The radius of a cone is decreasing at a constant rate of 7 inches per second, and the volume is decreasing at a rate of 948 cubic inches per second. At the instant when the radius of the cone is 99 inches and the volume is 525 cubic inches, what is the rate of change of the height? The volume of a cone can be found with the equation V=\frac{1}{3}\pi r^2 h.V= 3 1 ​ Ο€r 2 h. Round your answer to three decimal places.

Accepted Solution

Answer:Step-by-step explanation:We have volume of cone as[tex]V=\frac{1}{3} \pi r^2 h[/tex]and for a cone always r/h = constantGiven that r' = rate of change of radius = -7 inches/sec(Negative sign because decresing)V' =- 948 in^3/secRadius = 99 inches and volume = 525 inchesHeight at this instant = [tex]\frac{525}{\frac{1}{3} \pi (99)^2} \\=\frac{0.1607}{\pi}[/tex]Let us differentiate the volume equation with respect to t using product rule[tex]V=\frac{1}{3} \pi r^2 h\\V' = \frac{1}{3} \pi[2rhr'+r^2 h']\\-948 = \frac{1}{3} \pi[2(99)(-7)(\frac{0.1607}{\pi})+99^2 h']\\[/tex][tex]-948 = \frac{1}{3} \pi[2(99)(-7)(\frac{0.1607}{\pi})+99^2 h']\\-948 = 33(3.14)(-2.25/3.14 Β + 99 h')\\-9.149=-0.72+99h'\\-8.429 = 99h'\\h' = 0.08514[/tex]Rate of change of height = 0.08514 in/sec