Rotating Disk

A disk is rotating about a vertical axis in a liquid.  If the retardation due to friction of the liquid is proportional to the angular velocity, \omega, find \omega after t seconds if the initial angular velocity was \omega_o.

If the disk is rotating very rapidly, the retardation is proportional to \omega^{2}.  Find \omega after t seconds if the initial amgular velocity was \omega_o.

Source: March, Herman W., and Wolff, Henry C. (1917). Calculus. New York: McGraw-Hill. (Questions 4 & 5, p. 158)

Population of Bacteria

The number of bacteria per cubic centimeter of culture increases under proper conditions at a rate proportional to the number present.  Find an expression for the number present at the end of time t.  Find the time required for the number per cubic centimeter to increase from b_{1} to b_{2}.  Does this time depend on the number present at the time t=0?

Source: March, Herman W., and Wolff, Henry C. (1917). Calculus. New York: McGraw-Hill. (Question 3, p. 158)

 

Retardation of Boat

Assuming that the retardation of a boat moving in still water is proportional to the velocity, find the distanced pass over in time t after the engine was shut off, if the boat was moving at the rate of 7 miles per hour at that time.

Answer:

    \[s=\frac{7}{k}\left(1-e^{-kt}\right)\]

Source: March, Herman W., and Wolff, Henry C. (1917). Calculus. New York: McGraw-Hill. (Question 2, p. 158)

Velocity of Chemical Reactions

A law for the velocity of chemical reactions states that the amount of chemical change per unit of time is proportional to the mass of changing substances present in the system.  The rate at which the change takes place is proportional to the mass of the substance still unchanged.  If q denotes the original mass, find an expression for the mass remaining unchanged after a time t has elapsed.

Source: March, Herman W., and Wolff, Henry C. (1917). Calculus. New York: McGraw-Hill. (Question 1, p. 158)

Differential Equation (Application)

In a ‘simple’ epidemic, the rate at which the number of susceptible, s, in a population of total size n+1, become infected is given by \displaystyle \frac{ds}{dt}=-cs(n+1-s), where c is a positive constant.

Show that, if an epidemic is ‘simple’, the number of new cases of the disease recorded per unit time will be given by:

    \[\frac{cn\left(n+1\right)^{2}\,e^{\left(n+1\right)ct}}{\left[ n + e^{\left(n+1\right)ct}\right]^{2}}\]

Source: Rosenberg, R.L. (1984). Elementary Calculus: Course Notes. Ottawa, Canada: Holt, Rinehart and Winston. (Exercise 44: Question 7, p. 132)

Differential Equation (Application)

The rate at which enzyme A is converted to enzyme B is governed by the differential equation \displaystyle \frac{dq}{dt}=k\left(a-q\right)\left(b+q\right) where q denotes the concentration of enzyme B produced after t hours.

Here a and b are the initial concentrations of enzyme A and enzyme B respectively, while k is a positive constant.  If the initial concentrations, a and b, respectively, are 100 and 5 moles/litre, find the equation giving the concentration of enzyme B produced in the reaction after t hours when the concentration of enzyme B produced in the first hour is 10 moles/litre.

Source: Rosenberg, R.L. (1984). Elementary Calculus: Course Notes. Ottawa, Canada: Holt, Rinehart and Winston. (Exercise 44: Question 6, p. 132)

Differential Equation (Application)

A population of mice, initially numbering 30, is kept in conditions that can support a population size of 120.  The rate of increase in the population of size N is assumed to be given by the equation \displaystyle \frac{dN}{dt}=cN\left(K-N\right), where c and K are positive constants.  Find the formula giving the number of mice t months later, if, after one month, the population numbers 80.

Source: Rosenberg, R.L. (1984). Elementary Calculus: Course Notes. Ottawa, Canada: Holt, Rinehart and Winston. (Exercise 44: Question 5, p. 132)

Differential Equation (Application)

The growth of a cell depends on the flow of nutrients through its surface.  Let W(t) be the weight of the cell at time t.  Assume that for a limited time the growth rate \displaystyle {dW}{dt} is proportional to W^{\frac{2}{3}}.  (If the density remains constant, then W is proportional to D^{3}, where D is the diameter of the cell, and the surface area is proportional to D^{2}, or equivalently W^{\frac{2}{3}).

Hence \displaystyle \frac{dW}{dt}=kW^{\frac{2}{3}} for some positive constant k.  Find the general solution to this differential equation.

Source: Rosenberg, R.L. (1984). Elementary Calculus: Course Notes. Ottawa, Canada: Holt, Rinehart and Winston. (Exercise 44: Question 4, p. 132)