DESCRIBING HOW THINGS CHANGE

Problems from the subsection titled ``Velocity''

• 2.1:
  Compute the average velocity of the melon over the time interval 4 < t < 5 and over the interval 1 < t < 3.
  

• 2.2:
  What is the average velocity of the melon over the interval [0.99,1]? dots over the interval [1,1.01]? Use the following table to get average velocities over correspondingly shorter intervals: What value (to two decimal places, say) should we assign to the instantaneous velocity of the melon at time t=1 sec?

• 2.3 :
  
  • a.) Even if you did the above problems by hand, why not use Maple this time? Approximate the instantaneous velocity of the melon at time t=5 sec by first defining the function

     f:= t -> 6 + 100*t - 16*t^2; 

    and then calculating average velocities over short intervals which begin or end at t=5:

           (f(5.01) - f(5))/ (5.01-5); 

           (f(5)-f(4.99))/(5-4.99);  

    and so forth. What value would you assign for the instantaneous velocity at t=5?
  • b.) You can get fancier: Let's define the function

          vav := t -> (f(t+h)-f(t))/h;  

    This will give the average velocity over the time interval [t,t+h] once we have assigned a value to h. Do part (a) again using vav. Then use vav to find the instantaneous velocity at several (at least three) other values of t.
  • c.) It's even better if you graph it. Assign a (small) value to h, and then

            plot({f(t),vav(t)},t=0..6)   

    Do this several times with decreasing values of h (but only turn in \it one plot). What do the graphs of all the 'vav' s have in common? How do they differ? How is the graph of f(t) related to that of vav(t) . In particular how does f behave when vav is positive? negative? big? small? -- Is this behavior precise or only approximate?

  ---

  Problems from the subsection titled ``Derivatives''

• 2.4:
  
  • a.) (Use Maple to) plot the function on the interval [-1,2].
  • b.) Measure the net growth of f on the interval [-1,2].
  • c.) Measure the average growth rate of f on the interval [-1,2] ... on the interval [1,3/2].
  • d.) Define a function gav which measures the average growth rate of f on any interval [t,t+h]. By assigning smaller and smaller values to h, approximate the instantaneous growth rate of f at t=1.
  • e.) Now, unassign h using h:='h'; and, using gav, find the expression for the average growth rate of f over the interval [t,t+h]. The statement expand(gav(t)); should be just what you need. What happened to the h in the denominator of the definition of gav? (Write out an answer to this question!!)
  • f.) Based on part (e) and the definition of instantaneous rate of change, what value should be assigned to the instantaneous rate of change of f for any value of t? For t=1, how does this compare with your answer to part (d)?

• 2.5:
  Use D or diff to compute the derivative of the f from Exercise 2.4. Plot f and its derivative on the same axes. How might the derivative help answer the following questions:
  • a.) At what point(s) in [-1,2] is f increasing most rapidly?
  • b.) At what point(s) in [-1,2] is f decreasing most rapidly?
  • c.) At what point(s) in [-1,2] is f changing least rapidly?

• 2.6:
  Still using the same f from the last two problems: Let h=0.1 and have Maple plot the graphs of gav and of the derivative of f on the same axes. Then let h=0.01 and do the same thing. Explain what you see in the plots.

• 2.7:
  
  • a.) Before turning to the computer, make a sketch with pencil and paper of what the graph of f(t)=sin(t^2) over the t-interval [0,15] should look like. Keep in mind that t^2 increases faster and faster as t gets larger. How small should the size h of the t interval be chosen to get a good idea of the instantaneous growth rate for values of t near t=0? How does this change as t gets to be closer to 10? How small do you think it would be for t near 100?
  • b.) Now define a Maple function f:= t -> sin(t^2); and try plotting it over the interval t=0..15. Does the plot look like what you expected? If it does not, perhaps this is because of the algorithm Maple uses for plotting: it chooses a certain number of points (about 50) to compute exactly and then attempts to connect them with a reasonably smooth curve (not so different from what humans do) --- the problem with this particular function over this particular range is that we need to plot more points exactly to get a good picture of the graph. We can force Maple to choose more points with the numpoints option as follows:

     plot(f(t),t=0..15,numpoints=250);  

    This statement forces Maple to compute 250 points exactly before trying to ``connect the dots''. (How many points you want to use becomes an accuracy-versus-resources issue, since doing more points takes longer and uses more memory.)
  • c.) As you did before, define a gav function for this problem. Set h:=0.1; and make a plot of f(t) and gav(t) for t=10..15 using numpoints=250. Look very carefully at the graphs -- why is this version of gav misleading (consider where you should expect the growth rate to be positive, negative or zero)?
  • d.) As you can verify, the situation improves (on this interval) if you set h:=0.01; but will get worse again if we look at larger values of t.
  • e.) So, one advantage of the derivative is that it uses the smallest possible value of h (namely, h=0) to give uniformly good results. Make a plot of f(t) and diff(f(t),t) on the interval t=10..15 to verify this.

