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Sun 17 Jul, 2005 08:26 am
Mr. Buckley, our math teacher, loves to present the class with challenge math questions for the weekend. I was not able to start the following two questions dealing with THE TANGENT and THE SLOPE OF SECANT LINES. He went over a QUICK tangent/secant lesson and said the rest is for us to use our math skills.
What math skills? We are now learning what math is truly all about, right? I think Mr. Buckley is way off but in any case, he decided to proceed by asking two very tough questions in the "CHALLENGE QUESTIONS" section of his white board.
1) Find the equation of the tangent line to the parabola y = x^2 at the point
p(1, 1). What on earth is the tangent line? Is this the line the touches ONE point on the outside of the circle?
2) The point p(0.5, 2) lies on the curve y = 1/x. If Q is the point (x, 1/x), use your calculator to find the slope of the secant line PQ (correct to six decimal places) for the following values of x: 2 and 0.7. How do I use my calculator to answer question 2? What is the slope of the secant line?
What on earth is Mr. Buckley talking about? Why present the class with topics he BRIEFLY discussed in class? Does this make for good math teaching?
Tangents are found on every continuous curve. A parabola, like a circle, is a continuous curve.
In a simple sense your teacher is trying to incorporate several things that you already know. You are familiar with the slope intercept equation of a line right? y=mx+b.
If I look at an analytic form of the tangent trigonometric function then tan is the rise over the run of a triangle, which interestingly enough, is the same as the slope of a line. Connecting these factoids then a line that touches a continuous curve at a single point is the tangent line.
Now how do you find this tangent line on the parabola?
Think about how to find a line from two points. One point you know, its (1,1) and it's on the parabola y=x^2. Now look for another point that is slightly off of (1,1) sat x=1+0.1, and y=(1.1)^2. This gives (1.1, 1.21).
The slope of this line is then m=(1.21-1)/(1.1-1)
Now have the difference between x=1 and 1.1. X=1.05 and y=1.1025 & m=(1.1025-1)/(1.05-1)
Now think about a tangent, it's the line at a single point. So if I want to find the slope, I'd pick two points that are incredibly close to each other* on the parabola.
The first point is given (1,1) the second? x=1+del where del is almost 0. Then y=(1+del)^2=1+2del+del^2
So the slope is
m=(1+2del+del^2-1)/(1+del-1)=(2del+del^2)/del
Now since del is almost zero then del^2 is incredibly smaller than almost zero, so compared to del, del^2=0 and
m=2del/del=2
Now I have the slope of the tangent line to the parabola y=x^2 at x=1.
*at some point you have, or will be, introduced to the concept of a limit. This is an application of that concept.
As for your teacher--yes, it is a good teaching method. He is simply trying to get you to think about the conception of what you already have learned, and since all concepts of mathematics is cumulative he is effectively guiding you subtly toward the next concepts.
Rap
See here for the definition of the secant line:
http://mathworld.wolfram.com/SecantLine.html
This probem will be easier since you just have to plug x values into the function to determine y values. Then you can easily calculate the slope between the two points.
okay
I want to thank both of you for your tips and great help.
raprap
One question for raprap:
What do you mean by the letters del in your explanation of my question?
It's an abbreviation for the Greek letter delta. Delta traditionally signifies difference between two independent variable, that is, if I have an x and a want a nearby x a specific distance (del) from x (I'll call this second x---x0) then x0=x+del.
Rap