   Chapter 2.P, Problem 27P

Chapter
Section
Textbook Problem

# A container in the shape of an inverted cone has height 16 cm and radius 5 cm at the top. It is partially filled with a liquid that oozes through the sides at a rate proportional to the area of the container that is in contact with the liquid. (The surface area of a cone is π r l , where r is the radius and l is the slant height.) If we pour the liquid into the container at a rate of 2cm3/min, then the height of the liquid decreases at a rate of 0.3 cm/min when the height is 10 cm. If our goal is to keep the liquid at a constant height of 10 cm, at what rate should we pour the liquid into the container?

To determine

To find: at what rate should we pour the liquid into the container

Explanation

1) Concept:

Differentiating the volume of cone and equating it with difference of what is being added and what is oozing out and solving further to maintain height to 10 cm

2) Formula:

i) Volume of cone:

V=12πr2h

ii) Surface area of cone:

S=πrl

3) Given:

i. h=16 cm

ii. r=15 cm

iii. When height is 10 cm, dhdt= -0.3 cm/min

iv. Liquid oozing out = kπrl

v. Liquid added = 2 cm3/min

4) Calculation:

By similar triangle,

r5=h16

r=516h

The volume of inner cone is,

V=12πr2h=12π516h2h=25π768h3

Differentiate with respect to t,

dVdt=ddt25π768h3

By using constant multiple rule,

dVdt=25π768ddt(h3)

By using power rule and chain rule,

dVdt=25π7683h2dhdt

dVdt=25π256h2dhdt

Now the rate of change of the volume is also equal to the difference of what is being added and what is oozing out

Thus,

dVdt=2-kπrl

By equating both equations,

25π256h2dhdt=2-kπrl

Since, dhdt= - 0

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