Equation 2.6.1.
\begin{align*}
f\big(x_0+\De x\,,\,y_0+\De y\big)
&\approx f\big(x_0\,,\,y_0\big)
+ \pdiff{f}{x}\big(x_0\,,\,y_0\big)\,\De x
+ \pdiff{f}{y}\big(x_0\,,\,y_0\big)\,\De y
\end{align*}
Section 2.6 Linear Approximations and Error
A frequently used, and effective, strategy for building an understanding of the behaviour of a complicated function near a point is to approximate it by a simple function. The following suite of such approximations is standard fare in Calculus I courses. See, for example, §3.4 in the CLP-1 text.
\begin{align*}
g(t_0+\De t) &\approx g(t_0)
&&\text{constant approximation}\\
g(t_0+\De t) &\approx g(t_0) +g'(t_0)\,\De t
&&\text{linear, or tangent line, approximation}\\
g(t_0+\De t) &\approx g(t_0) +g'(t_0)\,\De t +\tfrac{1}{2} g''(t_0)\,\De t^2
&&\text{quadratic approximation}
\end{align*}
More generally, for any natural number \(n\text{,}\) the approximation
\begin{align*}
g(t_0+\De t) &\approx g(t_0) +g'(t_0)\,\De t + \tfrac{1}{2} g''(t_0)\,\De t^2
+\cdots+ \tfrac{1}{n!} g^{(n)}(t_0)\,\De t^n
\end{align*}
is known as the Taylor polynomial of order \(n\text{.}\) You may have also found a formula for the error introduced in making this approximation. The error \(E_n(\De t)\) is defined by
\begin{align*}
g(t_0+\De t)
&=
g(t_0)+g'(t_0)\De t+\tfrac{1}{2!}g''(t_0)\De t^2
+\cdots
+\tfrac{1}{n!}g^{(n)}(t_0) \De t^n +E_n(\De t)
\end{align*}
and obeys
1
You may have seen it written as \(E_n(x)=\tfrac{1}{(n+1)!}g^{(n+1)}(c) (x-a)^{n+1}\)
\begin{equation*}
E_n(\De t)=
\tfrac{1}{(n+1)!}g^{(n+1)}(t_0+c\De t) \De t^{n+1}
\end{equation*}
for some (unknown) \(0\le c\le 1\text{.}\)
It is a simple matter to use these one dimensional approximations to generate the analogous multidimensional approximations. To introduce the ideas, we’ll generate the linear approximation to a function, \(f(x,y)\text{,}\) of two variables, near the point \((x_0,y_0)\text{.}\) Define
\begin{equation*}
g(t) = f\big(x_0+t\,\De x\,,\,y_0+t\,\De y\big)
\end{equation*}
We have defined \(g(t)\) so that
\begin{equation*}
g(0) = f\big(x_0\,,\,y_0\big)\qquad\text{and}\qquad
g(1) = f\big(x_0 + \De x\,,\,y_0+\De y\big)
\end{equation*}
Consequently, setting \(t_0=0\) and \(\De t=1\text{,}\)
\begin{align*}
f\big(x_0+\De x\,,\,y_0+\De y\big)
&=g(1) = g(t_0+\De t)\\
&\approx g(t_0) + g'(t_0)\,\De t\\
&= g(0) + g'(0)
\end{align*}
We can now compute \(g'(0)\) using the multivariable chain rule of 2.4.2:
\begin{equation*}
g'(t) = \pdiff{f}{x}\big(x_0+t\,\De x\,,\,y_0+t\,\De y\big)\,\De x
+ \pdiff{f}{y}\big(x_0+t\,\De x\,,\,y_0+t\,\De y\big)\,\De y
\end{equation*}
so that,
Of course exactly the same procedure works for functions of three or more variables. In particular
Equation 2.6.2.
\begin{align*}
&f\big(x_0+\De x\,,\,y_0+\De y\,,\,z_0 + \De z\big)\\
&\hskip0.25in\approx f\big(x_0\,,\,y_0\,,\,z_0\big)
+ \pdiff{f}{x}\big(x_0\,,\,y_0\,,\,z_0\big)\,\De x
+ \pdiff{f}{y}\big(x_0\,,\,y_0\,,\,z_0\big)\,\De y\\
&\hskip2in
+ \pdiff{f}{z}\big(x_0\,,\,y_0\,,\,z_0\big)\,\De z
\end{align*}
While these linear approximations are quite simple, they tend to be pretty decent provided \(\De x\) and \(\De y\) are small. See the optional §2.6.1 for a more precise statement.
