To determine To calculate: The number of infected populations in 25 days into the epidemic. Explanation Given: Provided that epidemic in a community of 50 , 000 starts with 80 people infected and that 15 days later, 640 are infected. Concept used: Growth rate of sale gives us the idea that growth is with respect to time, which is proportional to the difference of amount of maximum population and the population at the current time. This makes the problem of an ordinary differential equation of variable separable form. Calculated: Let t be the time (in days) and y the number of individuals infected at time t . At any moment there are ( N − y ) y possible contacts between an uninfected individual and an infected individual. Thus, it is reasonable to assume that the rate of spread of the disease satisfies the bellow logestic equation d y d t = k ( 1 − y N ) y solve above diffrential equation by variable seperable, N ( N − y ) y d y = k d t ( 1 y + 1 N − y ) d y = k d t Take integral on both sides in above equation, ∫ 1 y d y + ∫ 1 N − y d y = k ∫ d t ln y − ln ( N − y ) = k t + c ln y N − y = k t + c y N − y = e k t + c [ Taking exponentials on both sides ] So, the required diffrential equation is y N − y = e k t + c (1) Now put y = 80 , N = 50 , 000 and t = 0 in above equation, 80 50 , 000 − 80 = e 0 + c 80 49920 = e c 0.00160 = e c Take natural logarithm both the side in above equation, ln ( 0

Calculus with Applications (11th Edition)

11th Edition

ISBN: 9780321979421

Author: Margaret L. Lial, Raymond N. Greenwell, Nathan P. Ritchey

Publisher: PEARSON

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Chapter 10.4, Problem 3YT

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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 10 Solutions

Calculus with Applications (11th Edition)

Ch. 10.1 - Find all solutions of the differential equation . Ch. 10.1 - Prob. 2YT Ch. 10.1 - Prob. 3YT Ch. 10.1 - Prob. 4YT Ch. 10.1 - Prob. 1WE Ch. 10.1 - Prob. 2WE Ch. 10.1 - Prob. 3WE Ch. 10.1 - Prob. 4WE Ch. 10.1 - Prob. 5WE Ch. 10.1 - Find the general solution for each differential...

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