sand is dumped such that the shape of the sandpile

questioned by administrator @ 12/07/2022 in Mathematics viewed by 286 People

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:)

Answered by administrator @ 12/07/2022

Answer:

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

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 radius
h 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 = d
Substitute: b = 2r

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

Define: b = 2r
Substitute: h = 2r

Now solve for height.

Divide 2 on both sides: h/2 = r
Rewrite 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.

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