The radius of a right circular cone is increasing at a rate of 5 inches per second and its height is decreasing at a rate of 3 inches per second. At what rate is the volume of the cone changing when the radius is 50 inches and the height is 20 inches

1700π in³/s

Step-by-step explanation:

Volume of a cone is expressed as

V = 1/3πr²h

Given

dr/dt = 5in/s

dh/dt = 3in/s

radius r = 50in

h = 20in

Required

dV/dt = ?

dV/dt = dV/dr * dr/dt + dV/dh*dh/dt

dV/dr = 1/3πh (5) + 1/3πr²(2)

dV/dt = 1/3π(20)(5)+1/3 π(50)²(2)

dV/dt =  100π/3 + 5000π/3

dV/dt = 5100π/3

dV/dt = 1700π in³/s

the direct variation says that:

then the equation is of the form is given by:

            where k is the constant variation.

from the given table:

consider any value of x and m(x)=y

x = 2 and m(x)=y = 3

substitute these value in [1]; we have;

divide both sides by 2 we get;

⇒

check:

put x = -1 and m(x) = -1.5

substitute in the equation:

-1.5 = -1.5                   true

therefore, the function m(x) represents a direct variation

graph of this function as shown below: