Un ballon sphérique est gonflé de sorte que son volume augmente à un rythme de 10 cm³/s. À quelle vitesse le rayon du ballon augmente-t-il lorsque le rayon est de 5 cm ?

["Titre : Comment le rayon d’un ballon sphérique augmente-t-il quand son volume croît à un rythme donné ?", "When a spherical balloon is inflated and its volume increases at a constant rate, a fundamental question arises: At what rate does the radius grow when the radius is 5 cm, if the volume increases at 10 cm³/s ? This article explores the relationship between volume and radius in a sphere and provides a precise, step-by-step solution using calculus.", "---", "### The Math Behind a Sphere’s Volume", "The volume $ V $ of a sphere with radius $ r $ is given by the formula:", "$$\nV = \frac{4}{3} \pi r^3\n$$", "This equation links the visible physical property—the volume—to the geometric property—the radius. To find how quickly the radius changes with time, we differentiate both sides with respect to time $ t $:", "$$\n\frac{dV}{dt} = \frac{d}{dt} \left( \frac{4}{3} \pi r^3 \right)\n$$", "Using the chain rule:", "$$\n\frac{dV}{dt} = 4\pi r^2 \cdot \frac{dr}{dt}\n$$", "This equation tells us that the rate of change of volume depends on both the square of the radius and the rate of change of the radius itself.", "---", "### Plugging in the Known Values", "We are given:\n- $ \frac{dV}{dt} = 10 $ cm³/s (volume increasing at 10 cubic centimeters per second)\n- $ r = 5 $ cm (the specific radius of interest)", "Substitute into the differentiated equation:", "$$\n10 = 4\pi (5)^2 \cdot \frac{dr}{dt}\n$$", "Simplify step by step:", "$$\n10 = 4\pi \cdot 25 \cdot \frac{dr}{dt}\n$$\n$$\n10 = 100\pi \cdot \frac{dr}{dt}\n$$", "Solve for $ \frac{dr}{dt} $:", "$$\n\frac{dr}{dt} = \frac{10}{100\pi} = \frac{1}{10\pi}\n$$", "---", "### Final Answer", "When the radius of the ballon is 5 cm, the radius increases at a rate of:", "$$\n\boxed{\frac{1}{10\pi} \ ext{ cm/s}}\n$$", "This precise value demonstrates how calculus connects changing volumes to real-time changes in shape—a concept crucial in physics, engineering, and everyday applications involving inflating objects.", "---", "### Summary", "- Volume of a sphere: $ V = \frac{4}{3}\pi r^3 $\n- Derivative: $ \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} $\n- At $ r = 5 $ cm and $ \frac{dV}{dt} = 10 $ cm³/s, $ \frac{dr}{dt} = \frac{1}{10\pi} $ cm/s", "Understanding this relationship helps in determining growth dynamics in many practical scenarios—from balloons to industrial pressure vessels.", "---", "Keywords: volume rate change, radius increase, calculus, sphere volume formula, related rates, math problem, rate of change, geometry, differential equations"]