Head start on Calculus: Differentiation, IntegrationDate: 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/ |
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