A spherical balloon a has a diameter 2 m when it is 1500m high . It is observed that the diameter increases at a constant rate of 4cm/min as it continues to rise. At what rate is the volume increasing when the diameter is 4m . At what rate is the surface area imcreasing then?

r = 1 meter

dr/dt = .02m/min
This is the same idea as your cone problem.

The rate of change of the volume is the rate of change of the surface area times the added thickness of the surface, dr/dt
In other words:
dV/dt = 4 pi r^2 dr/dt

surface area = A = 4 pi r^2
dA/dr = 8 pi r
dA/dt = 8 pi r dr/dt

T=3s

To find the rate at which the volume of the balloon is increasing when the diameter is 4m, we need to use related rates. Related rates involve finding the rate of change of one variable with respect to another variable. In this case, we are given the rate at which the diameter is increasing and asked to find the rate at which the volume is increasing.

Let's denote the diameter of the balloon as D and the volume of the balloon as V. We are given that D increases at a constant rate of 4 cm/min and we need to find the rate at which V is increasing.

From the given information, we can see that the radius (r) of the balloon is initially 1m (half of the diameter) and it is increasing at a rate of 2 cm/min (half of the rate of increase in diameter). Now we can express V in terms of r:

V = (4/3)πr^3

To find the rate at which V is increasing, we differentiate both sides of the equation with respect to time:

dV/dt = dV/dr * dr/dt

We know that dr/dt = 2 cm/min, so now we need to find dV/dr.

dV/dr = (d/dt)((4/3)πr^3) = 4πr^2 * dr/dt

Substituting the values we have, we get:

dV/dt = 4π(1^2)(2) = 8π cm^3/min

Therefore, the volume is increasing at a rate of 8π cm^3/min when the diameter is 4m.

To find the rate at which the surface area is increasing, we can differentiate the surface area formula of a sphere with respect to time.

The surface area (A) of a sphere is given by:

A = 4πr^2

Differentiating both sides of the equation with respect to time, we get:

dA/dt = dA/dr * dr/dt

We know that dr/dt = 2 cm/min, so now we need to find dA/dr.

dA/dr = (d/dt)(4πr^2) = 8πr * dr/dt

Substituting the values we have, we get:

dA/dt = 8π(1)(2) = 16π cm^2/min

Therefore, the surface area is increasing at a rate of 16π cm^2/min when the diameter is 4m.