Mathematics, 03.01.2021 01:30 sainijasdeep27

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.

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Answer from: iixyloa

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

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets Parenthesis Exponents Multiplication Division Addition 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.

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

Step 3: Differentiate

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

Step 4: Find Height Rate

Find dh/dt.

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

Answer from: Quest

yes

step-by-step explanation:

Answer from: Quest

a 360 degree rotation

step-by-step explanation:

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