Question

A container in the shape of a cone of height 20cm and radius 5 cm is held vertex downward and filled with water, which then drips out from the vertex at a rate of 5cm³s⁻¹.

a) Write down the value of the rate of change of the volume, dt dV
b) Write an expression for the radius of the cone in terms of the height and hence write an expression for the volume of the cone in terms of the height. (The formula for the volume of a cone with base radius r and height h is 1/3πr²h
c) Find dV/dh and using the formula dh/dV = 1/dv/dh, find dh/dV
d) Find the rate of change of the height of water in the cone when the height of the water is 4cm

A) a) -5, b) r = √(h² - 400), V = 1/3π(h² - 400)h, c) dV/dh = 1/3π(3h - 400/h), dh/dV = 3h - 400, d) -3
B) a) -5, b) r = √(h² - 400), V = 1/3π(h² - 400)h, c) dV/dh = 1/3π(3h - 400/h), dh/dV = 400 - 3h, d) 3
C) a) 5, b) r = √(h² - 400), V = 1/3π(h² - 400)h, c) dV/dh = 1/3π(3h - 400/h), dh/dV = 3h - 400, d) -3
D) a) 5, b) r = √(h² - 400), V = 1/3π(h² - 400)h, c) dV/dh = 1/3π(3h - 400/h), dh/dV = 400 - 3h, d) 3

Answer 1

a) dt dV = -5 cm³s⁻¹, b) r = √(h² - 400), V = 1/3π(h² - 400)h, c) dV/dh = 1/3π(3h - 400/h), dh/dV = 3h - 400, d) When the height of the water is 4cm, the rate of change of the height of water in the cone is -3.

Step-by-step explanation:

a) dt dV = -5 cm³s⁻¹

b) The expression for the radius of the cone in terms of the height is r = √(h² - 400), and the expression for the volume of the cone in terms of the height is V = 1/3π(h² - 400)h.

c) dV/dh = 1/3π(3h - 400/h). Using the formula dh/dV = 1/dv/dh, we can find that dh/dV = 3h - 400.

d) When the height of the water is 4cm, the rate of change of the height of water in the cone is -3.