Explanation Given information: The given function is, y = 2 x + 1 x 2 + x + 1 Formula used: The formula to calculate the differential of a function h ( x ) = [ g ( x ) ] n by General Power Rule is, h ′ ( x ) = n [ g ( x ) ] n − 1 ⋅ g ′ ( x ) The formula to calculate the differential of a function h ( x ) = [ g ( x ) f ( x ) ] by Quotient Rule is, h ′ ( x ) = f ( x ) g ′ ( x ) − f ′ ( x ) g ( x ) f 2 ( x ) Calculation: The given function is, y = 2 x + 1 x 2 + x + 1 (1) Differentiate equation (1) with respect to x on both sides to evaluate d y d x . d y d x = ( 2 x + 1 ) ′ ( x 2 + x + 1 ) − ( x 2 + x + 1 ) ′ ( 2 x + 1 ) ( x 2 + x + 1 ) 2 </

APPLIED CALCULUS FOR MGRL, LIFE, SOC SCI

10th Edition

ISBN: 9780357465400

Author: Tan

Publisher: CENGAGE L

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Question 1: If a barometer were built using oil (p = 0.92 g/cm³) instead of mercury (p = 13.6 g/cm³), would the column of oil be higher than, lower than, or the same as the column of mercury at 1.00 atm? If the level is different, by what factor? Explain. (5 pts) Solution: A barometer works based on the principle that the pressure exerted by the liquid column balances atmospheric pressure. The pressure is given by: P = pgh Since the atmospheric pressure remains constant (P = 1.00 atm), the height of the liquid column is inversely proportional to its density: Step 1: Given Data PHg hol=hgx Poil • Density of mercury: PHg = 13.6 g/cm³ Density of oil: Poil = 0.92 g/cm³ • Standard height of mercury at 1.00 atm: hμg Step 2: Compute Height of Oil = 760 mm = 0.760 m 13.6 hoil = 0.760 x 0.92 hoil = 0.760 × 14.78 hoil = 11.23 m Step 3: Compare Heights Since oil is less dense than mercury, the column of oil must be much taller than that of mercury. The factor by which it is taller is: Final…

Question 3: A sealed flask at room temperature contains a mixture of neon (Ne) and nitrogen (N2) gases. Ne has a mass of 3.25 g and exerts a pressure of 48.2 torr. . N2 contributes a pressure of 142 torr. • What is the mass of the N2 in the flask? • Atomic mass of Ne = 20.1797 g/mol • Atomic mass of N = 14.0067 g/mol Solution: We will use the Ideal Gas Law to determine the number of moles of each gas and calculate the mass of N2. PV = nRT where: • P = total pressure • V volume of the flask (same for both gases) n = number of moles of gas • R 0.0821 L atm/mol K • T = Room temperature (assume 298 K) Since both gases are in the same flask, their partial pressures correspond to their mole fractions. Step 1: Convert Pressures to Atmospheres 48.2 PNe = 0.0634 atm 760 142 PN2 = = 0.1868 atm 760 Step 2: Determine Moles of Ne nNe = mass molar mass 3.25 nNe 20.1797 nne 0.1611 mol Step 3: Use Partial Pressure Ratio to Find n

Chapter 3 Solutions

APPLIED CALCULUS FOR MGRL, LIFE, SOC SCI

Ch. 3.1 - State the following rules of differentiation in...Ch. 3.1 - Prob. 2CQ Ch. 3.1 - Prob. 3CQ Ch. 3.1 - Prob. 4CQ Ch. 3.1 - Prob. 1E Ch. 3.1 - Prob. 2E Ch. 3.1 - Prob. 3E Ch. 3.1 - Prob. 4E Ch. 3.1 - Prob. 5E Ch. 3.1 - Prob. 6E

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AUTO TRANSMISSIONS .AND FUEL ECONOMY In trying to...Ch. 3.5 - Prob. 32E Ch. 3.5 - Prob. 33E Ch. 3.5 - Prob. 34E Ch. 3.5 - Prob. 35E Ch. 3.5 - Prob. 36E Ch. 3.5 - Prob. 37E Ch. 3.5 - Prob. 38E Ch. 3.5 - Prob. 39E Ch. 3.5 - Prob. 40E Ch. 3.5 - Prob. 41E Ch. 3.5 - Prob. 42E Ch. 3.5 - Prob. 43E Ch. 3.5 - Prob. 44E Ch. 3.5 - Prob. 45E Ch. 3.5 - Prob. 46E Ch. 3.5 - Prob. 47E Ch. 3.5 - Prob. 48E Ch. 3.5 - Prob. 49E Ch. 3.5 - Prob. 50E Ch. 3.5 - Prob. 1TE Ch. 3.5 - Prob. 2TE Ch. 3.5 - Prob. 3TE Ch. 3.5 - Prob. 4TE Ch. 3.5 - In Exercises 1-8, find the value of the second...Ch. 3.5 - Prob. 6TE Ch. 3.5 - Prob. 7TE Ch. 3.5 - Prob. 8TE Ch. 3.6 - Prob. 1CQ Ch. 3.6 - Prob. 2CQ Ch. 3.6 - Prob. 3CQ Ch. 3.6 - Prob. 4CQ Ch. 3.6 - Prob. 1E Ch. 3.6 - Prob. 2E Ch. 3.6 - Prob. 3E Ch. 3.6 - Prob. 4E Ch. 3.6 - Prob. 5E Ch. 3.6 - Prob. 6E Ch. 3.6 - Prob. 7E Ch. 3.6 - Prob. 8E Ch. 3.6 - Prob. 9E Ch. 3.6 - Prob. 10E Ch. 3.6 - Prob. 11E Ch. 3.6 - Prob. 12E Ch. 3.6 - Prob. 13E Ch. 3.6 - Prob. 14E Ch. 3.6 - Prob. 15E Ch. 3.6 - 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Volume of a Cylinder The volume of a...Ch. 3.6 - Prob. 40E Ch. 3.6 - Prob. 41E Ch. 3.6 - Prob. 42E Ch. 3.6 - Prob. 43E Ch. 3.6 - Prob. 44E Ch. 3.6 - Prob. 45E Ch. 3.6 - Prob. 46E Ch. 3.6 - Prob. 47E Ch. 3.6 - PRODUCTION FUNCTION FOR A FURNITURE COMPANY The...Ch. 3.6 - Prob. 49E Ch. 3.6 - Prob. 50E Ch. 3.6 - Prob. 51E Ch. 3.6 - Prob. 52E Ch. 3.6 - Prob. 53E Ch. 3.6 - Prob. 54E Ch. 3.6 - Prob. 55E Ch. 3.6 - Prob. 56E Ch. 3.6 - Prob. 57E Ch. 3.6 - Prob. 58E Ch. 3.6 - Prob. 59E Ch. 3.6 - Prob. 60E Ch. 3.6 - Prob. 61E Ch. 3.6 - Prob. 62E Ch. 3.6 - Prob. 63E Ch. 3.6 - Prob. 64E Ch. 3.6 - Prob. 65E Ch. 3.6 - 66. 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The derivative of given function.

Solution Summary: The author calculates the derivative of a given function using the following formula: cdy

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Chapter 3 Solutions