(b]1. The problem statement, all variables and given/known data[/b] A spherical weather balloon has a radius of 1m when it is 1500m high. You observe that the radius increases at a rate of 2cm/min as it continues to rise. At what rate is the surface area increasing when the radius is 4m? 2. Relevant equations I thought volume of a sphere might be useful to help solve this question. V = 4/3pi(r^3) 3. The attempt at a solution I was not sure if the first sentence of the question could help me solve this problem or not, so i decided to try and solve it without using the info from there. It looks like from the given information that the following might be the case: if i let r be the radius then r = 4 the rate at which the radius increases with respect to time could be shown as: dr/dt = 2 Now, this is where i get stuck; i am not sure if i am correct in making my next step to differentiate the volume of a sphere function, but this is what i arrived at: if V=4/3pi(r^3) THEN dV/dt=4pi(r^2) I am stuck at this point, because i don't know how to take it further, help please?