  ---

  From the subsection titled ``Derivatives at Work and Play"

• 2.8:
  Use derivatives (computed by Maple or by hand) to answer the following:
  • a.) As x leaves x=2.6 and advances a little bit, does x^3-3x^2 go up or down?
  • b.)As x leaves x=1.7 and advances a little bit, does x^3-3x^2 go up or down?

• 2.9:
  
  • a.) Let z:= x*y + 8/(x^2*y^3) + 6*x; in Maple. Then set p:=diff(z,x); and q:=diff(z,y);. They are different. Describe carefully in words what each of them represents.
  • b.) Suppose you are sitting at the point in the plane where x=1 and y=2. What is the value of z there? How will z change if you increase x a little bit? What happens as you increase y a little bit?

• 2.10:
  Suppose f(x) is positive on the interval [a,b]. For x between a and b, let A(x) be the area of the region in the plane bounded by the vertical line x=a, the horizontal line y=0 (i.e., the x-axis), the vertical line x=x (you understand what I mean, don't you?) and the graph of y=f(x). What is the instantaneous growth rate of A(x)?

• 2.11:
  Find the highest and lowest points on the graph of the function y := sin(x)/(3 - 2x + x^2). First plot, then use a statement something like fsolve(diff(f(x),x)=0,x). This might only return one of the values you expect. You can tell Maple to look in a specific range by adding an optional argument, such as:

           fsolve(diff(f(x),x)=0,x,-5..0) 

  to look only for a solution in the interval [-5,0]. Explain in words what you did and why you did it.

• 2.12:
  
  • a.) Sketch with pencil and paper the graph of a function f(x) with the properties that f'(x)>0 for all for all x.
  • b.) Sketch the graph of a function g(x) with the properties that g'(x) < 0 for all

  ---

  From the subsection titled ``Differential Equations"

• 2.13 :
  
  • a.) With pencil and paper, sketch several examples of curves which might occur as graphs of solutions of y'(x)=ay(x) with a>0, based upon the above analysis.
  • b.) Do the corresponding analysis for a<0 and sketch some graphs for this case.

  Now think back to last week -- you've previously seen graphs like the ones you just drew. Your graphs should remind you of the graphs of exponential functions. One way to check this out is to see (using Maple) if the derivatives of exponential functions are proportional to themselves:

• 2.14:
  
  • a.) Choose several values of b>1. Use Maple to differentiate the function K*b^x with respect to x. You may have to evalf the constants that appear. Have we found solutions to our equation?
  • b.) What if you choose values of b < 1 (but still positive of course)?

    All right, we have found some (in fact all) solutions to differential equations of the form y'(x)=ay(x), but the relationship between the b in our solutions and the a in the differential equation might seem to be pretty obscure. Mathematicians have come up with a convention which saves us from the trouble of having to figure out the relationship between b and a each time. To do this, they have settled on the name e (in Maple, it's E) for the number with the property that the exponential function y=e^x (and in fact y=Ke^x for any constant K) satisfies our differential equation with proportionality constant a=1. In other words, the derivative of the function y=e^x is precisely y=e^x itself.

• 2.15:
  Use Maple to differentiate the function f:= x -> E^x; (this function is also known as the exponential function, and is denoted exp(x) both in Maple and in ``real life''). Then use Maple to differentiate g:= x -> K*E^(a*x); with respect to x. What do you conclude about solving the differential equation y'(x)=ay(x)? What is the precise relationship between a and b we were looking for above?

• 2.16:
  Find a formula for y(x) if it is known that the instantaneous growth rate y'(x) is proportional to y(x) and it is known that y(1)=2.7 and y(3.1)=5.4. (Hint: you know what the general form of the solution must look like. Use the two pieces of numerical data to determine the constants K and a.)

• 2.17:
  The half-life of a radioactive substance (or of a decomposing first-order chemical compound) is the amount of time it takes for half of the substance to change into something else. At 25 degrees C, the half-life of N_2O_5 is 4.03 x 10^4 sec. What differential equation does the concentration of N_2O_5 satisfy at this temperature? What percentage of N_2O_5 molecules present at the start of an experiment will have decomposed (careful!) at the end of one day?

