Calc 3

Taylor Polynomial

A function which easily allows us to approximate the value of a function at point when the original function makes this difficult or impossible. The Taylor polynomial and the original function will have the same derivative at a point ,to the nth degree of the taylor polynomial. The equation for the polynomial is p(x)= f(a)+ f'(a)(x−a)+ f''(a)(x−a)2 + f'''(a)(x−a)3++f(n)(a)(x−a)n 2! 3! n!

Translation

A translation is a function T:Rp →Rp such that T(x)=a+x,where a is a fixed vector in R n . A function that is the composition of a linear function followed by a translation is called an affine function. An affine function F thus has the form F(x) = a + L(x) , where L is a linear function.

Maxima and minima

All local max or min will result in a value of zero for the gradient at the point, but the converse is not necessarily true. To begin find all the zeroes of the partial derivatives and then use them in a system of equations to come up with the points that satisfy the gradient equaling zero.A point a in the domain of f is called a local minimum if there is an open ball B (a ; r ) centered at a such that f ( x) − f (a) ≥ 0 for all x ∈ B( a; r )

Lagrange multiplier

Allows us to find the maxima and minima of a function without actually finding a vector equation for the domain. For example if f is the transformation of a level set of a function g(x,y) to one dimension, we know that the domain on which we wish to find a minimum or maximum will be a composition function of f of the vector equation of g(x,y). In order to choose candidates for maximum or minimum we must set the derivative of this function equal to zero,and since from the chain rule we see that this derivative is the dot product of the gradient of f at r(t) and r'(t), we see that in order for this to occur the gradient of f at g must be orthogonal to to the tangent of g(x,y) . This means that since the gradient of f at g is orthogonal to the tangent to g(x,y), that the gradient must be the same as the normal to g(x,y). In order to solve, we find the gradients of the function and the curve, set them equal to each other using the Lagrange multiplier as a variable, and create a system of equations using the individual gradients for each variable and the original equation of the domain. Solve for the multiplier and resubstitute this value to come up with values that correspond to coordinate pairs that will be candidates for maximum and minimum.

Gradient

Derivative of a function that is real valued: f(x1,x2)=x3, the gradient is thus a row vector of the partial derivatives of a scalar valued function

Center of Mass

If the mass is continuously distributed over space, then the sum of the mass used in the equation for force becomes an integral. If we have a plane region in which the mass density is given by an equation p(x,y), we can calculate the mass over the region by dividing the region into equal rectangles of area A and choosing a point within each rectangle defined by the vector description of the region to insert into the mass density equation, and then use this value times the area of a rectangle as an approximation for the mass the rectangle, and finally summing the mass of the individual rectangles. As A becomes infinitely small, this sum becomes the mass of the plane region. Using the previous definition for the center of mass, we now see that the center of mass has coordinates x and y based on the vector components. This is equivalent to the double integral of (the density over the plane region times x)/(the double integral of the density over the plane region.) If mass is uniformly distributed, or p(x,y)=k, we see that the constant has no bearing on the center of mass, which is determined solely by geometry. In this case the center of mass is the centroid of the region.

Force

If we have a system of point masses whose position is defined by a vector r at a point "t" in time , and we have a force "fi" acting on the individual point masses, according to physics we say that this force is equal to the mass of the point times the acceleration(second derivative of r with respect to t) of the mass. If we sum this system we can see that the total force equals the total mass times the derivative( sum of the (masses*position in r)/sum of masses). If we define (sum of the (masses*position in r)/sum of masses) as R, then R becomes the center of mass.

Double Integral in Polar

If we integrate r first, the lower bound is always 0, while the upper bound depends on theta(e.g. r=1 +cos(theta)), and then the region of integration of theta is all values at which we have a slice of theta(e.g. 0-2pi)

Derivative

Let f : D → R p , where D ⊂ R n , and let x 0 ∈int D . Then f is differentiable at x0 if there is a linear function L such that lim1[f(x0 +h)−f(x0)−L(h)]=0. h→0 |h| The linear function L is called the derivative of f at x 0

Matrix Algebra

Letusconsiderthecomposition h=gfoftwolinearfunctions f:Rn →Rp and g:Rp → Rq . Suppose A is the matrix off and B is the matrix of g. Let's see about the matrix C of h. We know the columns of C are the vectors g( f (e j )), j = 1,2,,n , where, of course, the vectors e j are the coordinate vectors for R n . Now the columns of Aarejustthevectors f(ej),j=1,2,,n. Thusthevectors g(f(ej )) aresimplythe products Bf (ej ) . In other words, if we denote the columns of A by ki , i =1,2,,n, so that A = [k1 ,k2 ,,kn ], then the columns of C are Bk1 , Bk2 ,, Bkn , or in other words, C =[Bk ,Bk ,,Bk ].

principal unit normal vector

N = 1/κ*(dT/ ds)

Unit Tangent Vector

T = R'(t)/ = R'(t)/ = R'(t)dt/ = dR/ |R'(t)| ds / dt ds ds Has a length of one and is tangent to R

Torsion

Taking the derivative of the binormal vector results in a vector that is orthogonal to both the binormal and tangent vectors, and must thus have the direction of the normal vector. Torsion is the scalar multiplier by which you must multiply by the unit normal in order to obtain the derivative of the binormal vector. dB/ ds = −τ N .

