Lecture 13: Implicit and Logarithmic Differentiation

Transcription

1 8 6 Lecture 13: Implicit and Logarithmic Differentiation log(e^)= 4 log(a b) = log(a^b) = 2 b log(a) + log(b) b log(a) Drag To 0 1 log(a/b) = a log(e) = 1 Y=log(X) log(1) = 0 log(a) - log(b)

2 Chapter 3: Derivatives LECTURE TOPIC 10 DEFINITION OF THE DERIVATIVE 11 PROPERTIES OF THE DERIVATIVE 12 DERIVATIVES OF COMMON FUNCTIONS 13 IMPLICIT AND LOGARITHMIC DIFFERENTIATION Lecture 13 Implicit and Logarithmic Differentiation 2

3 Calculus Inspiration Brian Greene 3

4 IMPLICIT DIFFERENTIATION Up to now we have differentiated eplicit functions, described in Lecture 1 as separable functions where y is a function of, r is a function of and so forth. For this case we discovered that: dy d d d f ( ) The properties of derivatives were developed and applied to f() in varied forms. However, many interesting curves are implicit, that is, we cannot separate the variables to obtain all the s on the right and y s on the left. When this is the case we use a variation of the chain rule to find the derivative. For eample consider the implicit equation for the circle: 2 + y 2 - r 2 = 0 Sketching our solution we implicity differentiate both sides and use the chain rule to obtain: d/d ( 2 +y 2 - r 2 ) = 0 d/d ( 2 ) + d/d (y 2 ) - d/d (r 2 ) = 0 2 d/d + 2y dy/d = 0 + y dy/d = 0 dy/d = -/y Lecture 13 Implicit and Logarithmic Differentiation 4

5 IMPLICIT DIFFERENTIATION Lets repeat that symbolic calculation using vertical coherence, commenting each step: d/d ( 2 + y 2 - r 2 ) = 0 Take Derivative Termwise d/d ( 2 ) + d/d (y 2 ) = 0 r is Constant, Deriv is 0 2 d/d + 2y dy/d = 0 Chain Rule to Each Term d/d + y dy/d = 0 Divide Through By 2 + y dy/d = 0 d/d = 1 y dy/d = - Sub. from Both Sides dy/d = -/y Divide Both Sides By y This same process can be applied to any implicit equation. 5

6 25 Implicit Derivative of Circle: Lecture13-CircleImplicitDeriv.g y Þ ~ f(,y)=0 10 EXERCISES: 1) Drag f(,y) = 0 to sweep out circle. 2) Drag r to change circle size. 3) Drag T and observe the values of -/y. 4) Does the slope of the tangent line depend on the radius of the circle? 5 r T -z 0 dy/d = -/y Þ = z 1 I O Þ ~ Lecture 13 Implicit and Logarithmic Differentiation 6

7 Lecture13-ImplicitDerivBifolium.wm IMPLICIT DIFFERENTIATION OF BIFOLIUM Open and run this file in wmaima to find the slope of the implicit bifolium curve. The resulting slope is an implicit equation in and y! 7

8 IMPLICIT DERIVATIVE IN POLAR COORDINATES Lecture13-ImplicitDerivBifoliumPolar.wm Perhaps the derivative in polar coordinates will be eplicit, running the net file gives us: Again dy/d is a function of both r and, which is still implicit. Lecture 13 Implicit and Logarithmic Differentiation 8

9 BIFOLIUM TANGENT LINE AUTOMATICALLY Since we can define the bifolium as an eplicit function of we can draw it in Geometry Epressions. We can then select the curve, choose the tangent icon and the tangent line is computed and displayed automatically. Try it yourself. 9 For the general case, the interconversion of the three representions, eplicit, implicit and parmetric is a research topic in advance mathematics.

