A spherical balloon is being inflated. Given that the volume of a sphere in terms of its radius is v(r)=(4/3)(pi(r^3))
and the surface area of a sphere in terms of its radius is s(r)=4pi(r^2), estimate 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.
To estimate the rate at which the volume of the balloon is changing with respect to its surface area, we need to find the derivative of the volume function with respect to the surface area.
First, let's differentiate the volume function, v(r), with respect to the radius, r, using the chain rule.
v'(r) = d(v)/d(r) = d(v)/d(r) × d(r)/d(s)
The derivative of the volume function, (4/3)π(r^3), with respect to the radius, r, is given by:
v'(r) = 4π(r^2)
Now, we need to relate the derivative of the volume function with the surface area, s(r), in terms of radius, r.
We have the surface area function, s(r) = 4π(r^2), and we want to express s in terms of r, so we need to solve for r:
s(r) = 4π(r^2) = 50 cm^2
Now, we can find the value of r that corresponds to a surface area of 50 cm^2 and use it to calculate the rate at which the volume is changing with respect to the surface area.
4π(r^2) = 50 cm^2
r^2 = 50 cm^2 / (4π)
r^2 = 3.9894 cm^2
r ≈ 1.998 cm
Now, substitute this value of r into the derivative equation to find the rate at which the volume is changing with respect to the surface area:
v'(r) = 4π(r^2)
v'(1.998 cm) ≈ 4π(1.998^2)
v'(1.998 cm) ≈ 4π(3.992 cm^2)
v'(1.998 cm) ≈ 50.04 cm^3/cm^2
So, 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 approximately 50.04 cm^3/cm^2.
To find the rate at which the volume of the balloon is changing with respect to its surface area, we need to use the concept of related rates.
Given:
The volume of a sphere: V(r) = (4/3)πr^3
The surface area of a sphere: S(r) = 4πr^2
Surface area = 50 cm^2
First, let's find the derivative of the volume function with respect to the surface area.
dV/dS = (dV/dr)/(dS/dr)
To find dV/dr, we differentiate the volume function V(r) with respect to r.
dV/dr = d/dx [(4/3)π(r^3)]
= (4/3)π(3r^2)
= 4πr^2
To find dS/dr, we differentiate the surface area function S(r) with respect to r.
dS/dr = d/dx [4π(r^2)]
= 8πr
Now we can substitute the values into the derivative formula.
dV/dS = (4πr^2)/(8πr)
= r/2
Since we are given that the surface area measures 50 cm^2, we can substitute this value into the equation to find the rate of change.
dV/dS = r/2
= 50/2
= 25 cm^2
So, the rate at which the volume of the balloon is changing with respect to its surface area is 25 cm^2.
using the area, r = √(s/4π)
so, v = 4/3 π (s/4π)^(3/2) = 1/(6√π) s^(3/2)
dv/ds =1/(4√π) s^(1/2) = 1/4 √(s/π)
at s=50, that would be 5/4 √(2/π)