word problem?

Question

A distance of 43 ft from the pad, a man observes a helicopter taking off from a heliport. If the helicopter lifts off vertically and is rising at a speed of 40 ft/sec when it is at an altitude of 125 ft, how fast is the distance between the helicopter and the man changing at that instant?

Answer 1

We know that
y(t)=125+40*t
and x(t)=43

The distance from the man is
=sqrt(43^2+(125+40*t)^2)

or
=sqrt(17474+10000*t+1600*t^2)


for the rate of change of distance, take the first derivative

=.5*(10000+3200*t)/(sqrt(17474+10000*t+1600*t^2))

=200*(25+8*t)/(sqrt(17474+10000*t+1600*t^2))

We know t, since
y(t)=40*t =125
t=3.125
therefore the v(t) at t=3.125 is
39.42 ft/s


j

Answer 2

1st calculate distance from the observer to the helicopter using Pythagorean theorem
http://en.wikipedia.org/wiki/Pythagorean_theorem

"how fast is the distance between the helicopter and the man changing at that instant" is apparent speed

present height / distance to observer equal rising speed / apparent speed.

125 / distance = 40 / apparent speed.