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Expert-verified Found in: Page 133 ### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280 # A boat is pulled into a dock by a rope attached to the bow of the boat and passing through a pulley on the dock that is $$1m$$ higher than the bow of the boat. If the rope is pulled in at a rate of$$1m/s$$, how fast is the boat approaching the dock when it is $$8m$$ from the dock?

Rate of decrease of distance between boat and dock is $$\frac{{\sqrt {65} }}{8} \approx 1m/s$$

See the step by step solution

## Step 1: Given Information.

It is given that pulling rate of rope is $$1$$$$m/s$$and pulley is $$1$$$$m$$higher than bow of the boat. Rate of boat approaching the dock is to be found out.

## Step 2: Definition of Derivativx

In mathematics, the derivative of a function of a real variable measures the sensitivity to change of the function value (output value) with respect to a change in its argument (input value)..

## Step 3: Find derivative

Let $$x =$$distance of boat from dock

$$y =$$length of the rope

Now , as per the given condition

$${x^2} + {1^2} = {y^2}$$

Differentiate with respect to $$t$$

$$\begin{array}{l}2x\frac{{dx}}{{dt}} = 2y\frac{{dy}}{{dt}}\\x\frac{{dx}}{{dt}} = y\frac{{dy}}{{dt}}\end{array}$$

Given $$\frac{{dy}}{{dt}} = 1m/s$$

Now ,for the values computed earlier

$$\begin{array}{l}x = 8\\{y^2} = {8^2} + {1^2}\\y = \sqrt {65} \end{array}$$

Therefore,

$$\begin{array}{l}8\frac{{dx}}{{dt}} = \sqrt {65} \\\frac{{dx}}{{dt}} = \frac{{\sqrt {65} }}{8}\end{array}$$

Therefore, rate of decrease of distance between boat and dock is $$\frac{{\sqrt {65} }}{8} \approx 1m/s$$ ### Want to see more solutions like these? 