, etc. Let s work with the last one. We can graph a few points determined by this equation.
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1 1. Lines By a line, we simply mean a straight curve. We will always think of lines relative to the cartesian plane. Consider the equation 2x 3y 4 = 0. We can rewrite it in many different ways : 2x 3y = 4, 3y = 2x 4, y = 2 3 x 4 3, etc. Let s work with the last one. We can graph a few points determined by this equation. (0, 4/3), (2, 0), (5, 2), ( 3, 10/3), etc. Consider the graph below. Notice that the four points we determined lie on a line. Notice as relationship on this line every time we move 3 units in the positive x direction, we move two units in the positive y direction. Equivalently, for every change of one unit in the positive x direction, there is a movement of 2/3 in the positive y direction. So 2/3 measures the steepness of the line. We ll call it m = 2/3. y (5, 2) (2, 0) x (0, 4/3) ( 3, 10/3) Figure 1 1
2 2 In general, we can find the slope of a line by finding the ratio of the change in x direction to the change in y direction. This is simple if we know two points that lie on this line, say (x 1, y 1 ) and (x 2, y 2 ). Then the change in y direction between these points is y 2 y 1, and the change in x direction is x 2 x 1. So, we can say the slope of a line is m = y 2 y 1 x 2 x 1. Note two special cases. What is the slope of the line with the points (0, 0) and (5, 0) on it? Well, m = = 0. So it has a slope of 0, or rather, a change of one in the positive x direction yields a change of 0 in the y direction. This makes sense as the line between these two points is a horizonal line. Now ask, what is the slope of the line with the points (0, 0) and (0, 5) on it? Well, m = , but this is undefined. We can see that the line between the two points is a vertical line. So this line does not have a slope. Finally note that if two lines are parallel, then they have the same slope. A good mental exercise is to show that if two lines are parallel, then do not intersect at any point, and vice versa. Another good exercise is showing that if two lines intersect at two or more points, then they are the same line. 2. A few examples Example 1. Find the equation of the line through the point (1, 7) with slope 1 2. We use the point-slope form to get y + 7 = 1 2 (x 1). So y = 1 2 x Example 2. Find the equation of the line through the point ( 1, 2) and (3, 4). The slope is m = ( 1) = 3 2. So using point-slope form we get y 2 = 3 2 (x + 1). So y = 3 2 x Another way to look at tangent lines Let s consider a parabola, say y = x 2. Let s consider the point P = (2, 4). What does it mean to be a tangent line at this point? What makes a specific line tangent to the curve at this point, while other lines are not? Really, we would like
3 a decent characterization of tangent. Consider the figure below. It shows the graph y = x 2, with the point P = (2, 4) in bold and several lines going through P. The black line, which is in bold is different than all the others. Can you see how? Study all the lines carefully. y 3 x Figure 2 Notice that the bolded black line is the only line that intersects the curve exactly once. All the other lines through P intersect the curve twice. In fact, any other line through P other than a vertical line and the bolded line will intersect the curve twice. This makes sense if we tie this image back with our discussion of P and Q being pushed toward each other. If we imagine that Q is the point of intersection with the curve other than P, as Q gets closer to P, the slopes of the lines through P and Q approaches the slope of the bolded line. We can also think of the bolded line going through P and Q, but now P is on the same spot as Q. Perhaps this notion is a suitable definition for a tangent line, at least for a parabola. i.e. For a parabola, a tangent line at a point P is the line that intersects the parabola at exactly one point, and that point is P.
4 4 But what about for other functions? If we take y = x 3, it is easy to find a point and a line that intersects that the curve in only that spot. For example, take P = (1, 1) and y = 1. This characterization for tangent lines on parabolas doesn t seem to work. But consider a small piece of the curve y = x 3, say only for x values 0.9 x 1.1. Now use the idea of the two points P and Q. This creates lines that intersect this piece of the curve at two points. As Q is puched toward P, we see that the slopes of the lines through P and Q approach that of te tangent line. Also, if P = Q, the tangent line intersects the curve at exactly one spot. (At least on our small localized interval.) Consider the following characterization : A tangent line at a point P = (x, y) is the line that intersects the curve uniquely at the point P on arbitrarily small intervals around x. This seems like a decent and intuitive characterization for a tangent line, and it will suit our purposes nicely. Example 3. Find the equation of the line that intersects the curve y = x uniquely at the point ( 1, 3). Don t be confused by the language; we just need to find the equation of the tangent line through ( 1, 3). So we will just use Leibniz s limiting method. y + y = x 2 + 2x x + ( x) y = 2x + x x lim x 0 y x = 2x. Thus, the slope of the tangent line to x at the point ( 1, 3) is 2 ( 1) = 2. Using point-slope form, we get y 3 = 2(x + 1). So our line is y = 2x + 1. Example 4. Find the equation of the line with slope 3 that intersects the curve y = x at a unique point. Again, we are looking for a tangent line to the curve, but this time we do not know the unique point at which the line intersects the curve. We could use Leibniz s limiting method to find that the slope of the tangent line of a given point is 2x. We could then solve 2x = 3 to get x = 3/2 and continue from there. But let s pretend we do not have Leibniz s calculus. Well the equation of the line is y = 3x + b for some number b. Our job is to find that b.
5 5 We know that the line and curve will intersect, so we can solve x = 3x + b. We are looking for the b such that we get a unique solution to this equation. One way to do this is with the quadratic formula. Rearranging the equation, we get x 2 3x + (2 b) = 0. Remember that b is just a number, so the left side of the equation above is just a quadratic formula. Thus, the equation is just asking for us to find the roots. So we are looking for the b such that the equation above as a unique solution. Using the qudratic formula, we get x = 3 ± 9 4(2 b). 2 Recall, that when using the quadratic formula, a quadratic has 2 distinct real roots if the determinant (thing under the square root) is positive. It has 0 real roots if the determinant is negative. And it has exactly one real root when the determinant is 0. So the problem comes down to solving Solving for b, we get 9 4(2 b) = 0 9 = 8 4b b = 1 4. Finally, we get that our line is y = 3x 1 4.
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