Table of Common Derivatives By David Abraham

Transcription

1 Prouct an Quotient Rules: Table of Common Derivatives By Davi Abraham [ f ( g( ] = [ f ( ] g( + f ( [ g( ] f ( = g( [ f ( ] g( g( f ( [ g( ] Trigonometric Functions: sin( = cos( cos( = sin( tan( = sec ( sec( = sec( tan( csc( = csc( cot( cot( = csc ( Inverse Trigonometric Functions: sin ( = 1 1 cos ( = 1 tan 1 1 ( = 1+ sec ( = 1 csc ( = cot 1 ( = 1+ Eponential an Logarithmic Functions: a a ln(a) e = e = ln( ) = 1 log ( ) a = 1 ln( a)

2 Stanar Integration Techniques: Table of Common Integrals f ( g( = f ( g( [ f ( ] g( Integration by parts: a a Trig Substitution: a b = sin( θ ) b a = sec( θ ) b b Trigonometric Ientities: a + b = a b tan( θ ) Pythagorean Ientities: sin ( θ ) + cos ( θ ) = 1 1+ tan ( θ ) = sec ( θ ) 1+ cot ( θ ) = csc ( θ ) Sum-Difference Formulae: sin( u ± v) = sin( u) cos( v) ± cos( u)sin( v) cos( u ± v) = cos( u) cos( v) ± sin( u)sin( v) Half Angle an Power Reucing Formulae: 1 cos(θ ) 1+ cos(θ ) sin ( θ ) = cos ( θ ) = sin( θ ) = sin( θ ) cos( θ ) cos(θ ) = 1 sin ( θ) Trigonometric Functions: sin( u ) u = cos( u) + c cos( u ) u = sin( u) + c tan( u ) u = ln sec( u) + c sec ( u ) u = tan( u) + c csc ( u ) u = cot( u) + c sec( u ) tan( u) u = sec( u) + c csc( u ) cot( u) u = csc( u) + c sec( u ) u = ln sec( u) + tan( u) + c csc( u ) u = ln csc( u) cot( u) + c Eponential an Logarithmic Functions: u u u u a e u = e + c a u= + c a) ln( ln( u ) u = u ln( u) u+ c

3 Inverse Trigonometric Functions: 1 u u = sin + c a u a 1 u= u u a 1 1 u = tan a + u a 1 sec a u + c a u + c a Avance Calculus Double Integrals: The volume of a region D, boune above by the function f(,y) can be calculate via the following ouble integral in Cartesian coorinates: The equivalent in polar coorinates is then: =,=, =,=, Where: = + = = If the area of the region D is esire instea of the volume of the region, the integran f(,y) or f(r, ) = 1. Triple Integrals: Triple integrals over a three imensional region can be compute in the following way:,,=,, The volume of the region can be compute if the integran is mae to be 1.

4 The equivalent in Cylinrical Coorinates is then:,,=,, Where: = + = = = The equivalent in Spherical Coorinates is then:,,=,, sin Where: =sincos =sinsin =cos = + + Change of Variable: The Jacobian When performing a ouble or triple integral, sometimes the region D or V can be simplifie by use of a change of variable. In such a case, a transformation factor must be ae to the integration. This factor is equal to the eterminant of the Jacobian matri:,, = = Therefore: = Vector Fiels: A vector fiel is a function that assigns a vector, having both magnitue an irection to any point in or 3 imensional space. It has the general form in three imensions: =,,+,,+,,

5 This inicates that for every point in the 3 imensional space, there eists a vector whose components are epenent on the location. Vector fiels are commonly use in the secon an thir imensions. By this efinition, the graient (seen earlier in Intermeiate Calculus) is therefore a vector fiel: =,, An can be referre to as the graient vector fiel. In this case it is important to realize that the function f is not vector in nature. It is a normal function an is therefore referre to as a scalar function. Line Integrals: Before proceeing with the line integrals, it is important to summarize some common parametric equations for a few simple curves. This will help us with line integrals, since line integrals are heavily epenent on parameterize forms. Ellipse: + =1 Counter clockwise: Clockwise: =acost =bsint for 0 t π =acos = bsint for 0 t π Circle: + = Counter clockwise: Clockwise: = = for 0 t π = = for 0 t π y = f( = t = f(y) y = t y = f(t) = g(t) Line Segment Form: Any line segment from a point,, to a point,, can be parameterize as Or =1,, +,, 0 1 =1 + =1 + =1 +

