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,[tex]b[/tex] = [tex]52[/tex] ft
Vertical distance between the helipad and the helicopter,[tex]p[/tex] = [tex]123[/tex] ft
Helicopter lift off vertically and is raising at a speed of 45 ft/sec.
That can be written as, [tex]\frac{dp}{dt}[/tex] = 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:
⇒ [tex]h^2=b^2+p^2[/tex]
⇒ [tex]h=\sqrt{b^2+p^2}[/tex]
⇒ [tex]h=\sqrt{123^2+52^2}[/tex]
⇒ [tex]h=133.5[/tex] 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.
⇒ [tex]h^2=b^2+p^2[/tex]
⇒ [tex]2h\frac{dh}{dt} =2b\frac{db}{dt} + 2p\frac{dp}{dt}[/tex]
⇒ [tex]2h\frac{dh}{dt} =0+ 2p\frac{dp}{dt}[/tex]
⇒  [tex]h\frac{dh}{dt} = p\frac{dp}{dt}[/tex]    ....eliminating the common 2
⇒ [tex]\frac{dh}{dt} = \frac{p\frac{dp}{dt} }{h}[/tex]
⇒ Plugging the values.
⇒ [tex]\frac{dh}{dt} = \frac{123\times 45 }{133.5}[/tex]
⇒ [tex]\frac{dh}{dt} = 41.46[/tex] ft/sec
So,
41.46 ft/sec is the distance between the helicopter and the man changing at that instant.
