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Introduction to Differential Equations

Introduction to Differential Equations

– An Introduction

Concepts of primary interest: Definition: first Differential of a Left-side and right-side limits Higher derivatives Mean value and Rolle’s theorem L’Hospital Rule for indeterminate forms Implicit differentiation Inverse functions Method for Products Extrema and the Test for functions of two variables Lagrange Undetermined Multipliers (*** needs serious edit ****) Sample calculations: 2 Derivative of x Saddle-points and local extrema Derivative of the inverse function for Squared d(sinθ) /dθ = 1 for θ = 0 Applications Closest point of approach or CPA (nautical term) Tools of the Trade Derivatives of inverse functions

Contact: [email protected] 11/1/2012 Plotting inverse functions

***ADD a table of the derivatives of common functions See Tipler; Lagrange

Approximating Functions Many functions are complicated, and their evaluations require detailed, tortuous calculations. To avoid the stress, approximate representations are sometimes substituted for the actual functions of interest. For a f(x), it might be adequate to replace the function in a small interval around x0 by its value at that point f(x0). A better approximation for a function f(x) with a continuous derivative follows from the definition of that derivative.

df fx()− fx ( ) = 0 [DI.1] x0 xx→ dx 0 x− x0

df df Clearly, fx()≅+ fx ( ) ( x − x) where is the of the function at x0.. 00dx dx x0 x0

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Figure DI-1 This handout formalizes the definition of a derivative and illustrates a few basic properties and applications of the derivative. The derivative of the function at an general point x is to be defined as: The limit of the slope of the chord df() x fx()− fx () = Limit 21 xx, → x dx 12 (x21− x This relation requires that the limit exist and be the same for all limit

paths with x1 and x2 approaching x. This condition means that the slope of the chord must converge to the

same value no matter how x1 and x2 approach x. This includes arbitrary one-sided approaches

dfx() fx (+∆ x ) − fx () fx (+∆ x ) − fx () = Limit = Limit  [DI.2] dx ∆→xx00()x+∆ x − x ∆→ ∆x

11/1/2012 Physics Handout Series.Tank: Derivatives D/DX- 3 and the left-side (from smaller values) and right-side (from larger values) approaches which must exist and be equal. dfx() fx (+∆ x ) − fx () fx (+∆ x ) − fx () = Limit = Limit −+ [DI.3] dx ∆→xx00∆∆x ∆→ x left−− side right side At minimum, the function f(x) must be continuous at a point if its derivative exists at that point.

As before, the derivative is useful for representing the change in the value of the function corresponding to an change in its argument. df df= f( x +− dx ) f () x ≅ dx [DI.4] dx The infinitesimal change df is called the differential of f(x).

The function plotted to the right is smooth except in a neighborhood of x0. (The function has a cusp at x0.) The derivative of the function is not defined at that point. Which condition required for the derivative to be defined is not met at x0?

Smoothness: A function is smooth if it is continuous and continuously differentiable. If the function is continuous and continuously differentiable through order n, it is n- smooth. Differentiating an n-smooth function yields a result that is (n-1)-smooth while integration yields an (n+1)-smooth result.

Cauchy’s : If a function is continuous and has a continuous

11/1/2012 Physics Handout Series.Tank: Derivatives D/DX- 4 derivative in an interval [a, b], then there exists at least one point c in that interval such that: df f() b− f () a = dxxc= b− a That is: For a continuously , the derivative attains its average value over a closed interval somewhere in that interval. Rolle’s theorem is the special f(b) = f(a) case for which the derivative must vanish somewhere in the interval.

Linear Operation: The derivative operation is linear. d dd [afx()+= bgx ()] a[ f () x] + b[ gx ()] dx dx dx

Sample Calculation: Compute the derivative of x2. (We assume that x is the independent variable.)

dx()2  ( x+∆ x )22 − x 2x∆ x + ( ∆ x )2  =Limit  =Limit =Limit[22 x +∆ x] = x [DI.5] dx ∆→xxx000()x+∆ x − x ∆→ ∆x ∆→ Vital Rule: The Chain Rule (Know this one by heart!) Start: F(x) = f(u(x)) , a function of a function of x.

dfux(()) fux ((+∆ x )) − fux (()) fux((+∆ x )) − fux (()) ux (+∆ x ) − ux () = Limit = Limit  dx ∆→xx00∆x ∆→ ux(+∆ x ) − ux () ∆x

df(()) u x f ( u+∆ u ) − f () u u( x +∆ x ) − u () x  df  du = Limit * Limit  =  [DI.6] dx ∆→xx00()u+∆ u − u ∆→  ∆x du  dx The derivative of a function f of a function u(x) is the derivative of f with respect to its argument the derivative of its argument with respect to x. The of change of f(u(x)) with respect to x is the rate of change of f(u) with respect to u times the rate of change of u(x) with respect to x. Given the importance of the Chain Rule, it is to be df(()) u x stated one more . is the rate of change in the value of f(u(x)) with respect dx

11/1/2012 Physics Handout Series.Tank: Derivatives D/DX- 5 df() u to changes in the value of x which is the rate of change of the function f(u) with du respect to changes in the value of its argument multiplied by the rate of change of the df(()) u x df du argument u with respect to changes in the value of x. =  dx du dx

Sample Calculation: The Chain Rule f(u) = eu ; u(x) = ln(x). Examine: df(()) u x df du = . dx du dx

u df u d[ln(x)] -1 f(u) = e ⇒ /du = e ; u(x) = ln(x) ⇒ /dx = x . Hence

df(()) u x ln( x ))  1  1 =ex =  =1 dx x  x The result is expected as e ln(x) = x.

Sample Calculation: The Product Rule Start: F(x) = f(x) g(x) d( f () xgx ()) f ( x+∆ xgx ) ( +∆ x ) − f () xgx () = Limit  dx ∆→x 0 ∆x f( x+∆ xgx ) () − f () xgx () f ( x+∆ xgx ) ( +∆ x ) − f ( x +∆ xgx ) () = Limit + ∆→x 0 ∆∆xx f( x+∆ x ) − f () x  gx(+∆ x ) − gx () =Limit g() x+ Limit[ f ( x +∆ x )] Limit  ∆→x0∆∆xx ∆→xx00∆→ 

d( f () xgx ()) d ( f ()) x d( gx ()) ⇒=+gx() f () x [DI.7] dx dx dx The rule can be understood in terms of a rectangle representing the product f(x+∆x) g(x+∆x) – f(x g(x) = (f + ∆f) (g +∆g) – f g = f ∆g + ∆f g + ∆f ∆g

Full Area = (f + ∆f) (g +∆g) The increase in the product is f ∆g + ∆f g + ∆f ∆g where the piece ∆f ∆g is second order in small ∆g ∆f (∆f ∆g) things. The limit of f /∆x + /∆x g + /∆x dg df df becomes [ f /dx + /dx g + /dx (dg)] which

11/1/2012 Physics Handout Series.Tank: Derivatives D/DX- 6 dg df approaches f /dx + /dx g as dx and dg go to zero. Used: g(x) is a continuous function. ∆(fg) = f ∆g + ∆f g + ∆f ∆g ≈ f ∆g + ∆f g

Logarithmic Derivative Method for Products It can be tedious to apply the product rule to f(x), a product of several factors. One can take the derivative of the natural log of f(x).

d 1 df df d (ln[fx ( )]) = . It follows that: = fx( ) (ln[ fx ( )]) . dx f() x dx dx dx

df x2 x +1 Exercise: Compute forfx ( ) = using the product and quotient rules. dx (x2+ 3) 2/3 Repeat using the logarithmic derivative approach.

