60_005.q //0 :5 PM Page SECTION.5 Implicit Differentiation Section.5 EXPLORATION Graphing an Implicit Equation How coul ou use a graphing utilit to sketch the graph of the equation? Here are two possible approaches. a. Solve the equation for. Switch the roles of an an graph the two resulting equations. The combine graphs will show a 90 rotation of the graph of the original equation. b. Set the graphing utilit to parametric moe an graph the equations an t t t t t t. From either of these two approaches, can ou ecie whether the graph has a tangent line at the point 0,? Eplain our reasoning. Implicit Differentiation Distinguish between functions written in implicit form an eplicit form. Use implicit ifferentiation to fin the erivative of a function. Implicit an Eplicit Functions Up to this point in the tet, most functions have been epresse in eplicit form. For eample, in the equation 5 Eplicit form the variable is eplicitl written as a function of. Some functions, however, are onl implie b an equation. For instance, the function is efine implicitl b the equation. Suppose ou were aske to fin for this equation. You coul begin b writing eplicitl as a function of an then ifferentiating. Implicit Form This strateg works whenever ou can solve for the function eplicitl. You cannot, however, use this proceure when ou are unable to solve for as a function of. For instance, how woul ou fin for the equation where it is ver ifficult to epress as a function of eplicitl? To o this, ou can use implicit ifferentiation. To unerstan how to fin implicitl, ou must realize that the ifferentiation is taking place with respect to. This means that when ou ifferentiate terms involving alone, ou can ifferentiate as usual. However, when ou ifferentiate terms involving, ou must appl the Chain Rule, because ou are assuming that is efine implicitl as a ifferentiable function of. EXAMPLE Eplicit Form Derivative Differentiating with Respect to a. Variables agree: use Simple Power Rule. Variables agree u n nu n b. Variables isagree: use Chain Rule. Variables isagree u c. Chain Rule:. Prouct Rule Chain Rule Simplif.
60_005.q //0 :5 PM Page CHAPTER Differentiation Implicit Differentiation Guielines for Implicit Differentiation. Differentiate both sies of the equation with respect to.. Collect all terms involving on the left sie of the equation an move all other terms to the right sie of the equation.. Factor out of the left sie of the equation.. Solve for. EXAMPLE Implicit Differentiation Fin given that 5. NOTE In Eample, note that implicit ifferentiation can prouce an epression for that contains both an.,, 0, 0 Point on Graph (, ) The implicit equation has the erivative Figure.7 (, ) (, 0) + 5 = 5 8 0 Unefine 5 5. Slope of Graph. Differentiate both sies of the equation with respect to. 5 5 0. Collect the terms on the left sie of the equation an move all other terms to the right sie of the equation. 5. Factor out of the left sie of the equation. 5. Solve for b iviing b 5. 5 5 To see how ou can use an implicit erivative, consier the graph shown in Figure.7. From the graph, ou can see that is not a function of. Even so, the erivative foun in Eample gives a formula for the slope of the tangent line at a point on this graph. The slopes at several points on the graph are shown below the graph. TECHNOLOGY With most graphing utilities, it is eas to graph an equation that eplicitl represents as a function of. Graphing other equations, however, can require some ingenuit. For instance, to graph the equation given in Eample, use a graphing utilit, set in parametric moe, to graph the parametric representations t t 5t, t, an t t 5t, t, for 5 t 5. How oes the result compare with the graph shown in Figure.7?
60_005.q //0 :5 PM Page SECTION.5 Implicit Differentiation + = 0 (0, 0) It is meaningless to solve for in an equation that has no solution points. (For eample, has no solution points.) If, however, a segment of a graph can be represente b a ifferentiable function, will have meaning as the slope at each point on the segment. Recall that a function is not ifferentiable at (a) points with vertical tangents an (b) points at which the function is not continuous. EXAMPLE Representing a Graph b Differentiable Functions (a) (, 0) (, 0) (b) = = = (, 0) = (c) Some graph segments can be represente b ifferentiable functions. Figure.8 If possible, represent as a ifferentiable function of. a. 0 b. c. a. The graph of this equation is a single point. So, it oes not efine as a ifferentiable function of. See Figure.8(a). b. The graph of this equation is the unit circle, centere at 0, 0. The upper semicircle is given b the ifferentiable function, < < an the lower semicircle is given b the ifferentiable function, < <. At the points, 0 an, 0, the slope of the graph is unefine. See Figure.8(b). c. The upper half of this parabola is given b the ifferentiable function, < an the lower half of this parabola is given b the ifferentiable function, <. At the point, 0, the slope of the graph is unefine. See Figure.8(c). EXAMPLE Fining the Slope of a Graph Implicitl Determine the slope of the tangent line to the graph of at the point,. See Figure.9. + = Figure.9 (, ) 8 0 8 So, at,, the slope is. Write original equation. Differentiate with respect to. Solve for. Evaluate when an. NOTE To see the benefit of implicit ifferentiation, tr oing Eample using the eplicit function.
