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Q1E

Expert-verifiedFound in: Page 132

Book edition
2nd

Author(s)
James Stewart

Pages
830 pages

ISBN
9781133112280

**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 \(\left( {{\rm{3}}{x^{\rm{2}}}} \right)\frac{{{\rm{d}}x}}{{{\rm{dt}}}}\).

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

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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