next up previous
Next: About this document ... Up: lab_template Previous: lab_template


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 arguement 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 pot of soup taken from the stovetop at $96 \, ^{\circ}\mathrm{C}$ and a jug of apple juice taken from a refrigerator at $4 \, ^{\circ}\mathrm{C}$ are put on a dinner table. The temperature in the dining room is held at $22 \, ^{\circ}\mathrm{C}$. After 10 minutes, the soup temperature is $52 \, ^{\circ}\mathrm{C}$ and the juice temperature is $15 \, ^{\circ}\mathrm{C}$
    1. How much longer will it take the soup to lose its taste quality, i.e., to cool down to the temperature of $32 \, ^{\circ}\mathrm{C}$? What is the juice temperature at this moment?
    2. Plot the graphs $T$ (temperature) versus $t$ (time) for both products. From the graphs, make an estimate when the soup should be eaten if it is supposed to be best at $65 \, ^{\circ}\mathrm{C}$
    Joseph Feribach
  2. Copy every command from the background under the section titled Inverse Functions and paste them into Maple and execute each command. Be sure to shrink your graphs.

  3. Given the functions $\displaystyle f(x) = \frac{x}{x^2+1}$ and $\displaystyle g(x)= x^{\frac{3}{5}}$ (Hint: Use Maple's surd function when entering fractional exponents.)
    Plot the function $f$ over the interval $-10 \leq x \leq 10$ and plot the function $g$ over the same interval, but on a separate graph. Which function is not invertible and why?
    Find the inverse of the invertible function.
    Plot the function and its inverse along with the line $y=x$ on the domain of $-10 \leq x \leq 10$ and $-10 \leq y \leq 10$.
    Show that you have the correct inverse by using the composite definition. (If you come across a simplifying problem type the command assume(x>0).)

next up previous
Next: About this document ... Up: lab_template Previous: lab_template
Dina Solitro