Higher Order Derivatives: A of f(x) w.r.t. x is the first derivative of the first derivative of f(x) and so on.

Partial Derivative: A function of several variables requires some generalization of the concept of derivative. Consider a function of three variables: f(x,y,t). The is introduced to represent the rate of change of the value of the function when only one variable is incremented with respect to the variable incremented. To designate that only one argument of the list is to be incremented, the ‘straight-back’ derivative symbol is replaced by the ‘curly-back’ derivative symbol.

More formally, the symbol ∂ is known as Jacobi’s . It is pronounced as ‘curly dee’ or partial.

The definition of the partial derivative of f(x,y,t) with respect to x is:

11/1/2012 Physics Handout Series.Tank: Derivatives D/DX- 7 ∂fxyt(,,) fx (+∆ xyt ,,) − fxyt (,,) = Limit  [DI.8] ∂ ∆→x 0 +∆ − x y, t fixed ()x xx For our purposes the partial derivative with respect to x is defined just as the derivative with respect to x with the understanding that the other arguments are to be treated as constants. This definition leads to the partial differential w.r.t. x which is ∂f dx and the total differential of f(x,y,t): ∂x ∂∂∂fff df=++ dx dy dt ∂∂∂xyt The : Given a function of several variables such as f(x,y,t), the total rate of change of the with respect to one variable is needed when the other arguments have an implicit dependence on that variable. That is: for f(x(t),y(t), t), df∂∂∂ f dx f  dy f =++  [DI.9] dt∂∂∂ x dt y  dt t Note that this expression includes all the variation in the value of f(x,y,t) when t varies.

df ∂f Note: that the total time derivative dt and the partial time derivative ∂t of a function can only be assumed to be equal if the function f is a function of t only. See problem 4 for an example.

The Dot Notation: Total derivative w.r.t. time Time derivatives are ubiquitous leading physicists to develop a shorthand notation to represent the total derivative with respect to time, the dot notation. df(, x y ,) t ∂∂∂f f f →f ⇒= f xy + + [DI.10] dt ∂∂x y ∂ t

An "overdot" is a raised dot appearing above a symbol most commonly used in dx mathematics to indicate a derivative taken with respect to time (e.g., x = dt ).

11/1/2012 Physics Handout Series.Tank: Derivatives D/DX- 8 The expression is voiced " dot," and was Newton’s notation for derivatives (which he called “”) Eric W. Weisstein. "Overdot." From MathWorld – Wolfram Web Resource. http://mathworld.com/Overdot.html

Clairaut’s Theorem: Given functions with continuous partials through second ∂∂22ff order, the second order cross partials are equal. = ∂∂xy ∂∂ yx

Implicit Differentiation: Given a function of two variables that equals a constant, it follows that the variation (total differential) of that function is zero. F(x,y) = C implies that the total variation of F(x,y) = 0. ∂F ∂∂FF dy ∂x dF=+= dx dy 0 or that = −∂ . ∂∂xy dx F ∂y ∂F dy ∂x = −∂ given that F(x,y) = C. [DI.11] dx F ∂y

2 ∂F ∂F Suppose that F(x,y) = (x – 2 y) = 16. =2(xy − 2 ) and =−−4(xy 2 ) so ∂x ∂y dy2( x− 2 y ) 1 =−=. This result is verified by taking the square root of the original dx−−4( x 2 y ) 2

dy 1 yielding (xy− 2) =±⇒= 4 y1 x 2from which = follows trivially. 2 dx 2 dy Implicit differentiation allows one to compute without first solving to find the dx explicit form of y(x).

Exercise: Given that F(x,y) = C, the total variation of F(x,y) = 0; so any total dF derivative of F(x,y) must also be zero. Give the expression for dx and then use the

11/1/2012 Physics Handout Series.Tank: Derivatives D/DX- 9 result in equation [DI.11] to show that the total derivative with respect to x vanishes, dF dF dx = 0. The result demonstrates that the total derivative dx is not equivalent to the ∂F partial derivative ∂x .

The Slope of a Parameterized Curve: For a parameterized curve, x and y are given as functions of an independent parameter to be labeled t. Given x(t) and y(t),

dy dy dy dt   =dt  = dt  [DI.12] dxdt  dx  dx dt dt  Consider a circle of radius 1 centered on the origin. The parameterized representation

dy−cos( t ) is: x(t) = cos(t) and y(t) = sin(t). It follows that: = = −cot(t ) . dxsin( t ) Verify that the formula above correctly represents the slope at each point on the circle. Recall that t is measured in in the CCW sense from the +x-axis.

Logarithmic Differentiation: Use when the function to be differentiated contains an exponential or has an extremely large value: Take the log first; Take the derivative; Recover the desired result. d 1 df df d (ln(()fx) = ⇒=fx()( ln(() fx) [DI.13] dx f() x dx dx dx

x da( ) d For f(x) = ax: = ax ( xaln( )) = ax ln(a)  dx dx Another technique changes the representation the function f(x) = ax = (eln(a))x. The previous result is quickly verified using this alternative approach:

ddx dx( ln( a )) (eln(a ) ) =( exaln( )) = e xa ln( ) = eaxaln( ) ln( ) = ax ln(a) dx ( ) dx dx

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Special Function: ex.

dex The natural base e is defined by the property: = ex . This property is equivalent to dx

e∆x −1 choosing that value of e such that Limit =1. ∆→x 0 ∆x

x ∆x de x e −1 Exercise: Show that = e is equivalent to Limit =1. To verify, numerically dx ∆→x 0 ∆x evaluate the quantity in brackets for ∆x = {0.1, 0.01, 0.001, 0.0001}. [Use the definition of the derivative.]

N x The number e is an irrational equal to: Limit 1+≈1 2.718281828.... It follows that e N →∞ N

N 100,000 = Limit 1+ x . For comparison, [1 + 0.00001] = 2.718268… . N →∞ N

dd d d Sample Calculation: Afx()= e ln()Afx () = e ln() Afx () (ln( Afx ) ( )) = ln( Ae ) ln()Afx () ( fx()) dx dx dx dx dd Afx()= ln( AA )fx() ( fx ( )) dx dx

Symmetry: The derivative of an even function is an odd function. Assume that f(x) is an even function. ⇒ f(x) = f(-x) dfx() fx (+ ½)εε −− fx ( ½) = Limit  dx ε →0 ε

dfx(− ) fx (−+ ½)εε − fx ( −− ½) fx(+− ½)εε − fx ( ++ ½)  = → Limit  Use:fx ( )= f ( − x ) Limit  dx εε→→00εε   df()− x f( ++ x ½)εε − f ( +− x ½)  df () x = − Limit = − dx ε →0 ε dx

11/1/2012 Physics Handout Series.Tank: Derivatives D/DX- 11 It follows that the derivative of an even function is an odd function.

