Air is being pumped into a spherical balloon so that its volume increases at a rate of 100 {cm}^3}. How fast is the surface area of the balloon increasing when its radius is 16{cm}?
To find the rate at which the surface area of the balloon is increasing, we need to use the formulas for the volume and surface area of a sphere.
The volume of a sphere is given by the formula: V = (4/3)πr^3, where V is the volume and r is the radius.
To find the rate at which the volume is increasing, we can take the derivative with respect to time (t): dV/dt = (dV/dr)(dr/dt), where dV/dt is the rate of change of volume and dr/dt is the rate of change of the radius.
In this case, we are given that the volume is increasing at a rate of 100 cm^3/s, so dV/dt = 100. We need to find dr/dt, the rate at which the radius is changing.
We can rearrange the volume formula to solve for r: r = (3V / 4π)^(1/3).
Now, we can differentiate the equation for r with respect to time to find dr/dt:
dr/dt = (1/3)(3/4π)^(1/3) * [(dV/dt)] / [V^(2/3)].
Substituting the given values, we have:
dr/dt = (1/3)(3/4π)^(1/3) * (100) / (V^(2/3)).
Now we need to calculate the value of V at the given radius, r = 16 cm. Substituting this value into the volume formula:
V = (4/3)π(16)^3 = 4/3 * 3.14 * 4096 ≈ 21747 cm^3.
Now we can substitute this value of V into the equation for dr/dt:
dr/dt = (1/3)(3/4π)^(1/3) * (100) / (21747^(2/3)).
Evaluating this expression will give us the rate at which the radius is changing.
v = 4/3 pi r^3
dv/dt = 4pi r^2 dr/dt
a = 4pi r^2
da/dt = 8pi r dr/dt
= 8pi r (dv/dt / (4pi r^2))
Now just plug in dv/dt and r