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. Water flows into a tank having the form of a frustum of a right circular cone. The tank is 4 m tall with an upper radius of 1.5 m and a lower radius of 1 m. When the water in the tank is 2 m deep, the surface rises at a rate of 0.015 m/s. Calculate the discharge of water flowing into the tank in cu.m/s.

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Questioner:marky
Country:Philippines
Category:Calculus
Private:No
Subject:Calculus

Question:
Water flows into a tank having the form of a frustum of a right circular cone. The tank is 4 m tall with an upper radius of 1.5 m and a lower radius of 1 m. When the water in the tank is 2 m deep, the surface rises at a rate of 0.015 m/s. Calculate the discharge of water flowing into the tank in cu.m/s.
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This is a'related rates' problem.  Steps:
I. Draw a diagram -- a good,careful one.  I leave that to you.

II. Choose the variables -- the things that change.  Give them names.

IIA. Write the derivatives of them.

III. Use your knowledge to determine the relationship between them.  

IV. Use implicit differentiation, and at the end, plug in known values.

Now in this case you might conclude that the UNFRUSTRATED cone, with a lower radius of zero, has a height of 8 + 4 = 12 maters.

HOW, you ask?  A frustum is a difference between two cones.  Tour smaller cone has a radius that is 2/3 the larger one.  Therefore, its height is 2/3 the larger one.  So 4 m = 1/3 the larger cone.

SO, we are ready to proceed with this problem:
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Water flows into a tank having the form of a cone. The tank is 12 m tall with a radius of 1.5 m. When the water in the tank is (8 + 2) m deep, the surface rises at a rate of 0.015 m/s. Calculate the discharge of water flowing into the tank in cu.m/s.

or:

Water flows into a tank having the form of a cone. The tank is 12 m tall with a radius of 1.5 m. When the height of water is 10 m, The height increases at a rate of 0.015 m/s. What is the increase in volume at the instant the height (depth) = 10?

Variables:

h = depth of the water.  (at h = 10)
r = radius of the surface of the water.
V = Volume of the water.

Rates:

dh/dt = rate of increase of depth.  GIVEN; = 0.015
dV/dt = rate of increase of volume.  TO BE FOUND.

Relations:

The height of the tank is 8 * the radius.  THEREFORE, so is the height of the water.

h = 8r.  (use it to eliminate 'r')

V = (1/3) pi r^2 h  <<< You would look this up.

Since r = h/8,

V = (1/3) pi (h/8)^2 h

V = (1/192) pi h^3

NOW differentiate implicitly:

dV/dt = (1/192) pi 3h^2 dh/dt

OK, put h = 10, dh/dt = 0.015  and you are in business.  

Calculus

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Paul Klarreich

Expertise

All topics in first-year calculus including infinite series, max-min and related rate problems. Also trigonometry and complex numbers, theory of equations, exponential and logarithmic functions. I can also try (but not guarantee) to answer questions on Analysis -- sequences, limits, continuity.

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I taught all mathematics subjects from elementary algebra to differential equations at a two-year college in New York City for 25 years.

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(See above.)

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