4.2 Implicit Differentiation

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1 6 Chapter 4 More Derivatives 4. Implicit Differentiation What ou will learn about... Implicitl Define Functions Lenses, Tangents, an Normal Lines Derivatives of Higher Orer Rational Powers of Differentiable Functions an wh... Implicit ifferentiation allows us to fin erivatives of functions that are not efine or written eplicitl as a function of a single variable. Implicitl Define Functions The graph of the equation = 0 (Figure 4.7) has a well-efine slope at nearl ever point because it is the union of the graphs of the functions = f (), = f (), an = f (), which are ifferentiable ecept at O an A. But how o we fin the slope when we cannot convenientl solve the equation to fin the functions? The answer is to treat as a ifferentiable function of an ifferentiate both sies of the equation with respect to, using the ifferentiation rules for sums, proucts, an quotients, an the Chain Rule. Then solve for > in terms of an together to obtain a formula that calculates the slope at an point (, ) on the graph from the values of an. The process b which we fin > is calle implicit ifferentiation. The phrase erives from the fact that the equation = 0 f f f efines the functions,, an implicitl (i.e., hien insie the equation), without giving us eplicit formulas to work with. 9 0 ( 0, ) ( 0, ) f () A f () O 0 Figure 4.7 The graph of = 0 (calle a folium). Although not the graph of a function, it is the union of the graphs of three separate functions. This particular curve ates to Descartes in 68. ( 0, ) f () EXAMPLE Differentiating Implicitl Fin > if =. To fin >, we simpl ifferentiate both sies of the equation = with respect to, treating as a ifferentiable function of an appling the Chain Rule: Slope = = 4 P(4, ) Q(4, ) Slope = = 4 Figure 4.8 The erivative foun in Eample gives the slope for the tangent lines at both P an Q, because it is a function of. = = =. ( ) = ( ) # Now Tr Eercise. In the previous eample we ifferentiate with respect to, an et the erivative we obtaine appeare as a function of. Not onl is this acceptable, it is actuall quite useful. Figure 4.8, for eample, shows that the curve has two ifferent tangent lines when = 4: one at the point ( 4, ) an the other at the point (4, -). Since the formula for > epens on, our single formula gives the slope in both cases. Implicit ifferentiation will frequentl iel a erivative that is epresse in terms of both an, as in Eample.

2 Section 4. Implicit Differentiation 6 0 (, 4) Slope 4 Figure 4.9 The circle combines the graphs of two functions. The graph of is the lower semicircle an passes through (, -4). (Eample ) EXAMPLE Fining Slope on a Circle Fin the slope of the circle + = at the point (, -4). The circle is not the graph of a single function of, but it is the union of the graphs of two ifferentiable functions, an = - - = - (Figure 4.9). The point (, -4) lies on the graph of, so it is possible to fin the slope b calculating eplicitl: But we can also fin this slope more easil b ifferentiating both sies of the equation of the circle implicitl with respect to : The slope at (, -4) is ` - = - ` -6 = - = - = - 9 = 4. ( + ) = () + = 0 = -. - ` = - (, -4) -4 = 4. Differentiate both sies with respect to. The implicit solution, besies being computationall easier, iels a formula for > that applies at an point on the circle (ecept, of course, (, 0), where slope is unefine). The eplicit solution erive from the formula for applies onl to the lower half of the circle. Now Tr Eercise. To calculate the erivatives of other implicitl efine functions, we procee as in Eamples an. We treat as a ifferentiable function of an appl the usual rules to ifferentiate both sies of the efining equation. Ellen Ochoa (98 ) After earning a octorate egree in electrical engineering from Stanfor Universit, Ellen Ochoa became a research engineer an, within a few ears, receive three patents in the fiel of optics. In 990, Ochoa joine the NASA astronaut program, an, three ears later, became the first Hispanic female to travel in space. Ochoa s message to oung people is: If ou sta in school ou have the potential to achieve what ou want in the future. EXAMPLE Solving for / Show that the slope > is efine at ever point on the graph of = + sin. First we nee to know >, which we fin b implicit ifferentiation: = + sin () = ( + sin ) = ( ) + ( sin ) = + cos - (cos ) = ( - cos ) = = - cos. Differentiate both sies with respect to......treating as a function of an using the Chain Rule. Collect terms with > an factor out >. Solve for > b iviing. The formula for > is efine at ever point (, ), ecept for those points at which cos =. Since cos cannot be greater than, this never happens. Now Tr Eercise.