60_005.q //0 :5 PM Page CHAPTER Differentiation EXAMPLE 5 Fining the Slope of a Graph Implicitl Determine the slope of the graph of 00 at the point,. (, ) 00 00 00 00 00 00 00 00 5 5 ( + ) = 00 Lemniscate Figure.0 At the point,, the slope of the graph is 5 5 90 65 5 75 0 5 9 as shown in Figure.0. This graph is calle a lemniscate. EXAMPLE 6 Determining a Differentiable Function π, ( ) π π π sin = The erivative is. Figure. (, π ) Fin implicitl for the equation sin. Then fin the largest interval of the form a < < a on which is a ifferentiable function of (see Figure.). sin cos cos The largest interval about the origin for which is a ifferentiable function of is < <. To see this, note that cos is positive for all in this interval an is 0 at the enpoints. If ou restrict to the interval < <, ou shoul be able to write eplicitl as a function of. To o this, ou can use cos sin an conclue that., < <
60_005.q //0 :5 PM Page 5 SECTION.5 Implicit Differentiation 5 With implicit ifferentiation, the form of the erivative often can be simplifie (as in Eample 6) b an appropriate use of the original equation. A similar technique can be use to fin an simplif higher-orer erivatives obtaine implicitl. Fining the Secon Derivative Implicitl EXAMPLE 7 The Granger Collection Given 5, fin. Differentiating each term with respect to prouces ISAAC BARROW (60 677) The graph in Figure. is calle the kappa curve because it resembles the Greek letter kappa,. The general solution for the tangent line to this curve was iscovere b the English mathematician Isaac Barrow. Newton was Barrow s stuent, an the correspone frequentl regaring their work in the earl evelopment of calculus. 0. Differentiating a secon time with respect to iels 5. Quotient Rule Substitute for. Simplif. Substitute 5 for. Fining a Tangent Line to a Graph EXAMPLE 8 Fin the tangent line to the graph given b at the point,, as shown in Figure.. B rewriting an ifferentiating implicitl, ou obtain 0 (, ) At the point,, the slope is ( + ) = an the equation of the tangent line at this point is The kappa curve Figure. 0..
60_005.q //0 :5 PM Page 6 6 CHAPTER Differentiation Eercises for Section.5 See www.calcchat.com for worke-out solutions to o-numbere eercises. In Eercises 6, fin / b implicit ifferentiation.. 6. 6. 9. 8 5. 6. 7. 8. 9. 0. sin cos. sin cos. sin cos. sin tan. cot 5. sin 6. sec. Bifolium:. Folium of Descartes: Point:, 6 0 Point:, 8 In Eercises 7 0, (a) fin two eplicit functions b solving the equation for in terms of, (b) sketch the graph of the equation an label the parts given b the corresponing eplicit functions, (c) ifferentiate the eplicit functions, an () fin / an show that the result is equivalent to that of part (c). 7. 6 8. 6 9 0 9. 9 6 0. 9 9 In Eercises 8, fin / b implicit ifferentiation an evaluate the erivative at the given point..,,.. 0,,,, 0.,, 5. 5, 8, 6.,, 7. tan, 0, 0 8. cos,, Famous Curves In Eercises 9, fin the slope of the tangent line to the graph at the given point. 9. Witch of Agnesi: 0. Cissoi: 8 Point:, Point:, Famous Curves In Eercises 0, fin an equation of the tangent line to the graph at the given point. To print an enlarge cop of the graph, go to the website www.mathgraphs.com.. Parabola. Circle 8 6 0 6 8 5. Rotate hperbola 6. Rotate ellipse 7. Cruciform 8. Astroi 9 = 0 6 (, ) ( ) = ( ) (, 0) 6 6 = (, ) ( + ) + ( ) = 0 8 6 7 6 + 6 = 0 8 / + / = 5 (8, ) (, ) (, )
60_005.q //0 :5 PM Page 7 SECTION.5 Implicit Differentiation 7 9. Lemniscate 0. Kappa curve ( + ) = 00( ) 6. (a) Use implicit ifferentiation to fin an equation of the tangent line to the ellipse at,. 8 (b) Show that the equation of the tangent line to the ellipse at is 0 a 0 0, 0. a b b. (a) Use implicit ifferentiation to fin an equation of the tangent line to the hperbola at,. 6 8 (b) Show that the equation of the tangent line to the hperbola at is 0 a 0 0, 0 b. a b In Eercises an, fin / implicitl an fin the largest interval of the form a < < a or 0 < < a such that is a ifferentiable function of. Write / as a function of.. tan. cos In Eercises 5 50, fin / in terms of an. 5. 6 6. 7. 6 8. 9. 50. In Eercises 5 an 5, use a graphing utilit to graph the equation. Fin an equation of the tangent line to the graph at the given point an graph the tangent line in the same viewing winow. 