Answer to Question #248924 in Calculus for Sam

Question

A dock is 6 feet above water. Suppose you stand on the edge of the dock and pull a rope attached to a boat at the constant rate of 2 ft/s. Assume that the boat remains at water level. At what speed is the boat approaching the dock when it is 20 feet from the dock? 10 feet from the dock? Isn’t it surprising that the boat’s speed is not constant?

Answer 1

Apply the Pythagorean theorem:

$x^2 + h^2 = L^2$  where   $h = 6 ft$        

 Now apply implicit differentiation:

$2x \frac{dx}{dt} = 2L \frac{dL}{dt}\\ \Rightarrow \frac{dx}{dt} = \frac{L}{x} \frac{dL}{dt}$

For  $x = 20 \ ft, L = \sqrt{ ( 20^2 + 6^2 )}$   

$\Rightarrow\frac{dx}{dt} = \frac{\sqrt{436}}{20} \times (2) = 2.08 \ ft/sec$

For  $x = 10 \ ft, L = \sqrt{ ( 10^2 + 6^2 )}$     

$\Rightarrow \frac{dx}{dt} = \frac{\sqrt{136}}{10} \times (2) = 2.33 \ ft/sec$

 

Result:     Speed of boat increases as it gets closer to dock.