Functions #10 - Questions and Answers

Question 1 tan (t + π) =, and cot (t + π) = , so the tangent and cotangent functions are___functions.The period of each of these functions is. Answer Step 1 Given tan (t + π) =, and cot (t + π) =, so the tangent and cotangent functions are___functions. The period of each of these functions is. Step 2 Use the identity: tan(x) = sin(x) / cos(x) tan (t + π) = sin (t + π) / cos (t + π) Use the identities sin (s + t) = cos(s)sin(t) + cos(t)sin(s) cos (s+t) = cos(s)cos(t) + sin(s)sin(t) = (cos(t)sin(π) + cos(π)sin(t)) / (cos(t)cos(π) + sin(t)sin(π)) = (0 + (-1)sin(t)) / (cos(t)(-1) + 0) = -sin(t) / -cos(t) = tan(t) tan (t + π) = tan(t) tan (t + π) = tan (t) thus tangent function is periodic function with period π. Step 3 Use the identity: cot(x) = cos(x) / sin(x) cot (t + π) = cos(t + π) / sin (t + π) Use the identities: sin (s + t) = cos(s)sin(t) + cos(t)sin(s) cos (s+t) = cos(s)cos(t) + sin(s)sin(t) =(cos(t)cos(π) – sin(t)sin(π))/(cos(t)sin(π) + cos(π)sin(t)) =(-cos(t) - 0) / (0 + (-1)sin(t)) = -cos(t) / -sin(t) = cos(t) / sin(t) = cot(t) cot(t+π)=cot(t) Thus, cotangent is periodic with period π Final Answer: tan(t+π) = tan(t) , and cot(t+π) = cot(t) , so the tangent and cotangent functions are periodic functions. The period of each of these functions is π. Question 2 If X is uniformly distributed over (-1,1), find (a) P(|X|> 1/2) (b) the density function of the random variable |X|. Answer Step 1 Given that X ~ U[-1, 1], a uniform random variable on [-1,1] (a) P[|X| > 1/2] = P[X > 1/2] + P[X < -1/2] = ∫[1/2, 1] (1/2) dx + ∫[-1, -1/2] (1/2) dx = (1/2) [x] [1/2, 1] + (1/2) [x] [-1, -1/2] = (1/2) [1 - 1/2] + (1/2) [-1/2 + 1] = (1/2)(1/2) + (1/2)(1/2) = 1/4 + 1/4 = 2/4 = 1/2 Step 2 (b) To find the density function of |X|. Let Y = |X|. We first find the distribution function of Y. Fy(y) = P[Y ≤ y] = P[|X| ≤ y] = P[-y ≤ X ≤ y] = ∫[-y, y] (1/2) dx for 0 ≤ y ≤ 1 = (1/2)[2y] for 0 ≤ y ≤ 1 = y for 0 ≤ y ≤ 1 Thus, the density function of Y is given as fy(y) = d/dy Fy(y) = d/dy (y) = 1 for 0 ≤ y ≤ 1 Thus, fy(y) = {1 for 0 ≤ y ≤ 1 0 o.w Result (a) P[|X| > 1/2] = 1/2 (b) fy(y) = {1 for 0 ≤ y ≤ 1 0 o.w Question 3 Explain why the function is differentiable at the given point. Then find the linearization L(x, y) of the function at that point. f(x,y) = 1 + xln(xy – 5), (2, 3) Answer 1 Step 1 The Linearization L(x, y) for f(x, y) at (x₀, y₀) is L(x, y) = f(x₀, y₀) + fx(x₀, y₀) (x - x₀) + fy(x₀, y₀) (y - y₀) Step 2 f(x,y) = 1 + xln (xy - 5) f(2,3) = 1 + 2ln (6-5) = 1 + 2ln 1 = 1 fx = ln(xy-5) + xy/(xy-5) fx(2, 3) = ln(6 – 5) + 6/(6 - 5) = 0 + 6 = 6 fy= x²/(xy-5) fy(2,3) = 2²/(6 - 5) = 4 Step 3 Linearization at (2,3) is L(x,y) = f(2,3) + fx(2,3)(x - 2) + fy(2, 3)(y - 3) Substitute f(2,3) = 1, fx(2,3) = 6, fy(2,3) = 4 L(x,y) = 1 + 6(x - 2) + 4(y - 3) L(x,y) = 1 + 6x - 12 + 4y - 12 L(x, y) = 6x + 4y - 23 Result L(x,y) = 6x + 4y - 23 Answer 2 Step 1: Let's find the partial derivatives of f with respect to x and y: ∂f/∂x = ln(xy - 5) ∂f/∂y = x² / (xy - 5) Step 2: Next, let's evaluate these partial derivatives at the given point (2,3): ∂f/∂x (2,3) = ln(23-5) = ln(1) = 0 ∂f/∂y (2,3) = 2² / (23 - 5) = 4/1 = 4 Since both partial derivatives exist and are continuous at (2, 3), the function f(x, y) is differentiable at that point. Step 3: Now, let's find the linearization L(x, y) of the function f at the point (2,3). The linearization is given by: L(x,y) = f(2,3) + ∂f/∂x (2,3) (x - 2) + ∂f/∂y (2,3) (y - 3) Substituting the values we found earlier: L(x,y) = 1 + 0(x - 2) + 4(y - 3) = 1 + 4(y - 3) Therefore, the linearization of the function f(x,y) at the point (2, 3) is L(x,y) = 1 + 4(y - 3). Answer 3 To show why the function is differentiable at the given point (2, 3) and find its linearization L(x, y), we need to evaluate the partial derivatives and