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
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
To determine the rate at which the volume is decreasing, we can use the formula for the volume of a sphere:
V = (4/3)πr^3
where V is the volume and r is the radius.
To find the rate at which the volume is changing with respect to time, we need to take the derivative of the volume formula with respect to time (t), which can be denoted as dV/dt.
Given that the radius is decreasing at a rate of 0.4 cm/min, we can express this as dr/dt = -0.4 cm/min, taking the negative sign since the radius is decreasing.
Now, differentiate both sides of the volume formula with respect to time:
dV/dt = d/dt ((4/3)πr^3)
To derive this equation, we use the chain rule. The derivative of r^3 with respect to t is 3r^2 (dr/dt). Plugging in the given value of dr/dt = -0.4 cm/min, we have:
dV/dt = (4/3)π(3r^2)(dr/dt)
= (4/3)π(3)(13^2)(-0.4)
= (4/3)π(3)(169)(-0.4)
= (-16/3)π(169)
≈ -849.53 cm^3/min
Therefore, the rate at which the volume is decreasing when the radius is 13 cm is approximately -849.53 cm^3/min.
Please note that the negative sign indicates that the volume is decreasing over time.