Test for Extrema: The maximum and minimum values of functions provide clues as to the behavior of systems modeled with mathematics. If a continuous function is to have a maximum value at x0, its first derivative must vanish at xo. The derivative is, after all, the rate of change with respect a change in the argument. If the function has been increasing in value, it must quit increasing if it has reached its maximum value.

The First : A function can only have an extrema at a point if its derivative vanishes at that point. The first derivative must vanish at a point if the function is to have an extrema at that point. In df the case that = 0 , dx the second derivative determines the nature of the extrema.

df d2 f ≈ + −+1 −2 fx()()() fx00 x x 2 2 ()x x 0 dx x dx 0 x0

d2 f df =−≈1 −2 = df fx( ) fx (00 )2 2 (x x ) for 0 dx dx x x0 0 If the function satisfies the first derivative test at a point, then the nature of that extrema is determined by the second derivative. The value is a local minimum if the

11/1/2012 Physics Handout Series.Tank: Derivatives D/DX- 12 second derivative is positive and a local maximum if it is negative. The extreme value must still be computed. The first derivative test provides xo, the value of the argument at which the extrema occurs. You must substitute this value into the function to compute the extreme value f(xo).

The second derivative is positive at x1 as the slope is increasing for increasing x. The curve is concave upward so f(x1) is a local minimum.

The second derivative is negative at x2 as the slope is decreasing for increasing x. The curve is concave downward so f(x2) is a local maximum.

The situation is less clear in the case that both the first and second derivatives vanish 3 4 at x0. One should reflect of the behaviors of x and x around x = 0. Examining the series:

df df234df df ≈ + −+111 −+234 −+ −+ fx()()() fx00 xx2 234 () xx 03! ()xx 04! ()...xx 0 dx x dx dx dx 0 xxx000 The function fails to have a local maximum or a minimum if the first non-zero derivative is of odd order; such points are called saddled points. If the first non-zero derivative is of even order, the function attains a minimum at the point if it is positive and a maximum if it is negative.

A point at which the concavity (second derivative) changes is a point of inflection. If the first derivative condition is met at a point of inflection, that point is a

11/1/2012 Physics Handout Series.Tank: Derivatives D/DX- 13 saddle point.

End point extrema can occur in the case that the search is limited to an interval such as x∈[ ab, ] . The left endpoint is a relative minimum if df > 0 while the right endpoint dx a is a relative maximum if df > 0 . Always check the endpoint behavior in addition to dx b applying the first derivative test.

Extrema for functions of several variables: Consider f(x,y). The function f(x,y) ∂f ∂f point has an extremum at (x0,y0) only if ∂x and ∂y vanish at that point. The second derivative test must be satisfied by both second derivatives with respect to the same

∂2 f ∂2 f variable. That is, for a local minimum, both ∂x2 and ∂y2 must be positive. Unfortunately, one must do more than just apply the single argument test to each argument individually. The saddle example illustrates that an additional test is necessary.

Imagine that as you face forward, that the x-axis is horizontal and to the right across the horse. The y-axis runs horizontal and forward. The relative height of the saddle as compared to the center point can be represented as: h(x,y) = - 4 x2 + y2 The saddle curves down strongly to the left and right and curves upward only weakly along the saddle. At the center point (0,0), the first derivatives vanish and the second derivatives have opposite signs. The point is a maximum with respect to excursions in the x direction and a minimum with respect to excursions in the y direction. Such points are called saddle points.

11/1/2012 Physics Handout Series.Tank: Derivatives D/DX- 14 ∂2 f ∂2 f The question is: If both second derivatives ∂x2 and ∂y2 are negative, is the point necessarily a maximum? Return to the saddle example and change to a coordinate system (x/, y/) that is rotated by 450 in the CW sense as viewed from above. The coordinates of a point in the original system can be expressed in terms of its coordinates in the new system. xx=1'' + 1 y y =−+11 x'' y 2 2 22 These axes split the difference between the original x- and y-directions so it is expected that the strong downward in the across direction dominates the weaker upward curvature along the saddle for both of the new axes. Substituting into the height equation: 2 2 / / 3 / 2 3 / 2 / / h(x,y) = - 4 x + y => h(x ,y ) = - /2 (x ) - /2 (y ) – 5 x y The function h(x/,y/) meets the second derivative requirement for a maximum, but it is the same saddle so the point remains a saddle point, not a local maximum. The point is that there is a direction along which excursions lead to an increase rather than a decrease. Return to the Taylor’s series omitting the zero-valued first derivatives.

∂∂22f f ∂ 2f df 2 ≈+++1 −22−− −− − df 2 22()xx0 ()()yy00 xx ()()xx0 yy 0 () yy0 ∂x ∂∂y x ∂∂x y dy xy00,,xy00 xy00 , x0 Clearly, the troublesome direction must be between the x and y directions. Introducing

()yy− 0 the notations u for the direction’s and ∆ for (x – xo), ()xx− 0

∂∂22ff ∂ 2 ∂∂yx AB ();()()xx−=∆−= yy uxx −=∆ u x = 0 00 ∂∂22ff 2 BC ∂∂xy ∂y 

1 2 2 df = ( /2) [ A +2 B u + C u ] ∆ In the case of the saddle, it appeared that df was negative as the point appeared to be a local maximum. How can this conclusion fail to be true? If one could find a direction in which df is positive, it would fail. The critical distinction occurs when df changes

11/1/2012 Physics Handout Series.Tank: Derivatives D/DX- 15 sign; that is: when df = 0. df = 0 => [ A +2 B u + C u2 ] = 0 Solve to find the critical value of u.

−±2B 44 B2 − AC u = critical 2 A The additional condition for an extremum is that there be no real solutions for u 2 satisfying df = 0 which follows if (B – AC ) < 0. The determinant of the second partials must be positive as it equals AC - B2.

∂∂22ff 2 ∂∂ 2 AB ∂x yx ∂∂22ff ∂ 2 f = =22 −>∂∂ 0 ∂∂22ff ∂∂xy( xy) Additional Condition BC ∂∂xy ∂y2 For functions of two variables, the function has an extremum if the determinant of the ∂2 f < partials is positive (and is a maximum if ∂x2 0); the test is inconclusive if the determinant vanishes; and the point is a saddle point if the determinant is negative.

Exercise: Suppose that the second partial derivatives with respect to each of the two arguments of a function have opposite signs. What is the sign of the determinant of the second partials in this case? Does the function have an extremum at the point tested?

Summary: Finding and testing extrema for a function of two variables f(x,y) 1.) Apply the first derivative test with respect to each variable to locate points at which (local) extrema might exist. Solve the simultaneous

∂∂f(,) xy f(,) xy ∂∂xy= 0 and= 0 to find points (xe, ye) that are possible locations of an extremum.

11/1/2012 Physics Handout Series.Tank: Derivatives D/DX- 16 22 ∂∂fxy(,)ee fxy (,)ee 2.) Apply the second derivative tests. Evaluate ∂∂xy22and . If the second derivatives have opposite signs, the point is a saddle point. If both are positive (negative), the point may be a local minimum (maximum).

