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5. Sand is dumped off a conveyor belt into a pile at the rate of 2 cubic feet per minute. The sand pile is shaped like a cone whose height and base diameter are always equal. At what rate is the height of the pile growing when the pile is 5 feet high? (The volume of a cone is fr2h where r is the radius of the base and h is the height:)​

Respuesta :

Space

Answer:

[tex]\displaystyle \frac{dh}{dt} = \frac{8}{25 \pi} \ ft/min[/tex]

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

Equality Properties

  • Multiplication Property of Equality
  • Division Property of Equality
  • Addition Property of Equality
  • Subtract Property of Equality

Geometry

Volume of a Cone: [tex]\displaystyle V = \frac{1}{3} \pi r^2h[/tex]

  • r is radius
  • h is height

Diameter: d = 2r

Calculus

Derivatives

Derivative Notation

Differentiating with respect to time

Basic Power Rule:

  • f(x) = cxⁿ
  • f’(x) = c·nxⁿ⁻¹

Step-by-step explanation:

Step 1: Define

[tex]\displaystyle \frac{dV}{dt} = 2 \ ft^3/min\\b = h\\h = 5 \ ft[/tex]

Step 2: Rewrite Volume Formula

We need to rewrite the cone volume formula in terms of height h only.

Base b = diameter d of the circular base

  1. Define:                    b = d
  2. Substitute:              b = 2r

We are given that the base of the cone is the same as the height.

  1. Define:                    b = 2r
  2. Substitute:              h = 2r

Now solve for height.

  1. Divide 2 on both sides:                    h/2 = r
  2. Rewrite expression:                          r = h/2

Now find new volume formula.

  1. Define [VC]:                    [tex]\displaystyle V = \frac{1}{3} \pi r^2h[/tex]
  2. Substitute:                      [tex]\displaystyle V = \frac{1}{3} \pi (\frac{h}{2})^2h[/tex]
  3. Exponents:                     [tex]\displaystyle V = \frac{1}{3} \pi (\frac{h^2}{4})h[/tex]
  4. Multiply:                          [tex]\displaystyle V = \frac{1}{12} \pi h^3[/tex]

We now have the same volume formula in terms of height h only.

Step 3: Differentiate

  1. Basic Power Rule:                                                                                           [tex]\displaystyle \frac{dV}{dt} = \frac{1}{12} \pi \cdot 3 \cdot h^{3-1} \frac{dh}{dt}[/tex]
  2. Simplify:                                                                                                           [tex]\displaystyle \frac{dV}{dt} = \frac{1}{4} \pi h^{2} \frac{dh}{dt}[/tex]

Step 4: Solve for height rate

  1. Substitute:                     [tex]\displaystyle 2 \ ft^3/min = \frac{1}{4} \pi (5 \ ft)^{2} \frac{dh}{dt}[/tex]
  2. Isolate h rate:                [tex]\displaystyle \frac{2 \ ft^3/min}{\frac{1}{4} \pi (5 \ ft)^{2}} = \frac{dh}{dt}[/tex]
  3. Exponents:                    [tex]\displaystyle \frac{2 \ ft^3/min}{\frac{1}{4} \pi (25 \ ft^2)} = \frac{dh}{dt}[/tex]
  4. Multiply:                         [tex]\displaystyle \frac{2 \ ft^3/min}{\frac{25}{4} \pi \ ft^2} = \frac{dh}{dt}[/tex]
  5. Divide:                           [tex]\displaystyle \frac{8}{25 \pi} \ ft/min = \frac{dh}{dt}[/tex]
  6. Rewrite:                         [tex]\displaystyle \frac{dh}{dt} = \frac{8}{25 \pi} \ ft/min[/tex]

Here this tells us that the rate at which the height is moving at a rate of 0.101859 feet per minute.

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