RATES OF CHANGE QUESTION
A spherical balloon is being blown up so that its volume is increasing by 0.6 m^3 s^-1. Find the rate at which the radius is increasing when the radius is 0.1 m.
So what I did was that I determined dV/dt =0.6
and dV/dr = 4/3 r^3 phi (which is the volume of a sphere), and I should calculate dr/dt.
The answer should be 15 / phi but I figured I'll have to inverse the dV/dr part in order to get dr/dt.
But how do I do this???
We know
V = (4/3)πr^3
Just differentiate with respect to t
dV/dt = 4πr^2 dr/dt
now sub in our values
.6 = 4π(.1)^2 dr/dt
.6/(.04π = dr/dt
dr/dt = 15/π
Hmmm you just have to invert the answer that all
sure I did but I do not get 15/phi.
To find the rate at which the radius is increasing, you need to differentiate the volume equation with respect to time and then solve for dr/dt, the rate of change of the radius with respect to time.
Given:
dV/dt = 0.6 m^3 s^-1 -- Equation (1)
V = (4/3)πr^3 -- Equation (2)
To find dV/dr, the derivative of the volume equation with respect to r, apply implicit differentiation:
dV/dr = d((4/3)πr^3)/dr
= (4/3)π * (3r^2)
= 4πr^2 -- Equation (3)
Now, to find dr/dt, the rate at which the radius is increasing, we can use equation (1) and equation (3):
dV/dt = (dV/dr) * (dr/dt)
Substitute the values we have:
0.6 = (4πr^2) * (dr/dt)
Now, isolate dr/dt:
dr/dt = 0.6 / (4πr^2)
dr/dt = 0.15 / (πr^2)
To simplify further, rewrite π as φ, the symbol representing pi. So:
dr/dt = 0.15 / (φr^2)
Therefore, the rate at which the radius is increasing when the radius is 0.1 m is 0.15 / (φ * 0.1^2).
Simplifying this expression gives:
dr/dt ≈ 15 / φ
Hence, the answer is approximately 15 / φ.