V= k r^3 you know k
dV/dt= 3kr^2 dr/dt
dr/dt=(dV/dt)/3k * r^2
How fast does the radius of a spherical soap bubble change when you blow air into it at the rate of 15 cubic centimeters per second? Our known rate is dv/dt , the change in volume with respect to time, which is 15 cubic centimeters per second. The rate we want to find is dr/dt , the change in the radius with respect to time. Remember that the volume of a sphere is v=4/3Pir^3.
4 answers
The change of volume of a sphere with increasing radius is the surface area of the sphere times dr.
Draw a picture !
In other words:
dV = 4 pi r^2 dr
so
dV/dt = 4 pi r^2 dr/dt
so here
15 = 4 pi r^2 dr/dt
or
dr/dt = 15 /(4 pi r^2)
----------------------
note , I am a ship builder so usually do this sort of problem this way.
calculus experts would not draw a picture but take the derivative of the volume:
V = (4/3) pi r^3
so
dV/dr = 4 pi r^2 (which we knew)
so
dV/dt = dV/dr*dr/dt = 4 pi r^2 dr/dt
which we also knew :)
Draw a picture !
In other words:
dV = 4 pi r^2 dr
so
dV/dt = 4 pi r^2 dr/dt
so here
15 = 4 pi r^2 dr/dt
or
dr/dt = 15 /(4 pi r^2)
----------------------
note , I am a ship builder so usually do this sort of problem this way.
calculus experts would not draw a picture but take the derivative of the volume:
V = (4/3) pi r^3
so
dV/dr = 4 pi r^2 (which we knew)
so
dV/dt = dV/dr*dr/dt = 4 pi r^2 dr/dt
which we also knew :)
Ohhhh okay thanks so much.. your a ship builder??? that's pretty awesome. That's a very interesting job :)
You are welcome. Actually I am retired now :)