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Find the Taylor polynomial of degree 2 of the function $$ f\left(\begin{array}{l} x \\ y \end{array}\right)=\sqrt{x+y+x y} \text { at the point } \quad\left(\begin{array}{l} -2 \\ -3 \end{array}\right) $$

Short Answer

Expert verified
The Taylor polynomial is: \( 1 - x - 2 - 0.5y - 1.5 - 0.5(x^2 + 4x + 4) - 0.125(y^2 + 6y + 9) + 0.25(xy + 3x + 2y + 6) \).

Step by step solution

01

Find the function value at the point

We need to evaluate the function \( f(x, y) = \sqrt{x+y+xy} \) at the point \( (-2, -3) \). Substituting these values in the function: \[ f(-2, -3) = \sqrt{-2 + (-3) + (-2)(-3)} = \sqrt{-2 - 3 + 6} = \sqrt{1} = 1. \] Thus, the function value at the point is 1.
02

Calculate the partial derivatives

We need the first and second partial derivatives of \( f(x, y) \). **First Partial Derivatives:**1. \( f_x(x, y) = \frac{1}{2\sqrt{x+y+xy}}(1+y) \) 2. \( f_y(x, y) = \frac{1}{2\sqrt{x+y+xy}}(1+x) \) **Second Partial Derivatives:**1. \( f_{xx}(x, y) = \frac{-(1+y)^2}{4(x+y+xy)^{3/2}} \)2. \( f_{yy}(x, y) = \frac{-(1+x)^2}{4(x+y+xy)^{3/2}} \) 3. \( f_{xy}(x, y) = \frac{1}{4(x+y+xy)^{3/2}} \) These derivatives will be used to calculate the Taylor polynomial.
03

Evaluate the derivatives at the point

Next, we substitute \( x = -2 \) and \( y = -3 \) into the partial derivatives:1. \( f_x(-2, -3) = \frac{1}{2\sqrt{1}}(1 + (-3)) = \frac{-2}{2} = -1 \)2. \( f_y(-2, -3) = \frac{1}{2\sqrt{1}}(1 + (-2)) = \frac{-1}{2} = -0.5 \)3. \( f_{xx}(-2, -3) = \frac{-(1-3)^2}{4 \cdot 1^{3/2}} = -1 \)4. \( f_{yy}(-2, -3) = \frac{-(1-2)^2}{4 \cdot 1^{3/2}} = -0.25 \)5. \( f_{xy}(-2, -3) = \frac{1}{4 \cdot 1^{3/2}} = 0.25 \)
04

Write the Taylor polynomial of degree 2

The Taylor polynomial of degree 2 for a function \( f(x, y) \) centered at \( (a, b) \) is given by: \[ f(a, b) + f_x(a, b)(x-a) + f_y(a, b)(y-b) + \frac{1}{2}f_{xx}(a, b)(x-a)^2 + \frac{1}{2}f_{yy}(a, b)(y-b)^2 + f_{xy}(a, b)(x-a)(y-b). \]Substituting the calculated values:\[ 1 - (x+2) - 0.5(y+3) - \frac{1}{2}(x+2)^2 - \frac{0.25}{2}(y+3)^2 + 0.25(x+2)(y+3) \]Simplifying the polynomial, we have:\[ 1 - x - 2 - 0.5y - 1.5 - 0.5(x^2 + 4x + 4) - 0.125(y^2 + 6y + 9) + 0.25(xy + 3x + 2y + 6). \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Partial Derivatives
In multivariable calculus, partial derivatives are the generalization of ordinary derivatives for functions with more than one variable. They help to analyze how a function changes as one of its variables varies, while keeping the others constant.
To find a function's partial derivative with respect to a variable, treat all other variables as constants and differentiate as usual.
For our function, we have the first order partial derivatives:
  • \( f_x(x, y) = \frac{1}{2\sqrt{x+y+xy}}(1+y) \)
  • \( f_y(x, y) = \frac{1}{2\sqrt{x+y+xy}}(1+x) \)

