Air is being pumped into a spherical balloon so that its volume increases at a rate of 80cm^3/s. How fast is the surface area of the balloon increasing when its radius is 7cm?
V = (4/3)pi(r^3)
dV/dt = 4pi(r^2)dr/dt
when t=7
80 = 4pi(49)dr/dt
dr/dt = 80/(196pi)
SA = 4pi(r^2)
d(SA)/dt = 8pi(r)dr/dt
= 8pi(7)*80/(196pi)
etc
nmnm
22.85 cubic cm per second
To solve this problem, we can use the formulas for the volume and surface area of a sphere.
The volume of a sphere is given by the formula V = (4/3)πr^3, where r is the radius of the sphere.
Since we are given that the volume of the balloon is increasing at a rate of 80 cm^3/s, we can differentiate both sides of the volume equation with respect to time (t) to find the rate of change of the volume:
dV/dt = (4/3)π * 3r^2 * dr/dt
Here, dV/dt represents the rate of change of the volume, and dr/dt represents the rate of change of the radius.
We can rearrange this equation to solve for dr/dt:
dr/dt = (1/(4πr^2)) * dV/dt
Now, we know that the radius of the balloon is 7 cm, so we can substitute this into our equation to find dr/dt:
dr/dt = (1/(4π(7^2))) * 80
Calculate the value of (1/(4π(7^2))) * 80 to find the rate at which the radius is changing.
Once you have the value of dr/dt, you can find the rate at which the surface area of the balloon is increasing by using the formula for the surface area of a sphere:
A = 4πr^2
Differentiate both sides of this equation with respect to time (t) to find the rate of change of the surface area:
dA/dt = 8πr * dr/dt
Now, substitute the value of the radius (7 cm) and the rate of change of the radius (dr/dt) into the equation to find the rate at which the surface area of the balloon is increasing.