### Create a StudySoup account

#### Be part of our community, it's free to join!

Already have a StudySoup account? Login here

# Calculus III MATH 2210

The U

GPA 3.95

### View Full Document

## 23

## 0

## Popular in Course

## Popular in Mathematics (M)

This 51 page Class Notes was uploaded by Miss Noel Mertz on Monday October 26, 2015. The Class Notes belongs to MATH 2210 at University of Utah taught by Staff in Fall. Since its upload, it has received 23 views. For similar materials see /class/229950/math-2210-university-of-utah in Mathematics (M) at University of Utah.

## Reviews for Calculus III

### What is Karma?

#### Karma is the currency of StudySoup.

#### You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!

Date Created: 10/26/15

Quadric Surface Zoo Math 22 10 1 Monday September 27 2004 Quadric surfaces are collections of points in R3 which satisfy a quadratic equation in the Xyz variables Thus these surfaces are de ned implicitly and satisfy the next simplest implicit equations after the linear equations which de ne planes It is good to draw quadric surfaces for yourself the process of learning to draw them is related to your ability to visualize them This document is Maple output and is posted on our lecture page If you open it in Maple you can modify the surfaces and also manipulate them with your mouse You ll try to draw these surfaces in class by hand It turns out from linear algebra that even if your quadratic equation has cross terms xyXzyz there is a rotated coordinate system in which the solution set satsifies a quadric surface equation without cross terms In our section of 2210 we ll only worry about quotnocross termquot surfaces If you eventually take Math 2270 Linear Algebra you should study this question in detail Our text discusses these issues brie y in section 123 and 117 gt with plots the library of graphing commands 0 We already understand the solution set to equations axbyczd ie planes 1 Cylinders If one of the variables is missing from your equation then that variable is free to be anything as long as the other two variables satisify the equation Thus if a point lies on your surface the entire line of points you get by varying the missing variable is also on your surface we call such surfaces cylinders even if the cross section is not circular If you can recognize the curve in the plane of the two visible variables you can draw the cylindrical surface in 3space 39 gt implicitplothZA2yA24lx 3 3 y 3 3 z 2 2 axe s boxed title elliptic cylinder Thus all of your knowledge of conics ellipses parabolas hyperbolas translates into knowledge of cylindrical quadric surfaces 2 Ellipsoids An ellipsoid centered at the origin will be the solution set to 2 2 2 x y z a2 b2 c2 If you look in each coordinate plane by requiring one of your variables to be zero you get a trace curve which is an ellipse and this is the key to drawing the ellipsoid by hand 39 gt implicitplothxA24yA29ZA2lx 3 3 y 3 3 z 2 2 grid 2 O 2 O 2 O axesboxed title ellipsoidfootballTabernacle ellipsoidfo allTabernacle quot L wA VA a l g at Axe n VAVA 395 W V x lb V4quot EIIE WEE v 3 Paraboloids If one of your variables say quot2quot only appears linearly but the other two appear quadratically then your creature is a paraboloid Depending on whether the cross sections perpendicular to the linear variable 2 aXis are ellipses or hyperbolas you either have an elliptic or hyperbolic paraboloid 39 gt plot3d x 2y 2 x 2 2 y sqrt 4 xA2 sqrt 4 xA2 axesboxed title graph of zxA2yA2 elliptic paraboloidspotlight mirror graph of zxquot2yquot2 elliptic u raboloidspotlight mirror guu e 39 0 l WWdz39m ll gt plot3dx 2 y 2x 2 2y sqrt4 XA2 sqrt4 x 2 axesboxed title graph of zxA2 yA2 hyperbolic paraboloidpotato chip grapga63fg ygt hg oIic ll b v VW 9 6 my 4 Hyperboloids If all variables appear quadratically but not all with the same sign that was the case of ellipsoids then there are several interesting possibilities 4a Onesheeted hyperboloid x2 yz 22 1 a2 b2 02 Notice that the traces of this surface in horizontal planes are all nonempty and are ellipses Thus there is only one quotpiecequot to this surface 39 gt implicitplot3dx 2yA2 ZA2lx 2 2y 2 2 z 2 2 axesboxed title one sheeted hyperboloid onesheeted erboloid 2 1 0 1 2 2 2 1 1 0 0 x 1 1 Y 2 Alb Twosheeted hyperboloid x2 yz 22 1 2 b2 2 c Now when you look at