6 Up until this point all stanar efinite integrals are performe over some interval of, [a,b]. As such we were summing up the contribution of the function at all points along this line, representing in the en the area of the region. In the case of a line integral however, we will not be integrating over a linear region in which only varies, we will be integrating over a curve, with the interval efine by two points. Each of these points having an an y coorinate. The line integral following such a path over a curve for a function of two variables f(,y) is therefore:, Unlike the stanar integral, the ifferential here is s, representing the line integral of the function with respect to arc length. Remembering from Intermeiate Calculus, we know that arc length is efine as follows: = h = + Therefore, the line integral, which will sum up the function along the curve s length is efine as: f,ys= fht,gt t + y t t Assuming that the parameterization yiels = h(t) an y = g(t) Noticing now that the ifferential s is simply the magnitue of the velocity of the vector function r(t), we can write: + =,= h, We can notice from this that as long as the values a an b cause the curve to be trace out eactly once, the integral will be inepenent of the parameterization. Shoul we be given a piecewise continuous function, the final integral can be consiere the sum of the line integrals of the pieces ae together. Just as in single variable calculus we were able to a various integrals over regions.

7 When performing these types of integrals over space curves, we say that the curve has a certain orientation. The positive sense of the curve with bouns a an b, is in the irection of increasing values of t, that is from a to b. The opposite sense is the negative orientation. Using the properties of integrals an arc length, we can then arrive at the following conclusion:,=, This property will always hol when integrating with respect to arc length. We may naturally epan these ieas to also cover 3 imensional space:,,=,, + +,,=,, Suppose now we wishe to take the line integral of a function with respect to the y-ais or -ais rather than arc length. These integrals woul then be:,=,,=, We can now use mathematical shorthan to combine these two integrals: +=,+, With these types of integrals, the irection in which the curve is traverse oes in fact make a ifference, contrary to the case of arc length. As such, the following relationships can be shown:,=,,=, += +

8 Again we can etrapolate to three imensions:,,=,,,,=,,,,=,, ++=,,+,,+,, Line Integrals of Vector Fiels: Now that we know about line integrals an vector fiels, we are able to take the line integral of a vector fiel. This will essentially sum up all the components of the fiel that lie in the same irection as the curve: = It can then be shown that a relationship eists between this type of integral an the line integrals with respect to, y an z eplaine earlier: = ++ As such, any properties that apply to the RHS (iscusse above) also apply to line integrals of vector fiels. Note that the vector fiel must be parameterize accoring to the function r(t). Funamental Theorem for Line Integrals: Shoul we be performing a line integral of a vector fiel, where sai vector fiel is a graient one, a special case for computing the result applies: = We say that a vector fiel is a conservative one, if there eists a scalar function f such that =. If the vector fiel is a conservative one, it can be sai that the line integral through the conservative fiel is inepenent of path, via the funamental theorem of line integrals. We can therefore say that in a conservative fiel, the line integral aroun a close loop is zero.

9 In two imensions, we can etermine if a vector fiel is conservative in the following way: Then the fiel is conservative if: =+ = The potential scalar function f can then be foun by integrating as follows:,=,,=, An solving for the appropriate constants. Green s Theorem: Green s theorem can only be applie to close, simple curves. This means it must enclose a region entirely, without crossing itself. Shoul this conition be satisfie, we inicate that each curve has an orientation. This orientation is positive counter clockwise an negative clockwise. Green s Theorem for such a curve is then as follows: += Note that special notation is sometimes use for line integrals of curves that satisfy Green s theorem: + If you are face with a region that has a hole in it, split the region up into two parts, such that the ae line segments will cancel in the en, then apply Green s theorem. From Green s theorem, we can then erive the following relationship between line integrals an area of the enclose region D: = = = 1

10 Curl an Divergence: In orer to simplify efinitions in this section, let us first efine the el operator as follows: = + + We now efine the Curl an Divergence of a vector fiel as: = = = = + + The following are important properties of the above operators: If Then the vector fiel is conservative. =0 =0 This leas to two vector forms of Green s Theorem, which are as follows: = Parametric Surfaces: = Just as with the parameterization of functions when computing line integrals, the upcoming surface integrals will require the parameterization of surfaces in 3 imensional space.,=,+,+, If the surface is given having one term (such as written in terms of the other two, the form becomes easy to obtain: =,,=++,

11 The last important point to note for parametric surfaces has to o with the surface area of any one minute portion of the surface. Since the surface is efine in terms of the quantities u an v, we can use the nature of the cross prouct to obtain the following relationship: Surface Integrals: = Just as in line integrals we were able to compute the contribution of a function along a certain line or path, we can equality compute integrals over surfaces. The following case for a surface escribe by z = f(,y) but it is easily aaptable to the other cases as well:,,=,,, + +1 The alternative form is using the parameters u an v as note in the previous section. In such a case: It can in fact be shown that,,=, = + +1 An so either form can in fact be use, assuming proper algebraic manipulation, this is because a function of the form z = f(,y) can in fact be parameterize as: r(t) = <,y,f(,y)> One last important fact to note is that the surface area of a surface in three imensional space can be compute by setting the integran to 1. Surface Integrals of Vector Fiels: = = Just as with line integrals we were able to compute the contribution of a vector fiel along a certain curve; we now turn our attention to computing the contribution of a vector fiel through a given surface.

