constant rate of 20ft/sec while Bob ran North at15ft/sec. How fast was the distance b/w Amy & Bob changing at 10sec after Bob started to run?

Answer 1

Hope that this can help u. I'm not very sure about wat u mean by "changing". So here's my understanding.

With Amy running west and Bob running North, this will form a right-angled triangle. Hence, by Pythagoras Theorem, the speed (or distance) between them is the hypotenus of the triange formed.

Resultant speed between them
= Sqrt(20^2 + 15^2) = 25ft/sec

At 10secs, distance between them
= 25 * 10
= 250ft

Answer 2

More related rates!

You want to know the change in distance between them at an instant, that would be the derivitive of the distance between them. So you need a function for the distance between them. Because they are running at a right angle use the pathagoran theorm

d = sqrt ( a² + b²)

d = distance between
a = amy's distance from the start
b = bob's distance from the start

differentiate

dd/dt = (2a*da/dt + 2b*db/dt) / ( 2 * sqrt(a²+b²) )

their velocities are given
v_a = 20ft/s
v_b = 15ft/s

velocity is the derivitive of position
position a = 20ft/s *t
position b = 15ft/s *t

substitute in 10s for t
position a = 200 ft
position b = 150 ft

put all that into back into the equation for dd/dt

dd/dt = (2*200*20 + 2*150*15) / (2 * sqrt(200² + 150²))
dd/dt = 25ft/s

Answer 3

Assuming they started at the same time ,after 10 sec they will be 250 ft apart

As for the part of how fast the distance is changing it is 25 ft/sec and it is independent of time (after 1 sec it is 25ft/sec,after 10000 sec the same)

Answer 4

s^2 = x^2 + y^2
sds = xdx + ydy
ds/dt = (xdx/dt + ydy/dt)/(x^2 + y^2)^(1/2)
Taking West & North as positive
ds/dt = (200*20 + 150*15)/(200^2 + 150^2)^(1/2)
ds/dt = (4000 + 2250)/(40000 + 22500)^(1/2)
ds/dt = 6250/(62500)^(1/2)
ds/dt = 6250/250
ds/dt = 6250/250
ds/dt = 25 ft/sec.