Let v be the volume of water inside the conical funnel
h be the depth of the water
and r be the radius of the water surface
By the property of similar triangles,
r = 10h/15 = 2h/3 --- (*)
i.e. when h = 10cm, r = 10*10/15 = 20/3 cm
Then v = πr^2h/3 = 4πh^3/27--- (**)
a) Differentiate both sides of (**) with respect to t, we have
dv/dt = (4πh^2/9)*dh/dt
Subsitute dv/dt = -5cm^3/s, h = 10cm, we have
-5 = (4π*100/9)*dh/dt
⇒ dh/dt = -9π*5/400 = -9π/80 cm/s
b) Differentiate both sides of (*) with respect to t, we have
dr/dt = 2(dh/dt)/3
Subsitute dh/dt = 9π/80 cm/s, we have
⇒ dr/dt = 2(-9π/80)/3 = -3π/40 cm/s