traces in horizontal planes zconstant the value of the constant must make 22 l 2 c positive in order for the cross section to be an ellipse If the value is negative then there is NO trace Thus the surface is a union of two pieces on one of which 2 is at least c and on the other 2 is less than c gt implicitplot3dxA2yA2 ZA2 lx 3 3 y 3 3 z 3 3 axesboxed title two sheeted hyperboloid 2ltomm ma 3 395085 41L 11 Arlen r 4 v Ar 1 r 1 l h dk x 7 Wv 39 zylni A 1 a VA an A 11114 WW if WW I v AKV 2 AIquot bqbrlAZUJWJVVme M OV 000 v HB HHOH HOanAXgtNltgtNINgtNHOXHIN Nbxnlw w Nu mHHQn Sobobg manmmnwvoxmnw HHnHmnOOdm u v Math 2210 Section 125 Notes Dylan Zwick Fall 2008 1 Directional Derivatives and Gradients So far we39ve learned about partial derivatives with respect to z and y and also the general notion of differentiability We now want to extend these notions so that we can deal with partial derivatives in any direction If we have a function f z7 y this usually describes a surface 2 f z7 We can imagine this surface as a mountain If you39re standing on a moun tain and I ask you what the slope is at that point you39d have to tell me that you can39t answer the question The reason for this is that the slope is different depending on which direction you walk If you walk straight up the mountain the slope is probably quite large while if you walk along a trail the slope is usually much less So we can39t really talk about the slope at a point only the slope in a given direction at that point This is the idea behind the directional derivative and it39s this concept that we will formalize in this section 11 The Directional Derivative 111 De nition If we have a function f z7 y we can view the input z7 y as coming from a position vector p that points from the origin to the point z7 So we can view the function f z7 y as a function of the vector p Using this idea we can represent the partial derivative with respect to x as fp hi 7 fp Mp hmf haO Along these lines we can define the directional derivative in any direc tion to be De nition For a unit vector u the directional derivative of the function f at p in the direction of u is defined as fp hu 7 fp Dufp hm f haO We note something very important in this definition is that the vector u must be a unit vector This is often something that trips people up who see this for the first time 112 The Directional Derivative and the Gradient We recall that for a function f Ly allta f p that the gradient of the function was defined as the vector vfp fmpi fypi Well the nice thing about taking directional derivatives is that as you might expect we don39t need to always refer back to the limit definition and in fact once we39ve calculated the gradient at a given point p calculat ing the directional derivative in any direction 11 is easy Theorem Let f be differentiable at p Then f has a directional deriva tive at p in the direction of the unit vector u uli Uzi and Dufp u vfp which can be written more explicitly as DUfx7y u1f7y u2fy7y We will not have time to prove this in class but the proof is straight forward and is in the textbook Example Find the directional derivative of the function f L y y2 In x at the point 17 4 in the direction of a i 7 j Example Find the directional derivative of the function f L y 2x2 sin y yz at the point 17 7r2 in the direction of 2i j 113 Direction of Maximum Rate of Change For a given function f at a point p a natural question we may wish to ask is in what direction 11 will the directional derivative be maximized In other words if you39re standing on a mountain in what direction would you have the steepest ascent The answer to this can be derived pretty easily from the earlier theorem and it39s so straightforward that we39ll go through it39s proof DufP uV p HUHH VfltPgtH0039 H VfltPgtH0039 Now cos 6 is maximized and equal to 1 when 6 0 and minimized equal to 1 when 6 7r What this means is that the gradient vector points in the direction of maximum increase and points away from the direction of maximum decrease The magnitude of this maximum increase is H V f I while the magnitude of the maximum decrease is 7 H V f 12 Level Curves and Gradients We recall that for a function 2 f z y the level curves for a given con stant k are all the input values that have an output value k In other words all points Ly such that f Ly k These sets of points usually