- The rope is pulled by a winch at 4 feet per second. When 14 feet of rope is out, what will be the speed of the boat? As the boat inches closer to the dock, what happens to its speed?
- 4 feet per second is a constant speed at which the boat is moving. When 13 feet of rope is out, what will be the speed at which the winch pulls the rope? As the boat inches closer to the dock, what happens to the speed at which the winch pulls in the rope?
This problem aims to introduce two main concepts at the same time, that is, derivation and Pythagoras theorem, which are required to thoroughly understand the statement and the solution.
Expert Answer
Pythagoras theorem is valid when we require an unknown side of a right triangle formed by summing the areas of 3 similar squares. At the same time, the derivation helps find the rate of change in any quantity for another quantity.
We will start the solution by declaring some variables, let l be the length of the rope and x be the speed per second with which the boat is moving.
By applying Pythagoras theorem:
\[ l^2=12^2+x^2 \]
\[ l^2=144+x^2 \]
Part 1:
Taking the derivative with respect to $t$:
\[ 2l\dfrac{dl}{dt}=2x \dfrac{dx}{dt} \]
\[ \dfrac{dx}{dt}=\dfrac{l}{x}. \dfrac{dl}{dt} \]
Given $\dfrac{dl}{dt}$ as $-4$
\[ \dfrac{dx}{dt}=\dfrac{-4l}{x} \]
Given $l=13$,
\[13^2=144+x^2 \]
\[ x=5\]
\[ =\dfrac{-4(13)}{5} \]
\[ \dfrac{dx}{dt}=\dfrac{-52}{5} f \dfrac{t}{sec} \]
Part 2:
\[ \dfrac{dl}{dt}=\dfrac{x}{l}. \dfrac{dx}{dt} \]
Putting $l$ and $x$:
\[ =\dfrac{5}{13}. -4 \]
\[ \dfrac{dl}{dt}=\dfrac{-20}{13} f \dfrac{t}{sec} \]
$\dfrac{dl}{dt}$ increases, as the $l \rightarrow 0$.
Hence, the speed of the boat increases as the boat gets closer to the dock.
Numerical Answers
Part 1: \[ \dfrac{dx}{dt}=\dfrac{-52}{5} f \dfrac{t}{sec} \]
Part 2: \[ \dfrac{dl}{dt}=\dfrac{-20}{13} f \dfrac{t}{sec} \]
Example
(b) $6$ feet per second is a constant speed at which the boat is moving. When $15$ feet of rope is out, what will be the speed at which the winch pulls the rope? As the boat gets closer to the dock, what happens to the speed at which the winch pulls in the rope?
\[ l^2=144+x^2 \]
Part a:
Taking the derivative with respect to $t$:
\[ \dfrac{dx}{dt}=\dfrac{l}{x}. \dfrac{dl}{dt} \]
Given $\dfrac{dl}{dt}$ as $-6$
\[ \dfrac{dx}{dt}=\dfrac{-6l}{x} \]
Given $l = 15$
\[15^2 = 144+x^2 \],
\[ x= 9\]
\[ = \dfrac{-6(15)}{9} \]
\[ \dfrac{dx}{dt} = -10 f \dfrac{t}{sec} \]
Part b:
\[ \dfrac{dl}{dt} = \dfrac{x}{l}. \dfrac{dx}{dt} \]
Putting $l$ and $x$:
\[ = \dfrac{9}{15}. -6 \]
\[ \dfrac{dl}{dt}= \dfrac{-54}{15} f \dfrac{t}{sec} \]
Hence, the speed of the boat increases as the boat gets closer to the dock.