math 222 LATEST EDITION 2024 AID GRADE A

Jacobian How to use the chain rule? When you have a composition of functions (dw/ds, dw/dt)= (df/du, df/dv)(du/ds, dv/ds, du/dt, dv/dt) so it gives a vector Classify critical points 1. differentiate WRT each variable x and y. 2. set them equal to 0. then solve for the values (often take the equation that is easy to factorize and take its values. 3. replace the values obtained in the other partial equation set equal to 0 to get the points ***trick: you can add the equations sometimes te the second derivatives and Use Hessian at a critical point: - if det. is negative: saddle - if det. positive definite and the trace is positive: local min - if det. positive definite and trace negative: local max - if det is 0, then critical points depend on higher order terms remember the trace is the addition of the diagonal terms Things to Look Out for When Buying a Used Caravan Previous What is the criteria for the convergence of a sequence? Converges to L if as n gets large, a.n tends to L. limits laws Find a length of a vector sqrt(x^2+y^2+z^2) Equation of a sphere of radius r centered at... (x-xo)^2+(y-yo)^2+(z-zo)^2=r^2 Two ways of computing a dot product 1. (x1, y1, z1)dot(x2, y2, z2)=x1x2+y1y2+z1z2 2. also equal to = |(x1, y1, z1)|*|(x2, y2, z2)| cos(theta) where theta is the angle between the two vectors Adding vectors and multiply by a scalar (x1, y1, z1)+(x2, y2, z2)=(x1+x2, ...) scalar t(x1, y1, z1)=(tx1, ...) two vectors. How do you know if they are perpendicular? Do the dot product, if 0 then perpendicular What are acute and obtuse angles? Acute: aigu Obtuse: obtus What to watch for the cross product minus sign for the middle term symmetric form of a line parametric form of a line symm: (t=) (x-3)/2=(y+5)/3=(z-1)/3 para x=3+2t, y=-5+3t, z=1+4t equation form of a plane ax+by+cz=d Find equation of a line passing through two vectors 1. subtract vectors. 2. obtain v(t)= vector subtracted+ t(new vector obtained). find a line passing through a point and // to a line, WTD? 1. find a vector direction with the line given. 2. parametrize the line with the point and the vector obtained. Find the line of intersection of two planes. WTD? 1. Subtract two equations of planes 2. Parametrize with t. 3. Obtain a form (x1, y1, z1)=(i+t,j+t....) Find line of intersection of x+2y+3z=1 and x-y+z=1. isolate and find 3y+2z=0 so y=2t, z=-3t so x=1+2t-(-3t)=1+5t so (x,y,z)=(1+5t, 2t, -3t) Find the angle between two vectors in R^3. use cross product v1 X v2 = |v1| |v2| sin (theta) where theta is the angle between the two vectors. Find a vector orthogonal to two planes, WTD? Use cross-product Find the plane that contains a line and is // to another plane. 1. use the plane given ax+by+cz=d so that the desired plane has the form ax+by+cz=E 2.Replace the xyz in the equation of the plane with the the coordinate of the line (in parametric with t). 3. the t's should cancel out and you get a constant E. Find the plane that contains x=1+t, y=2-t, z=4-3t and is // to 5x+2y+z=1 5x+2y+z=13 Find the plane that contains the point (6,0,-2) and the line x=4-2t, y=3+5t, z=7+4t. Then answer the question with the equation of the plane ax+by+cz=D IMPORTANT! How to verify that you chose the correct line/plane/point? Replace the coordinates in the equations... SUBSTITUTION! What property of the cross-product is useful to inverse it? bXa=-aXb antisymmetric How to find the normal of a plane? in the form ax+by+cz=D, the normal is (a,b,c) Plane through (0,-2,5) and (-1,3,1) perpendicualr to 2z=5x+4y? 1. transform to obtain 5x+4y-2z=0 2. normal to the plane is (5,4,-2) 3. find the vector thru points: (0,-2,5) - (-1,3,1)= (1,-5,4) 4. cross-product: (5,4,-2)x(1,-5,4)=(6,-22,-29) 5. use the plane formula with one point: 6(x-0)-22(y-(-2))-29(z-5)=0 ...=-101 Validate with the other point... Find plane through (1,5,1) Perpend. to 2x+y-2z=2 and x+3z=4. Remember to complete the formula for the plane with 0 if coordinates are not there. 1. with the trick of the normal to the planes (coefficients!) we get normals: (2,1,-2) and (1,0,3) 2. Their cross-products gives the direction of the normal to the desired plane (which can be used then for the formula of the plane!!!) (3,-8,-1) 3. answer using equation of a plane (normal components times (x-xcoordinate of a point)): 3x-8y-z=-38 METHOD FOR LINE INTERSECTIONS Find plane thru line of intersection x-z=1 and y+2z=3 and perpend. to x+y-2z=1. 