Question

This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive any points for the skipped part, and you will not be able to come back to the skipped part. Tutorial Exercise The radius of a right circular cone is increasing at a rate of 1.7 in/s while its height is decreasing at a rate of 2.2 in/s. At what rate is the volume of the cone changing when the radius is 185 in. and the height is 134 in.? Part 1 of 4 The volume of a cone with base radius r and height h is given by V = 1 3 πr2h. To find the rate of change of the volume, we need to find dV dt . By the Chain Rule, we know that dV dt = ∂V ∂ Incorrect: Your answer is incorrect. dr dt + ∂V ∂ Incorrect: Your answer is incorrect. dh dt .

Answer 1

Answer:

Step-by-step explanation:

The volume of a cone is expressed as shown;

V = πr²h/3

The rate of change of volume is expressed as;

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

dV/dr is the rate of change of volume with respect to radius

∂V/dh is the rate of change of volume with respect to height

dr/dt is the rate of change in radius = 1.7 in/s

dh/dt is the rate of change in height = 2.2 in/s

∂V/∂r = 2πrh/3

if r = 185 and h = 134

∂V/∂r = 2π(185)(134)/3

∂V/∂r = 51, 893.73 in²

Also  ∂V/∂h =  πr²/3

if h = 134 in

 ∂V/∂h  =  π(134)²/3

∂V/∂h  =  π(134)²/3

∂V/∂h = 18,793.95in²

Substitute the given values into the differential equation above

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

dV/dt =  51, 893.73 *(1.7) + 18,793.95 *(2.2)

dV/dt =  88,219.341 + 41,346.68

dV/dt = 129,566.021 in³/s

Hence the volume of the cone is changing at 129,566.021 in³/s