10 Implicit 14Derivative of Bifolium: Lecture13-BifoliumImplicitDeriv.g EXERCISES: 1) Drag tangent T and observe the values of dy/d. 2) How is this curve different than the one in Lecture 2? 3) What impact does this have on the scale factor a? y Þ ~ r=4 a sin(t) cos(t) 2 T 4 2 a H z 4 dy/d = Þ z 3 I K J -2-2 O Þ ~16.4 Lecture 13 Implicit and Logarithmic Differentiation 10

11 2.0 Implicit Derivative of Atlas Curves: Cardioid: Eplicit Polar Form: Implicit Cartesian Form: B 0.5 θ A EXERCISES: For any four curves in Lecture 2 1) Add a tangent line. 2) Compute the derivative of the implicit form. 3) Add a control for parameters such as a above and drag it a (1+cos(θ)) 11

12 LOGARITHMIC DIFFERENTIATION Just as the chain rule provides a tool for differentiating implicit functions, logarithmic differentiation provides a tool for differentiating eponential functions which do not fit into any of the categories we have studed. One eample of this is finding the derivative of: y The power rule will not work for this because in the equation: n y The eponent n is considered constant. The solution is logarithmic differentation, but for that to work, we must first review the properties of logarithms and we will do that with Geometry Epressions. Note that natural log is written as log(). Other sources may use ln() for natural log. Lecture 13 Implicit and Logarithmic Differentiation 12

13 8 Properties of Logarithms EXERCISES: 1) Find log(1) = 0 property. 2) Find 6 log(e) = 1 property. 3) Find log(e ) = property. 4) Find log(a b ) = b log(a) property. 5) Find log(a b) = log(a) + log(b) property. 6) Find log(a/b) = log(a) - log(b) property. 7) Drag point b towards point a. 4 8) Drag the black dot to 0. log(e^)= log(a b) = Lecture13-LogProperties.g log(a^b) = 2 b log(a) + log(b) b log(a) Drag To 0 1 log(a/b) = a log(e) = 1 Y=log(X) log(1) = log(a) - log(b)

14 LOGARITHMIC DIFFERENTIATION OF AN EXPLODING FUNCTION Now that we have the properties of logarithms in hand, let s find the derivative of: y The trick is to take the log of BOTH SIDES, and use the property of logs: log y log( ) log( ) Now we differentiate both sides using the product rule to obtain: 1 y 1 log( ) y Multiplying both sides by y yields: y y(1 log( )) Back substituting the original function of y provides the eplicit result: y (1 log( )) Which we can now eamine in Geometry Epressions. Lecture 13 Implicit and Logarithmic Differentiation 14

15 Log Derivative of Eploding Function 2.5 Lecture13-LogDiffEploding.g EXERCISES: 1) Drag the blue dot to move the tangent line ) What are the values being computed? 3) At what is the slope of the tangent line zero? 4) What is the value of y at this? 5) Which increases faster, the function or its derivative? 1.5 6) How would you prove that? y 1.0 Y=X X ' Y=X X (1+log(X)) z 1 Þ ~ z 1 z 0 Þ O z 0 Þ ~

16 LOGARITHMIC DIFFERENTIATION OF A SLOW FUNCTION Functions like 1, and log() change slowly relative to linear functions for large values of. We need logarithms to differentiate increasing functions like: log( ) y log( ) Lecture13-LogDerivSlow.wm Using wmaima to manage epression compleity we run the problem and obtain: Which we can now eamine in Geometry Epressions. Lecture 13 Implicit and Logarithmic Differentiation 16

17 Log Derivative of Slow Function Lecture13-LogDiffSlow.g EXERCISES: 1) Drag the blue dot to move the tangent line. 2) What are the values being computed? 3) 3 Zoom out and assess how slow the function really is. 4) Which is slower the slow function or its derivative? 5) Is a slow function raised to a slow power still slow? 4 2 y 1 z 1 Þ ~1.12 O Y=log(X) log(x) z 1 z 0 Þ z 0 Þ ~ ' Y= 1 X + log(log(x)) X log(x) log(x) 17

18 LOGARITHMIC DIFFERENTIATION OF A PERIODIC FUNCTION We can logarithmically differentiate the periodic function: 1 y sin( ) Lecture13-LogDerivPeriodic.wm Using wmaima to manage epression compleity we run the problem and obtain: We now eamine the periodic function and its derivative in Geometry Epressions : Lecture 13 Implicit and Logarithmic Differentiation 18

19 Log Derivative of Periodic Function Lecture13-LogDiffPeriodic.g EXERCISES: 1) Drag the blue dot across different sections of the function. 2) What are the values of the function between sections? 3) How would you evaluate them? δ y δ Þ y O Y=sin(X) δ y Þ ~ X 2009 L. Van Warren / wdv.com / All Rights Reserved Y'= -log(sin(x)) X 2 + cos(x) 1 X sin(x) sin(x) X 19

20 End Lecture 13 Implicit and Logarithmic Differentiation 20