Questions and Answers #1 Multivariable Functions

Questions and Answers Sheet 1 Multivariable Functions Question #1 How to find expressions for 𝑓𝑥𝑥 and 𝑓𝑦𝑦 for the multivariable function 𝑓(𝑥, 𝑦) = ln(𝑥 2 𝑦) + 𝑦 3 𝑥 2 ? Answer: 𝑓(𝑥, 𝑦) = ln(𝑥 2 𝑦) + 𝑦 3 𝑥 2 For finding 𝑓𝑥 , we have to differentiate 𝑓(𝑥, 𝑦) with respect to x as follows: 2𝑥𝑦 𝑓𝑥 = 2 + 𝑦 3 𝑥 𝑦 Now, for finding 𝑓𝑥𝑥 , differentiate 𝑓𝑥 with respect to x . Try in the same way for 𝑓𝑦𝑦 . Question #2 I have the multivariable function log(𝑦 2 + 4𝑥 2 − 4) and I have found the maximal domain to be 𝑥2 + 𝑦2 >1 4 Answer: We have ln : (0, ∞) → 𝑅𝑅 and ln : (1, ∞) → 𝑅𝑅 + − {0} Question #3 Multivariable functions works when the partial derivatives are continous but if the function is just differentiable does the chain rule work ? I mean if 𝑧 = 𝑓(𝑥, 𝑦) is differentiable function and 𝑥 = 𝑔(𝑡), 𝑦 = ℎ(𝑡) are also differentiable functions can we write: 𝑑𝑧 𝑑𝑡 = 𝜕𝑓 𝜕𝑥 × 𝑑𝑥 𝑑𝑡 + 𝜕𝑓 𝜕𝑦 × 𝑑𝑦 𝑑𝑡 Answer: The derivative, if it exists, of a function 𝑓: 𝑅𝑅𝑛 → 𝑅𝑅𝑚 at a point x is a linear transformation 𝑓 ′ (𝑥): 𝑅𝑅𝑛 → 𝑅𝑅𝑚 that satisifies 𝑓(𝑥 + ℎ) − 𝑓(𝑥) = 𝑓 ′ (𝑥)ℎ + 𝑟(ℎ) where 𝑟 = 𝑜(|ℎ|) Now, suppose, 𝜙: 𝑅𝑅 → 𝑅𝑅2 : 𝑡 ↦ (𝑔(𝑡), ℎ(𝑡)) and 𝑓: 𝑅𝑅2 → 𝑅𝑅 . If 𝜙 is differentiable at 𝑡 = 𝑡0 and f is differentiable at (𝑔(𝑡0 ), ℎ(𝑡0 )) , then we can apply the chain rule to assert that 𝑓 ∘ 𝜙: 𝑅𝑅 → 𝑅𝑅 is differentiable at 𝑡 = 𝑡0 and (𝑓 ∘ 𝜙)′ (𝑡0 ): 𝑅𝑅 → 𝑅𝑅 exists and is equal to 𝑓′(𝜙)(𝑡_0)) ∘ 𝜙′ (𝑡_0) So far, nothing to do with partials. Of course, it turns out that if the derivative exists then it matches your formula. And if the relevant partial derivatives are continuous, the derivative exists and the formula is also satisfied. On the other hand, if one or more of the partials is not continuous at a given point, the derivative may or may not exist there, and you need to check directly to see if (1) is satisfied. Question #4 Prove multivariable function is surjective The function is 𝑓: 𝑁𝑁 2 → 𝑁𝑁 such that 𝑓(𝑎, 𝑏) = 𝑎𝑏 Answer: Take 𝑏 = 1 and a arbitrary. That shows any a in 𝑁𝑁 is hit by your function. Question #5 How do you define iterations of multivariable functions? To be clear(example): If 𝑓: 𝑅𝑅2 → 𝑅𝑅 How do you define 𝑓 ∘ 𝑓,or𝑓 ∘ … ∘ 𝑓? ? Answer: The composition is undefined as 𝑓 ∘ 𝑓 = 𝑅𝑅2 →𝑓 𝑅𝑅 →? ?. However if you have a function 𝑅𝑅2 →𝐹 𝑅𝑅2 , then we can easily form the composition. Question #6 Differentiation of multivariable function proof 𝑑 𝑢(𝑥) 𝑓(𝑡, 𝑥)𝑑𝑡 ∫ 𝑑𝑥 𝑣(𝑥) 𝑢(𝑥) 𝜕 = 𝑢′ (𝑥)𝑓(𝑢(𝑥), 𝑥) − 