calc3 cheat sheet onesheet - Math

1 DerivativesDxex=exDxsin(x)=cos(x)Dxcos(x )= sin(x)Dxtan(x)=sec2(x)Dxcot(x)= csc2(x)Dxsec(x)=sec(x)tan(x)Dxcsc(x)= csc(x)cot(x)Dxsin 1=1p1 x2,x2[ 1,1]Dxcos 1= 1p1 x2,x2[ 1,1]Dxtan 1=11+x2, 2 x 2 Dxsec 1=1|x|px2 1,|x|>1 Dxsinh(x)=cosh(x)Dxcosh(x)=sinh(x)Dxtanh (x)=sech2(x)Dxcoth(x)= csch2(x)Dxsech(x)= sech(x)tanh(x)Dxcsch(x)= csch(x)coth(x)Dxsinh 1=1px2+1 Dxcosh 1= 1px2 1,x >1 Dxtanh 1=11 x2 1<x<1 Dxsech 1=1xp1 x2,0<x<1 Dxln(x)=1xIntegralsR1xdx= ln|x|+cRexdx=ex+cRaxdx=1lnaax+cReaxdx=1a eax+cR1p1 x2dx= sin 1(x)+cR11+x2dx= tan 1(x)+cR1xpx2 1dx=sec 1(x)+cRsinh(x)dx= cosh(x)+cRcosh(x)dx= sinh(x)+cRtanh(x)dx= ln|cosh(x)|+cRtanh(x)sech(x)dx= sech(x)+cRsech2(x)dx= tanh(x)+cRcsch(x)coth(x)dx= csch(x)+cRtan(x)dx= ln|cos(x)|+cRcot(x)dx= ln|sin(x)|+cRcos(x)dx= sin(x)+cRsin(x)dx= cos(x)+cR1pa2 u2dx= sin 1(ua)+cR1a2+u2dx=1atan 1ua+cRln(x)dx=(xln(x)) x+cU-SubstitutionLetu=f(x)(canbemorethan onevariable).

2 Determine:du=f(x)dxdxand solve , if a definite integral, substitutethe bounds foru=f(x)ateachboundsSolve the integral using by PartsRudv=uv RvduFns and Identitiessin(cos 1(x)) =p1 x2cos(sin 1(x)) =p1 x2sec(tan 1(x)) =p1+x2tan(sec 1(x))=(px2 1ifx 1)=( px2 1if x 1)sinh 1(x)=lnx+px2+1sinh 1(x)=lnx+px2 1,x 1tanh 1(x)=12lnx+1+x1 x,1<x< 1sech 1(x)=ln[1+p1 x2x],0<x 1sinh(x)=ex e x2cosh(x)=ex+e x2 Trig Identitiessin2(x)+cos2(x)=11+tan2(x)=sec 2(x)1+cot2(x)=csc2(x)sin(x y)=sin(x)cos(y) cos(x)sin(y)cos(x y)=cos(x)cos(y) sin(x)sin(y)tan(x y)=tan(x) tan(y)1 tan(x) tan(y)sin(2x)=2sin(x)cos(x)cos(2x)=cos2( x) sin2(x)cosh(n2x) sinh2x=11+tan2(x)=sec2(x)

3 1+cot2(x)=csc2(x)sin2(x)=1 cos(2x)2cos2(x)=1+cos(2x)2tan2(x)=1 cos(2x)1+cos(2x)sin( x)= sin(x)cos( x)=cos(x)tan( x)= tan(x)Calculus 3 ConceptsCartesian coords in 3 Dgiven two points:(x1,y1,z1)and(x2,y2,z2),Distance between them:p(x1 x2)2+(y1 y2)2+(z1 z2)2 Midpoint:(x1+x22,y1+y22,z1+z22)Sphere with center (h,k,l) and radius r:(x h)2+(y k)2+(z l)2=r2 VectorsVector:~uUnit Vector: uMagnitude:||~u||=qu21+u22+u23 Unit Vector: u=~u||~u||Dot Product~u ~vProduces a Scalar(Geometrically, the dot product is avector projection)~u=<u1,u2,u3>~v=<v1,v2,v3>~u ~v=~0meansthetwovectorsarePerpendicular is the angle betweenthem.~u ~v=||~u||||~v||cos( )~u ~v=u1v1+u2v2+u3v3 NOTE: u v= cos( )||~u||2=~u ~u~u ~v= 0 when?

