a splenical ballon is being inflated @ 10 cubic inches per secound. how fast is the radius of the ballon increasing when the surface area of the ballon is 2 pi square inches( answer to 4 decimal places)?
I once had a splenical balloon. It was almost spherical.
With that in mind,
v = 4/3 pi r^3
dv/dt = 4 pi r^2 dr/dt
so, plug and chug
10 = 2 pi dr/dt
dr/dt = 10/2pi = 1.5915 in/s
Note: surface area is 4pi r^2
To find out how fast the radius of the balloon is increasing, we can use the related rates technique and differentiate the equation for the surface area of a sphere with respect to time.
The surface area of a sphere is given by the formula: A = 4πr², where A is the surface area and r is the radius.
Now, we have that the surface area of the balloon is 2π square inches, which gives us the equation: 2π = 4πr².
To find the rate at which the radius is increasing, let's differentiate the equation with respect to time. Since time is represented by "t", we use the chain rule:
d/dt (2π) = d/dt (4πr²)
0 = 8πr * dr/dt
Notice that we have the rate of change of the volume (10 cubic inches per second), but we are looking for the rate of change of the radius (dr/dt). So, we need to relate these two using the volume formula for a sphere.
The volume of a sphere is given by the formula: V = (4/3)πr³, where V is the volume and r is the radius.
Since we know that dV/dt = 10 cubic inches per second, we can differentiate the volume formula with respect to time:
dV/dt = d/dt [(4/3)πr³]
10 = 4πr² * dr/dt
Now, we have an equation that relates the rate of change of the volume and the rate of change of the radius.
Combining this equation with the earlier equation obtained from the surface area, we have the following system of equations:
8πr * dr/dt = 0
4πr² * dr/dt = 10
Solving the first equation, we find that either r = 0 or dr/dt = 0. Since the radius of a balloon cannot be zero, we conclude that dr/dt = 0.
Substituting this into the second equation, we get:
4πr² * 0 = 10
0 = 10
Since the equation is impossible, there is no solution. Therefore, the radius of the balloon is not changing when the surface area is 2π square inches.