3 64 Chapter 4 More Derivatives Normal line A Light ra P Tangent Point of entr B Curve of lens surface Implicit Differentiation Process. Differentiate both sies of the equation with respect to.. Collect the terms with > on one sie of the equation.. Factor out >. 4. Solve for >. Figure 4.0 The profile of a lens, showing the bening (refraction) of a ra of light as it passes through the lens surface. Normal (, ) Tangent 7 Figure 4. Tangent an normal lines to the ellipse - + = 7 at the point (-, ). (Eample 4) Lenses, Tangents, an Normal Lines In the law that escribes how light changes irection as it enters a lens, the important angles are the angles the light makes with the line perpenicular to the surface of the lens at the point of entr (angles A an B in Figure 4.0). This line is calle the normal to the surface at the point of entr. In a profile view of a lens like the one in Figure 4.0, the normal is a line perpenicular to the tangent to the profile curve at the point of entr. Profiles of lenses are often escribe b quaratic curves (see Figure 4.). When the are, we can use implicit ifferentiation to fin the tangents an normals. EXAMPLE 4 Tangent an normal to an ellipse Fin the tangent an normal to the ellipse - + = 7 at the point (-, ). (See Figure 4..) We first use implicit ifferentiation to fin > : - + = 7 ( ) - () + ( ) = (7) - a + b + = 0 ( - ) = - = - -. Differentiate both sies with respect to treating as a prouct an as a function of. Collect terms. Solve for >. We then evaluate the erivative at = -, = to obtain The tangent to the curve at (-, ) is ` = - (-, ) - ` (-, ) = - (-) () - (-) = 4. - = 4 ( - (-)) = continue

4 Section 4. Implicit Differentiation 6 The normal to the curve at (-, ) is - = - ( + ) 4 = Now Tr Eercise 7. Derivatives of Higher Orer Implicit ifferentiation can also be use to fin erivatives of higher orer. Here is an eample. EXAMPLE Fining a Secon Derivative Implicitl Fin > if - = 8. To start, we ifferentiate both sies of the equation with respect to in orer to fin =>. ( - ) = (8) 6-6 =0 - =0 We now appl the Quotient Rule to fin. =, when Z 0 = a b = - = - # Finall, we substitute = / to epress in terms of an. = - a b = - 4, when Z 0 Now Tr Eercise 9. EXPLORATION An Unepecte Derivative Consier the set of all points (, ) satisfing the equation - + = 4. What oes the graph of the equation look like? You can fin out in two was in this Eploration.. Use implicit ifferentiation to fin >. Are ou surprise b this erivative?. Knowing the erivative, what o ou conjecture about the graph?. What are the possible values of when = 0? Does this information enable ou to refine our conjecture about the graph? 4. The original equation can be written as ( - ) - 4 = 0. B factoring the epression on the left, write two equations whose graphs combine to give the graph of the original equation. Then sketch the graph.. Eplain wh our graph is consistent with the erivative foun in part.