4 Angle Between~uand~v: = cos 1(~u ~v||~u||||~v||)Projection of~uonto~v:pr~v~u=(~u ~v||~v||2)~vCross Product~u ~vProduces a Vector(Geometrically, the cross product is thearea of a paralellogram with sides||~u||and||~v||)~u=<u1,u2,u3>~v=<v1,v2,v3>~u ~v= i j ku1u2u3v1v2v3 ~u ~v=~0meansthevectorsareparalellLines and PlanesEquation of a Plane(x0,y0,z0) is a point on the plane and<A,B,C>is a normal vectorA(x x0)+B(y y0)+C(z z0)=0<A,B,C> <x x0,y y0,z z0>=0Ax+By+Cz=DwhereD=Ax0+By0+Cz0 Equation of a lineAlinerequiresaDirectionVector~u=<u1,u2,u3>and a point(x1,y1,z1)then,aparameterizationofa linecouldbe:x=u1t+x1y=u2t+y1z=u3t+z1 Distance from a Point to a PlaneThe distance from a point (x0,y0,z0)toa plane Ax+By+Cz=D can be expressedby the formula:d=|Ax0+By0+Cz0 D|pA2+B2+C2 Coord Sys ConvCylindrical to Rectangularx=rcos( )y=rsin( )z=zRectangular to Cylindricalr=px2+y2tan( )=yxz=zSpherical to Rectangularx= sin( )cos( )y= sin( )sin( )z= cos( )Rectangular to Spherical =px2+y2+z2tan( )=yxcos( )=zpx2+y2+z2 Spherical to Cylindricalr= sin( ) = z= cos( )Cylindrical to Spherical =pr2+z2 = cos( )=zpr2+z2 SurfacesEllipsoidx2a2+y2b2+z2c2=1 Hyperboloid of One Sheetx2a2+y2b2 z2c2=1(Major Axis.)

5 Z because it follows - )Hyperboloid of Two Sheetsz2c2 x2a2 y2b2=1(Major Axis: Z because it is the one notsubtracted)Elliptic Paraboloidz=x2a2+y2b2(Major Axis: z because it is the variableNOT squared)Hyperbolic Paraboloid(Major Axis: Z axis because it is notsquared)z=y2b2 x2a2 Elliptic Cone(Major Axis: Z axis because it s the onlyone being subtracted)x2a2+y2b2 z2c2=0 Cylinder1ofthevariablesismissingOR(x a)2+(y b2)=c(Major Axis is missing variable)Partial DerivativesPartial Derivatives are simply holding allother variables constant (and act likeconstants for the derivative) and onlytaking the derivative with respect to agiven z=f(x,y), the partial derivative ofzwithrespecttoxis:fx(x, y)=zx=@z@x=@f(x,y)@xlikewise for partial with respect to y.

6 Fy(x, y)=zy=@z@y=@f(x,y)@yNotationForfxyy,work insidetooutside fxthenfxy,thenfxyyfxyy=@3f@x@2y,For@3f@x @2y,workrighttoleftinthedenominatorGradi entsThe Gradient of a function in 2 variablesisrf=<fx,fy>The Gradient of a function in 3 variablesisrf=<fx,fy,fz>Chain Rule(s)Take the Partial derivative with respectto the first-order variables of thefunction times the partial (or normal)derivative of the first-order variable tothe ultimate variable you are looking forsummed with the same process for otherfirst-order variables this makes sense :let x = x(s,t), y = y(t) and z = z(x,y).zthenhasfirstpartialderivative:@z @xand@z@yxhasthepartialderivatives:@x@sa nd@x@tand y has the derivative:dydtIn this case (with z containing x and yas well as x and y both containing s andt), the chain rule for@z@sis@z@s=@z@x@x@sThe chain rule for@z@tis@z@t=@z@x@x@t+@z@ydydtNote: the use of d instead of @ withthe function of only one independentvariableLimits and ContinuityLimits in 2 or more variablesLimits taken over a vectorized limit justevaluate separately for each componentof the to show limit exists1.

