Suppose that the volume of a spherical raindrop is proportional to the 3/2 power of its surface area and the rate of reduction of the volume of the raindrop is proportional to its surface area, find a formula for the amount of time it takes for the raindrop to evaporate completely, giving your answer in terms of the constants you introduced and the initial surface area of the raindrop. This is a diferential equation problem I started with volume 0f raindrop = k(4pi r^2)^3/2 then not sure what to do?

The volume of a sphere is 4/3*pi*r^3, so using that you can solve for k \[k(4\pi r^{2})^{3/2} = (4/3)\pi r^{3}\] \[k(8\pi \sqrt{\pi} r^{3}) = (4/3)\pi r^{3}\] \[k =\frac{1}{ 6\sqrt{\pi}}\] Now define function S(t) as surface area of rain drop at time t, call it s \[V(t) = \frac{1}{6\sqrt{\pi}} s^{3/2}\] Differentiate using chain rule \[V'(t) = \frac{1}{4\sqrt{\pi}}\frac{ds}{dt}\sqrt{s}\] You know that the rate of change in volume (V'(t)) is proportional to s \[\rightarrow \frac{1}{4\sqrt{\pi}}\frac{ds}{dt}\sqrt{s} = k*s\] Rearrange equation to get s and ds on one side, dt on the other \[\frac{ds}{\sqrt{s}} = 4k \sqrt{\pi} dt\] integrate both sides \[\int\limits_{}^{} \frac{ds}{\sqrt{s}} =\int\limits_{}^{} 4k \sqrt{\pi} dt\] \[2\sqrt{s} = 4k \sqrt{\pi}t + C\] \[s = (2k \sqrt{\pi}t +C)^{2}\] Notice when t=0, initial surface area is C^2, replace C with sqrt( s_o) so its in terms of initial surface area \[s = (2k \sqrt{\pi}t+\sqrt{s_{o}})^{2}\] Substitute into volume function and solve for t when V =0 \[V(t) = \frac{1}{6 \sqrt{\pi}}[(2k \sqrt{\pi}t +\sqrt{s_{o}})^{2}]^{3/2} = \frac{1}{6 \sqrt{\pi}}(2k \sqrt{\pi}t +\sqrt{s_{o}})^{3} = 0\] \[\rightarrow 2k \sqrt{\pi}t +\sqrt{s_{o}} = 0\] \[t = -\frac{\sqrt{s_{o}}}{2k \sqrt{\pi}}\]

Join our real-time social learning platform and learn together with your friends!