Question

when a hemispherical bowl with a radius of 1 ft is filled with water to a depth of x inches, the volume of water in the bowl in cubic inches is V=(Pi/3)(36x^2-x^3). The water flows out a hole in the bottom of the bowl at a rate of 36Pi in^3/s. How fast is the height of the water decreasing when x=6 inch? when a hemispherical bowl with a radius of 1 ft is filled with water to a depth of x inches, the volume of water in the bowl in cubic inches is V=(Pi/3)(36x^2-x^3). The water flows out a hole in the bottom of the bowl at a rate of 36Pi in^3/s. How fast is the height of the water decreasing when x=6 inch? @Mathematics

Answer 1

you want
\[x'\] so put
\[V'=\frac{\pi}{3}(72xx'-3x^2x')\] you are told
\[V'=36\] so replace \[V'\] by 36 and x by 6 and solve for x'

Answer 2

so i got x' = 1/3

Answer 3

well - 1/3 decreasing