How do you solve this related rates problem?

Question

At a distance of 60 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 49 ft/sec when it is at an altitude of 114 ft, how fast is the distance between the helicopter and the man changing at that instant?

Answer 1

Let s be the distance between the man and the helicopter. Then:
s=√(x²+y²)
A simple application of the chain rule gives us:
ds/dt = 1/(2√(x²+y²)) (2x dx/dt + 2y dy/dt).
Since the man's horizontal position is not changing, dx/dt=0. The other values are given in the problem (x=60, y=114, dy/dt=49), so all that's left is to plug in the variables and simplify. Thus:

ds/dt = 1/(2√(3600+12996)) (228* 49)
=11172/(2√16596)
=11172/(2√(2^2 * 3^2 * 461)
=11172/(12√461)
=931/√461
≈43.36099 ft/sec

Answer 2

Take this with a grain of salt, because I'm rusty.

The first thing to do is draw a picture, with a right triangle, x feet between the helicopter and the ground, 60 feet between the man and the point directly beneath the helicopter (this is a constant), and z feet for the hypoteneuse.

Of course, using the Pathagorean Theorem, x^2 + y^2 = z^2.
Therefore, 2x (dx/dt) + 2y (dy/dt) = 2z (dz/dt)
But y=60, and it is constant, so 2y (dy/dt) = 120 (0) = 0.
So 2x (dx/dt) = 2z (dz/dt)
At this exact moment, x = 114, dx/dt = 49, and z = the hypoteneuse of your triangle, which is about 128.8
If you plug these numbers in, you should get dz/dt.

Hope this helps!

Answer 3

Draw the right angled triangle. ABC. AB is ground. BC is the vertical line from the pad to helicopter. AC is the observers line of sight to helicopter. so, A is the man. B is the pad, C is the helicopter.

If you manage to draw the triangle now it's obvious AB = 60 ft. BC = 114 ft. from pithogorus rule we get AC = 128.85 ft

take BC as h and AC as l. then l²=h²+60². we differentiate with respect to time. 2l dl/ dt = 2h dh/dt. from the data given dh/dt (the rising speed of helicopter) is 49 ft/sec. then we substitute:

2(128.85) dl/dt = 2(114) * 49
therefore dl/dt = 43.35 ft/sec. ##
(dl/dt is the speed of helicopter relative to man)

Answer 4

I solved a very similar problem yesterday for someone else!

Get time from heli speed and altitude.

Write equation (Pythagoras) of hypothenuse of triangle with a base of 60 feet and a height of 114 feet. Differentiate and evaluate at the time that you got first.

Position coords:
man (0,0)
heli (60, 49t)

When 49t = 114 feet, get t

You can do the rest, right?

I think that you can use trig, too. find
theta = arctan (60/114),
then the speed you want is 49*cos theta

I get 43.36

I think. You should do both. I'll check back.

Answer 5

h=49*t

d**2=(49*t)**2 + 60**2

take d( )/dt of both sides

2 d d' =2(49 t) * 49

d'= (49**2) (t)/ d

h=49 t

114/49 = t

d**2=(114)**2 + 3600

d=128.825

d'= (49**2) (114/49) / 128.825

d'= 43.361 ft/s

Answer 6

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