Answer:
-1.28 ft/s
Step-by-step explanation:
We are given that
The height of tip of fisherman's rod from the water surface=y=8 ft
[tex]\frac{dz}{dt}=-1ft/sec[/tex]
We have to find the rate at which the fish is approaching the base of the dock when x=10 ft
[tex]z=\sqrt{x^2+y^2}[/tex]
By Pythagoras theorem
[tex]Hypotenuse=\sqrt{base^2+(perpendicular\;side)^2}[/tex]
Substitute x=10 and y=8
[tex]z=\sqrt{(10)^2+8^2}=\sqrt{164}=2\sqrt{41}ft[/tex]
[tex]x^2+y^2=z^2[/tex]
Differentiate w.r.t t
[tex]2x\frac{dx}{dt}+2y\frac{dy}{dt}=2z\frac{dz}{dt}[/tex]
[tex]x\frac{dx}{dt}+y\frac{dy}{dt}=z\frac{dz}{dt}[/tex]
Substitute the values
[tex]10\frac{dx}{dt}+8(0)=2\sqrt{41}\times (-1)[/tex]
[tex]\frac{dy}{dt}=0[/tex]
Because he never moves the rod.
[tex]\frac{dx}{dt}=\frac{-2\sqrt{41}}{10}=-1.28 ft/s[/tex]
Hence, the fish is approaching the base of the dock at the rate of 1.28 ft/s
