Questions and Answers #2 Multivariable Functions

Questions and Answers Sheet 2 Multivariable Functions Question #1 How do you find critical points of multivariable function 𝑓(𝑥, 𝑦) = 𝑥 3 + 𝑥𝑦 − 𝑦 3 Answer: 𝑓(𝑥, 𝑦) = 𝑥 3 + 𝑥𝑦 − 𝑦 3 𝑓𝑥 = 3𝑥 2 + 𝑦 = 0 𝑓𝑦 = 𝑥 − 3𝑦 2 = 0 From the first equation: 𝑦 = −3𝑥 2 So, we need: 𝑥 − 3(−3𝑥 2 )2 = 0 𝑥 − 27𝑥 4 = 0 when 𝑥 = 0 , in which case 𝑦 = −3(0)2 = 0 and also when 𝑥 = 1/3 in which case 𝑦 = −1/3 The critical points are: (0,0) and (1/3, −1/3) Question #2 Let S denote the solid enclosed by 𝑥 2 + 𝑦 2 + 𝑧 2 = 2𝑧 and 𝑧 2 = 𝑥 2 + 𝑦 2 . What is the length of of the curve determined by (𝑥, 𝑦, 𝑧): 𝑥 2 + 𝑦 2 + 𝑧 2 = 2𝑧 and 𝑧 2 = 𝑥 2 + 𝑦 2 ? What is the surface area of S? Answer: We have two surfaces: 𝑥 2 + 𝑦 2 + 𝑧 2 = 2𝑧 and 𝑧 2 = 𝑥 2 + 𝑦 2 The loci of the intersection of the surfaces is thus that of the simultaneous solution, thus: (𝑧 2 ) + 𝑧 2 = 2𝑧 ⇒ 𝑧 2 − 𝑧 = 0 ⇒ 𝑧 = 0,1 Leading to two loci: (𝑧 = 0), (𝑧 = 1): ⇒ (𝑥 2 + 𝑦 2 = 0, a circle of radius 0), (𝑥 2 + 𝑦 2 = 1, a circle of radius 1): If we examine the surfaces, then the first solution is that of a single (tangency) point, and thus the second solution is the sought solution, as such the length of the curve (using 𝑃(circle) = 2𝜋𝑟 ) is: 𝐿 = 2𝜋 Question #3 How do you find critical points for function of two variables 𝑓(𝑥, 𝑦) = 8𝑥 3 + 144𝑥𝑦 + 8𝑦 3 ? Answer: For two-variables function, critical points are defined as the points in which the gradient equals zero, just like you had a critical point for the single-variable function f(x) if the derivative 𝑓 ′ (𝑥) = 0 . The matter is that you now can differentiate the function with respect to more than one variable (namely 2, in your case), and so you must define a derivative for each directions. The gradient is thus defined as the n-dimensional vector (again, in your case 𝑛 = 2 ), and its coordinates are the derivatives with respect to each variable. So, the gradient of a two-variable function 𝑓(𝑥, 𝑦) is the vector ( 𝜕𝑓 𝜕𝑓 , 𝜕𝑥 𝜕𝑦 ), where deriving with respect to a variable means to consider the other as a constant. Let's compute the two derivatives: 𝜕𝑓 𝜕𝑥 𝜕𝑓 𝜕𝑦 = 24𝑥 2 + 144𝑦 = 144𝑥 + 24𝑦 2 To find the critical points, we must find the values of x and y for which ( 𝜕𝑓 𝜕𝑓 , 𝜕𝑥 𝜕𝑦 ) = (0,0) holds. In other words, we must solve 24𝑥 2 + 144𝑦 = 0 24𝑦 2 + 144𝑥 = 0 Simplifying both expression, we have 𝑥 2 + 6𝑦 = 0 𝑦 2 + 6𝑥 = 0 An obvious solution is (𝑥, 𝑦) = (0,0) , which is thus our first critical point, while the other solution are (𝑥, 𝑦) = (−6, −6) , which is the second and last critical point. Question #4 How do you find the critical point and determine whether it is a local maximum, local minimum, or neither for 𝑓(𝑥, 𝑦) = 𝑥 2 + 4𝑥 + 𝑦 2 ? Answer: The first-order partial derivatives of 𝑧 = 𝑓(𝑥, 𝑦) = 𝑥 2 + 4𝑥 + 𝑦 2 are 𝜕𝑧 𝜕𝑥 = 2𝑥 + 4 and 𝜕𝑧 𝜕𝑦 = 2𝑦. Setting both of these equal to zero results in a system of equations whose unique solution is clearly (𝑥, 𝑦) = (−2,0) , so this is the unique critical point of f . The second-order partials are 𝜕2 𝑧 𝜕𝑥 2 = 2, 𝜕2 𝑧 𝜕𝑦 2 = 2 and 𝜕2 𝑧 𝜕𝑥𝜕𝑦 = 𝜕2 𝑧 𝜕𝑦𝜕𝑥 =0 This makes the discriminant for the (multivariable) Second Derivative Test equal to 𝐷= 𝜕2 𝑧 𝜕𝑥 2 ∗ 𝜕2 𝑧 𝜕𝑦 2 −( 𝜕2 𝑧 𝜕𝑥𝜕𝑦 2 ) = 2 ∗ 2 − 02 = 4 > 0 which means the critical point is either a local max or a local min (it's not a saddle point). Since 𝜕2 𝑧 𝜕𝑥 2 = 2 > 0 , the critical point is a local min. Question #5 A quick question; is it possible to say in a way analogous to the single variable case that a multivariable function is "asymptotically equivalent" to a second multivariable function? For example, consider the function of 𝑛1 , 𝑛2 ∈ 𝑅𝑅 given by Var(𝜇̂ ) = 𝜎 2 (𝑛1 +2𝑛2 ) (𝑛1 +𝑛2 )2 . where 𝜎 2 is a constant. Can we say that Var(𝜇̂ ) ≈ lim 1 𝑛1 +𝑛2 and then conclude that Var(𝜇̂ ) → 0 as 𝑛1 → ∞ and 𝑛2 → ∞ ? 