22 2 2 22 ∂∂ff ∂ f ∂∂fxy(,)ee fxy (,)ee 22−> 3.) If ∂∂xy( ∂∂xy) 0 and both ∂∂xy22and have the same sign,

2 ∂∂22ff ∂ 2 f 22−< the function has a local extremum at the point. If ∂∂xy( ∂∂xy) 0 and

2 2 ∂ fxy(,)ee ∂ fxy(,)ee both ∂x2 and ∂y2 have the same sign, the function has a saddle

point at (xe, ye).

In more dimensions, the minimum requirements for a local extremum are: (1) all first partials vanish, (2) all the repeated second partials (the second partials with respect to the same variable) have the same sign and (3) the determinant of second partials is non-zero and has the same sign as the product of the repeated second partials.

Lagrange Undetermined Multipliers: (Section under development MUCH ) http://www.slimy.com/~steuard/teaching/tutorials/Lagrange.html There are cases in which an extremum of a function is sought subject to the condition that its arguments satisfy a constraint equation. For example, one might need the maximum value of f(x,y,z) subject to the condition that φ(x,y,z) = x2 + y2 + z2 = R2. A condition required for an extremum is that the df, differential of f, vanishes for   infinitesimal variations consistent with the constraint. As dφφ=∇⋅ dr , and dφ must   vanish, only argument variations dr that are orthogonal to ∇φ are consistent with the constraint. The conclusion is that, in order to have an extrema consistent with the   constraint, it is not required that ∇f = 0, but only that the components of ∇f that are

11/1/2012 Physics Handout Series.Tank: Derivatives D/DX- 17   orthogonal to ∇φ must vanish. The requirement is that ∇+∇=f λφ 0 for some λ.   That λ is the values that makes ∇+∇f λφ orthogonal to ∇φ . In other words: The constraint (in this example) restricts the argument to points on the of a sphere,   and df=∇⋅ f dr . The constraint is to be specified as φ(x,y,z) = 0 so the to the   surface is ∇φ . The condition for an extremum is that the ∇f is parallel to (or  proportional to) ∇φ .  ∇=−∇f λφ. [DI.14] The scalar value λ is called the Lagrange undetermined multiplier.

In the low dimensional representation to the A  ∇f left, the red arrow represents the gradient of f B and the brown line is the constraint surface. At   df = ∇f ⋅ dr A, the gradient has a component parallel to the surface so the value of f can be increased by Constraint :gxy ( , ) ∇⊥g gxy(, ) = c displacing the point to the ‘right’ along the surface. At B, the gradient is perpendicular to the surface so df vanishes to first order for infinitesimal displacements about the point B that are consistent with the constraint. That is an extremum might exist at B. One does not exist at A.

11/1/2012 Physics Handout Series.Tank: Derivatives D/DX- 18 Consider the figure which represents the level

A surfaces of the function f(x,y) in blue and the constraint equation in red. The points A and C at   B which ∇f and ∇φ are parallel meet the requirements for a constrained extremum. At B on the other hand,

C displacements along the constraint line have  components along ∇f so the value of f changes indicating that the point is not a local maximum or minimum.

Extension to several constraints: Each independent constraint defines a direction (the direction of its gradient), and all allowed displacements must be orthogonal to that direction at that point. Each constraint gets its own multiplier and that multiplier times the gradient of the constraint subtracts the components of the gradient that are in directions orthogonal to displacements allowed by the constraint. If there is to be an extremum at the point, the components of the function’s gradient in the allowed directions must vanish.

Method Summary: Construct the function h(x,y) = f(x,y) + λ φ(x,y).  Compute ∇h and find the conditions that it vanishes.  The point at which ∇h vanishes can be the locations of local extrema. It is not always necessary to find the value of the .

Algebraic Development: If one wishes to find an extremum of a function f(x,y) subject to the constraint that f(x,y) = const., one cannot simply set the two partials of f(x,y) to zero because the variations in x and y are linked by the constraint equation.

11/1/2012 Physics Handout Series.Tank: Derivatives D/DX- 19 The requirements are that:

=+=∂∂ff φ =+=∂∂φφ df∂∂xy dx dy0 and d∂∂xy dx dy 0

∂φ ∂∂ =−=ff∂x or df∂∂xy dx ∂φ dx 0 ∂y which can be rearranged as:

∂∂φφ−1 ∂∂−1 ff= = −λ ( ∂∂xy) ∂∂xy( ) (,xy )

⇒ Conditions for an extrema of f(x,y) subject to a constraint φ(x,y) = constant.

∂∂ff+=λλ∂∂φφ += ∂∂xx0 and ∂∂ yy0 [DI.15] Along with the constraint equation φ(x,y) = constant, these generalizations of the first derivative test give sufficient information to find the at which the extremum occurs (x*,y*) and the undetermined multiplier λ(x,y). Note that equations [DI.14] and [DI.15] are consistent.

Sample Calculation LUM1: Find the ratio of r and h such that a right circular 2 cylinder has the greatest π r h subject to the constraint that the total surface 2 area of the cylinder (2π r h+ 2π r ) is a constant. 2 φ(r,h) = 2π r h+ 2π r = Ao

∂ ∂∂22 ∂ VA+=λ π + λ π + π = π + λπ + π = ∂r ∂∂ rr(r h ) ∂ r (2 rh 2 r ) 0; 2rh (2 h 4 r ) 0 ∂VA+=λ ∂∂ π2 + λ ∂ π + π22 = π + λπ = ∂h ∂∂ hh(r h ) ∂ h (2 rh 2 r ) 0; r(2 r ) 0

The second relation quickly yields λ = -½ r. Substituting in the first equation, we find

Ao ½ h = 2 r, and hence r = [ /6π] according to the constraint equation. Compare this solution to first solving for h in terms of r in the equation for the total area and then substituting the result for h in the expression for volume. The problem is thereby transformed to a simple first derivative test problem for a function of a single variable.

11/1/2012 Physics Handout Series.Tank: Derivatives D/DX- 20 The work around can be executed in this case, but it can be difficult to implement in many cases of interest.

Sample Calculation LUM2: Find the extrema of the function f(x,y,z) = x + y + z given that the points are constrained to lie of the surface of a sphere of radius centered on the origin. That is:

ˆˆˆ φ(x,y,z) = x2 + y2 + z2 = 1. The level surfaces of f(x,y,z) are planes with normal i++ jk. 3

The extreme values will correspond to the planes that are tangent to the sphere at (1,1,1) 3

−−− and at ( 1, 1, 1) where f(x,y,z) has values 3 and - 3 . 3 Employ the method: Construct the function h(x,y,z) = f(x,y,z) + λ φ(x,y,z). h(x,y,z) = (x + y + z) + λ (x2 + y2 + z2 – 1)  Compute ∇h and find the conditions that it vanishes.  ∇=+h(1 2 xiλλλ )ˆ ++ (1 2 y ) ˆj ++ (1 2 zk )ˆˆ = 0 iˆˆ + 0 j + 0 k We conclude that x = y = z.  The point at which ∇h vanishes can be the locations of local extrema. Invoking the constraint, the points are (1,1,1) and a (−−− 1, 1, 1) 3 3

where f(x,y,z) has values 3 and - 3 . It is not always necessary to find the value of the Lagrange multiplier.

