Question

Watching a Helicopter Take Off At a distance of 52 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 45 ft/sec when it is at an altitude of 123 ft, how fast is the distance between the helicopter and the man changing at that instant? (Round your answer to one decimal place.)

Answer 1

Answer:

The distance between the helicopter and the man changing at that instant is 41.46 ft/sec.

Explanation:

Given:

Distance between the man and the helipad,$b$ = $52$ ft

Vertical distance between the helipad and the helicopter,$p$ = $123$ ft

Helicopter lift off vertically and is raising at a speed of 45 ft/sec.

That can be written as, $\frac{dp}{dt}$ = 45 ft/sec

Now from the diagram (attached with) we can find the hypotenuse of the triangle that is the distance between the man and the helicopter.

Applying Pythagoras theorem:

⇒ $h^2=b^2+p^2$

⇒ $h=\sqrt{b^2+p^2}$

⇒ $h=\sqrt{123^2+52^2}$

⇒ $h=133.5$ ft

Now to find the changing distance between the helicopter and the man we have to  differentiate the Pythagoras theorem as dh/dt can be obtained from there.

And we know that the distance b is constant so db/dt=0.

Using chain rule and power rule of differentiation.

⇒ $h^2=b^2+p^2$

⇒ $2h\frac{dh}{dt} =2b\frac{db}{dt} + 2p\frac{dp}{dt}$

⇒ $2h\frac{dh}{dt} =0+ 2p\frac{dp}{dt}$

⇒  $h\frac{dh}{dt} = p\frac{dp}{dt}$       ....eliminating the common 2

⇒ $\frac{dh}{dt} = \frac{p\frac{dp}{dt} }{h}$

⇒ Plugging the values.

⇒ $\frac{dh}{dt} = \frac{123\times 45 }{133.5}$

⇒ $\frac{dh}{dt} = 41.46$ ft/sec

So,

41.46 ft/sec is the distance between the helicopter and the man changing at that instant.