Class Notes (807,462)
MATH 140 (141)
Ewa Duma (13)
Lecture 11

# Lecture 11 - 2.7.pdf

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School
McGill University
Department
Mathematics & Statistics (Sci)
Course
MATH 140
Professor
Ewa Duma
Semester
Summer

Description
y amounts to saying that the tangent line is the limiting position of the secant line PQ Q{x,▯ƒ } approaches . SPe Figure 1.) ƒ-f(a) P{a,▯f(a} 1 DEFINITION The tangent line to the curve y ▯ f▯x▯ at the point P▯a, f▯a▯▯ is x-a the line through P with slope MATH140 - Lecture 11 Notes f▯x▯ ▯ f▯a▯ m ▯ lixla x ▯ a SECTION 2.7 DERIVATIVES AND RATES OF CHANGE 0 a x x provided that this Q a+h,▯f(a+h)es and Rates of Change: { } (See Figure 3 where the cash ▯ 0 is illustrated aQdis to the right of . If it happened y t thath ▯ 0, however,Q would be to the left Pf .) y In our ﬁrst example we conﬁrm the guess we made in Example 1 in Section 2.1.es0p(becauseh ▯ x ▯ as the line through t P with slopesion for the slope of the tangent line in Deﬁnition 1 becomes P {,▯f(}) Q V 2 Q EXAMPLE 1 hind anf(a+h)-f(a)f the tm = lim (x > a) [(f(x) - f(a))/(x-a)] *provided that this limit exists point P▯1, 1▯. 2 m ▯ lim f▯a ▯ h▯ ▯ f▯a▯ h l 0 h P Q SOLUTION Here we haveh a ▯ 1plandFif▯x▯ ▯ xuat, so the slope ist line to the parabola y = x^2 at the point (1,1) FIGURE 3 EXf▯x▯ ▯ f▯1▯ind an equax ▯ 1of the tangent line to the hyperbola m ▯ lim point ▯3, 1▯ ▯ lim xl1 x ▯ 1 xl1 x ▯ 1 SOLUTION Letf▯x▯ ▯ 3▯x. Then the slope of the tangen▯3, 1▯ 0 x ▯ lim ▯x ▯ 1▯▯x ▯ 1▯ 3 y xl1 x ▯ 1 ▯ 1 m ▯ lim f▯3 ▯ h▯ ▯ f▯3▯ ▯ lim 3 ▯ h ▯ lim x+3y-6=0 y= 3 hl0 h hl0 h FIGURE 1 x ▯ lim ▯x ▯ 1▯ ▯ 1 ▯ 1 ▯ 2 xl1 ▯h 1 1 (3,▯1) ▯ lhl0 h▯3 ▯ h▯ ▯ hl0 ▯ 3 ▯ h ▯ ▯ 3 N Point-slope form for a line through the Using the point-slope form of the equation of a line, we ﬁnd that an equation of the 0 We may xefer to the slope of the tangent line to a curve at a point as the slope of the point▯1 ,1ywith slope : tangent line at ▯1, 1▯ is curve at the point.re an equation of the tangent at the p▯3, 1▯s y ▯ y1▯ m▯x ▯ x ▯1 y ▯ 1 ▯ ▯ 3x ▯ 3▯ This brings us to understanding how to calculate the derivative of a function where the FIGURE 4 which simpliﬁes to x ▯ 3y ▯ 6 ▯ 0 FIRST derivative allows us to ﬁnd out the slope of the function. We sometimes refer to the slope of the tangent line to a curve at a point as the slope of The hyperbola and its tangent are shown in Figure 4. the curve at the point. TheWhat we mean by this is that m = lim (h > 0) [f(a+h) - f(a)]/h will give us the slope looks almost like a straight line. Figure 2 illustrates this procedure for the curve y ▯ x time t=a time t=a+h VELOCITIES TEC Visual 2.7 shows an animation of Example 1. The more we zoom in, the morIn Section 2.1 we investigated the motion of a ball dropped from the CN Tower and deﬁned Figure 2. the curve becomes almost indistinguishable from its tangent line. 0 s its velocity to be the limiting value of average velocities over shorter and shorter time f(a+h)-f(a) periods. lim (h > 0) [(x+h)^2 - x^2]/h In general, suppose an object moves along a straight line according to an equation of 2 1.5 f(a) motion s ▯ f▯t▯ where ss the displacement (directed distance) of the object from the ori- lim (h > 0) 2x f(a+h) gin at timet. The functionf that describes the motion is called the position function of the object. In the time interval frot ▯ ato t ▯ a ▯ h FIGURE 5 Velocities: f▯a ▯ h▯ ▯ f▯a▯ (See Figure 5.) The average velocity over this time interval is (1,▯1) (1,▯1) (1,▯1) s Q{a+h,▯f(a+h)}on: Average velocity =average velocity ▯displacement ▯(af▯a ▯ h▯ ▯ f▯a▯ time P{a,▯f}a) This equation is the same as above in calculating the slope/derivative. 0 2 0.5 1.5 0.9 1.1 h which is the same as the slope of the secantPQinin Figure 6. Assume we are computing the average velocities over shorter and shorter timed shorter time intervals FIGURE 2 Zooming in toward the point (1,▯1) on the parabola y=▯ ▯a, a ▯ h▯. In other words, we heapproach 0 As in the example of the falling ball, we intervals, wdeﬁne the velocity (or instantaneous velocitv▯a▯at time instantaneous 0 a a+h velocity at
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