Question

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.

Answer 1

`\displaystyle (dh)/(dt) = (1)/(20 \pi) \ cm/s`

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:
  `\displaystyle V = (1)/(3) \pi r^2h`

Calculus

Derivatives

Derivative Notation

Differentiating with respect to time

Basic Power Rule:

• f(x) = cxⁿ
• f’(x) = c·nxⁿ⁻¹

Explanation:

Step 1: Define

`\displaystyle (r)/(h) = (2)/(3) \\(dV)/(dt) = 5 \ cm^3/s\\h = 15 \ cm`

Step 2: Rewrite Cone Volume Formula

Find the volume of the cone with respect to height.

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

Step 3: Differentiate

1. Basic Power Rule:
   `\displaystyle (dV)/(dt) = (4)/(27) \pi \cdot 3 \cdot h^(3-1) \cdot (dh)/(dt)`
2. Simplify:
   `\displaystyle (dV)/(dt) = (4)/(9) \pi h^(2) (dh)/(dt)`

Step 4: Find Height Rate

Find dh/dt.

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