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Section 5.2 Linear Approximations

In Section 5.1 we found that the equation of the tangent plane to the function \(f(x,y)\) at the point \((x_0,y_0,f(x_0,y_0))\text{,}\) which is a linear equation.

Definition 5.2.1.

We call the function

\begin{equation*} L(x,y) = f(x_0,y_0) + f_x(x_0,y_0)(x-x_0) + f_y(x_0,y_0)(y-y_0) \end{equation*}

the linearisation of \(f\) at the point \((x_0,y_0)\text{.}\)

Find the linearisation of \(z=\sin(x)\cos(y)\) at the point \((x,y) = \left(\dfrac{\pi}{4},\dfrac{\pi}{4} \right)\text{.}\)

Answer.

\(L(x,y) = \dfrac{1}{2} + \dfrac{1}{2} \left( x - \dfrac{\pi}{4} \right) - \dfrac{1}{2} \left(y - \dfrac{\pi}{4} \right)\)

Solution.

Begin by calculating the partial derivatives of \(z\text{,}\)

\begin{equation*} z_x = \cos(x)\cos(y) \: \text{ and } \: z_y = -\sin(x)\sin(y)\text{.} \end{equation*}

Thus

\begin{alignat*}{2} z \left(\dfrac{\pi}{4},\dfrac{\pi}{4} \right) \amp = \, \, \, \, \sin \left(\dfrac{\pi}{4} \right) \cos \left(\dfrac{\pi}{4} \right) \amp = \dfrac{1}{2}\\ z_x \left(\dfrac{\pi}{4},\dfrac{\pi}{4} \right) \amp = \, \, \, \, \cos \left(\dfrac{\pi}{4} \right) \cos \left(\dfrac{\pi}{4} \right) \amp = \dfrac{1}{2}\\ z_y \left(\dfrac{\pi}{4},\dfrac{\pi}{4} \right) \amp = -\sin \left(\dfrac{\pi}{4} \right) \sin \left(\dfrac{\pi}{4} \right) \amp = -\dfrac{1}{2}\text{.} \end{alignat*}

and so the linearisation is

\begin{equation*} L(x,y) = \dfrac{1}{2} + \dfrac{1}{2} \left( x - \dfrac{\pi}{4} \right) - \dfrac{1}{2} \left(y - \dfrac{\pi}{4} \right)\text{.} \end{equation*}

When we use the linearisation of \(f\) at the point \((x_0,y_0)\) to approximate the function near the point \((x_0,y_0)\) we call this the linear (or tangent plane) approximation of \(f\) at the point \((x_0,y_0)\text{.}\) Notice that if we let the independent variables change by the amounts \(\Delta x\) and \(\Delta y\) then the linearisation will change from \(L(x_0,y_0)\) to \(L(x_0+\Delta x, y_0 + \Delta y)\text{.}\) Thus we can approximate the change in the function value \(z=f(x,y)\) by

\begin{equation*} \Delta z \simeq L(x_0 + \Delta x, y_0 + \Delta y) - L(x_0,y_0). \end{equation*}

On using the linearisation formula given above, we end up with the following result.

Definition 5.2.3. The Linear Approximation Formula.

The linear approximation to the change, \(\Delta z\text{,}\) in the function \(z = f(x,y)\) when the independent variables change from \((x_0,y_0)\) to \((x_0+\Delta x, y_0 + \Delta y)\) is

\begin{equation*} \Delta z \simeq f_x (x_0, y_0) \Delta x + f_y (x_0, y_0) \Delta y\text{.} \end{equation*}

This result is sometimes called the “small change” formula for functions of two variables.

For the function \(f(x,y) = 3x^4+2y^4\text{,}\) \(f(1,2) = 35\text{.}\) Use a linear approximation to estimate \(f(1.01,2.03)\text{.}\)

Answer.

\(f(1.01,2.03) \simeq 37.04\)

Solution.

Via a linear approximation

\begin{equation*} f(1.01,2.03) \simeq f(1,2) + \Delta f\text{.} \end{equation*}

Here

\begin{equation*} f_x = 12x^3 \: \text{ and } \: f_y = 8y^3 \end{equation*}

and so

\begin{equation*} f_x(1,2) = 12 \: \text{ and } \: f_y(1,2) = 64\text{.} \end{equation*}

Thus, with \(\Delta x = 0.01\) and \(\Delta y = 0.03\text{,}\) via the linear approximation formula

\begin{equation*} \Delta f = 12 \times 0.01 + 64 \times 0.03 = 2.04 \end{equation*}

and hence

\begin{equation*} f(1.01,2.03) \simeq 35 + 2.04 = 37.04 \end{equation*}

A steel ball has a mass, \(m\text{,}\) of 6300 \(\pm\) 50 g and has volume, \(V\text{,}\) 800 \(\pm\) 10 cm3. Find the density of the ball, including an estimate of the error.