1. Find a direction of the line intersection: a) express the line with one variable: (1+z, 3-2z, z) b) an arbitrary point on line is (1,3,0) and a direction of the line using the coefficients is (1,-2,1) 2. take the cross product with the normal to the plane: (1,-2,1)X(1,1,-2)=(3,3,3)=(1,1,1) 3. Use the formula of the plane: and the point: x+y+z=4 SKEW LINES (lignes obliques) 1) find perpendicular line that joins them 2) find distance of x=6+2t, y=2+3t, z=-1+4t and x=-1+6s, y=-s, z=-5+2s 1. use the cross product to find their perpend. vector: gives (1,2,-2) 2. relate coordinates with the perpend. vector times r. x1= x2 +(xnormal*r) giving 6+2t=-1+6s+r ... solving: t=-11/15, s=13/15, r=1/3 3. find distance: d=|r*(1,2,-2)| => r=1 Equation of a cylinder Equation of a cone Equation of an Ellipsoid Equation of an elliptic parabola Equation of a hyperbolic paraboloid Equation of one sheeted hyperboloid Cyl: x^2+y^2=R but with z coord Cone: z^2=x^2+y^2 (trick: zizi en cercle) Ellipsoid: x^2/a^2+y^2/b^2+z^2/c^2=1 where abc are the distances from the origin (trick: trois foyers éligibles) Elliptic parabola: z=x^2/a^2+y^2/b^2 living in z>=0 (trick: ziab) Hyperbolic paraboloid (looks like a valley): z=x^2/a^2-y^2/b^2 (trick: zia mais bé) one sheeted hyperboloid x^2/a^2+y^2/b^2 - z^2/c^2=1 notice the minus sign. if a=b then x^2/a^2-z^2/c^2=1 (trick: ab mais c donne 1) One sheeted hyperboloid one sheet x^2=y^2+1 2-sheeted hyperboloid two sheets y^2=x^2+1 then rotate about the y axis. position vector of a particle r(t)=(x(t), y(t), z(t)) Velocity vector acceleration vector r(t)=(x(t), y(t), z(t)) and r*(t)=(x(t), y(t), z*(t)) Find the arctlength from a to b Integral of the speed where speed=absolute value of the velocity so 1. |r|=sqrt(x^2+y^2+z^2) 2. take the integral from a to be of |r*| Find arclength from 0 to 1 of (sqrt(2)t, e^t, e^-t) 1. find r* 2. |r*|=sqrt(2+e^2t+e^-2t) TRICK!!! REPLACE u=e^t becomes =sqrt(2+u^2+u^-2) and since (u+u^-1)^2= 2+u^2+u^-2 we get |r*|=e^t+e^-t 3. then evaluate the integral and you get e-e^-1 Chain rule formula d/dt(a(u(t)))= da/du*du/dt Arclength: what to check?? Check the sign so that it is POSITIVE!!! Find length of curve of intersection between x^2=2y and 3z=xy from (0,0,0) to (6, 18, 36) 1. Parametrize the curve: use x=t such that r=(t,0.5t^2, t^3/6) 2. Validate: at t=0 and t=6... 3. Find arclength =42 Arclenght formula ds/dt=sqrt((dx/dt)^2+(dy/dt)^2+(dz/dt)^2) such that you obtain s with integration reparametrize a helix r=(cost, sint, t) |r*|=sqrt(2) so that s=sqrt(2)*t+constant which is zero express then with r=(cos(s/sqrt(2)), sin (s/sqrt(2)), s/sqrt(2)) How to compute arclength? 1. find r* 2. find length |r*|=sqrt(... 3. Integrate from the boundaries of t Integral of sec(t) ln(sect+tant) (Intégrer du sec? La laine c'est sec plus tant) Binormal vector B= T X N Curvature 1. Kappa= |r X r|/|r|^3 TRICK: Curve Criss, eux rient à trois! 2. also with function kappa= |f*| / ((1+(f)^2)^(3/2)) 3. Kappa= | dT/ds | Unit normal vector 1. derivative of T/ norm of derivative of T Trick: c'est normal en ti ti!!! Unit tangent vector 1. tangent= velocity/norm of the velocity TRICK: TANT DE GENS RI RI 2. T=dr/ds Center of curvature point???????????? =r(t)- (1/k(t))*N(t)?????????????? Equation of an osculating plane (formed by the normal and the tangent) we get since binormal is perpend. to T and N: B dot ((x,y,z)-r(t))=0 If you have arclength, Tangent, Curvature, Normal... what equation? dT/ds= kappa*N acceleration vector of a particle (tangential and radial or normal components r**= (d^2s/dt^2)T+ Kappa(ds/dt)^2 N Tangential: =(d^2s/dt^2) Normal: =Kappa(ds/dt)^2 where ds/dt is the velocity! acc. tangential formula aTang.= r dot r / | r | Trick: Latrine dote eux rit acc. normal formula aNormal.= | r X r | / | r | Trick: Anorexique c'est normal unit binormal B=r X r /|r X r** | torsion formula (r X r) dot r*/ |r X r**|^2 Trick 2. dB/ds=- tau*N normal plane contains B and N find normal plane with Tangent=(1,t,0.5t^2) and r=(t,0.5t^2, t^3/6) plane is normal to T: 1*(x-t)+t(y-0.5t^2)+0.5t^2(z-t^3/6)=0 r* dot ((x,y,z)-r(t))=0 find osculating plane with Binormal=(0.5t^2,-t, 1) and r=(t,0.5t^2, t^3/6) plane is normal to B: 0.5t^2*(x-t)-t(y-0.5t^2)+1(z-t^3/6)=0 find center of curvature with Normal=((1+0.5t^2)^-1)(-t, 1-0.5t^2, t) and r=(t,0.5t^2, t^3/6) and Kappa=(1+0.5t^2)^-2 (t,0.5t^2, t^3/6)-(kappa^-1)N = (2t+0.5t^3, 0.5t^2-1+t^4/4, -t^3/3+t) find curvature of y=tanx use formula with function kappa= |f*| / ((1+(f)^2)^(3/2)) and use y*=sec^2(x) and d(secx)/dx= secxtanx Find normal and osculating plane at (1,1,1) of x=y^2 and z=x^2 1. Parametrize to get r(t^2, t, t^4) with y=t 2. find velocity r*