12 In orer to compute surface integrals we must have a two sie surfaces. As such there must eist two ifferent normal vectors, perpenicular to this surface. They are equal in magnitue but opposite in irection, with one being the negative of the other. If we have a close region, such as a sphere, a positive orientation is efine for all normal vectors pointing away from the enclose region an negative pointing into the region. Returning to the two forms in which we parameterize a surface earlier, there are two ways of fining the unit normal vector. Suppose Therefore: Or =,,,=,,,=0 = = = We can also compute the normal for surfaces parameterize by u an v: = We now efine the surface integral of the vector in the following way: Where is a unit normal vector to the surface. = We can simplify this again in both cases for the surfaces escription. In the case of z = f(,y) with vector fiel with i,j,k components respectfully being P,Q,R: An for u an v: = = = + = =

13 Stokes Theorem: Stokes theorem is a higher imensional version of Green s theorem. Before we can make use of it though we must efine an orientation as we i for line integrals. If we have a surface which has a bounary curve, C, then we can say that if you were to walk aroun the bounary with your hea pointing in the same irection as the unit normals of the surface, you are walking in the positive orientation. As such, Stokes theorem is then efine as: The Divergence Theorem: = The ivergence theorem is use to relate a 3 imensional enclose volume to the surface that encloses it. It allows for a irect relation between ouble an triple integrals. Full Range Fourier Series: = For any perioic function with perio L, there eists an infinite series approimation for the function compose entirely of orthogonal sine an cosine functions. This infinite series is calle a Fourier Series an has the following form: = + cos + sin Where the Fourier coefficients are given by: = 1 = 1 cos = 1 Since the function f( is perioic with perio L, the bouns of integration can equally be change to 0 to L, without moifying the equations.

14 Half Range Fourier Sine Series: Whereas the full range Fourier series uses both cosines an since to represent functions, half range epansions are equally possible. In a half range sine epansion, the Fourier coefficient of the cosine terms becomes zero an the resulting series is solely epenent on sine. As such, the Sine series epansion will prouce an o function. Half Range Fourier Cosine Series: = + = = This is the same iea as before, ecept using solely cosine components. As such it will prouce an even function. Comple Fourier Series: = = Owing to Euler s formula relate the comple eponential to cosine an sine functions, we may write a Fourier series with comple coefficients: = Where the comple value coefficients for a function with perio L are: = 1 / /

15 Solving the Heat Equation: The heat equation in a 1 imensional bar of material ictates the propagation of heat in the bar as a function of location an time. The heat equation is one of the form: = +, In orer to solve such an equation, with forcing function Q(,t) zero, we make use of the technique of separation of variables. That is, we assume that the solution of the prouct of a function t an a function of.,= We may then substitute into the ifferential equation an obtain the following: = It now stans that since the LHS is epenent solely on t an the RHS solely on, the only way these two can be equal is if they are both equal to some constant, non varying number. = = In orer to avoi trivial solutions to the equations, the value of cannot be negative or zero. A first orer ifferential equation will then nee to be solve for the t epenent solution, yieling an eponential. The epenent solution will return a superposition of sine an cosine terms. This will in turn lea to solving for Fourier coefficients, in orer to obtain the most general solution for the propose bounary value problem. Solving the Wave Equation: The wave equation ictates the propagation of a wave in a meium. In its simplest 1 imensional form, it takes on the following form with the letter c representing the spee of propagation. Or equivalently: = = 1 It can then be shown that the general solution for such a problem, known as D Alembert s solution, is:

16 ,= ++ The 1 imensional wave equation can be solve using the above metho or by separation of variables, which is similar to the heat equation solution. To solve the wave equation with D Alembert s solution, we perform the following steps. Assuming we are given the initial shape of the string,0= an its initial velocity,,0= : We assume the solution is of the form:,= ++ Taking the erivative: =+ = + + = + Now integrating both sies: = + We may now use this equation an the secon one involving v( as a system an obtain the following result: = 1 1 = Plugging this into the original problem, we obtain the following solution to the wave equation with the specifie initial conitions:,= Goo Luck