A spherical balloon is being inflated such that its radius increases at a constant rate of 0.5 cm/s. At what rate is the volume increasing when the radius is 10 cm?
["Title: How Fast Is the Volume of an Inflating Spherical Balloon Increasing?", "When a spherical balloon is being inflated, understanding how its volume changes with time is a classic and insightful problem in calculus and applied physics. In this article, we explore the precise rate at which the volume increases when the balloon’s radius reaches 10 cm, given that the radius grows at a constant rate of 0.5 cm per second.", "---", "Understanding the Problem", "The volume ( V ) of a sphere is given by the formula:", "[\nV = \frac{4}{3} \pi r^3\n]", "where ( r ) is the radius of the sphere. Since the radius increases at a constant rate of ( \frac{dr}{dt} = 0.5 , \ ext{cm/s} ), we want to find the rate of change of volume, ( \frac{dV}{dt} ), when ( r = 10 , \ ext{cm} ).", "---", "Differentiating Volume with Respect to Time", "We differentiate the volume formula with respect to time ( t ) using the chain rule:", "[\n\frac{dV}{dt} = \frac{dV}{dr} \cdot \frac{dr}{dt}\n]", "We first compute ( \frac{dV}{dr} ):", "[\n\frac{dV}{dr} = \frac{d}{dr} \left( \frac{4}{3} \pi r^3 \right) = 4 \pi r^2\n]", "Now substitute into the rate equation:", "[\n\frac{dV}{dt} = 4 \pi r^2 \cdot \frac{dr}{dt}\n]", "---", "Plugging in Known Values", "At ( r = 10 ) cm and ( \frac{dr}{dt} = 0.5 ) cm/s:", "[\n\frac{dV}{dt} = 4 \pi (10)^2 \cdot 0.5 = 4 \pi \cdot 100 \cdot 0.5 = 200 \pi , \ ext{cm}^3/\ ext{s}\n]", "---", "Final Answer", "Thus, when the radius of the spherical balloon is 10 cm, the volume is increasing at a rate of:", "[\n\boxed{200\pi , \ ext{cm}^3/\ ext{s}}\n]", "---", "Why This matters", "This result is not just mathematically elegant but also practical—whether inflating balloons for parties, studying atmospheric phenomena, or engineering applications, understanding how volume evolves over time helps in planning and control.", "Understanding related rates like this showcases the power of calculus in solving real-world problems involving changing shapes and sizes efficiently."]