• 2.18:
  The President and the Prime Minister order coffee and receive cups of equal temperature (hot) at the same time. The President adds a small amount of cool cream immediately, but does not drink his coffee until 10 minutes later. The Prime Minister waits 10 minutes, and then adds the same amount of cool cream and begins to drink. Who drinks the hotter coffee? (Hint: It would help to begin by showing that solutions to the differential equation derived from Newton's law of cooling, namely T'(t)= a(T(t)-A) are of the form )

  ---

  Problems from the section--"The Growth of Populations"

• 2.19:
  Learn about the properties of the logistic equation by producing plots of the function for changing values of the a and c constants. What is the effect of increasing the a constant? What is the effect of decreasing the c constant? How is the maximum attainable population affected by changing a? How is it affected by changing c?

• 2.20:
  Using the logistic equation and the data for the U.S. population between 1940 and 1970, try to adjust the values in the equation to get a better fit. Can you get a better fit? Explain in words, why it is or is not possible to do better. What is the projected maximum population of the U.S. using your fit and when will we reach it?

• 2.21:
  In July of 1993, the United Nation's population division released projections as to what the world's population would be in the year 2150. They reported that it took 123 years for the world's population to double from 1 billion in 1804 to 2 billion in 1927. By 1987, the global population was 5 billion. In 1993, it totaled 5.57 billion.
  • a.) What would the world's population be in the year 2150 if the current fertility rate continues? You will have to try various values of a and c to get a good fit to the data. Once you have your fit, sketch your curve showing the fit.
  • b.) The newspaper report of this news release stated that the world population in the year 2150 could be as high as 694 billion! On the basis of your fit could you reasonably justify a number that large? What is the largest population you could get by that time and still have a good fit to your data?

  ---

  Problems from the section--"What About the Force"

• 2.22:
  Normally, Physics 150 would not make use of the full power of Newton's Laws as he originally stated them. Maple gives us the ability to do far more realistic problems, however, so we will consider some of them here. For example, you should now know that the acceleration of an object depends upon the vector sum of all external forces acting on it. Typically, one of those forces will be air resistance or just friction induced by moving through the air. The dragging force of air resistance on an object can be approximately characterized (for speeds significantly slower than the speed of sound) as a force whose magnitude is proportional to the velocity of the object and whose direction is, of course, opposite to the direction of the fall. Therefore, if we consider a falling object, like a skydiver, her equation of motion would be

                              F_{net} = mg - bv
  

  where b is the constant of proportionality of air resistance on the skydiver. In general b will depend on the shape of the falling object, so you can reduce b by curling into a ball or increase it by going to the usual spread-eagle position to maximize your time in the air. As the skydiver speeds up, eventually, no matter what value of b is used, the skydiver will hit terminal velocity, the speed at which the drag force equals the gravitational force and no more acceleration occurs. For the spread-eagle position, the terminal velocity is about 130 miles/hour or roughly 60 meters/s. With this information, we can quickly solve for the acceleration at any given time from our previous equation:

                             a = g - b/m * v
  
  	At the terminal velocity, by definition, a = 0, so,
  
                            0 = g - b/m * v_t  ==> b/m = g/v_t
  
  	Thus, we can replace b/m in our original equation of
  	motion to get
                             a = g * (1 - v/v_t)
  
        
  
        
    a.) Now, we know that a = dv/dt,
  			so you should use Maple to solve for the
  			velocity as a function of time assuming
  			that g = 9.81 m/s^2 and
  			v_t = 60 m/s.  Plot your result
  			for the velocity as a function
  			of time and find the time (after the jump
  			begins) at which the skydiver's velocity
  			is within 5 m/s of the terminal velocity.
        
    b.) Find the distance fallen by the skydiver 
  			as a function of time.  If the skydiver
  			starts from an altitude of 10,000 feet and
  			needs to open the parachute at an altitude
  			no lower than 500 feet, what is the maximum
  			amount of time the jump can last?
        
  
    
    2.23:
  
  	A more accurate characterization of the air resistance is given
  	by the assumption that the drag force increases as the
  	square of the velocity of the object rather than
  	proportionally to the velocity.  Redo the above calculations
  	for this new assumption.
  
  	(Warning! this is not as easy as it looks!  Make good use
  	of Maple's Help facility and the instructors if you have trouble
  	doing this!).
  
  	Make a plot to display your result.  How long after starting the
  	jump does it take for the skydiver to reach a speed within 5 m/s
  	of the terminal velocity?  Just for fun, can you explain how to
  	find the distance fallen as a function of time?  Bonus points
  	will be awarded for those who can produce a correct plot of
  	position vs. time.
  
  

Large Lab II: The Death of Atoms

  The number of pennies always decreases when you
        	toss them (or at best does not change)   

 The number of pennies that ``decay'' on any given toss is 
	proportional to the number of pennies you tossed (you cannot
	lose more pennies than you tossed, but yet the more you
	toss, the more you tend to lose)  

 The number of pennies we started with was 100

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