# C4 Problems

I'm having a lot of trouble trying to solve the questions below, could someone please help me.

1.) A spherical balloon is allowed to deflate. The rate at which air is leaving the balloon is proportional to the volume V of air left in the balloon. When the radius of the balloon is 15cm, air is leaving at a rate of 8cm^3 s^-1. Find an expression for dV/dt.

2.) A tank is shaped as a cuboid with a square base of side 10cm. Water runs out through a hole in the base at a rate proportional to the square root of the height, h cm, of water in the tank. At the same time, water is pumped into the tank at a constant rate of 2cm^3 s^-1. Find an expression for dh/dt.
1)
dA/dt = k4pi r^3/3
r = 15, dA/dt = 8
dhokes
I'm having a lot of trouble trying to solve the questions below, could someone please help me.

1.) A spherical balloon is allowed to deflate. The rate at which air is leaving the balloon is proportional to the volume V of air left in the balloon. When the radius of the balloon is 15cm, air is leaving at a rate of 8cm^3 s^-1. Find an expression for dV/dt.

2.) A tank is shaped as a cuboid with a square base of side 10cm. Water runs out through a hole in the base at a rate proportional to the square root of the height, h cm, of water in the tank. At the same time, water is pumped into the tank at a constant rate of 2cm^3 s^-1. Find an expression for dh/dt.

1.) dV/dt = -kV, V = (4/3)Pi.r^3
dV/dt = -8 -> kV = 8. -> (8/k) = (4/3)Pi.15^3 -> k = [8]/[(4/3)Pi.15^3] = 2/(1125Pi)

-> dV/dt = -2V/(1125Pi)

2.) dV/dt = 2 - kSqrt(h)
V = Area Of Cross Section * Length = 10^2.h = 100h
-> dV/dh = 100
-> dh/dt = dh/dV * (dV/dt) = (1/100)*(2 - kSqrt(h)) = (1/50) - (k/100)Sqrt(h)
thanks a lot mates!