𝑥+2𝑦 (𝑥,𝑦)→(∞,∞) (𝑥+𝑦)2 does not exist. Am I wrong to think of Var(𝜇̂ ) as a function of two variables? Answer: The two previous answers are perfectly right, thus I just would like it emphasize the statistical perspective. Note that when you are talking about a variance of an estimator, n1 and n2 are sample sizes, hence they are (strictly) positive integers, i.e., 𝑛1 , 𝑛2 ∈ 𝑁𝑁 . As such, when you take the limit, you cannot consider any possible route (as Wolfram does) in 𝑅𝑅2 . The extreme cases in this variance are 𝑛1 >> 𝑛2 or 𝑛2 > > 𝑛1 , that don't change nothing asymptotically. Furthermore, you can conclude that your estimator 𝜇̂ 𝑛1 ,𝑛2 converges in probability to some constant 𝜇 , i.e., a consistent estimator of what-ever you are estimating (assuming that it is an unbiased estimator or 𝑝 the bias vanishes asymptotically). This stems from the fact that if lim 𝑀𝑆𝐸(𝜇̂) ̂𝑥 𝑛 = 0 , then 𝜇 𝑛 → 𝑛→∞ Question #6 𝑓(𝑥, 𝑦) = 𝑥 sin(𝑥) + cos(𝑥) − 𝑦 sin(𝑥) + 𝑦2 2 My task is to find the critical points of this multivariable function (Determine the set of critical points of the function). Now I found the partial derivative to be 𝑓𝑥′ = (𝑥 − 𝑦) cos(𝑥) 𝑓𝑦′ = 𝑦 − sin(𝑥) Wolfram Alpha says the critical point(s) is/are at 𝑥 = 𝑦 ? How do I get the above in that form? The best I could do was get it in the form 𝑥 − 𝑦 sec(𝑥) + tan(𝑥) = 𝑦 Answer: Only one critical point is located on the line 𝑦 = 𝑥 . There are an infinite number of critical points on the lines 𝑦 = ±1 Question #7 I have a multivariable function that is composed of six variables; four of these are constants 𝐶1 , 𝐶2 , 𝐶3 , 𝐶4 and the other two variables are x and y. The function is defined as follows Ξ(𝑥, 𝑦, 𝐶1 , 𝐶2 , 𝐶3 , 𝐶4 ) I would like to find the maximum value of Ξ over 𝑥 ∈ {−∞, ∞} and y ∈ {−∞, ∞} . What is the rigorous mathematical notation I should use to describe this. Is this fine: 𝑚𝑎𝑥{Ξ(𝑥, 𝑦, 𝐶1 , 𝐶2 , 𝐶3 , 𝐶4 )} Answer: If it uses four constants, isn't it just a function of 2 variables? Question #8 Suppose that 𝑥, 𝑦, 𝑧 ∈ Ⅎo𝑅 with 𝑎 > 0 𝑎 constant. Then, is the following true? 1 𝑓(𝑥, 𝑦 + 𝑐𝑧, 𝑧) − 𝑓(𝑥, 𝑦, 𝑧) = 𝑐 ∫ (𝜕𝑥 + 𝑧𝜕𝑦 )𝑓(𝑥, 𝑦 + 𝜃𝑧, 𝑧)𝑑𝜃 . 0 It looks like the mean value theorem for multivariable function but I don't know why 𝜕𝑥 appears. Answer: 1 That looks quite wrong. The right-hand side should be something like 𝑐𝑧 ∫0 𝜕𝑦 𝑓(𝑥, 𝑦 + 𝜃𝑐𝑧, 𝑧)𝑑𝜃 Question #9 I have the following multivariable function: 𝑓(𝑥, 𝑦) = 𝑥𝑦 2 , and I must prove it is continuous at (1,1) . I 3 3 have come to: |((𝑥 − 1)2 + (𝑦 − 1)2 )2 − 1| ≤ ((𝑥 − 1)2 + (𝑦 − 1)2 )2 + 1 (Triangle inequality) < 𝛿 3 + 1 = 𝜖. What can I do with this 1? Any tips are welcome and thank you in advance! Answer: One can pose this problem by considering the graph of the function as follow. From the definition of continuity the 𝜀 -neighbourhood of the image 𝑓(1,1) = 1 determines a couple of planes 𝑧 = 1 + 𝜀 and 𝑧 = 1 − 𝜀 The intersections of the planes 1 ± 𝜀 with the graph of f determine two hyperbolas ((𝑥), (𝑦), (1 ± 𝜀)) with 𝑥𝑦 2 = 1 ± . Also a little circle of radius 𝛿 , which is the border of a 𝛿 -neighbourhood centered at the preimage (1,1)⊤ , when is transformed gives a closed curve C viewed in the graph of f . Then there is a point in the curve C which is the farest away from 𝑝 = (1,1,1)⊤ , which will be determined by considering that for the curve C can be parametrized as 𝐶(𝑦) = ( distance function 𝑑(𝑝, 𝐶(𝑦)) = √( 1+𝜀 𝑦2 2 − 1) + (𝑦 − 1)2 + 𝜀 2 . 1+𝜀 𝑦2 , 𝑦, 1 + 𝜀) ⊤ and the