Remark 2.6.3.
\begin{align*}
f\big(x\,,\,y\big)
&\approx f\big(x_0\,,\,y_0\big)
+ \pdiff{f}{x}\big(x_0\,,\,y_0\big)\,(x-x_0)
+ \pdiff{f}{y}\big(x_0\,,\,y_0\big)\,(y-y_0)
\end{align*}
Looking at part (b) of Theorem 2.5.1, we see that this just says that the tangent plane to the surface \(z=f(x,y)\) at the point \(\big(x_0\,,\,y_0\,,\,f(x_0,y_0)\big)\) remains close to the surface when \((x,y)\) is close to \((x_0,y_0)\text{.}\)
Example 2.6.4.
Let
\begin{equation*}
f(x,y) = \sqrt{x^2+y^2}
\end{equation*}
Then
\begin{align*}
\pdiff{f}{x}(x,y)&=\frac{1}{2}\,\frac{2x}{\sqrt{x^2+y^2}} &
f_x(x_0,y_0)&=\frac{x_0}{\sqrt{x_0^2+y_0^2}}\\
\pdiff{f}{y}(x,y)&=\frac{1}{2}\,\frac{2y}{\sqrt{x^2+y^2}} &
f_y(x_0,y_0)&=\frac{y_0}{\sqrt{x_0^2+y_0^2}}
\end{align*}
so that the linear approximation to \(f(x,y)\) at \((x_0,y_0)\) is
\begin{align*}
f\big(x_0+\De x\,,\,y_0+\De y\big)
&\approx f\big(x_0\,,\,y_0\big)
+ f_x\big(x_0\,,\,y_0\big)\,\De x
+ f_y\big(x_0\,,\,y_0\big)\,\De y\\
&=\sqrt{x_0^2+y_0^2}
+ \frac{x_0}{\sqrt{x_0^2+y_0^2}}\,\De x
+ \frac{y_0}{\sqrt{x_0^2+y_0^2}}\,\De y
\end{align*}
We now formalise the above terminology and notation, and add some new terminology and notation (for functions of two variables — the variants for functions of three or more variables are obvious).
Definition 2.6.5.
- The linear approximation to the function \(f(x,y)\) at the point \((x_0,y_0)\) is\begin{equation*} f(x_0,y_0) + \pdiff{f}{x}\big(x_0\,,\,y_0\big)\,(x-x_0) + \pdiff{f}{y}\big(x_0\,,\,y_0\big)\,(y-y_0) \end{equation*}People often write \(\De f\) for the change, \(f\big(x_0+\De x\,,\,y_0+\De y\big) - f\big(x_0\,,\,y_0\big)\) in the value of \(f\text{,}\) \(\De x=x-x_0\) for the change in the value of \(x\text{,}\) and \(\De y=y-y_0\) for the change in the value of \(y\text{.}\) In this notation, the linear approximation gives\begin{align*} \De f &\approx \pdiff{f}{x}\big(x_0\,,\,y_0\big)\,\De x + \pdiff{f}{y}\big(x_0\,,\,y_0\big)\,\De y \end{align*}
- If they want to emphasize that \(\De x\) and \(\De y\) are really small (they may even say“infinitesimal”), they’ll write \(\dee{x}\) and \(\dee{y}\) instead and define
2
Don’t take the terminology “infinitesimal” too seriously. It is just intended to signal “very small”.\begin{align*} \dee{f} &=\pdiff{f}{x}\big(x_0\,,\,y_0\big)\,\dee{x} + \pdiff{f}{y}\big(x_0\,,\,y_0\big)\,\dee{y} \end{align*}That is, \(\dee{f}\) is the change of the value of the linear approximation to \(f\) when \(x\) is changed by \(\dee{x}\) and \(y\) is changed by \(\dee{y}\text{.}\) It approximates the corresponding change \(f\big(x_0+\dee{x}\,,\,y_0+\dee{y}\big) - f\big(x_0\,,\,y_0\big)\) in the value of \(f\text{.}\) People sometimes call \(\dee{x}\text{,}\) \(\dee{y}\) and \(\dee{f}\) “differentials” and sometimes \(\dee{f}\) is called the “total differential of \(f\)” to indicate that it includes the impact of small changes in both \(x\) and \(y\text{.}\)
Definition 2.6.6.