For second order partial derivatives, we derive the first partial derivatives with respect to each variable:
  • \( f_{xx}(x, y) = \frac{-(1+y)^2}{4(x+y+xy)^{3/2}} \)
  • \( f_{yy}(x, y) = \frac{-(1+x)^2}{4(x+y+xy)^{3/2}} \)
  • \( f_{xy}(x, y) = \frac{1}{4(x+y+xy)^{3/2}} \)
Evaluating these derivatives at a specific point gives us insights into the behavior of the function near that point.
Multivariable Calculus
Multivariable calculus extends the concepts of single-variable calculus to functions of two or more variables. This field is essential in understanding phenomena in higher dimensions.
In multivariable calculus, we often deal with functions like \( f(x, y) \) which depend on two independent variables. The tasks include studying the function’s rates of change, intersections, and behavior in a space, often using tools like partial derivatives and Taylor polynomials.
Functions can exhibit more complex behavior, such as non-linear interactions between variables, which simple derivatives cannot capture fully. That's why partial derivatives are fundamental in multivariable calculus as they allow us to examine the correlation between different variables. Additionally, concepts like gradients, Hessians, and Jacobians enhance our understanding of changes and directional behaviors in multivariable systems.
Degree 2 Polynomial
The Taylor polynomial of degree 2 approximates a multivariable function around a specific point by a quadratic expression. This polynomial provides a good balance between simplicity and accuracy for smoothly approximating functions within a particular range.
A degree 2 polynomial for a function \( f(x, y) \) centered at \((a, b)\) is written as: \[ f(a, b) + f_x(a, b)(x-a) + f_y(a, b)(y-b) + \frac{1}{2}f_{xx}(a, b)(x-a)^2 + \frac{1}{2}f_{yy}(a, b)(y-b)^2 + f_{xy}(a, b)(x-a)(y-b). \]
In the given exercise, this expression evaluates at the point \((-2, -3)\), providing the polynomial: \[ 1 - (x+2) - 0.5(y+3) - \frac{1}{2}(x+2)^2 - \frac{0.25}{2}(y+3)^2 + 0.25(x+2)(y+3). \]
This polynomial approximates \( f(x, y) \) near \((-2, -3)\), capturing the curvature of the function and providing a local model for study.
Function Evaluation
Evaluating a multivariable function at a given point helps to find the exact output of the function at that location in its domain. In this process, you substitute the specific values for each variable into the function.
For our function \( f(x, y) = \sqrt{x+y+xy} \), evaluating it at \((-2, -3)\) involves substituting these values into the expression. This leads to:\[ f(-2, -3) = \sqrt{-2 + (-3) + (-2)(-3)} = \sqrt{1} = 1. \]
The function evaluation not only provides the function’s value at a particular point but also acts as the first step in constructing more complex analyses, such as Taylor polynomials. This practice is critical in approximating functions and understanding their behavior near specific points in their domain.

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Most popular questions from this chapter

What is the volume of the largest rectangular parallelepiped contained in the ellipsoid $$ x^{2}+4 y^{2}+9 z^{2} \leq 9 ? $$

Let \(A, B, C, D\) be a convex quadrilateral in the plane, with the vertices free to move but with \(a\) the length of \(A B, b\) the length of \(B C, c\) the length of \(C D\) and \(d\) the length of \(D A\) all assigned. Let \(\varphi\) be the angle at \(A\) and \(\psi\) be the angle at \(C\) (a) Show that the angles \(\varphi\) and \(\psi\) satisfy the constraint $$ a^{2}+d^{2}-2 d \cos \varphi=b^{2}+c^{2}-2 b c \cos \psi $$ (b) Find a formula for the area of the quadrilateral in terms of \(\varphi, \psi\) and \(a, b, c, d\) (c) Show that the area is maximum if the quadrilateral can be inscribed in a circle. You may use the fact that a quadrilateral can be inscribed in a circle if the opposite angles add to \(\pi\).

(a) What is the Taylor polynomial of degree 2 of the function \(f\left(\begin{array}{l}x \\ y\end{array}\right)=\) \(\sin (2 x+y)\) at the point \(\left(\begin{array}{c}\pi / 6 \\ \pi / 3\end{array}\right) ?\) (b) Show that $$ f\left(\begin{array}{c} x \\ y \end{array}\right)+\frac{1}{2}\left(2 x+y-\frac{2 \pi}{3}\right)-\left(x-\frac{\pi}{6}\right)^{2} $$ has a critical poiut at \(\left(\begin{array}{c}\pi / 6 \\ \pi / 3\end{array}\right) .\) What kind of critical point is it?

(a) Find a parametrization for the union \(X\) of the lines through the origin and a point of the parametrized curve \(t \mapsto\left(\begin{array}{c}t \\\ t^{2} \\ t^{3}\end{array}\right)\). (b) Find an equation for the closure \(\bar{X}\) of \(X\). Is \(\bar{X}\) exactly \(X ?\) (c) Show that \(\bar{X}-\\{0\\}\) is a smooth surface. (d) Show that $$ \left(\begin{array}{l} r \\ \theta \end{array}\right) \mapsto\left(\begin{array}{c} r(1+\sin \theta) \\ r \cos \theta \\ r(1-\sin \theta) \end{array}\right) $$ is another parainetrization of \(\bar{X}\). In this form you should have no trouble giving a name to the surface \(\bar{X}\). (e) Relate \(\bar{X}\) to the set of non-invertible \(2 \times 2\) matrices.

(a) What is the Taylor polynomial of degree 3 of \(\sin \left(x+y^{2}\right)\) at the origin? (b) What is the Taylor polynomial of degree 4 of \(1 /\left(1+x^{2}+y^{2}\right)\) at the origin?

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