Double integral

The double integral can be pictured to be the volume of a shape constrained below by the domain D in the xy plane and above by the surface z=f(x,y). If the domain is bounded in the xy plane by y=h(x) and y=g(x), and to the right by x=a and to the left by x=b, and we partition the domain into slices parallel to the y-axis bounded by the surface z=f(x,y), we can obtain the volume of this shape by adding the volume of the partitions. The volume of one partition is the cross-sectional area times the height. When we make these partitions infinitely small and we sum them we arrive at the double integral of D. It is more simply defined as the integration of f(x,y) with respect to y from g(x) to h(x), integrated with respect to x from a to b.

u=(a,b,c)

The shorthand for associating points with vectors. Technically this makes no sense, but it establishes that that this point is at the end of a representative of u originating from the origin.

Maxima and minima on a closed domain

To evaluate maxima and minima we must have a domain that is both closed and bounded. For example if the domain is a circular region, we will first find any points within the domain at which the gradient is equal to zero, and then we must evaluate the boundary of the domain. In order to test the boundary we use a vector equation of the boundary and create a composition function of the original function and the vector equation of the boundary. Now using the composition function we identify candidates at the ends of the closed interval and we also find where the derivative of the composition equals zero, which from the chain rule is the dot product of the gradient evaluated at the vector equation of the boundary and the derivative of the vector equation. This will result in an equation that can be solved for x or y and these values can be used determine all the coordinate pairs that will become candidates, in addition to the original candidate found from the gradient.

Triple Integrals

Triple integrals find the volume of a three-dimensional region by defining the bounds of the first integration to be the planes parallel to the projection of D onto a coordinate plane, which will define the bounds for the next two integrals.

Level Set Gradient

Using the gradient from a curve described in r3, you can obtain an equation for a surface tangent to the curve at a point. The gradient is normal to the curve, and therefore using the point on the surface as coordinate vectors and the gradient at the point, the resulting dot product will give you the equation of a plane tangent to the curve at the point.

Acceleration

a(t) = dv/dt is the acceleration The derivative of the speed (d^2s/dt^2) is the tangential component of the acceleration, and κ ds/dt ^2 is the normal component of the acceleration.

Vector

an equivalence set of directed line segements, defined by having the same magnitude and direction.

scalar product of u = (a,b,c) and v = (x, y, z) :

ax + by + cz

arc length

b d = ∫| f'(t)|dt. a

Level Set

curve that results from slicing a curve with a plane. For example a function f(x,y,z) can be used to create a level set for the vertical axis by setting f(x,y)=k, where k is a constant representing the z value that every pairs of x and y correspond to. The graph will be in two dimensions and looks similar to a contour map.

Directional Derivative

dot product of gradient taken at a point and unit vector, which results in the magnitude of the derivative in the direction of the unit vector.

f is continuous

f (t) = x(t)i + y(t) j + z(t)k , then f is continuous at t0 if and only if each of the everyday scalar functions x(t),y(t), and z(t ) is continuous at t0 .

vector functions

functions f : X → Y in which Y is a set of vectors.

linear function

i) F ( x + y ) = F ( x) + F ( y ) for all x , y ∈ R n , and ii)F(ax)=aF(x) for all scalars and x∈Rn .

partial derivative of fi with respect to the j th variable.

in a matrix composed of L(h), each component is the partial derivative. f is differentiable at x only if the partials exist and are continuous.

Curvature

length of the second derivative, which is orthogonal to the tangent vector. κ = |dT / ds|

parametric equations

r(t)= f(t)i+g(t)j+h(t)k is equivalently described by the equations x = f (t) y = g(t) z = h(t)

coordinate vectors

represented by i,j, and k, these vectors are (1,0,0),(0,1,0), and (0,0,1) and allow any vector to be expressed as a linear combination of these vectors.

f (t) = ti + t^2 j

the nose will lie on the curve y = x^2, called a vector description of the curve.

Vector addition

the sum of adding two vectors, u+v, is the vector containing the directed line segment that connects the tail of v and the nose of u, or the diagonal resulting from the parallelogram formed by the vectors.

u=(a,b,c)=ai+bj+ck andv=(x,y,z)=xi+yj+zk.

u× v = (bz − cy)i + (cx − az) j + (ay −bx)k, or from the determinants of a 3x3 matrix.

vector product

u×v=|u||v||sinθ | n, where n is a vector of length 1 which is orthogonal to both u and v

scalar product

u⋅v =|u||v|cosθ, the dot product of u and v is the length of the projection of v onto u. Theta is the angle between u and v, which is only zero when the vectors are orthogonal.

Velocity

v(t) = dR/dt is the velocity, and we know that v(t) =(ds/dt)T and so the direction of the velocity is the unit tangent T.

vector derivative

vector derivative v is v =a'(t)i+b'(t)j +c'(t)k. The direction of the derivative vector is tangent to the vector at a point.

Binormal unit vector

vector that is orthogonal to both the unit tangent and principal unit normal B = T × N

Polar Coordinates

x = r cosθ, and y = r sinθ. When taking the double integral in cartesian coordinates when we partitioned the region of our plane it resulted in rectangles or delta A , in polar coordinates we now have regions defined by r and theta, thus delta A is approximately rdrdtheta.