Here's the problem:
As a spherical mothball evaporates, its volume decreases at a rate proportional to its surface area. Show that the rate of decrease of the radius is constant.
Here's what I've done so far:
Let r be the radius of the sphere, V its volume, and A its surface area. We know V = (4/3)?r[sup:3pja7b27]3[/sup:3pja7b27] and A = 4?r[sup:3pja7b27]2[/sup:3pja7b27]. We are given dV/dt = c(dA/dt), where c is a nonzero constant. We want to show that dr/dt is a constant.
V = (4/3)?r[sup:3pja7b27]3[/sup:3pja7b27] ? dV/dt = (4/3)? ? d(r[sup:3pja7b27]3[/sup:3pja7b27])/dt = (4/3)? ? 3r[sup:3pja7b27]2[/sup:3pja7b27] ? dr/dt
A = 4?r[sup:3pja7b27]2[/sup:3pja7b27] ? dA/dt = 4? ? d(r[sup:3pja7b27]2[/sup:3pja7b27])/dt = 4? ? 2r ? dr/dt
dV/dt = c(dA/dt) ? (4/3)? ? 3r[sup:3pja7b27]2[/sup:3pja7b27] ? dr/dt = c[4? ? 2r ? dr/dt] ? r = 2c
I'm not sure how this result connects with dr/dt being a constant. Am I on the right track, and where do I go from here?
As a spherical mothball evaporates, its volume decreases at a rate proportional to its surface area. Show that the rate of decrease of the radius is constant.
Here's what I've done so far:
Let r be the radius of the sphere, V its volume, and A its surface area. We know V = (4/3)?r[sup:3pja7b27]3[/sup:3pja7b27] and A = 4?r[sup:3pja7b27]2[/sup:3pja7b27]. We are given dV/dt = c(dA/dt), where c is a nonzero constant. We want to show that dr/dt is a constant.
V = (4/3)?r[sup:3pja7b27]3[/sup:3pja7b27] ? dV/dt = (4/3)? ? d(r[sup:3pja7b27]3[/sup:3pja7b27])/dt = (4/3)? ? 3r[sup:3pja7b27]2[/sup:3pja7b27] ? dr/dt
A = 4?r[sup:3pja7b27]2[/sup:3pja7b27] ? dA/dt = 4? ? d(r[sup:3pja7b27]2[/sup:3pja7b27])/dt = 4? ? 2r ? dr/dt
dV/dt = c(dA/dt) ? (4/3)? ? 3r[sup:3pja7b27]2[/sup:3pja7b27] ? dr/dt = c[4? ? 2r ? dr/dt] ? r = 2c
I'm not sure how this result connects with dr/dt being a constant. Am I on the right track, and where do I go from here?