Air is being pumped into a spherical balloon so that its volume increases at a rate of 90{cm}^3/s. How fast is the surface area of the balloon increasing when its radius is 7{cm}? Recall that a ball of radius r has volume \displaystyle V={4}/{3}pie r^3 and surface area S=4\pi r^2
dV/dt= 4/3 PI 3 r^2 dr/dt
solve for dr/dt given dv/dt
Then dS/dt=4PI 2r dr/dt, knowing dr/dt solve for dS/dt
and it is done.
ok thanks alot for clearing it up
To find the rate at which the surface area of the balloon is increasing, we first need to express the surface area in terms of the radius. Given that the surface area of a sphere is given by \(S = 4\pi r^2\).
Let's differentiate both sides of the equation with respect to time t using implicit differentiation, as we want to find the rate of change with respect to time:
\(\frac{dS}{dt} = \frac{d}{dt}(4\pi r^2)\)
Since we are given the rate at which the volume is increasing \(\frac{dV}{dt} = 90 \, \text{cm}^3/\text{s}\), and the relationship between the volume and radius is given by:
\(V = \frac{4}{3}\pi r^3\)
We can differentiate this equation with respect to time t using implicit differentiation:
\(\frac{dV}{dt} = \frac{d}{dt}\left(\frac{4}{3}\pi r^3\right)\)
Now we can solve for \(\frac{dS}{dt}\) in terms of \(\frac{dV}{dt}\):
\(\frac{dS}{dt} = \frac{d}{dt}(4\pi r^2) = 8\pi r \frac{dr}{dt}\)
We are given that the radius is 7 cm, so substituting the value of r = 7:
\(\frac{dS}{dt} = 8\pi (7) \frac{dr}{dt}\)
Now we need to find \(\frac{dr}{dt}\). We are given that \(\frac{dV}{dt} = 90 \, \text{cm}^3/\text{s}\), which represents the rate of change of volume. Rewriting the volume equation, we have:
\(\frac{dV}{dt} = \frac{d}{dt}\left(\frac{4}{3}\pi r^3\right)\)
\(\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}\)
Now we can solve for \(\frac{dr}{dt}\):
\(90 = 4\pi (7)^2 \frac{dr}{dt}\)
\(\frac{dr}{dt} = \frac{90}{4\pi (7)^2}\)
Substituting this value into the equation for \(\frac{dS}{dt}\):
\(\frac{dS}{dt} = 8\pi (7) \left(\frac{90}{4\pi(7)^2}\right)\)
Calculating this expression will give us the rate at which the surface area of the balloon is increasing when its radius is 7 cm.