substitute the point's coordinates into the linearization formula. Let's start by calculating the partial derivatives of f(x, y) with respect to x and y using the product rule and chain rule: ∂f/∂x = ∂/∂x (1 + xln(xy - 5)) = ln(xy - 5) + x ∂/∂x (ln(xy-5)) = ln(xy-5) + x (1/(xy-5))∂/∂x (xy-5) = ln(xy-5) + x (1/(xy-5))y = ln(xy - 5) + xy/(xy - 5) Now, let's find the partial derivative with respect to y: ∂f/∂y = ∂/∂y (1 + xln(xy - 5)) = x ∂/∂y (ln(xy - 5)) = x (1/(xy-5))∂/∂y (xy-5) = x (1/(xy-5))x = x²/(xy-5) Now, let's substitute the coordinates (2,3) into the partial derivatives: ∂f/∂x (2,3) = ln(23 - 5) + (23)/(23-5) = ln(1) + 6/1 = 0 + 6 = 6 ∂f/∂y (2,3) = 2²/(23-5) = 4/1 = 4 Now, we can write the linearization L(x,y) using the point-slope form of a linear equation: L(x,y) = f(2, 3) + ∂f/∂x (2,3)(x - 2) + ∂f/∂y (2,3)(y - 3) L(x,y) = f(2, 3) + 6(x - 2) + 4(y - 3) L(x,y) = f(2,3) + 6x -12 + 4y-12 L(x,y) = f(2, 3) + 4x + 4y - 20 Since we're given the point (2,3), we substitute these values into the original function to find f(2,3): f(2,3) = 1 + 2ln(23-5) = 1 + 2ln(6-5) = 1 + 2ln(1) = 1 + 20 = 1 + 0 = 1 Therefore, the linearization L(x,y) of the function at the point (2, 3) is: L(x,y) = 1 + 6x + 4y - 20 L(x,y) = 4x + 4y - 19. Answer 4 To determine the differentiability of the function f(x, y) = 1 + xln(xy-5) at the point (2, 3), we need to check if the partial derivatives of f exist and are continuous in a neighborhood of that point. First, let's find the partial derivatives of f with respect to x and y: ∂f/∂x = ∂/∂x (1+xln(xy-5)) Using the product rule, we have: ∂f/∂x = 0 + ln(xy - 5) + x(∂/∂x (ln(xy - 5))) Now, let's find ∂/∂x (ln(xy – 5)): ∂/∂x (ln(xy-5)) = (1/(xy-5)) * (∂/∂x (xy - 5)) Simplifying further: ∂/∂x (ln(xy-5)) = (1/(xy-5)) * y Therefore, we have: ∂f/∂x = ln(xy-5) + x ((1/(xy-5)) * y ) = ln(xy-5) + xy/(xy-5) Next, let's find ∂f/∂y: ∂f/∂y = ∂/∂y(1 + xln(xy-5)) Applying the chain rule: ∂f/∂y = x ∂/∂y(ln(xy – 5)) = x (1/(xy - 5)) * ∂/∂y (xy - 5) Simplifying further: ∂f/∂y = x (1/(xy-5)) * x = x²/(xy-5) Now, let's evaluate the partial derivatives at the point (2, 3): ∂f/∂x(2,3) = ln(23 - 5) + (23)/(23-5) = ln(1) + 6/1 = 0 + 6 = 6 ∂f/∂y(2,3) = 2²/(23-5) = 4/1 = 4 Since both partial derivatives exist and are continuous at the point (2,3), the function f is differentiable at that point. Next, let's find the linearization L(x,y) of the function f(x,y) at the point (2, 3). The linearization can be expressed using the equation of a plane: L(x,y) = f(2, 3) + ∂f/∂x (2,3) (x - 2) + ∂f/∂y (2,3) (y - 3) Substituting the values we found earlier: L(x,y) = f(2,3) + 6(x - 2) + 4(y - 3) Therefore, the linearization of the function f(x,y) at the point (2, 3) is: L(x,y) = f(2, 3) + 6(x - 2) + 4(y - 3) Question 4 Find a function whose square plus the square of its derivative is 1. Answer We want to find a function y(x) that satisfy the equation y² + (dy/dx)² = 1 (dy/dx)² = 1 - y² dy/dx = ±√(1 - y²) Which is a separable DE, thus we multiply both sides by dy / √1 - y² to get dy/√(1 - y²) = ± dx From appendix I in textbook we find that ∫ du/√(1 - u²) = sin⁻¹u + C Applying this formula we get sin⁻¹y = ± x + C → y = ±sin(x + C) Thus, any solution of the form y = ± sin (x + C) satisfy the required condition, from the DE, we note that y = 1 and y = -1 are singular solutions which make the denominator √(1-y²) = 0 If the negative sign was used we get y = - sin(x + C) Result: y = ±sin (x + C) Question 5 Write the polynomial as the product of factors that are irreducible over the rationals, f(x) = x⁴ + 6x² - 27 Answer Let us factor the given polynomial function as the product of factors that are irreducible over the rationals f(x) = x⁴ + 6x² - 27 = x⁴ + 9x² - 3x² - 27 = (x⁴ + 9x²) - (3x² + 27) Let us factor out x² and -3: = x²(x² + 9) - 3(x² + 9) = (x² - 3) (x² + 9) Result: (x²-3) (x²+9)