Suppose that we wish to approximate a quantity \(Q\) and that the approximation turns out to be \(Q+\De Q\text{.}\) Then
- the absolute error in the approximation is \(|\De Q|\) and
- the relative error in the approximation is \(\left|\frac{\De Q}{Q}\right|\) and
- the percentage error in the approximation is \(100\left|\frac{\De Q}{Q}\right|\)
In Example 3.4.5 of the CLP-1 text we found an approximate value for the number \(\sqrt{4.1}\) by using a linear approximation to the single variable function \(f(x)=\sqrt{x}\text{.}\) We can make similar use of linear approximations to multivariable functions.
Example 2.6.7.
Find an approximate value for \(\frac{(0.998)^3}{1.003}\text{.}\)
Solution.
Set \(f(x,y) = \dfrac{x^3}{y}\text{.}\) We are to find (approximately) \(f(0.998\,,\,1.003)\text{.}\) We can easily find
\begin{align*}
f(1,1) &= \frac{1^3}{1}=1
\end{align*}
and since
\begin{equation*}
\pdiff{f}{x}=\frac{3x^2}{y}\qquad \text{and}\qquad
\pdiff{f}{y}=-\frac{x^3}{y^2}
\end{equation*}
we can also easily find
\begin{align*}
\pdiff{f}{x}(1,1) &= 3\frac{1^2}{1}=3\\
\pdiff{f}{y}(1,1) &= -1\frac{1^3}{1^2}=-1
\end{align*}
So, setting \(\De x=-0.002\) and \(\De y=0.003\text{,}\) we have
\begin{align*}
\frac{0.998^3}{1.003}
&=f(0.998\,,\,1.003)
=f(1+\De x\,,\,1+\De y)\\
&\approx f\big(1,1\big)
+ \pdiff{f}{x}\big(1,1\big)\,\De x
+ \pdiff{f}{y}\big(1,1\big)\,\De y\\
&\approx 1 +3(-0.002)-1(0.003)
=0.991
\end{align*}
By way of comparison, the exact answer is \(0.9910389\) to seven decimal places.
Example 2.6.8.
Find an approximate value for \((4.2)^{1/2} + (26.7)^{1/3} + (256.4)^{1/4}\text{.}\)
Solution.
Set \(f(x,y,z) = x^{1/2} + y^{1/3} + z^{1/4}\text{.}\) We are to find (approximately) \(f(4.2\,,\,26.7\,,\,256.4)\text{.}\) We can easily find
\begin{align*}
f(4,27,256) &= (4)^{1/2} + (27)^{1/3} + (256)^{1/4} = 2+3+4 =9
\end{align*}
and since
\begin{equation*}
\pdiff{f}{x}=\frac{1}{2x^{1/2}}\qquad
\pdiff{f}{y}=\frac{1}{3y^{2/3}}\qquad
\pdiff{f}{z}=\frac{1}{4z^{3/4}}
\end{equation*}
we can also easily find
\begin{align*}
\pdiff{f}{x}(4,27,256) &= \frac{1}{2(4)^{1/2}} =\frac{1}{2}\times\frac{1}{2}\\
\pdiff{f}{y}(4,27,256) &= \frac{1}{3(27)^{2/3}}=\frac{1}{3}\times\frac{1}{9}\\
\pdiff{f}{z}(4,27,256) &= \frac{1}{4(256)^{3/4}}=\frac{1}{4}\times\frac{1}{64}
\end{align*}
So, setting \(\De x=0.2\text{,}\) \(\De y=-0.3\text{,}\) and \(\De z=0.4\text{,}\) we have
\begin{align*}
&(4.2)^{1/2} + (26.7)^{1/3} + (256.4)^{1/4}
=f(4.2\,,\,26.7\,,\,256.4)\\
&\hskip0.5in=f(4+\De x\,,\,27+\De y\,,\,256+\De z)\\
&\hskip0.5in\approx f\big(4,27,256\big)
+ \pdiff{f}{x}\big(4,27,256\big)\,\De x
+ \pdiff{f}{y}\big(4,27,256\big)\,\De y\\
&\hskip1in + \pdiff{f}{z}\big(4,27,256\big)\,\De z\\
&\hskip0.5in\approx 9 +\frac{0.2}{2\times2}-\frac{0.3}{3\times9}
+\frac{0.4}{4\times64}
= 9+\frac{1}{20}-\frac{1}{90}+\frac{1}{640}\\
&\hskip0.5in=9.0405
\end{align*}
to four decimal places. The exact answer is \(9.03980\) to five decimal places.
