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2. Initially, there are 500 strands of bacteria in a culture. If the growth rate of the population of the bacteria in the culture is proportional to the population itself, the strands of the bacteria after 5 minutes are 750 strands. Determine the growth rate factor of the population growth and determine when will there be 1500 strands of the bacteria in the culture.

2. Initially, there are 500 strands of bacteria in a culture. If the growth rate of the population of the bacteria in the culture is proportional to the population itself, the strands of the bacteria after 5 minutes are 750 strands. Determine the growth rate factor of the population growth and determine when will there be 1500 strands of the bacteria in the culture.

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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NATURAL GROWTH AND DECAY These problems involve time rate of change of variable that is proportional to the same variable being referred to. If x represents this variable, then the differential equation is dx = kx dt Solving this equation dx k dt In x = kt + c x = ekt+c x = cekt At initial condition, time t = 0, x = xo Xo = cek(0) c = Xo Then the general solution or formula for this type of problems is x = x,ekt The following summarizes the formulas and their variables for each problem Natural Population Growth Continuous Compounding Interest P = Poekt A = Aoet P – population at any time, t A – amount at any time, t Po – population at t = 0, initial condition Ao – amount deposited at t = 0 k – growth rate factor, k > 0 i- annual rate of continuously compounded interest Radioactive Decay Half-Life of a Radioactive Material x = x,e-kt x = x,e-kt x – amount of unconverted substance present Att = 1 (half-life), x =÷xo 0, initial condition In 2 T=- k Xo - amount present at t = k – decay constant, k > 0
Transcribed Image Text:NATURAL GROWTH AND DECAY These problems involve time rate of change of variable that is proportional to the same variable being referred to. If x represents this variable, then the differential equation is dx = kx dt Solving this equation dx k dt In x = kt + c x = ekt+c x = cekt At initial condition, time t = 0, x = xo Xo = cek(0) c = Xo Then the general solution or formula for this type of problems is x = x,ekt The following summarizes the formulas and their variables for each problem Natural Population Growth Continuous Compounding Interest P = Poekt A = Aoet P – population at any time, t A – amount at any time, t Po – population at t = 0, initial condition Ao – amount deposited at t = 0 k – growth rate factor, k > 0 i- annual rate of continuously compounded interest Radioactive Decay Half-Life of a Radioactive Material x = x,e-kt x = x,e-kt x – amount of unconverted substance present Att = 1 (half-life), x =÷xo 0, initial condition In 2 T=- k Xo - amount present at t = k – decay constant, k > 0
2. Initially, there are 500 strands of bacteria in a culture. If the growth rate of the population of the bacteria in the culture is proportional to the population itself, the strands of the bacteria after 5 minutes are 750 strands. Determine the growth rate factor of the population growth and determine when will there be 1500 strands of the bacteria in the culture.
Transcribed Image Text:2. Initially, there are 500 strands of bacteria in a culture. If the growth rate of the population of the bacteria in the culture is proportional to the population itself, the strands of the bacteria after 5 minutes are 750 strands. Determine the growth rate factor of the population growth and determine when will there be 1500 strands of the bacteria in the culture.
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