𝑣 ′ (𝑥)𝑓(𝑣(𝑥), 𝑥) + ∫𝑣(𝑥) 𝜕𝑥 𝑓(𝑡, 𝑥)𝑑𝑡 Differentiation of multivariable function proof 𝑑 𝑢(𝑥) 𝑓(𝑡, 𝑥)𝑑𝑡 ∫ 𝑑𝑥 𝑣(𝑥) 𝑢(𝑥) 𝜕 = 𝑢′ (𝑥)𝑓(𝑢(𝑥), 𝑥) − 𝑣 ′ (𝑥)𝑓(𝑣(𝑥), 𝑥) + ∫𝑣(𝑥) 𝜕𝑥 𝑓(𝑡, 𝑥)𝑑𝑡 Answer: Start with 𝑢 𝐼(𝑥) = ∫𝑣 𝑓(𝑡, 𝑥)𝑑𝑡 = 𝐹(𝑢, 𝑥) − 𝐹(𝑣, 𝑥) Where F is the antiderivative of f 𝑑 𝑑𝑥 𝐼= 𝑑 𝑑𝑥 𝐹(𝑢, 𝑥) − 𝑑 𝑑𝑥 𝐹(𝑣, 𝑥) Now for a function 𝑤(𝑎(𝑥), 𝑏(𝑥)) its derivative with respect to x can be written as 𝜕𝑎 𝜕𝑥 (𝑥) 𝜕𝑤 𝜕𝑥 (𝑎, (𝑥), 𝑏(𝑥)) + 𝜕𝑏 𝜕𝑥 (𝑥) 𝜕𝑤 𝜕𝑥 (𝑎, (𝑥), 𝑏(𝑥)) In terms of F we have 𝐹(𝑢, 𝑥) and 𝐹(𝑣, 𝑥) which remember u and v are functions of x. Therefore for each of them we can write this by plugging into the above formula. 𝑤(𝑥) = 𝐹(𝑥), 𝑎(𝑥) = 𝑣(𝑥) or 𝑢(𝑥) and 𝑏(𝑥) = 𝑥 . because 𝑑 𝑑𝑥 𝜕𝑥 𝜕𝑥 𝐹(𝑢, 𝑥) = Therefore = 1 Now the above also holds for 𝐹(𝑢, 𝑥) so 𝜕𝑢 𝜕𝑥 𝑓(𝑢, 𝑥) + 𝜕𝑥 𝜕𝑥 𝑓(𝑢, 𝑥) 𝐼 ′ = 𝑓(𝑣, 𝑥) + 𝜕𝑣 𝜕𝑥 𝑓(𝑣, 𝑥) − 𝑓(𝑢, 𝑥) − 𝜕𝑢 𝜕𝑥 𝑓(𝑢, 𝑥) Now rearranging you can see that it starts to take on your form. 𝐼 ′ = 𝑣 ′ 𝑓(𝑣, 𝑥) − 𝑢′ 𝑓(𝑢, 𝑥) + 𝑓(𝑣, 𝑥) − 𝑓(𝑢, 𝑥) Now the last two terms can be written in terms of an integral. 𝑣 𝑓(𝑣, 𝑥) − 𝑓(𝑢, 𝑥) = ∫ 𝑓 ′ (𝑡, 𝑥)𝑑𝑡 𝑢 Which then can all come together to give 𝑣 𝜕 𝐼 ′ = 𝑣 ′ 𝑓(𝑣, 𝑥) − 𝑢′ 𝑓(𝑢, 𝑥) + ∫𝑢 ⇒ 𝑢 ∫ 𝑓(𝑡, 𝑥)𝑑𝑡 𝑑𝑥 𝑣 𝑑 ′ 𝜕𝑥 𝑓(𝑡, 𝑥)𝑑𝑡 ′ 𝑢 𝜕 = 𝑣 𝑓(𝑣, 𝑥) − 𝑢 𝑓(𝑢, 𝑥) + ∫𝑣 𝜕𝑥 𝑓(𝑡, 𝑥)𝑑𝑡 Which is the formula you want to prove. Question #7 First of all, I'm new to multivariable calculus... in a multivariable function, by assuming that its domain is going to be 𝑅2 and its image is going to be all real numbers, the graph of that function is defined as a subset of 𝑅3 in which the x and y axis are going to receive the inputs, and the output is going to be in z (𝑥, 𝑦, 𝑓(𝑥, 𝑦))? Is that correct? Will its graph, in this example, be some kind of surface? Answer: Yes. If such a 𝑓(𝑥, 𝑦) is defined for all real pairs (x,y), for each such pair corresponds a single real value, which could be taken as the z coordinate of a point. This interpretation leads to a surface, as you suppose. Using Wolfram Alpha, for instance, you can get the plots of the following functions. Guessing the surfaces before looking them up would be an interesting exercise. 𝑓(𝑥, 𝑦) = 𝑥 𝑓(𝑥, 𝑦) = 𝑦 𝑓(𝑥, 𝑦) = 𝑥 + 𝑦 𝑓(𝑥, 𝑦) = 𝑥 − 𝑦 Question #8 let x be a differentiable function 𝑥: [𝑎, 𝑏] → 𝑅 which satisfies: 𝑑𝑥(𝑡) 𝑑𝑡 = 𝑓(𝑡, 𝑥(𝑡)) 𝑥(𝑎) = 𝑥𝑎 In particular I am trying to understand what 𝑓(𝑡, 𝑥(𝑡)) means, I understand this represents a multivariable function with parameters t and x(t) , but I can't think of what this would mean in the given