3 λ =  2 , but in some cases its value has an interpretation. Add a complication: LUM2B. Note this complication has not be checked!!! Add the constraint 2 x – y = 0 to favor larger y. Construct the function h(x,y,z) = f(x,y,z) + λ φ(x,y,z) + µ ψ(x,y,z). h(x,y,z) = (x + y + z) + λ (x2 + y2 + z2 – 1) + µ (2 x - y)

11/1/2012 Physics Handout Series.Tank: Derivatives D/DX- 21  Compute ∇h and find the conditions that it vanishes.  ∇=+h(1 2 xλ + 2 µ ) iˆ ++ (1 2 y λµ − ) ˆj ++ (1 2 zk λ )ˆˆ = 0 iˆˆ + 0 j + 0 k We find λ = - 1/ , y = z - µ, x = 2 µ - z, µ = - z 2z The point at which ∇h vanishes can be the locations of local extrema. Invoking the constraint, we find the point (1,2,1) at which f(x,y,z) = 4 . 6 6

We note that 4 ≈ 1.633 < 1.732 ≈ 3 .The additional constraint did 6 limit the maximum value.

NOTE: The operational gouge is that you take the function V(...) for which an extremum is sought and add to it an undetermined multiplier λ times the constraint that the arguments of V must satisfy. Next apply the normal routine to find an extremum. The standard first derivative test finds the point (argument values) at which the extremum of a function occurs. Add a constant times the constraint and apply the standard method to find an extremum Interpret the multiplier. http://www.slimy.com/~steuard/teaching/tutorials/Lagrange.html

An alternative, and sometimes more direct, method to find the position at which the extremum occurs is to invert the constraint to find y(x). The problem may then be ∂ solved using the first derivative test [ /∂x{f(x,y(x))} = 0]. This technique may be less desirable than using undetermined multipliers if the constraints are difficult to invert or if there are several constraints that must be satisfied simultaneously.

Use the Euler-Lagrange equations to find the extremum for a path . The constraint equation supplies the additional condition needed to find ¬ (if you care). ******* Editing in progress

11/1/2012 Physics Handout Series.Tank: Derivatives D/DX- 22

L’Hospital Rule: If the limit of the ratio of two functions appears to be indeterminate, that limit, if it exists, is the limit of the ratio of the derivatives of those same two functions. Suppose that the limiting values of the two functions are both zero or . The ratio of the values of the two functions may approach a limit nonetheless.

fx()  f'() x  Limit→ Limit   [DI.16] → xx→ 0 xx0 gx()  g'() x  sin(x) A favorite physics example arises in the study of single-slit diffraction: /x. Limit[cos( x )] sin()x fx'() cos()x x→0 1 Limit →=Limit Limit   == x→0x xx →→ 00g'( x ) 1 Limit[1] 1 x→0 sin(x) The combination /x is called the ‘sink’ function sinc(x).

Another application confirms that ln(x) diverges more slowly that any positive power ln(x) of x as x becomes large. Equivalently, /xn approaches zero as x grows for all n > 0. The function ln(x) is as near to x0 as you can get and not be there!

ln()x fx'()  x−−1   xn →===> Limitnn Limit Limit−1  Limit 0( forn 0) x→∞ x xxx→∞ g'( x ) →∞  nx →∞ n No general proof is to be presented, but L'Hospital's rule can be understood easily in the case that the limits are approaching zero and the functions are analytic (have expansions).

[1] 1 [2] 2 fx() fx()+ f ()( x xx −+ )2! f ()( x xx − ) + ... Limit→ Limit 0 00 00 xx→→xx [1] 1 [2] 2 00gx() gx ()0+ g ()( x 00 x −+ x )2! g ()( x 00 x − x ) + ...

If both f(x0) and g(x0) vanish, they are removed and (x – x0) is factored out.

[1] 1 [2] fx() f( x )+2! f ( x )( xx −+ ) ... ⇒→Limit Limit 0 00 xx→→xx [1] 1 [2] 00gx()  g( x0 )+2! g ( x 00 )( x −+ x ) ...

11/1/2012 Physics Handout Series.Tank: Derivatives D/DX- 23 And so it continues. If the limits of the function diverge, the argument can be made with the inverse ratio of their inverses. The general proof including the undefined limit case is in the appendix along with an argument that the procedure can be applied when xo = ∞.

df  dx fx() xo Applying L’Hospital’s rule: Limit= Limit xx→→00xx dg gx() dx xo

1.) Apply this formula when the limits Limit[ f() x ] and Limit[ g() x ] are both either 0 xx→ 0 xx→ 0 or undefined (infinite). 0 ∞ 2.) A limit of the /0 form can be converted to the /∞ form by considering it as (1/g) /(1/f) and vice versa. Try converting forms if the L’Hospital’s Rule method is not productive using the original form. 3.) L’Hospital’s Rule can be used recursively to reach a defined limit. Use the

dfn n fx() dx x form: Limit= Limit o beginning with n =1, and increasing n →→ dgn xx00gx() xx  dxn xo integer by integer until the procedure yields a defined limit. Once a defined limit is found, n cannot be increased further.

L’Hospital, Guillaume de (1661-1704) French mathematician who, at age 15, solved a difficult problem about cycloids posed by Pascal. He published the first book ever on differential , L'Analyse des Infiniment Petits pour l'Intelligence des Lignes Courbes (1696). In this book, l'Hospital included l'Hospital's rule. l'Hospital's name is commonly seen

11/1/2012 Physics Handout Series.Tank: Derivatives D/DX- 24 spelled both "l'Hospital" and "l'Hôpital" (e.g., Maurer 1981, p. 426), the two being equivalent in French spelling. © Eric W. Weisstein http://scienceworld.wolfram.com/biography/LHospital.h tml

sin∆θ Geometric evaluation of the limit of /∆θ as ∆θ goes to zero. The standard limits are to be assumed. Limit[sinθ ] = 0 and Limit[cosθ ] = 1 θ →0 θ →0

1 2 The area of a circle is ( 2 )2π r so the area

1 2 of a pie wedge with angle θ is ( 2 )θ r .

Working with the unit circle below, we conclude that the area of the ∆θ pie wedge Ocd of radius cos(∆θ) is less than the area of the triangle Ocb which is less than the area of the unit radius pie ∆θ wedge Oab. Area(Ocd) < Area(Ocb) < Area(Oab)

112 1 ( 22)cos (∆θθ ) ∆< cos( ∆ θ )sin( ∆ θ ) <( 2) ∆ θ

∆θ 1 sin( ) 1 Divide by ( 2 ) cos(∆∆θθ ) : cos(∆≤θ ) ≤ ∆∆θθcos( )

11/1/2012 Physics Handout Series.Tank: Derivatives D/DX- 25 sin(∆∆θθ ) 1 sin( ) Limit cos(∆θ ) ≤ ≤ →≤1 Limit ≤1 ∆→θ 0 ∆∆θθcos( ) ∆→θ 0  ∆ θ

sin(∆θ ) ⇒=Limit 1 ∆→θ 0 ∆θ

sin(∆θ ) Exercise: Use L’Hospital’s Rule to evaluate Limit  ∆→θ 0 ∆θ

Exercise: Use L’Hospital’s Rule to evaluate Limit xnx e− . Read the three application x→∞  rules!

