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An ice sculpture in the form of a sphere melts in such a way that it maintains its spherical shape. The volume of the sphere is decreasing at a constant rate of 2π2\pi cubic meters per hour. At what rate, in square meters per hour, is the surface area of the sphere decreasing at the moment when the radius is 55 meters? (Note: For a sphere of radius rr, the surface area is 4πr24\pi r^{2} and the volume is 43πr3\frac{4}{3}\pi r^{3}.)\newline(A) 4π5\frac{4\pi}{5}\newline(B) 40π40\pi\newline(C) 80π280\pi^{2}\newline(D) 100π100\pi

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Q. An ice sculpture in the form of a sphere melts in such a way that it maintains its spherical shape. The volume of the sphere is decreasing at a constant rate of 2π2\pi cubic meters per hour. At what rate, in square meters per hour, is the surface area of the sphere decreasing at the moment when the radius is 55 meters? (Note: For a sphere of radius rr, the surface area is 4πr24\pi r^{2} and the volume is 43πr3\frac{4}{3}\pi r^{3}.)\newline(A) 4π5\frac{4\pi}{5}\newline(B) 40π40\pi\newline(C) 80π280\pi^{2}\newline(D) 100π100\pi
  1. Identify values and formulae: Identify the known values and the formulae needed.\newlineWe know the volume of the sphere is decreasing at a constant rate of 2π2\pi cubic meters per hour. The formula for the volume of a sphere is V=43πr3V = \frac{4}{3}\pi r^3, and the formula for the surface area of a sphere is A=4πr2A = 4\pi r^2. We need to find the rate of change of the surface area with respect to time when the radius is 55 meters.
  2. Rate of volume change: Find the rate of change of the volume with respect to the radius. The rate of change of the volume with respect to time dVdt\frac{dV}{dt} is given as −2π m3/hour-2\pi\, \text{m}^3/\text{hour} (negative because the volume is decreasing). We can relate this to the rate of change of the volume with respect to the radius dVdr\frac{dV}{dr} using the chain rule: dVdt=dVdr⋅drdt\frac{dV}{dt} = \frac{dV}{dr} \cdot \frac{dr}{dt}.
  3. Differentiate volume formula: Differentiate the volume formula with respect to the radius.\newlineDifferentiate V=43πr3V = \frac{4}{3}\pi r^3 with respect to rr to get dVdr=4πr2\frac{dV}{dr} = 4\pi r^2.
  4. Solve for dr/dt: Solve for dr/dt.\newlineWe have dVdt=−2π\frac{dV}{dt} = -2\pi and dVdr=4πr2\frac{dV}{dr} = 4\pi r^2. Plugging these into the chain rule equation gives us −2π=4πr2×(drdt)-2\pi = 4\pi r^2 \times \left(\frac{dr}{dt}\right). Now we solve for drdt\frac{dr}{dt} when r=5r = 5 meters.\newline−2π=4π(5)2×(drdt)-2\pi = 4\pi(5)^2 \times \left(\frac{dr}{dt}\right)\newline−2π=100π×(drdt)-2\pi = 100\pi \times \left(\frac{dr}{dt}\right)\newlinedrdt=−2π100π\frac{dr}{dt} = \frac{-2\pi}{100\pi}\newlinedrdt=−150\frac{dr}{dt} = -\frac{1}{50} m/hour
  5. Rate of surface area change: Find the rate of change of the surface area with respect to the radius. Differentiate the surface area formula A=4πr2A = 4\pi r^2 with respect to rr to get dAdr=8πr\frac{dA}{dr} = 8\pi r.
  6. Rate of change with respect to time: Find the rate of change of the surface area with respect to time.\newlineNow we use the chain rule again to relate dAdr\frac{dA}{dr} and drdt\frac{dr}{dt} to find dAdt\frac{dA}{dt}: dAdt=(dAdr)⋅(drdt)\frac{dA}{dt} = \left(\frac{dA}{dr}\right) \cdot \left(\frac{dr}{dt}\right).
  7. Calculate dAdt\frac{dA}{dt}: Calculate dAdt\frac{dA}{dt} when r=5r = 5 meters.\newlineWe have dAdr=8πr\frac{dA}{dr} = 8\pi r and drdt=−150\frac{dr}{dt} = -\frac{1}{50}. Plugging in r=5r = 5 meters gives us:\newlinedAdt=8π(5)×(−150)\frac{dA}{dt} = 8\pi(5) \times (-\frac{1}{50})\newlinedAdt=40π×(−150)\frac{dA}{dt} = 40\pi \times (-\frac{1}{50})\newlinedAdt=−4π5\frac{dA}{dt} = -\frac{4\pi}{5} m2^2/hour
  8. Conclude final answer: Conclude with the final answer.\newlineThe rate at which the surface area of the sphere is decreasing when the radius is 55 meters is −4π5-\frac{4\pi}{5} square meters per hour. The negative sign indicates that the surface area is decreasing. The correct answer is (A) 4π5\frac{4\pi}{5}.

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