5 66 Chapter 4 More Derivatives Rational Powers of Differentiable Functions We know that the Power Rule n = n n- hols for an integer n (Rules an 7). We can now prove that it hols when n is an rational number. RULE 9 Power Rule for Rational Powers of If n is an rational number, then n = n n-. If n 6, then the erivative oes not eist at = 0. Proof Let p an q be integers with q 7 0 an suppose that = q p = p>q. Then Since p an q are integers (for which we alrea have the Power Rule), we can ifferentiate both sies of the equation with respect to an obtain If Z 0, we can ivie both sies of the equation b to solve for >, obtaining = pp- q q- = p q # p- ( p>q ) q- = p q # p- p-p>q = p q # (p-)-(p-p>q) = p q # (p>q)-. q = p. q- q = pp-. = p>q p q q- q (q - ) = p - p q A law of eponents This proves the rule. B combining this result with the Chain Rule, we get an etension of the Power Chain Rule to rational powers of u: n If n is a rational number an u is a ifferentiable function of, then u is a ifferentiable function of an un n- u = nu, provie that u Z 0 if n 6. The restriction that u Z 0 when n 6 is necessar because 0 might be in the omain of u n but not in the omain of u n-, as we see in the first two parts of Eample 6.

6 Section 4. Implicit Differentiation 67 EXAMPLE 6 Using the Rational Power Rule (a) () = (> ) = -> = Notice that is efine at = 0, but >( ) is not. (b) (> ) = (-> ) = > The original function is efine for all real numbers, but the erivative is unefine at = 0. Recall Figure., which showe that this function s graph has a cusp at = 0. (c) (cos )-> = - (cos # )-6> (cos ) = - (cos )-6> (-sin ) = sin (cos )-6> Now Tr Eercise. Quick Review 4. (For help, go to Section. an Appeni A..) Eercise numbers with a gra backgroun inicate problems that the authors have esigne to be solve without a calculator. In Eercises, sketch the curve efine b the equation an fin two functions an whose graphs will combine to give the curve.. - = = = = 9. + = + In Eercises 6 8, solve for in terms of an = 4-7. sin - cos = + 8. ( - ) = ( - ) In Eercises 9 an 0, fin an epression for the function using rational powers rather than raicals. 9. ( - ) 0. + Section 4. Eercises In Eercises 8, fin >.. + = 6. + = 8. = - 4. = = tan 6. = sin 7. + tan () = sin = In Eercises 9, fin > an fin the slope of the curve at the inicate point =, (-, ) 0. + = 9, (0, ). ( - ) + ( - ) =, (, 4). ( + ) + ( + ) =, (, -7) In Eercises 6, fin where the slope of the curve is efine.. - = 4 4. = cos. + = = 6 In Eercises 7 6, fin the lines that are (a) tangent an (b) normal to the curve at the given point =, (, ) 8. + =, (, -4) 9. = 9, (-, ) = 0, (-, ) = 0, (-, 0). - + =, (, ). + p sin = p, 4. sin = cos,. = sin (p - ), (, p>) (p>4, p>) (, 0) 6. cos - sin = 0, (0, p) In Eercises 7 0, use implicit ifferentiation to fin > an then. > 7. + = 8. > + > = 9. = = + In Eercises 4, fin >.. = 9>4. = ->. = 4. = 4. = ( + ) -> 6. = ( - 6) > 7. = + 8. = + 9. = = ( + ) -> 4. = (csc ) > 4. = [sin ( + )] >4

