A paper cup has the shape of a cone with height 10 cm and radius 3 cm (at the top. if water is poured into the cup at a rate of 2cm3/s, how fast is the water level rising when the water is 5 cm deep?
Accepted Solution
A:
Let h: height of the water r: radius of the circular top of the water V: the volume of water in the cup. We have: r/h = 3/10 So, r = (3/10)*h the volume of a cone is: V = (1/3)*π*r^2*h Rewriting: V (t) = (1/3)*π*((3/10)*h(t))^2*h(t) V (t) =(3π/100)*h(t)^3 Using implicit differentiation: V'(t) = (9π/100)*h(t)^2*h'(t) Clearing h'(t) h'(t)=V'(t)/((9π/100)*h(t)^2) the rate of change of volume is V'(t) = 2 cm3/s when h(t) = 5 cm. substituting: h'(t) = 8/(9π) cm/s Answer: the water level is rising at a rate of: h'(t) = 8/(9π) cm/s