Air is being pumped into a spherical balloon so that its volume increases at a rate of 80 \mbox{cm}^3\mbox{/s}. How fast is the surface area of the balloon increasing when its radius is 14 \mbox{cm}? Recall that a ball of radius r has volume \displaystyle V=\frac{4}{3}\pi r^3 and surface area S=4\pi r^2.
just use the formula
v = 4/3 pi r^3
a = 4pi r^2, so
dv/dt = 4pi r^2 dr/dt = a dr/dt
da/dt = 8pi r dr/dt
v(14) = 11494
a(14) = 2463
So, we have
80 = 2463 dr/dt
da/dt = 8pi(14)(80/2463) = 11.43
what is an mbox?
V = (4/3) pi r^3
S = dV/dr = 4 pi r^2
dS/dr = 8 pi r
so
dS/dt = 8 pi r dr/dt
now find dr/dt
if dV/dt = 80 cm^3/s
dV = S * dr = 4 pi r^2 dr
so
dV/dt = 4 pi r^2 dr/dt
so
dr/dt = ( 1/4 pi r^2 )dV/dt = (20/pi 14^2)
so
dS/dt = 8 pi 14 (20/14^2 pi)
= 80/7
=
To find the rate at which the surface area of the balloon is increasing, we first need to find an equation that relates the surface area to the radius and the volume. The volume of a sphere is given by the equation:
V = (4/3)πr^3
To find an equation relating the surface area to the radius and the volume, we can differentiate both sides of the volume equation with respect to time (t):
dV/dt = (4/3)d/dt(πr^3)
Since we are given the rate at which the volume is increasing, dV/dt, we can substitute that into the equation:
80 cm^3/s = (4/3)d/dt(πr^3)
Now we need to find an equation relating the surface area to the radius, and differentiate both sides with respect to time:
S = 4πr^2
dS/dt = d/dt(4πr^2)
Now we can substitute the given radius of 14 cm into the equation:
dS/dt = d/dt(4π(14)^2)
Simplifying:
dS/dt = d/dt(784π)
Since there are no variables in the equation, the derivative of a constant is zero:
dS/dt = 0
Therefore, the surface area of the balloon is not changing with time, so the rate at which it is increasing is zero.
To find the rate at which the surface area of the balloon is increasing, we need to relate the change in the surface area to the change in the volume.
Given that the volume of the balloon is increasing at a rate of 80 cm³/s, we can find how the volume is changing with respect to time. This means we need to find dV/dt.
We are also given that the radius of the balloon is 14 cm, so we can find the value of the radius, r.
We can differentiate the volume formula V = (4/3)πr³ with respect to time, using the chain rule:
dV/dt = dV/dr * dr/dt
The first term dV/dr represents the derivative of the volume with respect to the radius, which is given by:
dV/dr = 4πr²
The second term dr/dt represents the derivative of the radius with respect to time, which is not given. However, we can find it using the information about the rate of volume increase.
We know that dV/dt = 80 cm³/s, and we are looking for dr/dt. We need to rearrange the previous equation to solve for dr/dt:
dV/dt = dV/dr * dr/dt
80 = 4πr² * dr/dt
dr/dt = 80 / (4πr²)
Now we can substitute the given radius of 14 cm into the equation to find dr/dt:
dr/dt = 80 / (4π(14)²)
dr/dt ≈ 0.0346 cm/s
Now that we have the value of dr/dt, we can use it to find how the surface area of the balloon is changing with respect to time.
The formula for surface area S = 4πr² can be differentiated with respect to time to find dS/dt:
dS/dt = dS/dr * dr/dt
The first term dS/dr represents the derivative of the surface area with respect to the radius, which is given by:
dS/dr = 8πr
Now we can substitute the given radius of 14 cm and the value of dr/dt into the equation:
dS/dt = 8π(14) * 0.0346
dS/dt ≈ 15.33 cm²/s
Therefore, the surface area of the balloon is increasing at a rate of approximately 15.33 cm²/s when its radius is 14 cm.