A spherical balloon of radius r cm has volume Vcm^3 , where V =4/3 * pi * r^3. The balloon is inflated at a constant rate of 10 cm^3 s^-1 . Find the rate of increase of r when r = 8.

We are being asked to find the rate of change of radius, dr/dt. We will need to use the chain rule to do this: dV/dt = dV/dr * dr/dt.

We are given that dV/dt is 10cm^3 per second, and differentiating V = 4/3 * pi *r^3 with respect to r gives us dV/dr = 4 * pi *r^2 which at r= 8 gives us dV/dr = 804.25

Now rearranging the chain rule equation we find that dr/dt = (dV/dt) / (dV/dr) = 10/804.25 = 0.0124 cm per second.

MA

Related Maths A Level answers

All answers ▸

Express (9x^2 + 43x + 8)/(3+x)(1-x)(2x+1) in partial fractions.


What are radians and what are they used for?


Integrate 2x^5 + 7x^3 - (3/x^2)


Find the turning point of y = x + 1 + 4/x2 and describe the nature of the turning point