1. A rate of change situation.

One way of changing the freezing point of some liquid (call it A) is to add another liquid (call it B). If we have a mixture that is 95% A and 5% B, then we say that the mole fraction of A is 0.95. Let T(m) be the freezing point (measured in degrees C) of a mixture of A and B for which the mole fraction of A is m. So we have a function . For certain substances A and B, chemists model this freezing point function by

You can see from this model that the mixture goes through some sort of phase transition when m =0.76.  Use piecewise to define T in Maple, and then reproduce this plot. Use help to review how to give a title to a graph, and how to label the vertical axis.

> T:=m->piecewise(m<.76,5*m^4-23.6*m^3-146*m^2+129*m-17,45.056*(m-.76)^(1/3)-11.98132480);

> plot(T(m),m=0..1,labels=[m,`Degrees C`], title=`Freezing point a function of the mole fraction of A`);

(a) Compute the average rate of change of T with respect to m between m=0 and at m=0.001 (so we take )  Now use D to find the instantaneous rate of change when m=0.

> D

(b) Next compute the average rate when m=0.4, again taking .  Then use D to find the instantaneous rate of change when m=0.4.

> D

(c) Next compute the average rate when m=0.6, again taking .  Also compute the average rate when .  Then use D to find the instantaneous rate of change when m=0.6.

(d) Next compute the average rate when m=0.76, again taking .  Also compute the average rate when Then try to use D to find the instantaneous rate of change when m=0.76.

(e) Finally plot the instantaneous rate of change of the freezing point with respect to m.  Use the syntax plot(f(m),m=0..1, c..d, discont=true) where you need to experiment to find decent values of c and d .  Title your graph, and label the vertical axis.

The left-hand and right-hand derivatives of a function are defined by (D-) and (D+) respectively.

Submission:

What are the left-hand and right-hand derivatives for the freezing point function T at the phase transition? Write in sentences, and explain your reasoning.

Submission worksheet:

In a certain experiment, a bactericide was added to a nutrient broth in which bacteria were growing. The population of bacteria continued to increase for a while, and then began to decline. Suppose that the population b of bacteria is given by the following function of the time t (in hours).

> b:=t->10^6+10^4*t-10^3*t^2;

Let us first find the rate db/dt, at which the population of bacteria was changing.

> D(b)(t);

The point in time at which the rate of change of the bacteria population came to a stop is found by solving db/dt=0. We can do this easily by hand, but illustrate how to use Maple to do it.

> solve (D(b)(t)=0,t);

The solution means that at t=5 hours, the rate of change of the population was zero. Afterwards the population began to decrease.  

Let us graph the population curve to understand it better. In order to plot, it makes sense to choose a domain on which the value of the function is not negative, because the physical model makes no sense when the value is negative. So let us solve for when .

> solve(b(t)=0,t);

These exact solutions are not so useful in graphing. Let us approximate them.

> evalf(5-5*sqrt(41)),evalf( 5+5*sqrt(41));

Supposing that we added the bactericide at time t=0, a reasonable domain to choose for our plot is 0..40.

> plot(b,0..40);

Just from this picture alone, it is clear that the rate of change of the population is the largest value (but negative) at the time when the population of bacteria reaches zero. So the rate of change increases after the population levels off. We could also have plotted the derivative D(b) on the same plot, and looked for its extreme values.

> plot([b,D(b)],0..40,color=[blue,red]);

The derivative reaches its maximum positive value when t=0, and decreases from then on, reaching its minimum value when t is about 37 hours. Note that when D(b)(t) is positive, the population of bacteria is increasing, and that the population reaches its maximum value exactly at the point where the derivative vanishes.

Submission:

It takes 12 hours to drain a storage tank by opening the valve at the bottom. The depth of the fluid in the tank hours after the valve is opened is given by the formula  meters.

(a) Find the rate (meters per hour) at which the tank is draining at time .

(b) When is the fluid level in the tank falling fastest? Slowest? What are the values of at these times.

(c) Graph y and together and discuss the behavior of in relation to the signs and values of . (Be sure that your discussion is in complete and coherent sentences.)

The volume of a growing spherical cell is , where the radius r is measured in micrometers (a micrometer is one millionth of a meter).

Submission:

(a) Find the average rate of change of V with respect to r when r changes from

1. 5 to 8

2. 5 to 6

3. 5 to 5.1

(b) Find the instantaneous rate of change of V with respect to r when . Be sure to include units in your answers and answer this question.  What is the relationship between the answers for (a) and the answers for (b)?

Submission worksheet: