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The exponential and logarithmic functions

Exponential growth

The simple model for growth is exponential growth, where it is assumed that $y'(t)$ is proportional to $y$. That is,

\begin{displaymath}\frac{dy}{dt} = ky\quad\hbox{where $k$\ is a positive constant.}\end{displaymath}

Separating the variables and integrating, we have

\begin{displaymath}\int\frac{dy}{y} = \int k\,dt \end{displaymath}

so that

\ln \mid y \mid = kt + C

In the case of exponential growth, we can drop the absolute value signs around $y$, because $y$ will always be a positive quantity. Solving for $y$, we obtain

\begin{displaymath}\mid y \mid = e^{kt + C} \end{displaymath}

which we may write in the form $y = Ae^{kt}$, where $A$ is an arbitrary positive constant. The same formula is used for exponential decay, except the decay constant $k$ is negative.

Newton's law of cooling

What is usually called Newton's law of cooling is a simple model for the change in temperature of an object that is in contact with an environment at a different temperature. It says that the rate of change of the temperature of an object is proportional to the difference between the object's temperature and the temperature of the environment. Mathematically, this can be expressed as the differential equation

\begin{displaymath}\frac{dT}{dt} = -k \left( T-T_{\mathrm{out}} \right) \end{displaymath}

where $k$ is the constant of proportionality and $T_{\mathrm{out}}$ is the temperature of the environment. Using a technique called separation of variables it isn't hard to derive the solution

\begin{displaymath}T(t) = T_{\mathrm{out}} + \left( T_0 - T_{\mathrm{out}} \right)
e^{-kt} \end{displaymath}

where $T_0$ is the temperature of the object at $t=0$.

Inverse Functions

Consider the functions $f,g$ defined by


To be able to get an inverse the function must be one-to-one. You can plot the functions to get a hint as to whether they are invertible or not.
> f:=x->exp(x)+exp(-x);
> plot(f(x),x=-5..5);
> g:=x->exp(x)-exp(-x);
> plot(g(x),x=-5..5);
Both satisfy the vertical-line test but $f(x)$ is not invertible since it does not satisfy the horizontal-line test. Indeed $f$ is not one-to-one, for instance $f(0.5)=f(-0.5)$. From the plot it seems that the function $g$ is one-to-one. In order to determine its inverse we solve for x.
> solve(g(x)=y,x);
We observe that one of the solutions is not defined since the argument of the logarithm can only be positive. Thus:


> ginv:=y->ln(y/2+sqrt(y^2+4)/2);
Let's look at the plot along with the line $y=x$ to see if our functions seem to make sense.
> plot({x,g(x),ginv(x)},x=-20..20,y=-20..20,scaling=constrained);
Let's check that we have computed the right inverse. By definition the composot ion of the functions should be the line $y=x$ since an inverse is the reflectio n about this line.

(g\circ g^{-1})(y)=y=(g^{-1}\circ g)(x)=x

> g(ginv(y));
> simplify(%);
> ginv(g(x));
> simplify(%);
We are having difficulty getting $x$ for the last composition. Think about what issue the computer has in dealing with this simplification as you will come across this in the exercises.


  1. A 5 lb. roast having initial temperature of $50\,^{\circ}\mathrm{F}$ is put into a pre-heated oven at $375\,^{\circ}\mathrm{F}$. How much longer will it take for the roast to be medium rare (a temperature of $150\,^{\circ}\mathrm{F}$ if the roast is $125 \,^{\circ}\mathrm{F}$ after 75 minutes.

  2. The loudness of sound is measured in decibels in honor of Alexander Graham Bell (1847-1922), inventor of the telephone. If the variation in pressure is $P$ pounds per square inch, then the loudness in decibels is

    \begin{displaymath}L = 20 \log_{10} (121.3P) \end{displaymath}

    Find how many times greater is the variation in pressure caused by a rock band at 115 decibels than a soft rock concert at 85 decibels.

  3. For each function below, use the background example under the section titled Inverse Functions to help answer each of the following. Be sure to shrink your graphs.

    \begin{displaymath}f(x) = \frac{x}{x^2+1}\end{displaymath}

    \begin{displaymath}g(x)= x^{\frac{3}{5}} \end{displaymath}

    \begin{displaymath}h(x) = \frac{\ln(x-2)}{x} \end{displaymath}

    \begin{displaymath}k(x)=\arctan(\ln(x)) \end{displaymath}

    Plot each funtion over a range that gives a good representation of the function. Which functions are not invertible and why?
    Find the inverse of the invertible function(s).
    For at least one of the invertivle functions, plot the function and its inverse along with the line $y=x$. Use a domain that shows that the functions are inverses.
    Show that you have the correct inverse(s) in part $B$ by using the composite definition. (If you come across a simplifying problem type the command assume(x>0).)

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Next: About this document ... Up: lab_template Previous: lab_template
Dina J. Solitro-Rassias