Head start on Calculus: Differentiation, Integration

Date: 07/08/97 at 16:25:36
From: Diane Truesdale
Subject: Calculus

Dr. Math,
I'm soon to be a Calculus student in high school.  Could you please 
teach me some Calculus over the internet?  I'm enrolled in Calculus 
BC, the harder of the two Calculus classes. Could you recommend a 
teach yourself book that I could use?

Thank you very much for your time and consideration!
Kelly Truesdale

Date: 07/09/97 at 18:51:39
From: Doctor Barney
Subject: Re: Calculus

Hey Kelly, 

You should congratulate yourself for your initiative in wanting to get 
a head start over the summer.  I'd be glad to help!

I would also recommend a Schaum's outline series, or similar "class 
notes" type of handbook.  Just ask at any full-service bookstore, or a 
college campus bookstore might be a good bet.  These self-study books 
are not a substitute for a real class with a complete text book and a 
live teacher, but they can be good for supplemental material or for a 
fast learner to get a head start.(They're also very good for review 
after you've had the class.)

You can teach yourself some calculus by graphing some functions. Let's 
start with f(x)=x.  Go ahead and graph that on some graph paper. It's 
a straight line, right? Now let's define a function f'(x) as the 
"derivative" of f(x).  The derivative of f(x) is the function whose 
value at any given point is the slope of f(x) at that point.  What is 
the derivative of f(x) in our example?  Well, since f(x)=x is a 
straight line with slope equal to 1 for all values of x, f'(x)=1.  

That was calculus.  No big deal. The process of finding derivative 
functions is called differentiation.  Let's try some other functions:

   g(x)=x+7
   h(x)=(1/2)x+4
   j(x)=mx+b

Can you find the derivative functions g'(x) h'(x) and j'(x) where the 
value of the derivative function at each point is the slope of the 
original function at that point? Plot the original function on graph 
paper, find the slope at various points, and then plot the derivative 
function. From looking at the graph try to find an algebraic 
expression for the derivative function.  

Now some harder (but funner) ones:

   k(x)=x^2    (x squared)
   l(x)=2x^2
   m(x)=3x^2

Again, plot the original function on graph paper, find the slope at 
various points (guess, if you have to) and then plot the derivative 
function. From looking at the graph try to find an algebraic 
expression for the derivative function. Do you start to see a pattern?

Some really fun ones:

   n(x)=x^3
   p(x)=e^x
   q(x)=sin(x)

Well, that should get you started. Don't get frustrated; it can be 
VERY difficult to teach this type of thing to yourself. Just be 
patient, keep trying, and if it's really hard set it aside for a while 
and try again later. It is, after all, summer vacation!  If you need 
help or you want the answers or you're ready for more, write back.  

The other big concept in calclus is integration. Where differentiation 
is finding a function that represents the slope of some other 
function, integration is finding a function that represents the area 
bounded by some other function. Integration is an extremely important 
tool in almost all branches of science.  

-Doctor Barney,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   

Date: 07/10/97 at 17:43:06
From: Diane Truesdale
Subject: Re: Calculus

Dr. Barney,

Derivatives seem easy enough.  Can you give me the answers to those 
last examples you gave me and teach me some more funthings about 
Calulus?  Is Calculus fun?  Do you think I'll like it?  Is it 
challenging?  Is it true that girls usually have more trouble in 
Geometry than boys do?  

Thanks for giving me some interesting and fun Calculus examples!
Kelly Truesdale

Date: 07/14/97 at 16:59:49
From: Doctor Barney
Subject: Re: Calculus

Wow! What a lot of questions in one little message!  I'm glad you 
thought it was "easy", but keep in mind that we started with some easy 
ones, and some functions can be very difficult to differentiate.  
Besides, if you don't have the answers yet, you don't know if you got 
them all right, so how do you know it's easy?

Okay, here are the answers.  How many did you figure out correctly?

   g(x) = x+7       g'(x) = 1
   h(x) = (1/2)x+4  h'(x) = 1/2
   j(x) = mx+b      j'(x) = m

   k(x) = x^2       k'(x) = 2x
   l(x) = 2x^2      l'(x) = 4x
   m(x) = 3x^2      m'(x) = 6x

   n(x) = x^3       n'(x) = 3x^2
   p(x) = e^x       p'(x) = e^x
   q(x) = sin(x)    q'(x) = cos(x)

Calculus, like everything else, is fun for some people, but not for 
others. I think that it is fun. It's fun to see how it relates to 
scientific phenomena in the real world.  I don't know if you will like 
it, but I suspect that if you are playing on the computer asking 
people to teach you some calculus during your summer vacation that 
there is a good chance that you will like it.  

It is not accurate to say that "girls usually have more trouble than 
boys" in geometry or in math generally, or in any other subject, for 
that matter.  It is true that some people do well in math and science, 
and that some people do not.  But this is probably because the people 
who do well simply like it more, and not because those who "have 
trouble" couldn't learn these subjects well if they really wanted to.  

Now, it is also true that of the people who enjoy it more and tend to 
do well that more of them tend to be boys (or men) BUT *this is the 
important part* the girls and women who choose to study math and 
science do just as well as the boys and men. In fact, some of my best 
math teachers have been women. I hope this answers that question.  