CPA Application: At 0230, an unidentified ship is 14 nautical miles (14 nmi.) due east of you current location. It is steady on course 3400T traveling at 23 knots. You (the Black Goo Maru) are steering 000T and making a steady 22 knots. Compute the time at which the separation of the two ships will be a minimum. Let τ be the time elapsed since 0230. Give the distance and bearing to the contact at that time. When should you wake the CO?  =−+00ττˆˆ r? 14 23sin(20 ) EN 23cos(20 )  ˆ rNBGM = 22τ

=−=−   00ττˆˆ + −  rrel rr? BGM 14 23sin(20 )E 23cos(20 ) 22 N

22 =−+− 0ττ02  rrel 14 23sin(20 ) 23cos(20 ) 22 

00 02 dr −−14 23sin(20 )ττ  23sin(20 )+ 23cos(20 ) − 22  rel = = 0 dτ 02202 14− 23sin(20 )ττ +− 23cos(20 ) 22 

(14)(23)sin 200 τ = = cpa 22 01.775408hrs . 23+− 22 (44)(23)cos 20

11/1/2012 Physics Handout Series.Tank: Derivatives D/DX- 26  ττ= − 00ˆ+ −  τˆˆ = +− ˆ rrel( cpa ) 14 23sin(20 )cpa E 23cos(20 ) 22 cpa NE[ 0.034] [ 0.687] N

 0 rrel(τ cpa )= [ 0.688] @ 177 T Do not wake the CO! Add turns to make good 25 knots. Your cpa distance will exceed 10 thousand yards which, according to the standing night orders, is distant enough that the no further action is required. You should inform the 0400 watch of the situation and recommend dropping turns to return to the base speed 22 knots at 0415.

 =00 ˆˆ +−  Exercise CPA1: The contact has vErel 23sin(20 )  23cos(20 ) 22  Nas its  relative . Compute vrrel⋅ rel()τ cpa .

Exercise CPA2: In the scenario above, you add turns to make good 25 knots at 0230. Compute the time at which the separation of the two ships will be a minimum. Let τ be the time elapsed since 0230. Give the distance and bearing to the contact at that time assuming that it maintains course and speed. (156.70 T)

Tools of the Trade: Derivatives of Inverse Functions

Define arcf(u) as the inverse function for f(x). That means arcf(f(x)) = x. It follows that f(arcf(f(x)) ) = f( x ) or that f(arcf(u) ) = u for all u in the range of f(x). That is: If arcf(u) is the inverse function for f(x) then f(x) is the inverse function for arcf(u).

The symbol arcf(x) is read as ‘Ark-f of x’ or as ‘inverse f of x’.

Use the chain rule to conclude: (See the next exercise for an alternative form.)

11/1/2012 Physics Handout Series.Tank: Derivatives D/DX- 27 −1 d(arc f ( f ( x ))) dx d(arc f ( u ))  df ==⇒==1 whereu fx ( ) [DI.17] dx dx du dx df To complete the evaluation, /dx is expressed in terms of f(x) and then f(x) is replaced by u.

Sample Calculation: Consider the function Sq(x) = x2. The inverse function is arcSq(x).

−1 d(arc Sq ( Sq ( x ))) dx d(arc Sq ( u )) dSq ( x ) 1 ==⇒====1 whereu Sq ( x ) x2 dx dx du dx2 x Express x in terms of u = x2: x = [Sq(x)]1/2

−1 d(arc Sq ( u )) dSq ( x ) 1 1 1 −1/2 = = = = 2 u du dx2[ Sq ( x )]1/2 2 u 1/2

1/2 d(arc Sq ( u )) 1 −1/2 It is known that arcSq(u) = u . Hence: = 2 u as advertised. The function du arcSq(u) is commonly represented as Sqrt(u).

−1 d( f ( arcf ( x ))) dx d( arcf ( x ))  df Exercise: Begin with = =1 and show that = . dx dx dx du u= arcf() x

d(Sq(u)) Using the alternative form from the previous exercise, f(u) = Sq(u), /du = 2u so

d(arcSq(x)) 1 1/2 d( arcSq ( x )) 1 /dx = /2u with u = arcf(x) = x . It follows that = . dx2 x1/2

Tools of the Trade: Plotting Inverse Functions All physics students should be familiar with the graphs of common function such as powers, trig functions, logs and exponentials. The graph of the inverse function [y = f(x)] can be constructed by inverting ‘x-y’ pairs point by point or by reflecting each point across a line at 450 between the x and y axes. The reflecting each point across a

11/1/2012 Physics Handout Series.Tank: Derivatives D/DX- 28 line at 450 procedure assumes that the graph is drawn with 1:1 scaling between the axes.

As a trivial example, let y(x) = 2 x. This represents a line at tan-1(2) = 63.4350. Reflecting across a line at 450 yields a line at 26.5650. Not surprisingly, this line has a slope of ½. Arcy(u) = ½ u. Arcy(y(x)) = ½ y(x) = ½ (2 x) = x

Geometry of the reflection process: Reflection is accomplished by moving the point along a line perpendicular to the reflection (plane) line. In moving from

(xi, yi) to the reflection line, x decreases by s and y increases by s. On the line x =

y or xi – s = yi + s; so

s = ½(xi – yi). The displacement up and to

the left is continued to (xi - 2s, yi + 2s) =

(xf, yf).

Substituting, (xf, yf) = (xi - (xi – yi), yi + (xi

– yi)) or (xf, yf) = (yi xi). Reflection across the line at 450 inverts the roles of y and x.

11/1/2012 Physics Handout Series.Tank: Derivatives D/DX- 29 3 Plot[{Sin[x],ArcSin[x],x},{x,-

2 Pi, Pi}, AspectRatio  1]

1 A plot of sin[x], arcsin[x] and x -3 -2 -1 1 2 3 from - π to π. Note that arcsin -1 plots as the reflection of sin[x] -2 across the line at 450 over the -3 range (-1, +1). What happens outside that range? What is the arcsin[1.1] ?

6 Plot[{Tan[x],ArcTan[x],x},{x,-2

4 Pi,2 Pi},PlotRange  {- 2 Pi, 2  2 Pi}, AspectRatio 1]

A plot of tan[x], arctan[x] and x -6 -4 -2 2 4 6

-2 from – 2 π to 2 π. Note that arctan plots as the reflection of -4 tan[x] across the line at 450 for x -6 in the range (-½ π, ½ π). What does tan do outside that range?

Leibnitz Rule for Differentiating a product: dn n dnk− u  dv k  (uxvx()()) = n! n ∑ (nk− )! k ! nk−  k  dx k=0 dx  dx 

Appendix: L’Hospital’s Rule not Proven

11/1/2012 Physics Handout Series.Tank: Derivatives D/DX- 30

No proof is to be presented. The arguments that follow are flawed and at best support the plausibility of the claims.

