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Chapter One 1.1 Linear Functions & the Tangent Problem

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Chapter One 1.1 Linear Functions & the Tangent Problem Chapter One Limits and Rates of Change 1.1 Linear Functions & the Tangent Problem Gavin Denham Tuesday, January 28, 2014 2:27:17 PM MT A linear function is a function f of the form f (x) = mx + b, m and b are constants This equation has a slope of m & a y-intercept of b. Gavin Denham Tuesday, January 28, 2014 2:27:17 PM MT The slope of a non-vertical line that passes through the points P1 (x1, y1 ) and P2 (x2 , y2 ) is defined by Since the slope is the ratio of the change in y (∆y) to the change !y y " y in x (∆x). m = = 2 1 !x x2 " x1 In calculus, we interpret this to be The rate of change of y with respect to x. Gavin Denham Tuesday, January 28, 2014 2:27:17 PM MT Find a linear equation whose graph passes through the points (-2,-2) and (3,4) ! !y y2 " y1 4 ! 2 4 + 2 6 m = = = ! = = !x x2 " x1 3! 2 3+ 2 5 In order to find an equation of any line we need the slope of the line and one point. We now have both of these: 6 m = and either point (-2,-2) or (3,4) 5 Gavin Denham Tuesday, January 28, 2014 2:27:17 PM MT Find a linear equation whose graph passes through the points (-2,-2) and (3,4) Equation of a line is defined as: y ! y1 = m(x ! x1 ) where m is slope and x1 & y1 are fixed co-ordinates on the line 6 6 m = and point (3,4) m = and point (-2,-2) 5 5 6 6 y ! 2 x ! 2 y ! 4 = (x ! 3) ! = ( ! ) 5 5 6 18 6 12 y ! 4 = x ! y + 2 = x + 5 5 5 5 20 10 +4 + -2 - 5 5 6 2 6 2 y = x + y= x + 5 5 5 5 Gavin Denham Tuesday, January 28, 2014 2:27:17 PM MT The concept of rate of change is critical to our understanding of calculus! Rate of change explains several concepts in mathematics and science. The simplest and easiest example is slope. How does x change as y changes… A linear function is given by y = 6 - 5x. If x increases by 2, how does y change? !y We are given that = " 5 !x " !y% ( !x$ ' = 5 !x The rate of change is # !x& ( ) !y " slope = = " 5 !y = 5!x !x Since !x = 2 Therefore we can state: + " !x = 2 !y = 5(2) When x increases by 2 !y = "10 y will decrease by 10. Gavin Denham Tuesday, January 28, 2014 2:27:17 PM MT The Tangent Problem The word tangent comes from the latin word tangens, which means touching. For a simple curve, such as a circle, a tangent is a line that intersects the circle once and only once. Things become more complicated when we introduce other curves. The curve C at right has a single point P that has two lines through it. Line t is tangent at P but intersects the curve at another point (still a tangent line at P). Line l intersects at P but it is not considered tangent because of its intersection (not touching) Gavin Denham Tuesday, January 28, 2014 2:27:17 PM MT Find the equation of the tangent line to the parabola y=x2 at the point P(1,1). We can find the equation of any line as long as we have slope m and a point (x,y). Our problem is we only have a point. In order to find slope, by definition, we need to find the rise over run between two points. The process we go through is to use a set of second points, Q(x,y) on the same curve, close to P(1,1). By calculating the slopes of such lines, known as secant lines, we can approximate the slope of the tangent line. Gavin Denham Tuesday, January 28, 2014 2:27:17 PM MT Find the equation of the tangent line to the parabola y = x2 at the point P(1,1). We choose a point x≠1 so that our slope is defined and P≠Q. But since Q lies on the curve y = x2, we can substitute x2 for y y !1 x2 !1 mmPPQQ == m = x !1 PQ x !1 For instance, for the point Q, where x = 1.1, the slope would be 2 2 x !1 1.1 !1 1.21!1 .21 Slope of m = = = = = 2.1 PQ x !1 1.1!1 1.1!1 .1 secant line Gavin Denham Tuesday, January 28, 2014 2:27:17 PM MT Find the equation of the tangent line to the parabola y = x2 at the point P(1,1). The following table uses the previous method to determine several values of x close to 1 for mPQ x > 1 mPQ x < 1 mPQ 2 3 0 1 1.5 2.5 .5 1.5 1.1 2.1 .9 1.9 1.01 2.01 .99 1.99 1.001 2.001 .999 1.999 We can now determine that the slope of secant lines Gavin Denham Tuesday, January 28,close 2014 2:27:17 PMto MT x = 1 seem to have a slope that approaches 2 Find the equation of the tangent line to the parabola y=x2 at the point P(1,1). We can now say that the slope of the y = 2x ! 1 line tangent at x=1 has a slope of 2. m = 2 and point (1,1) y ! y1 = m(x ! x1 ) y !1 = 2(x !1) The equation of the tangent line y !1 = 2x ! 2 at the point (1,1) on the curve y = x2 +1 +1 is y = 2x - 1 y = 2x !1 Gavin Denham Tuesday, January 28, 2014 2:27:17 PM MT Page 9 Exercises 1-10 odd letters (a,c,e…) Gavin Denham Tuesday, January 28, 2014 2:27:17 PM MT.
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