Determine how fast the length of an edge of a cube is changing at the moment when the length of the edge is $5 c m$ $5 cm$ and the volume of the cube is decreasing at a rate of $100 \frac{c m^{3}}{sec}$ $100 (cm^3)/sec$ ?

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Answer 1 · 2016-07-18T10:55:29

$- \frac{4}{3}$ $- 4/3$ cm/sec

$V = x^{3}$ $V = x^3$

$\frac{d V}{d t} = \frac{d}{d t} x^{3} = 3 x^{2} \frac{d x}{d t}$ $(dV)/(dt) = d/dt x^3 = 3x^2 dx/dt$

$\frac{d x}{d t} = \frac{d V}{d t} \frac{1}{3 x^{2}}$ $dx/dt = (dV)/(dt) 1/ (3x^2)$

$\frac{d x}{d t} = - 100 \cdot \frac{1}{3 {(5)}^{2}} = - \frac{4}{3}$ $dx/dt = - 100 * 1/ (3(5)^2) = - 4/3$ cm/sec

Determine how fast the length of an edge of a cube is changing at the moment when the length of the edge is 5cm5 cm and the volume of the cube is decreasing at a rate of 100cm3sec100 (cm^3)/sec?