context? Answer: Think of it as if you have any first order differential equation, say: (3𝑥 2 + 1)𝑡𝑥 = −𝑥 3 + 𝑥 What you are looking for is an equation 𝑥 = 𝑥(𝑡) that satisfies the given differential equation, because the analytical solution may not exist or is simply to hard or tedious to obtain. The numerical techniques give an approximation of this function in various values in an interval [𝑡𝑎 , 𝑡𝑏 ] . To calculate the approximation, we write the differential equation in the form 𝑥(𝑡) = 𝑓(𝑡, 𝑥(𝑡)) So, what we are doing is simply isolating in the ode the x term: −𝑥 3 +𝑥 𝑥 = (3𝑥 2 +1)𝑡 And then we have an equation that we can use numerical methods to calculate some approximations for x(t) in [𝑡𝑎 , 𝑡𝑏 ] . Question #9 Derivative of a multivariable function evaluate the derivative of a function 𝑅𝑅 → 𝑅𝑅 defined as 𝑔(𝑡) = 𝑓(𝑥 + 𝑡(𝑦 − 𝑥)) where 𝑓: 𝑅𝑅𝑛 → 𝑅𝑅 is a multivariable function and 𝑥, 𝑦 ∈ 𝑅𝑅𝑛 . Prove that 𝑔′ (𝑡) = (𝑦 − 𝑥)𝑇 ∇𝑓(𝑥 + 𝑡(𝑦 − 𝑥)) Answer: Let 𝑥 = (𝑥1 , 𝑥2 , … , 𝑥𝑛 ) and 𝑦 = (𝑦1 , 𝑦2 , … , 𝑦𝑛 ). Then: 𝑥 + 𝑡(𝑦 − 𝑥) = [𝑥1 + 𝑡(𝑦1 − 𝑥1 ), … , 𝑥𝑛 + 𝑡(𝑦𝑛 − 𝑥𝑛 )]. So, 𝑔(𝑡) = 𝑓(𝑥1 + 𝑡(𝑦1 − 𝑥1 ), … , 𝑥𝑛 + 𝑡(𝑦𝑛 − 𝑥𝑛 )). Define 𝑧𝑖 (𝑡) = 𝑥𝑖 + 𝑡(𝑦𝑖 − 𝑥𝑖 ). So, 𝑔(𝑡) = 𝑓(𝑧1 (𝑡), . . , 𝑧𝑛 (𝑡)). Now, 𝑔′ (𝑡) = ∑ 𝜕𝑓 𝑑𝑧1 𝜕𝑧1 𝑑𝑡 which equals the desired product you have written. Question #10 Sketch the level curves for 𝑓(𝑥, 𝑦) = 𝑥 2 − 𝑦 together with the gradient ∇𝑓 at a few typical points. Write an equation for the tangent line of the level curve 𝑓(𝑥, 𝑦) = 1 at the point (√2, 1) . Answer: Step 1 If 𝑦 = 𝑓(𝑥) , then the independent variable x is called the input of the function and the dependent variable y is called the output of the function. The derivative of the function is the measure of how fast the function is changing with respect to x. A multi variable function 𝑓(𝑥, 𝑦, 𝑧) contains more than one inputs. Step 2 The gradient of a multivariable function 𝑓(𝑥, 𝑦) is defined as ∇𝑓 − 𝑓(𝑥, 𝑦) = 𝑥 2 − 𝑦 , substitute it in ∇𝑓 − 𝜕𝑥 𝑛 𝜕𝑥 = 𝑛𝑥 𝑛−1 𝑎𝑛𝑑 ∇𝑓 = =𝑖 𝜕 𝜕𝑥 𝜕𝑓 𝜕𝑥 𝑖+ 𝜕𝑓 𝜕𝑦 𝜕𝑦 𝑛 𝜕𝑥 𝜕𝑓 𝜕𝑥 𝑖+ 𝜕𝑓 𝜕𝑦 𝜕𝑓 𝜕𝑥 𝑖+ 𝜕𝑓 𝜕𝑦 𝑗. To find gradient of 𝑗. According to standard results in partial differentiation = 𝑛𝑦 𝑛−1 , where n is an integer. 𝑗 (𝑥 2 − 𝑦) + 𝑗 𝜕 𝜕𝑦 (𝑥 2 − 𝑦) = 2𝑥𝑖 − 𝑗 To find the gradient at a particular point (𝑥0 , 𝑦0 ), substitute it in ∇𝑓 = 2𝑥𝑖 − 𝑗 and find the corresponding vectors. Thus the value of gradient corresponding to the points that lies on the y axis can be calculated by substituting 𝑦 = 0 in ∇𝑓 = 2𝑥𝑖 − 𝑗 ∇𝑓 = 2𝑥𝑖 − 𝑗 = 0𝑖 − 𝑗 = −𝑗