Air is being pumped into a spherical balloon so that its volume increases at a rate of 80 cubic centimeters per second. How fast is the surface area of the balloon increasing when its radius is 8 cm?
A = surface area = (4/3) pi r^2
so
d A/dt = (8/3) pi r dr/dt
now get dr/dt from:
D Volume = surface area*dr=(4/3) pi r^2 dr
so
dV/dt = (4/3)pi r^2 dr/dt
To find how fast the surface area of the balloon is increasing, we can differentiate the surface area formula with respect to time. The formula for the surface area of a sphere is given by:
A = 4πr^2
Where A is the surface area and r is the radius of the sphere.
Differentiating both sides of the equation with respect to time (t) gives us:
dA/dt = d(4πr^2)/dt
To find dA/dt, the rate of change of surface area, we need to differentiate the equation and substitute the given values.
First, let's differentiate the equation:
dA/dt = 8πr(dr/dt)
Now we need to find the value of dr/dt, the rate of change of the radius. We are given that the volume of the balloon is increasing at a rate of 80 cubic centimeters per second. The formula for the volume of a sphere is:
V = (4/3)πr^3
Differentiating both sides of the equation with respect to time (t), we get:
dV/dt = d[(4/3)πr^3]/dt
Simplifying the equation and substituting the given value, we have:
80 cm^3/s = 4πr^2(dr/dt)
Now, we need to calculate dr/dt when the radius is 8 cm. Substituting the values into the equation:
80 cm^3/s = 4π(8^2)(dr/dt)
Simplifying further:
80 = 256π(dr/dt)
Now, we can solve for dr/dt:
dr/dt = 80 / (256π)
dr/dt ≈ 0.098 cm/s
Finally, substitute this value back into the equation for dA/dt:
dA/dt = 8πr(dr/dt)
dA/dt = 8π(8)(0.098)
dA/dt ≈ 19.89 cm^2/s
Therefore, when the radius is 8 cm, the surface area of the balloon is increasing at a rate of approximately 19.89 square centimeters per second.