A spherical balloon is being inflated. Given that the volume of a sphere in terms of its radius is V(r)= 4/3 πr^3 and the surface area of a sphere in terms of its radius is S(r)= 4πr^2, estimate the rate
at which the volume of the balloon is changing with respect to its surface area when the surface area
measures 50cm cubed.
dV/dt = 4πr^2 dr/dt
dS/dt = 8πr dr/dt
dV/dS = (dV/dt) / (dS/dt) = r/2
So now just find r when A = 50
To estimate the rate at which the volume of the balloon is changing with respect to its surface area, we can use calculus. Specifically, we can use the chain rule to find the derivative of the volume with respect to the surface area.
Given that the volume of the sphere is V(r) = (4/3)πr^3 and the surface area is S(r) = 4πr^2, we need to find dV/dS, the derivative of the volume with respect to the surface area.
First, we need to find the derivative of the volume with respect to the radius, dV/dr. Using the power rule, we differentiate V(r) = (4/3)πr^3 as follows:
dV/dr = 3(4/3)πr^2
= 4πr^2
Next, we need to find the derivative of the surface area with respect to the radius, dS/dr. Again, using the power rule, we differentiate S(r) = 4πr^2 as follows:
dS/dr = 2(4πr)
= 8πr
Finally, we can find dV/dS by dividing dV/dr by dS/dr:
dV/dS = (dV/dr) / (dS/dr)
= (4πr^2) / (8πr)
= (1/2)r
Given that the surface area measures 50 cm^2, we can substitute this value into our equation to find the rate at which the volume changes:
dV/dS = (1/2)r
= (1/2)(50)
= 25 cm^2
Therefore, the rate at which the volume of the balloon is changing with respect to its surface area when the surface area measures 50 cm^2 is 25 cm^2.