Applications of calc to physical world (1 Viewer)

mahli · Apr 21, 2012

surface area of spherical bubble is increasing at a constant rate of 1.9mm^2s^-1. Find the rate of increase of its volume when its radius is 0.6mm

nightweaver066 · Apr 21, 2012

Use:
$V = \frac{4}{3}\pi r^3$

and

$SA = 4\pi r^2$

$\frac{d(SA)}{dt} = \frac{dV}{dt} \times \frac{d(SA)}{dV}$

$\frac{dV}{dt} = 1.9 \times \frac{dV}{d(SA)}$

$\frac{V}{SA} = \frac{1}{3}r$

$V = \frac{rSA}{3}$

$\frac{dV}{d(SA)} = \frac{r}{3}$

$= 0.2$

$\therefore \frac{dV}{dt} = 1.9 \times 0.2 = 0.38mm^3/s$

Applications of calc to physical world (1 Viewer)

mahli

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nightweaver066

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