Now for some more calculus!  Once you know how to differentiate, you 
will want to know how to use this tool.  On of the most common uses of 
differentiation is to find the minima and maxima of various functions.  
These are the smallest points of a function.  y=x^2, for example, has 
a minimum at x=0.  One interesting thing about derivatives is that the 
derivative of a function will be zero at the points where the function 
has a minima or a maxima. That makes sense when you think about it, 
because at a maxima, for example, the function has been getting 
bigger, stops getting bigger at one point, and then starts getting 
smaller. So the derivative of that function (the slope) is first 
positive, goes to zero at one point, and then becomes negative, right?  
So now we can use this tool to find the maxima and minima of simple 
functions. This can be very useful in real life.  

For example, suppose you are told that the fuel efficiency of a car is 
given by the function mpg = -(1/160)x^2 + (1/2)x  where mpg is the 
distance in miles that the car will travel using one gallon of gas and 
x is the speed the car is driven in miles per hour.  How fast should 
you drive to get the best mileage? 

Well, the mpg will be a maximum where its derivative is zero. 
 
     mpg' = -(1/80)x + 1/2   
     
This derivative function will be zero when

  (1/80)x = 1/2
        x = 40 miles per hour.  

Go back to the original function for mpg, plot it on graph paper for 
values of x from  0 to 60 mph in 5 mph increments, and see if 40 mph 
is really the speed that gives the best gas mileage.  

Now your turn. Suppose you are a graphic artist designing posters, and 
you discover that the price people are willing to pay for a square 
poster is proportional to the length one side of the poster and is 
given by the equation price = five dollars per foot.  However, the 
cost of square posters is proportional to the amount of material used, 
and is given by the equation cost = 83 and 1/3 cents per square foot.  
What size poster will generate the most profit for your company?  
Hint: profit = price - cost.

Write back if you get stuck, but don't just ask me for the answer; 
show me how far you got.

-Doctor Barney,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   

Date: 07/14/97 at 19:38:13
From: Diane Truesdale
Subject: Re: Calculus

Dr. Barney,

I think I've got it figured out!

Okay, here are my steps. Please correct them if they're wrong.

     x = the number of feet
   x^2 = the nimber of square feet
 price = 5x
  cost = (83 and 1/3)x^2/100 or 5x^2/6
 profit = price - cost
 profit = 5x- 5x^2/6
        = 5-5x/3
   5x/3 = 5
     5x = 15
      x = 3 feet
    x^2 = 9 square feet
and a profit of $7.50.

Am I right on this problem or just partially right? Did I leave out
some steps? 

Do you have any tips for me that my help me with the SAT? Could you 
help me prepare with some more questions?

Thank you for all your help!  I really appreciate it.

Yours truly,
Kelly Truesdale

P.S. I did get all those equations from before correctly. Thanks 
again.

Date: 07/16/97 at 18:54:53
From: Doctor Barney
Subject: Re: Calculus

You got the answer right! You didn't really leave out any steps, but 
if it had been a test you probably should have explained that you were 
taking the derivative of the profit function, and that you were 
setting that derivative equal to zero to find the value of x at the 
maximum profit.
  
I don't think there's very much calculus on the SAT. If you want to 
study calculus you should really get a study guide or a surplus 
textbook, as I suggested before.  You can write to us with specific 
questions when you get stuck.

Meanwhile, here's a little more to keep you busy:

Double derivatives: if you take the derivative of a derivative 
function, you will be calculating the rate of change of the slope of 
the original function.  One classic example is acceleration.  If an 
object (like your car, say) is at position x(t) as a function of time, 
the rate of change of its position is called velocity, and is the 
derivative of its position with respect to time: v(t)=x'(t).  

Now, the rate at which your car's velocity is changing is called 
acceleration, and it is just the time derivative of the velocity 
function, or the double derivative of position:  a(t)=v'(t)=x''(t).  
v' is read "v prime" and x'' is read "x double prime".

But you can calculate double derivatives of any function. Wherever the 
double derivative is positive, the slope of the original function is 
increasing, and whenever the double derivative is negative the slope 
of the original function is decreasing.  (Note that the slope can be 
decreasing even when the original function is increasing, and vice 
versa.)  

Where the derivative is an indication of the slope of a function, the 
double derivative is an indication of the concavity of a function.  
Wherever concavity is positive, the original function is curved 
upwards, and wherever it is negative the original function is curved 
downward. Where the double derivative is zero what does that tell you 
about the original function?  You could try to find the double 
derivatives of the functions you found the first derivatives of 
earlier, graph them, and see how they show you the slope of the first 
derivatives, and how they relate graphicaly to the original functions.

When you set the derivative equal to zero in a problem like the one 
you just did with the posters, how do you know if you found a maximum 
or a minimum or some other point where the function is flat? 

   Answer: find the concavity (the double derivative) of the function 
at that point. Where concavity is positive and slope is zero, you have 
found a minimum. Where concavity is negative and slope is zero, you 
have found a maximum. Where concavity and slope are both zero, this is 
called an "inflection point". What does the original function look 
like at an inflection point? Graph the function  y=x^3 to find out. 

There are also higher derivatives, each of which shows the rate of 
change of the next lower derivative, but they are not used very often.  
In mechanics, for instance, "jerk" is defined as the rate of change of 
acceleration. You could think of that as the triple derivative of 
position.  

Have fun! 

-Doctor Barney,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/