form curves Now we note that if we move along one of these level curves then by definition f z y is not changing and so it must be that along these level curves the directional derivative is 0 This means that the angle between the direction of the level curve at a point and the direction of the gradient vector at that point must be 7r 2 In still other words the level curves are perpendicular to the gradient vector field Theorem The gradient of f at a point P is perpendicular to the level curve of f that goes through P Example Find the vector in the direction of most rapid increase for the function fz7 y 6 sin x at the point 57r67 0 Then find the rate of change in that direction 121 Higher Dimensions We note here finally that just as functions of two variables f z7 y give us level curves functions of three variables f z7 y 2 give us level surfaces although these can be harder to visualize for obvious reasons As an ex ample the surface defined by wf7y7z2y2722 would have a level surface for w 1 that would be a hyperboloid of one sheet Example Graph the level surface just described Math 2210 Section 128 Notes Dylan Zwick Fall 2008 1 Maxima and Minima 11 Criteria for Maxima and Minima If we have a function f z y with a domain S we define 1 07110 all x y 2 07110 all Ly is a global maximum value of f on S if fz0y0 2 fzy for in S is a global minimum value of f on S if fz0y0 fzy for in S VV 3 f zo7 yo is a global extreme value of f on S if it is either a global maxi mum value or a global minimum value Now a point 0110 6 S is a local maximum or minimum value if it is a global maximum or minimum on S N where N is any neighborhood Of 07y0 in These are just the definitions you39d expect Now the theory of maxima and minima begins with a theorem that is rather difficult to prove but intuitively obvious MaxMin Existence Theorem If f is continuous on a closed and bounded set S then f attains both a global maximum value and a global mini mum value of S We wont39 be proving this but it is proven in most texts on advanced calculus It39s a pain to prove Now we want to figure out a criteria for determining which points are local maxima or minima To aid us in this task we have another big theorem this time easier to prove it39s proven in the textbook but not in these notes but perhaps less intuitively obvious Critical Point Theorem Let f be defined on a set S containing 0110 If f zo7 yo is an extreme value then 0110 must be a critical point that is either 0110 is 1 a boundary point of S or 2 a stationary point of f or 3 a singular point of f Now recall that a stationary point is a point in S at which the functin f is differentiable and V f 0 A singular point is a point where f is not differentiable 12 Critical Point Test Now if we have a critical point a boundary stationary or singular point then the point may or may not be a local maximum or minimum We39ll discuss what to do with boundary points later and singular points you just have to check individually we won39t be dealing with situations where there are an infinite number of singular points yuck but there39s a very useful test for stationary points that is the big brother of the second deriva tive test from single variable calculus Second Partials Test Suppose that fz7y has continuous second partial derivatives in a neighborhood of 07110 and that V f zo7 yo 0 Let D DWOWO fwzlt07yofyy07yo zz0790 Then H if D gt 0 and fmz07 yo lt 0 then f zo7 yo is a local maximum Value if D lt 0 and fmz07 yo gt 0 then fz07 yo is a local minimum Value if D lt 0 then f zo7 yo is a saddle point not an extreme Value P93 if D 0 then the test is inconclusive This number D is actually the determinant of a matrix called the Hes sian Hm i fyw fy Example For f L y zyz 7 6x2 7 312 find all critical points Indicate whether each is a min max 0r saddle point Example For f L y e 2y2 4y find all critical points Indicate whether each is a min max 0r saddle point 13 Dealing with Boundary Points Dealing with boundary points can be a major pain because usually the boundary is a curve and it39s not always easy to parameterize the curve The basic idea is that to find the maxima and minima on a boundary you break the boundary down into curves and then use methods from cal culus 1 to find the maxima and minima on those curves It isn39t always obvious or even remotely obvious how to do this