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Essential Calculus: Early Transcendentals
Found in: Page 133
Essential Calculus: Early Transcendentals

Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

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Short Answer

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 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

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


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

Therefore, rate of decrease of distance between boat and dock is \(\frac{{\sqrt {65} }}{8} \approx 1m/s\)

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