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Beads are dropped to create a conical pile such that the ratio of its radius to the height of the pile is constant at 2:3 and the volume is increasing at a rate of 5 cm^3/s. Find the rate of change of height at h = 15cm.

Sagot :

Space

Answer:

[tex]\displaystyle \frac{dh}{dt} = \frac{1}{20 \pi} \ cm/s[/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

Algebra I

  • Equality Properties

Geometry

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

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{r}{h} = \frac{2}{3} \\\frac{dV}{dt} = 5 \ cm^3/s\\h = 15 \ cm[/tex]

Step 2: Rewrite Cone Volume Formula

Find the volume of the cone with respect to height.

  1. Define ratio:                         [tex]\displaystyle \frac{r}{h} = \frac{2}{3}[/tex]
  2. Isolate r:                               [tex]\displaystyle r = \frac{2}{3} h[/tex]
  3. Substitute in r [VC]:             [tex]\displaystyle V = \frac{1}{3} \pi (\frac{2}{3}h)^2h[/tex]
  4. Exponents:                          [tex]\displaystyle V = \frac{1}{3} \pi (\frac{4}{9}h^2)h[/tex]
  5. Multiply:                               [tex]\displaystyle V = \frac{4}{27} \pi h^3[/tex]

Step 3: Differentiate

  1. Basic Power Rule:                                                                                          [tex]\displaystyle \frac{dV}{dt} = \frac{4}{27} \pi \cdot 3 \cdot h^{3-1} \cdot \frac{dh}{dt}[/tex]
  2. Simplify:                                                                                                           [tex]\displaystyle \frac{dV}{dt} = \frac{4}{9} \pi h^{2} \frac{dh}{dt}[/tex]

Step 4: Find Height Rate

Find dh/dt.

  1. Substitute in known variables:                                                                      [tex]\displaystyle 5 \ cm^3/s = \frac{4}{9} \pi (15 \ cm)^{2} \frac{dh}{dt}[/tex]
  2. Isolate dh/dt:                                                                                                   [tex]\displaystyle \frac{5 \ cm^3/s}{\frac{4}{9} \pi (15 \ cm)^{2} } = \frac{dh}{dt}[/tex]
  3. Rewrite:                                                                                                           [tex]\displaystyle \frac{dh}{dt} = \frac{5 \ cm^3/s}{\frac{4}{9} \pi (15 \ cm)^{2} }[/tex]
  4. Evaluate Exponents:                                                                                      [tex]\displaystyle \frac{dh}{dt} = \frac{5 \ cm^3/s}{\frac{4}{9} \pi (225 \ cm^2) }[/tex]
  5. Evaluate Multiplication:                                                                                  [tex]\displaystyle \frac{dh}{dt} = \frac{5 \ cm^3/s}{100 \pi cm^2 }[/tex]
  6. Simplify:                                                                                                          [tex]\displaystyle \frac{dh}{dt} = \frac{1}{20 \pi} \ cm/s[/tex]
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