and even when it is fairly straightforward it can still be a major pain I think the best way to get a feel for how to do this is just to see it in the context of some examples Example Find the global max Value and min Value for the elliptical paraboloid fzy 2 112 on S E 713Ly E 7174 and points that yield these extreme Values Example Find the global max and min points for the function ay 9627696y278y7 on S mWZ 92 1 m m 552 s 23 5 W 3 M 5 is sag w K 1 g 555 151 Q a 1 3 6 x w mzrmW w wwumwmwvm f ax gavx a im 5 32 S g SEEK a MMWWWS if Ezagama w m i M w ng 6534 a g gzgg a MT 5 s 5 jggg39m M 22 agiigi 39 a 3 i V ig 3 z x G f5 4r wmm w 39 m 2 7 Q a a x w w w E E 3 2559 2 w W m i H x g k 2 H N i I w m MM W M 4m w xdwix siz 2 f ii f f w N Math 2210 Section 127 Notes Dylan Zwick Fall 2008 1 Tangent Planes and Approximations 11 Implicitly De ned Surfaces and their Tangent Planes We recall from calculus l and brie y from our lecture on the chain rule that we can define curves in R2 implicitly by an equation of the form f 7 y k where k is a constant or without loss of generality by f 7 y 0 An example is the unit circle 2 y2 1 which can also be written as 2 112 71 0 Now the same idea applies to surfaces in three dimensions as we39ve already seen in our study of quadric surfaces A surface in R3 can be defined by an equation of the form Fz7 y7 2 k where k is a constant or again without loss of generality as Fz7 y7 2 0 For example the unit sphere is 2 y2 22 1 which can be rewritten as 2 y2 22 7 1 0 Now if we consider a curve on this surface zt 7 yt 7 2t where the component functions all must satisfy the surface equation Fz7 y7 27 k then we get the relation Ft7yt72t k which says that the function F when Viewed as a function of the Vari able t is a constant Well here we can apply the chain rule to get dFicBFdx 9de 6Fdzid7k70 dtiaxdt aydt azdt dt 739 If we View the curve as a vector Valued function rt lt zt 7 yt7 2t gt then we see that this is equivalent to 1 dr F 7 0 V dt dr Now 7 1s tangent to the curve rt and as the above equatlon 1s valid for any curve on the surface we see that the gradient vector of the func tion F at a point on the surface is perpendicular to all tangent lines to the surface at that point This provides the following definition De nition If a surface is defined implicitly as F x y 2 k then if F is differentiable at a point Pzo7 yo 20 on the surface with VF207 yo 20 7 0 then the plane through P perpendicular to VF 20710720 is called the tangent plane to the surface at the point P Now the equation for this tangent plane is 1 For a surface defined as Fzy7 2 k FzWOJJOVZOXz 0 Fy07yo720y 90 Fz07yo7202 20 0 which just comes from the above discussion and our earlier equa tions about tangent planes N For a surface defined as 2 f z y 2 i 20 fwlt07110gtlt 0 nyO ley 90 which is the equation we derived earlier in this chapter but it can also be derived from the previous equation with F x y 2 f z y 7 2 0 Example Find the equation of the tangent plane to 8z2y2822 16 at 17 27 2 12 Differentials and Approximations If we have a surface defined as 2 f Ly and if zo7y0720 is a fixed point on the surface we can introduce a new coordinate system dz7 dy7 d2 parallel to the old coordinate system but with 071107 20 as the origin In other words we just change Lg7 2 to dz7 dy7 d2 and change zo7y0720 to 07 07 0 If we make this change then our equation for the tangent plane to the surface at the point zo7 yo7 20 changes from Z i 20 fwlt07110gtlt 0 nyOWOXy 90 d2 fwlt07110gtd fy07yody which is much simpler We can actually use this to define a quantity d2 which we call the di erential of the function f L y at the point 07110 De nition Let 2 f 27 y where f is a differentiable function and let dz and dy called the differentials of z and y be Variables The di erential of the dependent Variable d2 also called the total di erential of f and written dfz7 y is defined as d2 dfz7y fwltx7yd fy967ydy vf lt duly gt This differential derives from the concept of a tangent plane and how for a function that is differentiable at a point the tangent plane is a good approximation of the function around that point Consequently we can use the differential d2 to obtain an estimate for the change in the function A2 when we change 2 and y by a small amount Example Use d2 to approximate the change in 2 as 27y moves from 27 3 to 2037 