when the water is 5cm deep, the radius of the surface will be 1/2 cm.
V = 1/3 πr^2 h = 1/2 πr^2 (10r) = 5πr^3
dV/dt = 15πr^2 dr/dt
1/10 = πr(1/2)^2 dr/dt
dr/dt = 2/(5π)
A = πr^2
dA/dt = 2πr dr/dt = 2πr * 2/(5π) = 4/5 cm^2/s
. A container in the shape of a right circular cone of height 10cm and base radius 1cm is
catching water from a tap leaking at the rate of 0.1cm3
s
-1
. Find the rate at which the
surface area of water is increasing when the water is half-way up the cone
2 answers
how did 1/3 turn into 1/2?