5., 9, 5. (, ) 6 6 6 ( + ) =, 5, 5 In Eercises 5 an 5, fin equations for the tangent line an normal line to the circle at the given points. (The normal line at a point is perpenicular to the tangent line at the point.) Use a graphing utilit to graph the equation, tangent line, an normal line. (, ) In Eercises 57 an 58, fin the points at which the graph of the equation has a vertical or horizontal tangent line. 57. 5 6 00 60 00 0 58. 8 0 Orthogonal Trajectories In Eercises 59 6, use a graphing utilit to sketch the intersecting graphs of the equations an show that the are orthogonal. [Two graphs are orthogonal if at their point(s) of intersection their tangent lines are perpenicular to each other.] 59. 6 60. 6. 0 6. sin 5 9 Orthogonal Trajectories In Eercises 6 an 6, verif that the two families of curves are orthogonal where C an K are real numbers. Use a graphing utilit to graph the two families for two values of C an two values of K. 6. C, K 6. C, K In Eercises 65 68, ifferentiate (a) with respect to ( is a function of ) an (b) with respect to t ( an are functions of t). 65. 0 66. 0 67. cos sin 68. sin cos Writing About Concepts 69. Describe the ifference between the eplicit form of a function an an implicit equation. Give an eample of each. 70. In our own wors, state the guielines for implicit ifferentiation. 7. Orthogonal Trajectories The figure below shows the topographic map carrie b a group of hikers. The hikers are in a wooe area on top of the hill shown on the map an the ecie to follow a path of steepest escent (orthogonal trajectories to the contours on the map). Draw their routes if the start from point A an if the start from point B. If their goal is to reach the roa along the top of the map, which starting point shoul the use? To print an enlarge cop of the graph, go to the website www.mathgraphs.com. 5. 5 5.,,, 9 0,,, 5 67 800 55. Show that the normal line at an point on the circle r passes through the origin. 56. Two circles of raius are tangent to the graph of at the point,. Fin equations of these two circles. A B 99 800
60_005.q //0 :5 PM Page 8 8 CHAPTER Differentiation 7. Weather Map The weather map shows several isobars curves that represent areas of constant air pressure. Three high pressures H an one low pressure L are shown on the map. Given that win spee is greatest along the orthogonal trajectories of the isobars, use the map to etermine the areas having high win spee. H L H H 76. Slope Fin all points on the circle 5 where the slope is. 77. Horizontal Tangent Determine the point(s) at which the graph of has a horizontal tangent. 78. Tangent Lines Fin equations of both tangent lines to the ellipse that passes through the point, 0. 9 79. Normals to a Parabola The graph shows the normal lines from the point, 0 to the graph of the parabola. How man normal lines are there from the point 0, 0 to the graph of the parabola if (a) (b) 0 0,, an (c) 0? For what value of 0 are two of the normal lines perpenicular to each other? 7. Consier the equation. (a) Use a graphing utilit to graph the equation. (b) Fin an graph the four tangent lines to the curve for. (c) Fin the eact coorinates of the point of intersection of the two tangent lines in the first quarant. 7. Let L be an tangent line to the curve c. Show that the sum of the - an -intercepts of L is c. 75. Prove (Theorem.) that n n n for the case in which n is a rational number. (Hint: Write pq in the form q p an ifferentiate implicitl. Assume that p an q are integers, where q > 0. ) 80. Normal Lines (a) Fin an equation of the normal line to the ellipse 8 (, 0) = at the point,. (b) Use a graphing utilit to graph the ellipse an the normal line. (c) At what other point oes the normal line intersect the ellipse? Section Project: Optical Illusions In each graph below, an optical illusion is create b having lines intersect a famil of curves. In each case, the lines appear to be curve. Fin the value of / for the given values of an. (a) Circles: C (b) Hperbolas: C,, C 5,, C (c) Lines: a b,, a, b () Cosine curves: C cos,, C FOR FURTHER INFORMATION For more information on the mathematics of optical illusions, see the article Descriptive Moels for Perception of Optical Illusions b Davi A. Smith in The UMAP Journal.