Air is pumped into a balloon such that its volume increases at the rate of 75cm^3 per second. It is assumed that the balloon is spherical all the time. Find in term of pi the rate at which the radius of the balloon is increasing when the radius is 5cm. Given the balloon was initially empty show that one minute after the pumping begins the readius is increasing at the rate of 1/12pi r^-1/3 per second. Help please!!
surface area = 4 pi r^2
so
dv/dt = 4 pi r^2 dr/dt
dv/dt = 75 cm^3/s
so
dr/dt = 75/(4 pi r^2)
when r = 5
dr/dt = 3/(4 pi) = .75/pi cm/s
after 60 seconds volume = 75*60 = 4500 cm^3
(4/3)pi r^3 = 4500
pi r^3 = 3375
r = 15/pi^(1/3)
r^2 = 225/pi^(2/3)
4 pi r^2 = 900 pi^(1/3)
dr/dt = 75/(4 pi r^2)
= (75/900)pi^-1/3)
= (1/12) pi^(-1/3)
So I disagree
So did I - thank you so much!! I have spent hours on this one and I am sure you are right.
Well, let's hope we are right :)
To find the rate at which the radius of the balloon is increasing, we can use the formula for the volume of a sphere and differentiate it with respect to time.
Let's start by writing down the formula for the volume of a sphere:
V = (4/3) * π * r^3
Where V represents the volume and r represents the radius.
Now, we need to differentiate both sides of the equation with respect to time (t):
dV/dt = d/dt[(4/3) * π * r^3]
The left side gives us the rate at which the volume is changing, which is given in the problem as 75 cm^3/s:
dV/dt = 75
On the right side, we need to apply the product rule for differentiation:
d/dt[(4/3) * π * r^3] = (4/3) * π * d/dt(r^3)
Using the power rule for differentiation, we can find d/dt(r^3):
d/dt(r^3) = 3r^2 * dr/dt
Now we can substitute this back into the equation:
75 = (4/3) * π * 3r^2 * dr/dt
Simplifying, we get:
25 = 4πr^2 * dr/dt
Now, we can solve for dr/dt, the rate at which the radius is changing:
dr/dt = 25 / (4πr^2)
To find the rate at which the radius is increasing when the radius is 5 cm, we can substitute r = 5 into the equation:
dr/dt = 25 / (4π * 5^2)
dr/dt = 25 / (4π * 25)
dr/dt = 1 / (4π)
Therefore, when the radius is 5 cm, the rate at which the radius is increasing is 1/(4π) cm/s.
Now, let's show that one minute after the pumping begins, the radius is increasing at the rate of 1/(12π * r^(1/3)) per second.
To find the rate of change when one minute has passed, we'll need to convert one minute into seconds. Since there are 60 seconds in a minute, one minute is equal to 60 seconds.
Now we can substitute t = 60 into the equation for dr/dt:
dr/dt = 25 / (4π * r^2)
dr/dt = 25 / (4π * r^2) * t, where t is in seconds
dr/dt = 25 / (4π * r^2) * 60
dr/dt = 1500 / (4π * r^2)
Next, we'll substitute r = (1/12π) into the equation:
dr/dt = 1500 / (4π * (1/12π)^2)
dr/dt = 1500 / (4π * 1/(144π^2))
dr/dt = 1500 / (4/144π^3)
dr/dt = 1500 / (1/36π^3)
dr/dt = 36π^3 * 1500
dr/dt = 54000π^3
So, one minute after the pumping begins, the rate at which the radius is increasing is 54000π^3 cm/s, which can also be written as 1/(12π * r^(1/3)) per second.