7 Plug in Numbers, Everything is Fine2. Algebraic Manipulation-factoring/dividing out-use trig identites3. Change to polar coordsif(x, y)!(0,0),r!0 Strategies to show limit DNE1. Show limit is di erent if approachedfrom di erent paths(x=y,x=y2,etc.)2. Switch to Polar coords and show thelimit ,z=f(x, y), is continuous at (a,b)iff(a, b)=lim(x,y)!(a,b)f(x, y)Which means:1. The limit exists2. The fn value is defined3. They are the same valueDirectional DerivativesLet z=f(x,y) be a fuction, (a,b) ap pointin the domain (a valid input point) and uaunitvector(2D).The Directional Derivative is then thederivative at the point (a,b) in thedirection of uor:D~uf(a, b)= u rf(a, b)This will return :D~uf(a, b, c)= u rf(a, b, c)Tangent Planeslet F(x,y,z) = k be a surface and P =(x0,y0,z0) of a Tangent Plane:rF(x0,y0,z0) <x x0,y y0,z z0>Approximationsletz=f(x, y)beadi erentiablefunction total di erential of f = dzdz=rf <dx,dy>This is theapproximatechange in zThe actual change in z is the di erencein z values: z=z z1 Maxima and MinimaInternal Points1.

8 Take the Partial Derivatives withrespect to X and Y (fxandfy)(Canusegradient)2. Set derivatives equal to 0 and use tosolve system of equations for x and y3. Plug back into original equation for Second Derivative Test for whetherpoints are local max, min, or saddleSecond Partial Derivative Test1. Find all (x,y) points such thatrf(x, y)=~02. LetD=fxx(x, y)fyy(x, y) f2xy(x, y)IF (a) D>0 ANDfxx<0,f(x,y) islocal max value(b) D>0 ANDfxx(x, y)>0f(x,y)islocal min value(c) D<0, (x,y,f(x,y)) is a saddle point(d) D = 0, test is inconclusive3. Determine if any boundary pointgives min or max. Typically, we have toparametrize boundary and then reduceto a Calc 1 type of min/max problem following only apply only if aboundary is given1.

9 Check the corner points2. Check each line (0 x 5wouldgive x=0 and x=5 )On Bounded Equations, this is theglobal min and derivativetest is not MultipliersGiven a function f(x,y) with a constraintg(x,y), solve the following system ofequations to find the max and minpoints on the constraint (NOTE: mayneed to also find internal points.):rf= rgg(x, y)=0(orkifgiven)Double IntegralsWith Respect to the xy-axis, if taking anintegral,RRdydxis cutting in vertical rectangles,RRdxdyis cutting in horizontalrectanglesPolar CoordinatesWhen using polar coordinates,dA=rdrd Surface Area of a Curvelet z = f(x,y) be continuous over S (aclosed Region in 2D domain)Then the surface area of z = f(x,y) overS is:SA=RRSqf2x+f2y+1dATriple IntegralsRRRsf(x, y, z)dv=Ra2a1R 2(x) 1(x)R 2(x,y) 1(x,y)f(x, y, z)dzdydxNote.

10 Dvcan be exchanged fordxdydzinany order, but you must then chooseyour limits of integration according tothat orderJacobian MethodRRGf(g(u, v),h(u, v))|J(u, v)|dudv=RRRf(x, y)dxdyJ(u, v)= @x@u@x@v@y@u@y@v Common Jacobians:Rect. to Cylindrical:rRect. to Spherical: 2sin( )Vector Fieldsletf(x, y, z)beascalarfieldand~F(x, y, z)=M(x, y, z) i+N(x, y, z) j+P(x, y, z) kbeavectorfield,Grandient of f =rf=<@f@x,@f@y,@f@z>Divergence of~F:r ~F=@M@x+@N@y+@P@zCurl of~F:r ~F= i j k@@x@@y@@zMN P Line IntegralsCgivenbyx=x(t),y=y(t),t2[a, b]Rcf(x, y)ds=Rbaf(x(t),y(t))dswhereds=q(dxdt)2+( dydt)2dtorq1+(dydx)2dxorq1+(dxdy)2dyTo evaluate a Line Integral, get a paramaterized version of the line(usually in terms of t, though inexclusive terms of x or y is ok) evaluate for the derivatives needed(usually dy, dx, and/or dt) plug in to original equation to get interms of the independant variable solve integralWorkLet~F=M i+ j+ k(force)M=M(x, y, z),N=N(x, y, z),P=P(x, y, z)(Literally)d~r=dx i+dy j+dz kWorkw=Rc~F d~r(Work done by moving a particle overcurve C with force~F)