Answer.

\(\rho = \) 7.875 \(\pm \) 0.1609375 g⁄cm3

Solution.

The density, \(\rho\text{,}\) is given by

\begin{equation*} \rho = \dfrac{m}{V}\text{.} \end{equation*}

Thus

\begin{equation*} \rho = \dfrac{6300}{800} = 7.875 \: g/cm^3. \end{equation*}

Using a linear approximation to estimate the error

\begin{equation*} \Delta \rho \simeq \dfrac{\partial \rho}{\partial m} \Delta m + \dfrac{\partial \rho}{\partial V} \Delta V\text{.} \end{equation*}

Now

\begin{equation*} \dfrac{\partial \rho}{\partial m} = \dfrac{1}{V} \: \text{ and } \: \dfrac{\partial \rho}{\partial V} = -\dfrac{m}{V^2} \end{equation*}

and so at \(m=6300\text{,}\) \(V=800\) and with \(\Delta m = 50\) and \(\Delta V = -10\) (to get the maximum value of \(\Delta \rho\))

\begin{equation*} \Delta \rho \simeq \frac{1}{800} \times 50 + \frac{6300}{800^2} \times 10 = 0.1609375\text{.} \end{equation*}

Thus

\begin{equation*} \rho = 7.875 \pm 0.1609375 \: g/cm^3\text{.} \end{equation*}

Use a linear approximation to estimate the value of \(z\) at \((x,y) = (1.1,-0.02)\) for surface \(z = f(x,y)\) defined implicitly by \(z-yz^3 = x+2\text{.}\)

Answer.

\(z(1.1,-0.02) \simeq 2.56\)

Solution.

Firstly notice that when \(x=1\) and \(y=0\text{,}\) \(z=3\text{.}\) Thus \(z(1,0)=3\text{.}\) Now, via a linear approximation

\begin{equation*} z(1.1,-0.02) \simeq z(1,0) + \Delta z \end{equation*}

where

\begin{equation*} \Delta z \simeq z_x(1,0) \Delta x + z_y(1,0) \Delta y\text{,} \end{equation*}

and \(\Delta x = 0.1\text{,}\) \(\Delta y = -0.02\text{.}\) To find the partial derivatives we need to use implicit differentiation. Differentiating with respect to \(x\text{:}\)

\begin{align*} \dfrac{\partial}{\partial x} (z - yz^3) \amp = \dfrac{\partial}{\partial x} (x+2)\\ z_x - 3y z^2 z_x \amp = 1\\ z_x \amp = \dfrac{1}{1-3yz^2}\text{.} \end{align*}

Thus

\begin{equation*} z_x(1,0) = \dfrac{1}{1-3 \times 0 \times 3^2} = 1\text{.} \end{equation*}

Differentiating with respect to y:

\begin{align*} \dfrac{\partial}{\partial y} (z - yz^3) \amp = \dfrac{\partial}{\partial y} (x+2)\\ z_y - (3y z^2 z_y + z^3) \amp = 0\\ z_y \amp = \dfrac{z^3}{1-3yz^2}\text{.} \end{align*}

Thus

\begin{equation*} z_y(1,0) = \dfrac{3^3}{1-3 \times 0 \times 3^2} = 27\text{.} \end{equation*}

Putting this together gives

\begin{equation*} \Delta z \simeq 1 \times 0.1 + 27 \times (-0.02) = -0.44 \end{equation*}

and hence

\begin{equation*} z(1.1,-0.02) \simeq 3-0.44 = 2.56. \end{equation*}

Exercises Example Tasks

1.

Use a linear approximation to find the value of \(z(2.96,-0.95)\) when

\begin{equation*} z=x^2-xy+3y^2\text{.} \end{equation*}

2.

Use a linear approximation to estimate the value of \(\sqrt{(2.01)^2-0.98}\text{.}\)

3.

A right angled triangle \(ABC\) with right angle at \(B\) is measured with \(AB = \) 10 \(\pm\) 0.02 cm and \(BC = \) 3.4 \(\pm\) 0.02 cm. What is the angle at \(A\text{,}\) including the error?

4.

In the figure below a rectangle initially with sides \(x\) and \(y\) has been made larger so that the sides are now \(x + \Delta x\) and \(y + \Delta y\text{.}\)

Figure 5.2.7.
Shade on the diagram the regions that represent:

  1. The increase in area.

  2. The linear approximation to the increase in area. Explain your answer.