Book edition 2nd

Author(s) James Stewart

Pages 830 pages

ISBN 9781133112280

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Short Answer

The altitude of a triangle is increasing at a rate of \(1cm/min\)while the area of the triangle is increasing at a rate of \(2cm/min\). At what rate is the base of the triangle changing when the altitude is \(10cm\) and the area is\(100c{m^2}\) ?

The base of the triangle is changing at the rate of \(\frac{{db}}{{dt}} = - 1.6cm/min\)

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given Information.

It is given that increasing rate of altitude of triangle is \(1cm/min\)and that of the area is \(2cm/min\). Rate change of base of triangle is to be found out.

Step 2: Definition of Derivative

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 \(h = \)altitude of the triangle

\(A = \)area of the triangle

\(b = \)area of the triangle

Now , area of the triangle

\(A = \frac{{bh}}{2}\)

Differentiate with respect to \(t\)

\(\begin{array}{l}\frac{d}{{dt}}\left( A \right) = \frac{d}{{dt}}\left( {\frac{{bh}}{2}} \right)\\\frac{{dA}}{{dt}} = \frac{{dh}}{{dt}}.\frac{b}{2} + \frac{h}{2}.\frac{{db}}{{dt}}\\\frac{h}{2}.\frac{{db}}{{dt}} = \frac{{dA}}{{dt}} - \frac{{dh}}{{dt}}.\frac{b}{2}\\\frac{{db}}{{dt}} = \frac{2}{h}.\frac{{dA}}{{dt}} - \frac{{dh}}{{dt}}.\frac{b}{h}\end{array}\)

Now,for the values

\(\begin{array}{l}h = 10cm\\A = 100c{m^2}\\b = \frac{{2.100}}{{10}}\\b = 20cm\end{array}\)

\(\begin{array}{c}\frac{{db}}{{dt}} = \frac{2}{h}.\frac{{dA}}{{dt}} - \frac{{dh}}{{dt}}.\frac{b}{h}\\ = \frac{2}{{10}}.2 - 1.\frac{{20}}{{10}}\\ = - \frac{8}{5}\\ = - 1.6\end{array}\)

Therefore, the base of the triangle is changing at the rate of \(\frac{{db}}{{dt}} = - 1.6cm/min\)

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Essential Calculus: Early Transcendentals

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Short Answer

The altitude of a triangle is increasing at a rate of \(1cm/min\)while the area of the triangle is increasing at a rate of \(2cm/min\). At what rate is the base of the triangle changing when the altitude is \(10cm\) and the area is\(100c{m^2}\) ?

Step by Step Solution

Step 1: Given Information.

Step 2: Definition of Derivative

Step 3: Find derivative

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Essential Calculus: Early Transcendentals

Answers without the blur.

Short Answer

The altitude of a triangle is increasing at a rate of \(1cm/min\)while the area of the triangle is increasing at a rate of \(2cm/min\). At what rate is the base of the triangle changing when the altitude is \(10cm\) and the area is\(100c{m^2}\) ?

Step by Step Solution

Step 1: Given Information.

Step 2: Definition of Derivative

Step 3: Find derivative

Most popular questions for Math Textbooks

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