8. Gas is pumped into a spherical balloon at \( 500 \mathrm{~cm}^{3} / \mathrm{s} \). Find the rate of increase in surface area when the radius is 15 cm . The side of an equilateral triangle is increasing at \( \frac{2}{\sqrt{0}} \mathrm{~cm} / \mathrm{s} \). Calculate the rate of ch

To find the rate of increase in surface area of the balloon when the radius is 15 cm, we first recognize that the volume \( V \) of a sphere is given by \( V = \frac{4}{3} \pi r^3 \). The rate of volume change \( \frac{dV}{dt} = 500 \, \mathrm{cm}^3/\mathrm{s} \) leads us to use related rates. We differentiate \( V \) with respect to \( t \), yielding \( \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} \). Plugging in \( r = 15 \) cm, we can solve for \( \frac{dr}{dt} \) and subsequently find the surface area \( S = 4\pi r^2 \); thus, we can find \( \frac{dS}{dt} \). Now for the equilateral triangle, the relationship between the side length \( s \) and the area \( A \) is \( A = \frac{\sqrt{3}}{4} s^2 \). Differentiating with respect to time \( t \), we get \( \frac{dA}{dt} = \frac{\sqrt{3}}{2} s \frac{ds}{dt} \). With \( \frac{ds}{dt} = \frac{2}{\sqrt{0}} \) cm/s (that's an exhilaratingly odd rate), we need to keep in mind that as \( s \) increases, it should still be defined (so let’s pretend \( s > 0 \)). We can substitute to find the rate of change of the area.