Question

The tip of a fisherman’s rod is 8 feet above the surface of the water when he catches a fish. If he reels in a fish at a rate of 1 foot per second, and never moves the position of the rod, at what rate is the fish approaching the base of the dock when 10 feet of fishing line is out?

Answer 1

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

$\frac{dz}{dt}=-1ft/sec$

We have to find the rate at which the fish is approaching the base of the dock when x=10 ft

$z=\sqrt{x^2+y^2}$

By Pythagoras theorem

$Hypotenuse=\sqrt{base^2+(perpendicular\;side)^2}$

Substitute x=10 and y=8

$z=\sqrt{(10)^2+8^2}=\sqrt{164}=2\sqrt{41}ft$

$x^2+y^2=z^2$

Differentiate w.r.t t

$2x\frac{dx}{dt}+2y\frac{dy}{dt}=2z\frac{dz}{dt}$

$x\frac{dx}{dt}+y\frac{dy}{dt}=z\frac{dz}{dt}$

Substitute the values

$10\frac{dx}{dt}+8(0)=2\sqrt{41}\times (-1)$

$\frac{dy}{dt}=0$

Because he never moves the rod.

$\frac{dx}{dt}=\frac{-2\sqrt{41}}{10}=-1.28 ft/s$

Hence, the fish is approaching the base of the dock at the rate  of 1.28 ft/s