This general proof is less intuitive than the one given previously using Taylor’s series 0 for the case of a /0 limit. It is presented here for completeness. One can push the ‘I believe’ button, and skip this appendix.

Earlier the validity of the approach was established in the case that the limit 0 approached the /0. A few steps are recopied to prompt your memory.

[1] 1 [2] 2 fx() fx()+ f ()( x xx −+ )2! f ()( x xx − ) + ... Limit→ Limit 0 00 00 xx→→xx [1] 1 [2] 2 00gx() gx ()0+ g ()( x 00 x −+ x )2! g ()( x 00 x − x ) + ...

If both f(x0) and g(x0) approach zero, they are removed and (x – x0) is factored out.

[1] 1 [2] fx() f( x )+2! f ( x )( xx −+ ) ... ⇒→Limit Limit 0 00 xx→→xx [1] 1 [2] 00gx()  g( x0 )+2! g ( x 00 )( x −+ x ) ...

∞ In the case that the limit approaches the indeterminate form /∞, the two functions are 0 inverted to yield the /0 form to which we have shown L’Hospital’s method applies. (That is: While the functions g(x) and f(x) almost certainly do not have convergent power series expansions, their inverses may so that the Taylor’s series development 0 for the /0 case may be valid.)

2 1 d 1  1 dg fx() ∞ gx() 0 ( gx()) ( gx()) Limit ≈→Limit  ≈ → Limit dx =Limit dx xx→ xx →1 LHospitalxx' →→d 1 xx 2 df 0gx() ∞ 00 00( fx())  1 fx() for0/0 case dx  ( fx()) dx

11/1/2012 Physics Handout Series.Tank: Derivatives D/DX- 31 Comparing the first and last terms:

2 1 dg dg f() x ( gx()) dx gx() dx =2 ⇒= gx() 1 df after cross multiplying f() x df ( fx()) dx dx

∞ fx()  f'() x  which is an /∞ form so the L’Hospital method, Limit→ Limit  , can be → xx→ 0 xx0 gx()  g'() x 

∞ 0 used to attack the /∞ and the /0 indeterminate forms. The steps above played the limit game fast and loose, but they should constitute a plausibility argument for the procedure.

The use of L’Hospital’s rule in the case that xo is ∞ can be justified by considering

fx() ff()11yy  '()  Limit→= Limit Limit    →→11 x→∞ gx()yy00  g ()yy  g '() 

Problems 1.) Using ln(ez) = z, (The natural log is the inverse of using the base e.)

d[ln(u)] -1 show that the derivative /du = u .

sin(∆θ ) d(sinθ ) d(cosθ ) 2.) Using Limit =1, show that = cosθ and = −sinθ . Use the ∆θ dθ dθ ∆→θ 0  trig identities for sine and cosine of sums of angles.

dx(arctan( )) 1 da(arctan(x )) 3.) Show that = and that a = . dx1+ x2 dx a22+ x Recall that arctan(x) is an alternative notation for tan-1(x).

4.) An exercise in blindly computing partial and total derivatives: Mechanics can be cast into the Lagrangian alternative to the Newton form. For our purposes, the

11/1/2012 Physics Handout Series.Tank: Derivatives D/DX- 32 Lagrangian function is defined as a function of the coordinates, their times derivatives (the coordinate ) and time. The function often assumes the form of T, the kinetic , minus V, the potential energy, but, in some cases, its form is more general. For central problems, ones in which the potential depends on r only, conservation of angular restricts the motion to a plane so one can adopt polar coordinates {r, φ} leading to the associated coordinates velocities {r,φ}. Without

1 2 22 explanation, the Lagrangian is Lr(,φφ ,, r )=( 2) m r +− rφ V() r = T - V.

dL∂∂ L In this formalism, the are: = where q represents first r dt∂∂ q q and then φ and q represents first r and thenφ .

∂∂∂LLL ∂ LdL ∂ d ∂ L d Compute: , , , , and . What is (φ) ? Express it in ∂∂rφφ ∂ r ∂ dt ∂ r dt  ∂ φ dt

‘shorthand’. What does r represent? Contrast r and ar.

dL∂∂ L dL∂∂ L ∂ V 2 Partial Answers: = ⇒mr(20 rφφ += r ) ; = ⇒=−+mr mφ r. dt ∂∂φφ dt∂∂ r r ∂r

Note that the rate of change with respect to distance moved in a direction of a potential function is the negative of the component of the force in that direction associated with that potential. The rate of change with respect to an angular coordinate change of a potential function is the negative of the torque associated with that potential about that angular axis. A torque is the generalized force expected when the generalized coordinate is an angle. In polar coordinates,

 2   ˆˆ a=−++( rφ r) rˆˆ( φ r2 r φφ) =+ arr aφ φ.

5.) Compute the first derivative of the following functions. The variable argument is x. 1 b x a.) xbc222++ , b.) ,c.) and d.) . 222 2223/2 2223/2 xbc++ xbc++ xbc++ 6.) Compute the partial derivative with respect to x of the following expressions: 1 y x a.) xyz2++ 22, b.) , c.) and d.) 2 22 2 223/2 2 223/2 xyz++ xyz++ xyz++

11/1/2012 Physics Handout Series.Tank: Derivatives D/DX- 33  ∂∂∂xiˆˆ++ y j zkˆ 7.) Compute pppxyz++ 3/2 where px, py andpz are the constant ∂∂∂xyz2 22 xyz++  components of a constant vector p .

ˆˆˆ ˆ 2   pixy++ p j pk z xiˆˆ++ yjzk rp −3( prr ⋅ ) Answer: −3( pxxyz ++ py pz) = 2 223/2 2 225/2 r5 xyz++ xyz++ 8.) Once the derivatives of sin(θ), cos(θ), sinh(x) and cosh(x) are known, the derivatives for the remaining trigonometric and hyperbolic function can be developed

d tanθ d secθ dxtanh ( ) using the product and quotient rules. Compute and . Compute dθ dθ dx dsec hx( ) and . dx

ex dx(ln( ln )) de( ) 9.) Compute and . dx dx

10.) Prepare a drawing that illustrates geometrically a sequence of chords approaching a tangent line for a left-side limit. Extend each chord beyond its intersection points with the curve representing the function as a straight, dashed line. Prepare a sketch of a function with a cusp to illustrate that a left-side limit and right-side limit need not be equal.

11.) Suppose that a function is discontinuous at the point xo. What problem arises in evaluating the derivative of that function at xo? Reference the definition of the derivative.

x ∆x de x e −1 12.) Show that = e is equivalent to Limit =1. [Use the definition of the dx ∆→x 0 ∆x

11/1/2012 Physics Handout Series.Tank: Derivatives D/DX- 34 derivative.] a.) Numerically evaluate the quantity in brackets for the sequence of values ∆x = {0.1, 0.01, 0.001, 0.0001, … }. b.) The value of each term in the sequence ∆x 2 is approximately 1+ /2. Deduce the form of the sequence accurate to order (∆x) ?

