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\[\boxed{\frac901,771}\] - Aurero

📅 March 14, 2026 👤 scraface

Mar 14, 2026

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📌 \tan 45^\circ = 1, \quad \tan 30^\circ = \frac{1}{\sqrt{3}}

📌 Rationalize the denominator:

📌 \]Question: A weather balloon is modeled as a sphere with radius $ r $ meters, and it ascends such that its volume increases at a rate proportional to $ \sin \theta $, where $ \theta $ is the angle of elevation from the ground. If the volume increase rate is given by $ \frac{dV}{dt} = k \sin \theta $, and $ \theta = 60^\circ $, compute the rate of volume increase in terms of $ k $ and $ r $.

📌 Solution: The volume of a sphere is given by $ V = \frac{4}{3} \pi r^3 $. Differentiating with respect to time $ t $, we get:

📌 We are given that $ \frac{dV}{dt} = k \sin \theta $, and $ \theta = 60^\circ $, so $ \sin \theta = \frac{\sqrt{3}}{2} $. Substituting:

📌 k \cdot \frac{\sqrt{3}}{2} = 4\pi r^2 \frac{dr}{dt}

📌 Solving for $ \frac{dr}{dt} $:

📌 \frac{dr}{dt} = \frac{k \sqrt{3}}{8\pi r^2}

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