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.
= 4pi r^2(15)??
Oh okay, thanks
To find the rate of change of the radius with respect to time (dr/dt), given the known rate of change of volume with respect to time (dv/dt), you can use the relationship between volume and radius of a sphere.
The volume (v) of a sphere is given by the formula:
v = (4/3) * π * r^3
Taking the derivative of both sides of the equation with respect to time (t) gives us:
dv/dt = d/dt[(4/3) * π * r^3]
Now, to find dr/dt, the rate of change of the radius with respect to time, we need to solve for it.
First, rearrange the equation to solve for r:
v = (4/3) * π * r^3
(3/4)v = π * r^3
(3/4v)^(1/3) = r
Next, differentiate both sides of the equation with respect to time (t):
d/dt[(3/4v)^(1/3)] = d/dt[r]
Now, substitute dv/dt into the equation, as it represents the change in volume with respect to time:
d/dt[(3/4v)^(1/3)] = dr/dt
Finally, plug in the given value of dv/dt (15 cubic centimeters per second) and solve for dr/dt:
dr/dt = d/dt[(3/4(15))^(1/3)]
Calculating the above expression will give you the rate of change of the radius with respect to time (dr/dt).
hmm. They gave you the formula, so just use it:
v = 4/3 pi r^3
dv/dt = 4pi r^2 dr/dt
So, just solve for dr/dt
dV/dr = area of surface = 4 pi r^2
so
dV/dt = 4 pi r^2 dr/dt
so
dr/dt = (dV/dt) / (4 pi r^2)