Air is being pumped into a spherical balloon so that its volume increases at a rate of 50cm3/s. How fast is the surface area of the balloon increasing when its radius is 11cm? Recall that a ball of radius r has volume V=43πr3 and surface area S=4πr2.
dS/dt = 8πr dr/dt
dV/dt = 4πr^2 dr/dt
so, dV/dt = r/2 dS/dt
dS/dt = dV/dt * 2/r
The given data thus mean that
dS/dt = 500*2/11 cm^2/min
To find the rate at which the surface area of the balloon is increasing, we can use the time derivative of the surface area equation.
Given:
dV/dt = 50 cm^3/s (rate of change of volume)
r = 11 cm (radius)
First, let's find the rate of change of volume with respect to time, dV/dt, using the equation for volume of a sphere:
V = (4/3)πr^3
Differentiating both sides with respect to time (t), we get:
dV/dt = d/dt[(4/3)πr^3]
= (4/3)π * d/dt(r^3)
Since we only have the rate of change of volume, we need to find the rate of change of radius with respect to time, dr/dt, to solve the problem.
Let's differentiate the volume equation with respect to r:
V = (4/3)πr^3
Differentiating both sides with respect to r, we get:
dV/dr = d/dr[(4/3)πr^3]
= 4π * d/dr(r^3)
= 12πr^2
Now, we can find dr/dt:
dV/dt = dV/dr * dr/dt
Substituting the given values for dV/dt and r:
50 cm^3/s = 12π(11 cm)^2 * dr/dt
To find dr/dt, rearrange the equation:
dr/dt = (50 cm^3/s) / [12π(11 cm)^2]
Now, we can calculate dr/dt.
To find how fast the surface area of the balloon is increasing, we can use the chain rule of differentiation.
Let's start by differentiating the given formulas for volume and surface area with respect to time.
The volume of the balloon, V, is given by the formula V = (4/3)πr^3.
The surface area of the balloon, S, is given by the formula S = 4πr^2.
Differentiating both sides of the equations with respect to time (t), we get:
dV/dt = (d/dt)(4/3)πr^3
dS/dt = (d/dt)(4πr^2)
Since we are given that the volume of the balloon is increasing at a rate of 50 cm^3/s, we can substitute dV/dt = 50 cm^3/s.
Now, let's find dS/dt when the radius of the balloon is 11 cm. We can substitute r = 11 cm into our equation for dS/dt.
dS/dt = (d/dt)(4π(11)^2)
To simplify this, we can calculate (d/dt)(121) first. Since the derivative of a constant with respect to time is always zero, we get:
(d/dt)(121) = 0
Therefore, dS/dt = (d/dt)(4π(11)^2) = 4π(2)(11)(0) = 0
So, when the radius of the balloon is 11 cm, the surface area of the balloon is not changing (dS/dt = 0).