That’s a difference of about
\begin{equation*}
100\frac{9.0405-9.0398}{9}\%
=0.008\%
\end{equation*}
Note that we could have used the single variable approximation techniques in the CLP-1 text to separately approximate \((4.2)^{1/2}\text{,}\) \((26.7)^{1/3}\) and \((256.4)^{1/4}\) and then added the results together. Indeed what we have done here is equivalent.
Example 2.6.9.
A triangle has sides \(a=10.1\)cm and \(b=19.8\)cm which include an angle \(35^\circ\text{.}\) Approximate the area of the triangle.
Solution.
The triangle has height \(h=a\sin\theta\) and hence has area
\begin{equation*}
A(a,b,\theta) = \frac{1}{2} bh =\frac{1}{2} ab\sin\theta
\end{equation*}
The \(\sin\theta\) in this formula hides a booby trap built into this problem. In preparing the linear approximation we will need to use the derivative of \(\sin\theta\text{.}\) But the standard derivative \(\diff{}{\theta}\sin\theta
=\cos\theta\) only applies when \(\theta\) is expressed in radians — not in degrees. See Warning 3.4.23 in the CLP-1 text.
So we are obliged to convert \(35^\circ\) into
\begin{equation*}
35^\circ = (30+5) \frac{\pi}{180}\ \text{radians}
=\Big(\frac{\pi}{6} + \frac{\pi}{36}\Big)\ \text{radians}
\end{equation*}
We need to compute (approximately) \(A(10.1\,,\,19.8\,,\,\frac{\pi}{6}+\frac{\pi}{36}\big)\text{.}\) We will, of course, choose
3
There are other choices possible. For example, we could write \(35^\circ=45^\circ-10^\circ\text{.}\) To get a good approximation we try to make \(\De\theta\) as small as possible, while keeping the arithmetic reasonably simple.
\begin{align*}
a_0&=10 & b_0&=20 & \theta_0&=\frac{\pi}{6}\\
\De a&=0.1 & \De b&=-0.2 & \De\theta&=\frac{\pi}{36}
\end{align*}
By way of preparation, we evaluate
\begin{align*}
A\big(a_0,b_0,\theta_0\big)
&=\frac{1}{2}a_0b_0\sin\theta_0 =\frac{1}{2}(10)(20)\frac{1}{2}=50\\
\pdiff{A}{a}\big(a_0,b_0,\theta_0\big)
&=\frac{1}{2}b_0\sin\theta_0 =\frac{1}{2}(20)\frac{1}{2} =5\\
\pdiff{A}{b}\big(a_0,b_0,\theta_0\big)
&=\frac{1}{2}a_0\sin\theta_0 =\frac{1}{2}(10)\frac{1}{2} =\frac{5}{2}\\
\pdiff{A}{\theta}\big(a_0,b_0,\theta_0\big)
&=\frac{1}{2}a_0b_0\cos\theta_0 =\frac{1}{2}(10)(20)\frac{\sqrt{3}}{2}
= 50\,\sqrt{3}
\end{align*}
So the linear approximation gives
\begin{align*}
\text{Area} & = A(10.1\,,\,19.8\,,\,\frac{\pi}{6}+\frac{\pi}{36}\big)
= A(a_0+\De a\,,\,b_0+\De b\,,\,\theta_0+\De\theta\big)\\
&\approx A\big(a_0,b_0,\theta_0\big)
+ \pdiff{A}{a}\big(a_0,b_0,\theta_0\big)\De a
+ \pdiff{A}{b}\big(a_0,b_0,\theta_0\big)\De b\\
&\hskip1in + \pdiff{A}{\theta}\big(a_0,b_0,\theta_0\big)\De\theta\\
&=50 +5\times 0.1 +\frac{5}{2}\times (-0.2) +50\sqrt{3}\frac{\pi}{36}\\
&=50 +\frac{5}{10} -\frac{5}{10} +50\sqrt{3}\frac{\pi}{36}\\
&=50\left(1+\sqrt{3}\frac{\pi}{36}\right)\\
&\approx 57.56
\end{align*}
to two decimal places. The exact answer is \(57.35\) to two decimal places. Our approximation has an error of about
\begin{equation*}
100\ \frac{57.56-57.35}{57.35}\%
=0.37\%
\end{equation*}
Another practical use of these linear approximations is to quantify how errors made in measured quantities propagate in computations using those measured quantities. Let’s explore this idea a little by recycling the last example.
