   Chapter 2.8, Problem 36E

Chapter
Section
Textbook Problem

# A faucet is filling a hemispherical basin of diameter 60 cm with water at a rate of 2 L/min. Find the rate at which the water is rising in the basin when it is half full. [Use the following facts: 1 L is 1000 c m 3 . The volume of the portion of a sphere with radius r from the bottom to a height h is V = π ( r h 2 − 1 3 h 3 ) , as we will show in Chapter 5.]

To determine

To find:

The rate at which the water is rising in the basin when it is half full.

Explanation

1. Concept:

We use volume function and differentiate it to find the rate of change.

2. Formulae:

i. The volume of the portion of a sphere with radius r from the bottom to a height h is, V= π(rh2-13h3)

ii. Volume of hemisphere, Vh=23*π*r3

iii. Power rule of differentiation: ddxxn=n*xn-1

iv. Chain rule of differentiation: ddtfx=ddxfx*dxdt

3. Given:

Diameter of hemispherical basin is 60 cm.

dvdt=2 L/min=2000 cm3/min

4. Calculations:

V= π(rh2-13h3)

Differentiate both sides with respect to t.

dVdt= π 2rh-h2dhdt

Diameter is 60 cm

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