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Consider a tank in the shape of an inverted right circular cone that is leaking water. The dimensions of the conical tank are a height of 7 ft and a radius of 10 ft. How fast does the depth of the water change when the water is 5 ft high if the cone leaks at a rate of 10 cubic feet per minute?



Answer :

xero099

By using the volume formula for a cone and the definition of rate of change, the depth of the cone is changing at a rate of approximately 0.062 feet per minute.

How much is the depth changing in a leaking cone?

The volume of a cone (V), in cubic feet, is represented by the following formula:

V = (π / 3) · R² · h    (1)

Where:

  • R - Radius of the cone, in feet.
  • h - Height of the cone, in feet.

In addition, the radius and the height have following relationship:

h / R = 7 / 10

h = (7 / 10) · R      (2)

h' = (7 / 10) · R'     (3)

The rate of change of the cone is found by means of first derivatives:

V' = (2π / 3) · R · h · R' + (π / 3) · R² · h'

V' = (π / 3) · R · (2 · h · R' + R · h')        (4)

Where:

  • R' - Rate of change of the radius, in feet per minute.
  • h' - Rate of change of the height, in feet per minute.
  • V' - Rate of change of the volume, in cubic feet per minute.

By (2) and (3) in (4):

V' = (π / 3) · (10 / 7) · h · [2 · h · (10 / 7) · h' + (10 / 7) · h · h']

V' = (10π / 21) · h · [(30 / 7) · h · h']

V' = (300π / 147) · h² · h'

If we know that h = 5 ft and V' = 10 ft³ / min, then the rate of change for the depth of the cone is:

10 = (300π / 147) · 5² · h'

h' ≈ 0.062 ft / min

The depth of the cone is changing at a rate of approximately 0.062 feet per minute.

To learn more on rates of change: https://brainly.com/question/11606037

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