298 for 2 22 75zyy Compare this estimate to the actual change in Value 13 Taylor Polynomials for Multivariable Functions We won39t go too deep into Taylor polynomials for multivariable functions because to be honest they just get too big and too nasty too fast and are best left to computers However the idea is basically the same as the idea for single variable functions For a single variable function we can get an approximation for the func tion by using its first derivative and finding a tangent line f 96 x f 960 f 96o96 7 960 but we can get a better approximation using an approximating parabola whose equation we get from using the second derivative me me f ltzogtltz e goo ltz0gtltz 7 mo Well the same idea applies to surfaces We can get an approximation with a tangent plane but we can get a better approximation with a tan gent elliptic paraboloid The equation is 1 711 flt07yO fz07yo 0 fy07yoy 110 ilmemloX 02 2fzylt 07110gtlt 0y 110 fyylt07110gtlty yo Now this can extend to third or higher order approximations and we can further extend this to functions of three or more variables but as I said it gets messy and is best left to a computer x2y2 Example For the function f L y tan lt find the second or der Taylor polynomial based at 07 0 Math 2210 Section 126 Notes Dylan Zwick Fall 2008 1 The Chain Rule 11 The Calculus I Chain Rule ln calculus l we learned that if we have a composite of two functions f then the derivative of the composite was the derivative of the outside function multiplied by the derivative of the inside function Example What is the derivative of In sin 2 e 111 The First Version of the Multivariable Chain Rule lf 2 f L y is a function of two variables and both of those variables are in turn functions of a single parameter t then we can view the function 2 as a function of the single parameter t The idea behind this sentence is much easier to understand than it ap pears For example suppose we have the function 2 sin 2 y with z t2 and y t3 then we could write 2 as a function of just t namely 2 sin t2 t3 Well 2 when expressed like this is just a single variable function and so if the functions f z and y are differentiable then it makes sense to talk about the derivative of 2 with respect to t The relationship between the derivative of 2 with respect to t and the other derivatives of f z and y are Theorem Let x 2t and y yt be differentiable at t and let 2 f 27y be differentiable at 7 Then 2 fzzt7 is differentiable at t and d2 7 62dz 32dy dt T azdt aydt This is the first version of the chain rule for multivariable functions Basically it39s just saying that the amount that 2 changes when we change It is how much 2 changes when we change 2 multiplied by how much 2 changes when we change It added to how much 2 changes when we change y multiplied by how much y changes when we change It Again that39s a long sentence but walk through it and you39ll see it39s really just logic The proof is pretty straightforward Proof If we simplify notation and let p Ax7 Ag and A2 f pAp 7 f p then since f is differentiable we have A2 fp AP fp vfp AP 619 AP fmpAz fyPAy 6AP AP where 61 a 0 as Ap a 0 Now if we divide both sides by At and take the limit as At a 0 we get 2 dt dy d up mm 2 which is what we want to prove Example d Find 71 given w y 7 yzx z cos t y sin t 12 The Second Version of the Multivariable Chain Rule This is a natural extension of the concepts we just discussed Suppose that we have a function 2 f 27 y and z and y are themselves functions of two other parameters 5 and t say 2 257 t and y gs7 t Then 2 itself can be viewed as a function of s and t and if everything is differentiable we can takes its partial derivative with respect to s or t The corresponding relations are Theorem Let x 257 t and y gs7 t have first partial derivaties at at and let 2 fz7y be differentaible at 2s7t7ys7t Then 2 f 257 t 7 1157 t has first partial derivatives given by 62 61 6f 62 61 3ny g aas asandaampat aw Example Find 2 given w 1nz y 71nz 7 y withz tes andy 59 We note that we can naturally extend these ideas to functions of three or more dimensions Example Ifw z2y222zywherez sty sitandz s2tcalcu1ate 310 a 13 The Implicit Function Theorem We may remember from calculus I that it is possible to define a curve im plicitly as all points z and y that satisfy a given relation F Ly 0 The unit circle for example