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TAMU MATH 151 - lec3_6-9

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Lecture for Week 6 (Secs. 3.6–9)Derivative Miscellany I1Implicit differentiationWe want to answer questions like this:1. What is the derivative of tan−1x ?2. What isdydxifx3+ y3+ xy2+ x2y − 25x −25y = 0 ?2x3+ y3+ xy2+ x2y − 25x − 25y = 0.Here we don ’t know how to sol ve for y as a f unc-tion of x, but we expect that the formula definesa f unction “implicitl y ” if we consider a smallenough “window” on the graph (to pass t he “ver-tical line test”).xy.....................................................................................................................................................................................................................................................................................................................................................3Temporarily assuming this is so, we differe n-tiate the equation with respect to x, re me mber-ing that y is a fu nction of x.0 =ddx(x3+ y3+ xy2+ x2y − 25x − 25y)= 3x2+ 3y2y′+ y2+ 2xyy′+ 2xy + x2y′− 25 − 25y′= (3x2+ y2+ 2xy − 25) + y′(3y2+ 2xy + x2− 25).y′=25 − 3x2− y2− 2xy3y2+ 2xy + x2− 25.4To use th i s formula, you need to k now apoint (x, y) on the curve. You can check that(3, 4) does satisfyx3+ y3+ xy2+ x2y − 25x − 25y = 0.Plug those nu mbe rs intoy′=25 − 3x2− y2− 2xy3y2+ 2xy + x2− 25to get y′= −34.5But (x, y) = (3, −3) also satisfies the equa-tion, and it gives y′= −1. And (3, −4) s at i s fiesthe equation and gives y′= +34. Three differentfunct i ons are defined near x = 3 by our equation,and each has a different slope.3−3−44...................................6The curve in t his problem is the un ion of acircle and a line:0 = x3+ y3+ xy2+ x2y − 25x − 25y= (x2+ y2− 25)(x + y).3−3−44.........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................••7We can cl early see the three points of intersec-tion with the line x = 3. Two other int erestingpoints are:1. x = 5, y = 0 (ve rti cal tangent): The denom-inator of the f ormula for y equals 0, but thenumerator does not.2. x = −5/√2, y = 5/√2 (intersection): Bothnumerator and denominator vanish, becausethe slope is finite but not unique.8An inportant applicati on of implicit differ-entiation is to find f ormulas for derivatives ofinverse functions, such as u = tan−1v. Thisequation just means v = tan u, together with the“branch condition ” that −π2< u <π2(withoutwhich u would not be unique ly defined). So1 =dvdv=ddvtan u =ddutan ududv.Butddutan u = sec2u = 1 + tan2u = 1 + v2.9Putting those two equations together, we getddvtan−1v =dudv=ddutan u−1=11 + v2.Generally speaking, the derivative of an in-verse trig function is an algebraic function! Wewill see more of this in Sec. 4.2.10Exercise 3.6.39Show th at the cu rve familiesy = cx2, x2+ 2y2= kare orthogon al traje ctories of each other.(That means that every curve in one family (eachcurve labeled by c) is orthogonal to every curve inthe other family (labeled by k).11For th e c urve s y = cx2we havedydx= 2cx.(No implicit differentiation was needed in th i scase.) For t he curves x2+ 2y2= k we have2x + 4ydydx= 0 ⇒dydx= −x2y.If t he curves are orthogonal , the prod uct of theslopes must be −1 (and vice ve rsa). We ll, theproduct is(2cx)−x2y= −cx2y= −1.12Derivatives of vector functionsNo big surprise here: Conceptually, we aresubtracting the “arrows” for two nearby valuesof the parameter, t hen dividin g by the parameterdifferenc e and t ak ing the limit.................................................................................................................................................................................................................................................................................................................r(t)r(a)∆rr′(a) = limt→ar(t) − r(a )t − a≡ limh→0∆rh13And calculati onally, since our basis vectorsdo not depend on t, we just differenti at e eachcomponent:ddtht2ˆi + 3tˆj + 5ˆki= 2tˆi + 3ˆj.14Second (and higher) derivativesThis is fairly obv i ous, too: The secondderivative is t he derivative of the first derivative.s(t) = At2+ Bt + C ⇒ s′(t) = 2At + B⇒ s′′(t) = 2A.(This was essentially Exercise 3.8.37.)15The most important ap plication of secondderivatives is acceleration , t he derivative of veloc-ity, which is the derivative of position.Exercise 3.8.49A satellite completes one orbit of Earth at an al-titude 1000 km every 1 h 46 min. Find the veloc-ity, speed, and acceleration at each time. (Earthradius = 6600 km.)16The period is 14660= 1.767 h r. Therefore,the angular speed i s 2π/1. 767 = 3.557 radiansper hour. The radius of the ci rc le is 7600, so thespeed in the orbit i s 7600 × 3.557 = 27030 km/hat all times. To represent the velocity we mustchoose a c oord i nate sys t em; say that the satelli t ecrosses the x axi s when t = 0 and moves counter-clockwise (so it crosses the y axi s after a quarterperiod). Thenv(t) = 27030h−sin(3.557t), cos(3.557t)i.17(When t = 0, v is in the positive y dire ction;after a quarter period, it is in the negative x di-rection.) The acce l eration is the derivative ofthat,a(t) = 27030 ×3.557h−cos(3.557t), −sin(3.557t)i.Finally, l et’s find t he position function. Its deriv-ative must be v, so a good fi rst guess isr(t) =270303.557hcos(3.557t), sin(3.557t)i.18To this we could add any constant vec t or, but aquick ch eck shows that r(0) is in the positive xdirection as we wanted, and this orbi t is centeredat the origin as it s hould be. So this is the rightanswer. Notice that a poi nts in the direction op-posite to r (i.e., toward the center of the orbit),as always for uniform circ ular motion.19Slopes and tangents of parametric cu


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