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3 years ago

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An accident at an oil drilling platform is causing a circular oil slick. The slick is 0.07 foot thick, and when the radius of th

e slick is 110 feet, the radius is increasing at the rate of 0.9 foot per minute. At what rate (in cubic feet per minute) is oil flowing from the site of the accident? (Round your answer to two decimal places.)

1 answer:

ololo11 [35]3 years ago

7 0

Answer: $43.54 ft^3/min$

Step-by-step explanation:

Given

Thickness of oil slick=0.07 foot

radius of slick=110 ft

$\frac{\mathrm{d} r}{\mathrm{d} t}=0.9 ft/s$

Let V be the volume of oil slick

so, $V=0.07\pi \cdot r^2$

rate of oil flowing is

$\frac{\mathrm{d} V}{\mathrm{d} t}=0.07\times 2\pi \cdot r\frac{\mathrm{d} r}{\mathrm{d} t}$

$\frac{\mathrm{d} V}{\mathrm{d} t}=0.07\times 2\pi \times 110\times 0.9=43.54 ft^3/min$

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$\displaystyle \int {\frac{2x}{3x + 1}} \, dx = \frac{-2(ln|3x + 1| - 3x)}{9} + C$

General Formulas and Concepts:

<u>Algebra I</u>

Terms/Coefficients
Factoring

<u>Algebra II</u>

Polynomial Long Division

<u>Calculus</u>

Differentiation

Derivatives
Derivative Notation

Derivative Property [Multiplied Constant]: $\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)$

Derivative Property [Addition/Subtraction]: $\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]$

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Integration

Integrals
Integration Constant C
Indefinite Integrals

Integration Rule [Reverse Power Rule]: $\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C$

Integration Property [Multiplied Constant]: $\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx$

Integration Property [Addition/Subtraction]: $\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx$

Logarithmic Integration

U-Substitution

Step-by-step explanation:

*Note:

You could use u-solve instead of rewriting the integrand to integrate this integral.

<u>Step 1: Define</u>

<em>Identify</em>

$\displaystyle \int {\frac{2x}{3x + 1}} \, dx$

<u>Step 2: Integrate Pt. 1</u>

[Integrand] Rewrite [Polynomial Long Division (See Attachment)]: $\displaystyle \int {\frac{2x}{3x + 1}} \, dx = \int {\bigg( \frac{2}{3} - \frac{2}{3(3x + 1)} \bigg)} \, dx$
[Integral] Rewrite [Integration Property - Addition/Subtraction]: $\displaystyle \int {\frac{2x}{3x + 1}} \, dx = \int {\frac{2}{3}} \, dx - \int {\frac{2}{3(3x + 1)}} \, dx$
[Integrals] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle \int {\frac{2x}{3x + 1}} \, dx = \frac{2}{3}\int {} \, dx - \frac{2}{3}\int {\frac{1}{3x + 1}} \, dx$
[1st Integral] Reverse Power Rule: $\displaystyle \int {\frac{2x}{3x + 1}} \, dx = \frac{2}{3}x - \frac{2}{3}\int {\frac{1}{3x + 1}} \, dx$

<u>Step 3: Integrate Pt. 2</u>

<em>Identify variables for u-substitution.</em>

Set <em>u</em>: $\displaystyle u = 3x + 1$
[<em>u</em>] Differentiate [Basic Power Rule]: $\displaystyle du = 3 \ dx$

<u>Step 4: Integrate Pt. 3</u>

[Integral] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle \int {\frac{2x}{3x + 1}} \, dx = \frac{2}{3}x - \frac{2}{9}\int {\frac{3}{3x + 1}} \, dx$
[Integral] U-Substitution: $\displaystyle \int {\frac{2x}{3x + 1}} \, dx = \frac{2}{3}x - \frac{2}{9}\int {\frac{1}{u}} \, du$
[Integral] Logarithmic Integration: $\displaystyle \int {\frac{2x}{3x + 1}} \, dx = \frac{2}{3}x - \frac{2}{9}ln|u| + C$
Back-Substitute: $\displaystyle \int {\frac{2x}{3x + 1}} \, dx = \frac{2}{3}x - \frac{2}{9}ln|3x + 1| + C$
Factor: $\displaystyle \int {\frac{2x}{3x + 1}} \, dx = -2 \bigg( \frac{1}{9}ln|3x + 1| - \frac{x}{3} \bigg) + C$
Rewrite: $\displaystyle \int {\frac{2x}{3x + 1}} \, dx = \frac{-2(ln|3x + 1| - 3x)}{9} + C$

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Integration

Book: College Calculus 10e

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3 years ago