would be a curve of this form 2 y2 7 1 0 This is a more general concept that a function y f z in that neither of the Variables must be a function of the other one It is possible to talk about the slope of a curve defined in this way around a point on the curve We learned in calculus l a rather long and laborious way of solving this type of problem Here we39ll learn a short cut If we have a relation F z7 y 0 then if we differentiate both sides with respect to x we get M Edi 3c dx 3y dx 7 dx Now if we note that g 1 then after some algebra we get CL d 7 BF which makes implicit differentiation much easier Example d For the curve defined by F z7 y 3 y 7 10y4 0 calculate dig as x a function of z and y We also get similar relations for surfaces defined by F 27 y7 2 0 3F am nd 5395 Ea ay 32 32 Example If Fz7 y7 2 2351 7y sin 2 7 2 0 defines 2 implicitly as a function of x and y find i 32 3 g fggmf 5 m tww Wi gxf u 31 lt wav g 7 05 Vi raw S m z 3 y m 39 gaa 5315 7651 31 53 Iquot v i a lt 2M I 5 52am 622 w a Eggfggg 59232 if i i baggage agggmz 55 m nggi a A A i gamm ga gag ixxizggmfw w 39 Mia A gig 52g 5 a m 7 f N A 39 2 x W 7quot fig 3 wg Ag a 22 w x 39 g wz 6er If E i 0 g 3 h 6 21am Q ag i Kiwi w sz s 5 3f 3 5 at g r f m Cquot m W iii sh quotK M m x a 4 Wm M M Wk me H WWWmm w u MM r My 9 v M x E E 35 mewwm wwwm v 5 WW L 39 E Ag W x3 quot321 W m A wax324 32 aim u 5 1 AW E xi g 5 a M 3 A3 a I f i a ajg A m r W W ms Siamg i 2 Ci g 3 awwxzx i 3 A L V x r w 22232si zisffNW is Amy 1 f lt N sizirfzfi ex 6 153523 2 5 a mi a M M W W W M W M w H W M y if m i g we Q 53 5 a g fquot x V w 5 a f 3 NJ u I W if fig gt YltFf7 f i a if 3 he f 524 3511 23534 5 11 13 s eggi v xgz Quag zgz 25 a The order of integration in triple integrals Fernando Guevara Vasquez November 16 2007 There are many ways of expressing a triple integral as an iterated integral all of them should give the same result but some are easier to evaluate by hand than others So it pays to try different orders of integration before embarking on a long calculation This preliminary exploratory work is done below but it is a good exercise to rewrite the iterated integrals by yourselvesi The following integration example comes from exercise 13721 in the Varberg Purcell and Rigdon text book We have to determine the volume of the solid 5 in the rst octant that is delimited by y 212 and the plane y 8 7 42 see Figure l The volume is given by VolS SldV If we integrate over 2 rst ie we sum the areas of slices of the body at 2 constant we get 2 8742 1y2 2 M 8742 VolS ldzdydz or VolS ldydzdz 0 0 0 0 0 2x2 depending on the order of the inner integrationi Because of the square roots in the integration bounds evaluating these integrals seems error prone So let us try to integrate over y rsti Unfortunately we also get square roots in the integration bounds 8 Mi2 27114 3 2y4 NJ2 VolS ldzdzdy or VolS ldzdzdyi 0 0 0 0 0 0 If the integration is done over I rst the expressions we get should be much easier to evaluate we only have polynomials to integrate 2 8 27y4 2 27122 8742 VolS ldzdydz or VolS ldydzdz 0 2x2 0 0 0 202 You can verify that VolS 12815 Figure 1 Solid from exercise 13i7i21i

### BOOM! Enjoy Your Free Notes!

We've added these Notes to your profile, click here to view them now.

### You're already Subscribed!

Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'

## Why people love StudySoup

#### "Knowing I can count on the Elite Notetaker in my class allows me to focus on what the professor is saying instead of just scribbling notes the whole time and falling behind."

#### "I made $350 in just two days after posting my first study guide."

#### "Knowing I can count on the Elite Notetaker in my class allows me to focus on what the professor is saying instead of just scribbling notes the whole time and falling behind."

#### "Their 'Elite Notetakers' are making over $1,200/month in sales by creating high quality content that helps their classmates in a time of need."

### Refund Policy

#### STUDYSOUP CANCELLATION POLICY

All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email support@studysoup.com

#### STUDYSOUP REFUND POLICY

StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here: support@studysoup.com

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to support@studysoup.com

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.