Example 2.6.10. Example 2.6.9, continued.
Suppose, that, as in Example 2.6.9, we are attempting to determine the area of a triangle by measuring the lengths of two of its sides together with the angle between them and then using the formula
\begin{equation*}
A(a,b,\theta) = \frac{1}{2} ab\sin\theta
\end{equation*}
Of course, in the real world , we cannot measure lengths and angles exactly. So if we need to know the area to within 1%, the question becomes: “How accurately do we have to measure the side lengths and included angle if we want the area that we compute to have an error of no more than about 1%?”
4
Of course in our “real world” everyone uses calculus.
Let’s call the exact side lengths and included angle \(a_0\text{,}\) \(b_0\) and \(\theta_0\text{,}\) respectively, and the measured side lengths and included angle \(a_0+\De a\text{,}\) \(b_0+\De b\) and \(\theta_0+\De\theta\text{.}\) So \(\De a\text{,}\) \(\De b\) and \(\De\theta\) represent the errors in our measurements. Then, by 2.6.2, the error in our computed area will be approximately
\begin{align*}
\De A &\approx \pdiff{A}{a}\big(a_0,b_0,\theta_0\big)\,\De a
+ \pdiff{A}{b}\big(a_0,b_0,\theta_0\big)\,\De b
+ \pdiff{A}{\theta}\big(a_0,b_0,\theta_0\big)\,\De\theta\\
&=\frac{\De a}{2} b_0\sin\theta_0
+\frac{\De b}{2} a_0\sin\theta_0
+\frac{\De \theta}{2} a_0b_0\cos\theta_0
\end{align*}
and the percentage error in our computed area will be
\begin{align*}
100\frac{|\De A|}{A(a_0,b_0,\theta_0)}
&\approx \left| 100\frac{\De a}{a_0} + 100\frac{\De b}{b_0}
+100\De\theta\frac{\cos\theta_0}{\sin\theta_0} \right|
\end{align*}
By the triangle inequality, \(|u+v|\le |u|+|v|\text{,}\) and the fact that \(|uv|=|u|\ |v|\text{,}\)
\begin{align*}
&\left| 100\frac{\De a}{a_0} + 100\frac{\De b}{b_0}
+100\De\theta\frac{\cos\theta_0}{\sin\theta_0} \right|\\
&\hskip1in\le 100\left|\frac{\De a}{a_0}\right| + 100\left|\frac{\De b}{b_0} \right|
+100|\De\theta|\ \left|\frac{\cos\theta_0}{\sin\theta_0} \right|
\end{align*}
We want this to be less than \(1\text{.}\)
Of course we do not know exactly what \(a_0\text{,}\) \(b_0\) and \(\theta_0\) are. But suppose that we are confident that \(a_0\ge 10\text{,}\) \(b_0\ge 10\) and \(\frac{\pi}{6}\le \theta_0 \le \frac{\pi}{2}\) so that \(\cot\theta_0\le \cot\frac{\pi}{6}=\sqrt{3}\le 2\text{.}\) Then
\begin{align*}
100\left|\frac{\De a}{a_0}\right|&\le 100\left|\frac{\De a}{10}\right|
= 10\,|\De a|\\
100\left|\frac{\De b}{b_0} \right|&\le 100\left|\frac{\De b}{10} \right|
= 10\,|\De b|\\
100|\De\theta|\ \left|\frac{\cos\theta_0}{\sin\theta_0} \right|
&\le 100|\De\theta|\ 2
=200\,|\De\theta|
\end{align*}
and
\begin{gather*}
100\frac{|\De A|}{A(a_0,b_0,\theta_0)}
\lesssim 10\,|\De a| + 10\,|\De b| +200\,|\De\theta|
\end{gather*}
So it will suffice to have measurement errors \(|\De a|\text{,}\) \(|\De b|\) and \(|\De\theta|\) obey
\begin{equation*}
10\,|\De a| + 10\,|\De b| +200\,|\De\theta| \lt 1
\end{equation*}
Example 2.6.11.
