If the volume of a spherical balloon is changing at a rate of 10 cm3
/s, how fast is the surface area
changing when the radius is 6.5 cm?
since v = 4/3 pi r^3
and a = 4 pi r^2,
v = 1/3 ar
so,
dv/dt = 4pi r^2 dr/dt = a dr/dt
so, dr/dt = (dv/dt)/a
dv/dt = 1/3 (r da/dt + a dr/dt)
= 1/3 (r da/dt + a*(dv/dt)/a)
= 1/3 (r da/dt + dv/dt)
= 1/2 r da/dt
So,
10 = 1/2 (6.5) da/dt
da/dt = 3.077 cm^2/s
To find the rate at which the surface area is changing, we need to use the formulas for the volume and surface area of a sphere. The formulas are as follows:
Volume of a sphere: V = (4/3)πr^3
Surface area of a sphere: A = 4πr^2
Given that the volume is changing at a rate of 10 cm3/s, we need to find how fast the surface area is changing with respect to time.
We can express the relationship between the volume (V), radius (r), and time (t) as follows:
V(t) = (4/3)πr(t)^3
Since we are given that the volume is changing at a rate of 10 cm3/s, we can differentiate the volume function with respect to time to find the rate of change of volume:
dV/dt = d/dt((4/3)πr(t)^3)
= 4πr(t)^2(dr/dt)
We can rewrite this as:
10 = 4πr(t)^2(dr/dt)
Now, we need to find the rate at which the surface area is changing when the radius is 6.5 cm. We can substitute this value into the equation and solve for (dr/dt):
10 = 4π(6.5)^2(dr/dt)
(dr/dt) = 10 / (4π(6.5)^2)
Now, we can calculate the value of (dr/dt) to find how fast the surface area is changing when the radius is 6.5 cm.