−1 d(arc f ( u ))  df 13.) For inverse function, it was found that: = whereu= fx ( ) . Express du dx this result in prose. Is the result reasonable?

14.) Find the limiting value of x -2 (1 – cos(x)) as x approaches zero using L’Hospital’s rule. For comparison, replace cos(x) by its Taylor’s series expansion about zero including terms through x4 and take the limit.

2 15.) Given f(x) = x (1 – x ), find the location xex of any extremum between 0 and 1.

Apply the second derivative test to decide if the f(xex) is a local maximum or 4 minimum. Compute f(xex). Answer: f(xex) = /27.

2 π/ 16.) Given f(θ) = sin θ cosθ, find the location θex of any extremum between 0 and 2.

Begin by sketching the f(θ). Apply the second derivative test to decide if the f(θex) is a local maximum or minimum. Compute f(θex). Complete the characterization of f(θ) by

π/ checking the endpoints 0 and 2. Is either point the location of a local extremum? If θ 2 so, what type? Answer: f( ex) = 33.

17.) Given f(x, y) = 2 x y – 4 x2 + 20 x – 6 y2 + 18 y – 54, find the location of a local extremum: (xex, yex). Compute the second derivatives and evaluate them at that point. Form and evaluate the determinant of the second partials at that point. Discuss the nature of the extremum.

11/1/2012 Physics Handout Series.Tank: Derivatives D/DX- 35 Note that the second partials are all constant values. This is always the case for a function that is a second order polynomial in two variables. Added a third order term or a trig function would lead to second partials that are non-constant.

18.) Given f(x, y) = 32 (x + y) – 3(x2 + y2) - 10 x y – 64, find the location of a point where the first derivative test is satisfied (xS, yS). Compute the second derivatives and evaluate them at that point. Form and evaluate the determinant of the second partials evaluated at that point. Discuss the nature of the ‘extremum’. Compute: f(xS, yS), f(xS+1, yS), f(xS, yS+1), f(xS+1, yS+1) and f(xS+1, yS-1).

sinh(x ) π 2 π 19.) Compute: Limit , Limit( x− 2) tan( x ) and Limit( x− 2) tan( x ) . x→0 x x→π /2 x→π /2

π 20.) Consider sinθ in the interval [0, /2 ]. According to the mean value theorem, what π value must the derivative of sinθ have for some θ between 0 and /2? For what value of θ does sinθ have that value? 50.460

21.) The tracks of a rail line rise from sea level as they run west to reach Denver. The rails continue west to return to sea level in San Francisco. What value must the slope of the track have at some point? What value must the slope have at the point along the tracks that is highest above sea level?

x n x n dfn 22. Given fn(x) = [1 + /n] and f(x) = Lim( fn () x) = Lim [ 1 + n] , compute /dx. Show nn→∞ →∞ ( )

x dfn that fn(x) = [1 + /n] /dx. What form does this equation approach in the limit that n becomes very large?

11/1/2012 Physics Handout Series.Tank: Derivatives D/DX- 36 x n 23. Give the first five or six terms of the binomial expansion for fn(x) = [1 + /n] . Take the limit of your result as n becomes very large. Identify f(x) = Lim( fn () x ). n→∞

d d d n 24. Compute dx [cosh(x )] and dx [sinh(x )]. Develop expressions for dxn [cosh(x )] and d n dxn [sinh(x )] for all positive integer values: n = 0, 1, 2, 3, … .

25. Compute the limit of x-n ln(x) as x becomes very large. Conclude that ln(x) diverges more slowly that any positive power of x as x → ∞.

26. Feynman’s Differentiation Rule: Given f(t) = k [u(t)]a [v(t)]b[w(t)]c … where k, du dv dw df a, b, c, … are constants, =ft()  adt ++ b dt c dt +... . For example, dt u v w 

sin()tt cos() (−+ sin()) t cos()22 tt sin() sin(t) 1 -1 −= f(t) = /cos(t) = [sin(t)] [cos(t)] = 11 2 cos(tt ) sin( ) cos( t ) cos ( t )

6 (1+−ttt232 ) ( ) 12+ t Apply this method to Feynman’s examples: ft()= 3 and gt()= . tt+ 52 (4 t ) 2 tt++1 2

27. Leibnitz Rule for Differentiating a Product: dn n dnk− u  dv k  (uxvx()()) = n! n ∑ (nk− )! k ! nk−  k  dx k=0 dx  dx 

d n Compute the explicit form of (uxvx()()) for n = 0, 1, 2 and 4. Discuss the dxn comparison with the result found using the formula above for each case. Recall that 0! = 1. In the Expansions handout, see Taylor’s Series of a Product of Functions.

28. Prove the Leibnitz Rule for Differentiating a Product:

11/1/2012 Physics Handout Series.Tank: Derivatives D/DX- 37 dn n dnk− u  dv k  (uxvx()()) = n! n ∑ (nk− )! k ! nk−  k  dx k=0 dx  dx 

d n Compute the explicit form of (uxvx()()) for n = 0 and 1. Assume the result is dxn valid for n and prove that it holds for n + 1. HINT: In the Expansions handout, see the proof of the .

29. Symmetry: Prove that the derivative of an odd function is an even function. Assume a function of a single variable, f(x).

d 30.) Compute (ax ) where a is assumed to be a real positive constant. dx

31.) Extremum Problem: A vertical fence 8 feet high is parallel to a much higher wall one foot from it. What is the

L minimum length of a ladder that can just 1' get over the wall and lean against the wall? 8' Answer: (125)½ feet not checked!

32.) A force FA is applied at a given angle α above the x axis. A second force FB is applied at an angle β below the x axis. What value of β yields the smallest FB given

11/1/2012 Physics Handout Series.Tank: Derivatives D/DX- 38 that the resultant of the two is to be Ftotal = 1000 N in the x direction? Find FA and FB. Prepare a vector addition diagram that illustrates the configuration. π β = /2 - α; FA = Ftotal cosα; Fb = Ftotal sinα Hint: As you seek apply the condition to find a minimum, list all the symbols that appear in the quantity to which the procedure is to be applied and list the symbols that are to be treated as constants.

1/ x 1/ x 33.) Compute Limit[1+ ax] . Consider ln[ 1+ ax] . x→0 ( )

1/ x 1/ x 34.) Compute Limit1++ ax bx2 . Compute Limit ex + ax x→0  x→0 

d d x 35.) Compute ( xx ) and xx . dx dx ( )

36.) Minimize the function x2 + y2 + z2 subject to the constraint that 2 x + y – z = 5. Basically, you are finding the minimum distance between the origin and points in the plane specified. For this problem, identify the value of the multiplier λ.

References:

1. K. F. Riley, M. P. Hobson and S. J. Bence, Mathematical Methods for Physics and Engineering, 2nd Ed., Cambridge, Cambridge UK (2002).

2. The Wolfram web site: mathworld.wolfram.com/

3. Donald A. McQuarrie, Mathematical Methods for Scientists and Engineers, University Science Books, Sausalito, CA (2003).

11/1/2012 Physics Handout Series.Tank: Derivatives D/DX- 39