Math Question: A spherical snowball is being made so that its volume is increasing at the rate of 8 cubic feet per minute. Find the rate at which the radius is increasing when the snowball is 4 feet in diameter?

I set these up this way. What do we know? dV/dt = 8 ft^3/min What do we want? dr/dt = ??? When: when d = 4 feet, r = 2 feet Equation relating V and r? V = 4pir^3/3 Rate equation - implicitly differentiate: d/dt(V) = d/dt(4pi*r^3/3) dV/dt = 4pi r^2 dr/dt Plug in the known values and solve for dr/dt. 8 = 4pi (2)^2 dr/dt 8 = 16pi dr/dt dr/dt = 1/(2pi) = 0.159 ft/min
Answered On 04/25/2020 01:20

Since the formula for the volume of a sphere is: V=4?/3r³ And the given is: dV/dt=8ft³/min d=4ft Required: dr/dt Since: d=2r 4=2r r=2ft To find the rate of increase of the radius with respect to time(dr/dt), we need to take the derivative of the formula of the volume of the sphere with respect to time. d/dt(V=4?/3r³)d/dt Using implicit differentiation: dV/dt=4?/3(3r²)dr/dt Substitute all the given to the formula to get dr/dt 8=4?(2²)dr/dt 8/16?=16? dr/dt /16? dr/dt=1/2? ft/min or 0.159 ft/min
Answered On 04/09/2023 14:51