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Essential Calculus: Early Transcendentals
Found in: Page 132
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

If \(V\) is the volume of a cube with edge length \(x\) and the cube expands as time passes, find \(\frac{{dV}}{{dt}}\) in terms of \(\frac{{dx}}{{dt}}\).

The value of \(\frac{{dV}}{{dt}}\) in terms of \(\frac{{dx}}{{dt}}\) is \(\left( {{\rm{3}}{x^{\rm{2}}}} \right)\frac{{{\rm{d}}x}}{{{\rm{dt}}}}\).

See the step by step solution

Step by Step Solution

Step 1: Definition of the Chain rule

If \(y = f\left( u \right)\) and \(u = g\left( x \right)\) are differentiable functions, then the Chain rule is given by \(\frac{{dy}}{{dx}} = \frac{{dy}}{{du}} \cdot \frac{{du}}{{dx}}\)

Step 2: Find \(\frac{{dV}}{{dt}}\) in terms of \(\frac{{dx}}{{dt}}\).

It is given that a cube of a volume V with edge length \(x\).

Then, the volume of the cube is given by, \(V\left( x \right) = {x^3}\).

Differentiate both sides of \(V\left( x \right) = {x^3}\) with respect to \(t\) as follows:

\(\frac{{dV}}{{dt}} = \frac{d}{{dt}}\left( {{x^3}} \right)\)

Apply the chain rule of differentiation as follows:

\(\begin{array}{c}\frac{{dV}}{{dt}} = \frac{d}{{dx}}\left( {{x^3}} \right)\frac{{dx}}{{dt}}\\\frac{{dV}}{{dt}} = 3{x^2}\frac{{dx}}{{dt}}\end{array}\)

Therefore, \(\frac{{dV}}{{dt}}\) in terms of \(\frac{{dx}}{{dt}}\) is equal to \(3{x^2}\frac{{dx}}{{dt}}\).

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