At a certain moment the surface area of a sphere is 8 m2
and increasing at a rate of 10m2/sec. Determine how fast the volume of the sphere is changing at this moment
and increasing at a rate of 10m2/sec. Determine how fast the volume of the sphere is changing at this moment
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Use the given surface area to calculate the radius.
Take the time derivative of area and volume and use the common dr/dt to create dV/dt as a function of dA/dt.
Calculate dV/dt when dA/dt = 10 m^2/sec
A = 4 pi r^2 = 8 pi
r^2 = 2
r = sqrt(2)
V = (4/3) pi r^3
dV/dt = (4/3) pi (3r^2) dr/dt = 4 pi r^2 dr/dt
dA/dt = 4 pi (2r) dr/dt
dr/dt = [1/(8 pi r)] dA/dt
Substitute dr/dt into dV/dt.
dV/dt = [(4 pi r^2)/(8 pi r)]dA/dt
Simplify.
dV/dt = [r/2] dA/dt
Plug in the values of r and dA/dt.
dV/dt = [sqrt(2)/2][10] = 5sqrt(2) cubic meters per second.
Take the time derivative of area and volume and use the common dr/dt to create dV/dt as a function of dA/dt.
Calculate dV/dt when dA/dt = 10 m^2/sec
A = 4 pi r^2 = 8 pi
r^2 = 2
r = sqrt(2)
V = (4/3) pi r^3
dV/dt = (4/3) pi (3r^2) dr/dt = 4 pi r^2 dr/dt
dA/dt = 4 pi (2r) dr/dt
dr/dt = [1/(8 pi r)] dA/dt
Substitute dr/dt into dV/dt.
dV/dt = [(4 pi r^2)/(8 pi r)]dA/dt
Simplify.
dV/dt = [r/2] dA/dt
Plug in the values of r and dA/dt.
dV/dt = [sqrt(2)/2][10] = 5sqrt(2) cubic meters per second.