Mathematics, 10.01.2021 04:00 josephmelichar777

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:)

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Answer from: juvie65

$\displaystyle \frac{dh}{dt} = \frac{8}{25 \pi} \ ft/min$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets Parenthesis Exponents Multiplication Division Addition 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: $\displaystyle V = \frac{1}{3} \pi r^2h$

r is radiush 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

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

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

Define: b = dSubstitute: b = 2r

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

Define: b = 2rSubstitute: h = 2r

Now solve for height.

Divide 2 on both sides: h/2 = rRewrite expression: r = h/2

Now find new volume formula.

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

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

Step 3: Differentiate

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

Step 4: Solve for height rate

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

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

Answer from: Quest

d≈7.07in

step-by-step explanation:

Answer from: Quest

the two may have that, even a breakdown is probably offered