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

Answer:

$\displaystyle \frac{dh}{dt} = \frac{1}{20 \pi} \ cm/s$

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: $\displaystyle V = \frac{1}{3} \pi r^2h$

Calculus

Derivatives

Derivative Notation

Differentiating with respect to time

Basic Power Rule:

• f(x) = cxⁿ
• f’(x) = c·nxⁿ⁻¹

Step-by-step explanation:

Step 1: Define

$\displaystyle \frac{r}{h} = \frac{2}{3} \\\frac{dV}{dt} = 5 \ cm^3/s\\h = 15 \ cm$

Step 2: Rewrite Cone Volume Formula

Find the volume of the cone with respect to height.

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

Step 3: Differentiate

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

Step 4: Find Height Rate

Find dh/dt.

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