im so confused :( Willie Makit blows up a spherical balloon. He recalls that the volume is V=4/3Pi*r^3. In order for the radius to increase at 2cm/sec, how fast must Willie blow air into the balloon when the radius is 3 cm?
given:
r'=2cm/sec
r=3cm
find V'
look at example 1:
http://tutorial.math.lamar.edu/Classes/CalcI/RelatedRates.aspx
i just looked at it and tried to figure it out but i just don't get it. in the example you gave me there was already a rate that the air was being pumped, but in mine there is none. it's asking for the rate. I just don't get it
forget it :) i figured it out! thanks!!!!
To find out how fast Willie must blow air into the balloon when the radius is 3 cm, we need to differentiate the volume formula with respect to time and then substitute the given values.
Let's start by differentiating the volume formula:
V = (4/3) * π * r^3
Now, differentiate both sides of the equation with respect to time (t):
dV/dt = d/dt[(4/3) * π * r^3]
To differentiate the right side of the equation, we can use the chain rule of differentiation. Let's break it down:
dV/dt = (4/3) * π * d/dt[r^3]
Next, differentiate r^3 with respect to time, keeping in mind that the derivative of r with respect to time is given as dr/dt:
dV/dt = (4/3) * π * 3r^2 * dr/dt
Now, we can substitute the given values into the equation:
r = 3 cm (radius)
dr/dt = 2 cm/sec (rate of increase in radius)
Substituting these values into the equation:
dV/dt = (4/3) * π * 3(3 cm)^2 * (2 cm/sec)
Simplifying further:
dV/dt = (4/3) * π * 3(9 cm^2) * (2 cm/sec)
dV/dt = (4/3) * π * 54 cm^2/sec
Finally, evaluate the expression:
dV/dt = 72π cm^2/sec
Therefore, Willie must blow air into the balloon at a rate of 72π cm^2/sec when the radius is 3 cm.