7 68 Chapter 4 More Derivatives 4. Which of the following coul be true if f () = ->? (a) (c) f() = > - f () = - -4> 0, 4, The Cissoi of Diocles (ates from about 00 B.C.E.) (a) Fin equations for the tangent an normal to the cissoi of Diocles, ( - ) =, at the point (, ) as picture below. (b) Eplain how to reprouce the graph on a grapher. (b) () ( ) f() = 9 0 > - 7 f () = > Which of the following coul be true if g t t 4? (a) g t 44 t 4 (b) g t 4> 4 t (c) g t t 7 6 t > 4 () g t 4 t > 4 4. The Eight Curve (a) Fin the slopes of the figure-eightshape curve 4 = - at the two points shown on the graph that follows. (b) Use parametric moe an the two pairs of parametric equations (t) = t - t 4, (t) = t, (t) = -t - t 4, (t) = t, to graph the curve. Specif a winow an a parameter interval. > 47. (a) Confirm that (-, ) is on the curve efine b = cos (p). (b) Use part (a) to fin the slope of the line tangent to the curve at (-, ). 48. Group Activit (a) Show that the relation - = - cannot be a function of b showing that there is more than one possible -value when =. (b) On a small enough square with center (, ), the part of the graph of the relation within the square will efine a function = f(). For this function, fin f () an f (). 49. Fin the two points where the curve + + = 7 crosses the -ais, an show that the tangents to the curve at these points are parallel. What is the common slope of these tangents? 0. Fin points on the curve + + = 7 (a) where the tangent is parallel to the -ais an (b) where the tangent is parallel to the -ais. (In the latter case, > is not efine, but > is. What value oes > have at these points?). Orthogonal Curves Two curves are orthogonal at a point of intersection if their tangents at that point cross at right angles. Show that the curves + = an = are orthogonal at (, ) an (, -). Use parametric moe to raw the curves an to show the tangent lines.. The position of a bo moving along a coorinate line at time t is s = (4 + 6t) >, with s in meters an t in secons. Fin the bo s velocit an acceleration when t = sec.. The velocit of a falling bo is v = 8s - t + feet per secon at the instant t (sec) the bo has fallen s feet from its starting point. Show that the bo s acceleration is ft/sec. 4. The Devil s Curve (Gabriel Cramer [the Cramer of Cramer s Rule], 70) Fin the slopes of the evil s curve 4-4 = 4-9 at the four inicate points. (, ) (, ) (, ) (, ) 0 (, ). The Folium of Descartes (See Figure 4.7 on page 6) (a) Fin the slope of the folium of Descartes, = 0 at the points (4, ) an (, 4). (b) At what point other than the origin oes the folium have a horizontal tangent? (c) Fin the coorinates of point A in Figure 4.7, where the folium has a vertical tangent.

8 Section 4. Implicit Differentiation The line that is normal to the curve + - = 0 at (, ) intersects the curve at what other point? 7. Fin the normals to the curve + - = 0 that are parallel to the line + = Show that if it is possible to raw these three normals from the point ( a, 0) to the parabola = shown here, then a must be greater than >. One of the normals is the -ais. For what value of a are the other two normals perpenicular? 0 (a, 0) Stanarize Test Questions 9. True or False The slope of + = at (>, ) is. Justif our answer. 60. True or False The erivative of = is. Justif our answer. > In Eercises 6 an 6, use the curve - + =. 6. Multiple Choice Which of the following is equal to >? - + (A) (B) (C) - - (D) (E) 6. Multiple Choice Which of the following is equal to (A) ( - ) + (B) ( - ) (C) (D) (E) ( - ) ( - )? 6. Multiple Choice Which of the following is equal to > if = >4? 4 (A) (B) (C) (D) (E) 4 4 > 64. Multiple Choice Which of the following is equal to the slope of the tangent to - = at (, )? (A) - (B) - (C) (D) (E) 0 Etening the Ieas 6. Fining Tangents (a) Show that the tangent to the ellipse a + b = at the point (, ) has equation. a + b = (b) Fin an equation for the tangent to the hperbola a - b = at the point (, ). 66. En Behavior Moel Consier the hperbola Show that 4 >4. a - b = (a) = b. a - a (b) g() = (b>a) ƒ ƒ is an en behavior moel for f() = (b>a) - a. (c) g() = -(b>a) ƒ ƒ is an en behavior moel for f() = -(b>a) - a. >4 >4 4 >4 Quick Quiz for AP* Preparation: Sections Multiple Choice Which of the following gives for = sin 4 ()? (A) (C) (E) 4 sin () cos () sin () cos () - sin () cos () (B) (D) sin () cos () sin (). Multiple Choice What is the slope of the line tangent to the curve - = - 6 at the point (, )? (A) 0 (B) (C) (D) (E) Multiple Choice Which of the following gives for the parametric curve = sin t, = cost t? (A) - (B) (C) - cos t cos t tan t (D) (E) tan t tan t 4. Free Response A curve in the -plane is efine b - = 6. (a) Fin. (b) Fin an equation for the tangent line at each point on the curve with -coorinate. (c) Fin the -coorinate of each point on the curve where the tangent line is vertical.