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Section 5.3 Quadratic Approximations

We have seen previously that for functions of one variable the idea of the linearisation of the function could be extended by considering the Taylor polynomial for the function.

For the function \(f(x) = e^x\) the Maclaurin polynomial of degree \(n\) is

\begin{equation*} \sum_{k=0}^{n} \dfrac{x^k}{k!} = 1+x+\dfrac{x^2}{2!}+\ldots+\dfrac{x^n}{n!}\text{.} \end{equation*}

We saw that the linearisation of \(e^x\) at \(x=0\) was the Maclaurin polynomial of degree 1, i.e.

\begin{equation*} T_1 (x) = 1+x\text{.} \end{equation*}

The Maclaurin polynomial of degree 2, i.e.

\begin{equation*} T_2 (x) = 1 + x + \dfrac{x^2}{2} \end{equation*}

gives a quadratic approximation to \(e^x\) about \(x=0\) and so on.

For functions of one variable we derived the Taylor series by trying to find a power series in \((x-a)\) that matched the function and all its derivatives at \(x=a\text{.}\) For functions of two variables we can use a similar idea to derive the Taylor series about the point \((x,y)=(a,b)\text{.}\) This series will be a power series in \((x-a)\) and \((y-b)\) that matches the function and its partial derivatives at the point \((x,y)=(a,b)\text{.}\) The formula for the Taylor series of two variables is quite lengthy to write and so we will not reproduce it here. However, as with functions of one variable truncations of this series are called Taylor polynomials and the Taylor polynomial of degree 1 is the linearisation of the function. Similarly, the Taylor polynomial of degree 2 will give us a quadratic approximation to the function.

Definition 5.3.2. Taylor Polynomial of Degree 2.

The Taylor Polynomial of degree 2 for the function of two variables \(f(x,y)\) about the point \((x,y) = (a,b)\) is
\begin{align*} T_2(x,y) = \amp f(a,b) + f_x(a,b)(x-a) + f_y(a,b)(y-b)\\ \amp + \dfrac{f_{xx}(a,b)}{2!} (x-a)^2 + f_{xy}(a,b)(x-a)(y-b) + \dfrac{f_{yy}(a,b)}{2!}(y-b)^2 \end{align*}

The question of how good an approximation this polynomial is goes beyond what we will cover in this course but if \(f(x,y)\) has continuous partial derivatives and if \((x,y)\) is “sufficiently close to” \((a,b)\) then the approximation should be useful.

Find the quadratic approximation to the function \(f(x,y) = x^2y\) about the point \((x,y) = (1,2)\text{.}\)

Answer.

\(T_2(x,y) = 2+4(x-1)+(y-2)+2(x-1)^2+2(x-1)(y-2)\)

Solution.

First calculate the partial derivatives:

\begin{equation*} f_x = 2xy, \: \: f_y = x^2, \: \: f_{xx} = 2y, \: \: f_{xy} = 2x, \: \: f_{yy} = 0. \end{equation*}

Now evaluate these at \((x,y) = (1,2)\)

\begin{equation*} f_x(1,2) = 4, \: \: f_y(1,2) = 1, \: \: f_{xx}(1,2) = 4, \: \: f_{xy}(1,2) = 2, \: \: f_{yy}(1,2) = 0. \end{equation*}

Thus, using \(f(1,2) = 2\text{,}\) we have

\begin{equation*} T_2(x,y) = 2+4(x-1)+(y-2)+2(x-1)^2+2(x-1)(y-2)\text{.} \end{equation*}

Using both a linear and a quadratic approximation, estimate the difference in the volume between a box with a square base of side length 1 m and height 2 m and a box with square base of side length 1.1 m and height 2.05 m.

Answer.

Via a linear approximation, \(\Delta V = 0.45\text{.}\)

Via a quadratic approximation, \(\Delta V = 0.48\)

Solution.

If we let the side length of the base of a box be \(x\) and the height be \(y\) then volume \(V\) of the box is given by the formula

\begin{equation*} V(x,y) = x^2y\text{.} \end{equation*}

Thus the difference in the volume between the boxes will be the change in \(V\) when \(x\) changes by 0.1 and \(y\) changes by 0.05. Using the results obtained in the example above, via a linear approximation

\begin{equation*} \Delta V = 4 \Delta x + \Delta y = 4 \times 0.1 + 0.05 = 0.45\text{.} \end{equation*}

Via a quadratic approximation

\begin{align*} \Delta V \amp = 4 \Delta x + \Delta y + 2 \Delta x^2 + 2\Delta x \Delta y \\ \amp = 4 \times 0.1 + 0.05 +2 \times 0.1^2 + 2 \times 0.1 \times 0.05 \\ \amp = 0.48\text{.} \end{align*}

Exercises Example Tasks

1.

Find the Taylor polynomial of degree 2 for \(f(x,y) = e^{x+y^2}\) about \((x,y) = (0,0)\text{.}\)

2.

Find the Taylor polynomial of degree 2 for \(f(x,y)=xy\) about \((x,y)=(2,3)\text{.}\)

3.

If \(x=10 \pm 0.5\text{,}\) \(y=15 \pm 0.05\) use a linear approximation and a quadratic approximation to find the value of the dependent variable z and an associated error bound when

\begin{equation*} z = y\ln(x)\text{.} \end{equation*}