Asked by Peopsh8
A spherical snowball is melting in such a way that its radius is decreasing at a rate of 0.4cm/min. At what rate is the volume decreasing when the radius is 13 cm ?
I got dV/dt= -849.5 but its seems too large
I got dV/dt= -849.5 but its seems too large
Answers
Answered by
oobleck
v = 4/3 pi r^3
dv/dt = 4 pi r^2 dr/dt = 4 pi * 13^2 * -0.4 = -849.48 cm^3/min
the engineers here will tell you a simpler way. Consider the surface of the sphere as a thin sheet of thickness dr. Then the volume will decrease by A*dr
That is, dv/dt = 4 pi r^2 dr/dt
as above
dv/dt = 4 pi r^2 dr/dt = 4 pi * 13^2 * -0.4 = -849.48 cm^3/min
the engineers here will tell you a simpler way. Consider the surface of the sphere as a thin sheet of thickness dr. Then the volume will decrease by A*dr
That is, dv/dt = 4 pi r^2 dr/dt
as above
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