A Question
Suppose that three variables are measured with percentage error \(\veps_1,\ \veps_2\) and \(\veps_3\) respectively. In other words, if the exact value of variable number \(i\) is \(x_i\) and measured value of variable number \(i\) is \(x_i+\De x_i\) then
\begin{equation*}
100\ \left|\frac{\De x_i}{x_i}\right|=\veps_i
\end{equation*}
Suppose further that a quantity \(P\) is then computed by taking the product of the three variables. So the exact value of \(P\) is
\begin{equation*}
P(x_1,x_2,x_3)=x_1x_2x_3
\end{equation*}
and the measured value is \(P(x_1+\De x_1\,,\,x_2+\De x_2\,,\,x_3+\De x_3)\text{.}\) What is the percentage error in this measured value of \(P\text{?}\)
Solution.
The percentage error in the measured value \(P(x_1+\De x_1\,,\,x_2+\De x_2\,,\,x_3+\De x_3)\) is
\begin{equation*}
100\ \left|\frac{P(x_1+\De x_1\,,\,x_2+\De x_2\,,\,x_3+\De x_3)-P(x_1,x_2,x_3)}
{P(x_1,x_2,x_3)}\right|
\end{equation*}
We can get a much simpler approximate expression for this percentage error, which is good enough for virtually all applications, by applying
\begin{align*}
P(x_1+\De &x_1\,,\,x_2+\De x_2\,,\,x_3+\De x_3)\\
&\approx P(x_1,x_2,x_3) +P_{x_1}(x_1,x_2,x_3)\,\De x_1
+P_{x_2}(x_1,x_2,x_3)\,\De x_2\\
&\hskip1in+P_{x_3}(x_1,x_2,x_3)\,\De x_3
\end{align*}
The three partial derivatives are
\begin{alignat*}{4}
P_{x_1}(x_1,x_2,x_3)&=\pdiff{}{x_1}\big[x_1x_2x_3\big]
&=x_2x_3\cr
P_{x_2}(x_1,x_2,x_3)&=\pdiff{}{x_2}\big[x_1x_2x_3\big]
&=x_1x_3\cr
P_{x_3}(x_1,x_2,x_3)&=\pdiff{}{x_3}\big[x_1x_2x_3\big]
&=x_1x_2
\end{alignat*}
So
\begin{align*}
&P(x_1+\De x_1\,,\,x_2+\De x_2\,,\,x_3+\De x_3)\\
&\hskip1in\approx P(x_1,x_2,x_3) +x_2x_3\,\De x_1+x_1x_3\,\De x_2+x_1x_2\,\De x_3
\end{align*}
and the (approximate) percentage error in \(P\) is
\begin{align*}
&100\ \left|
\frac{P(x_1+\De x_1,x_2+\De x_2,x_3+\De x_3)-P(x_1,x_2,x_3)}{P(x_1,x_2,x_3)}
\right|\\
&\hskip0.5in
\approx
100\ \left|
\frac{x_2x_3\De x_1+x_1x_3\De x_2+x_1x_2\De x_3}{P(x_1,x_2,x_3)}
\right|\\
&\hskip0.5in
=100\ \left|\frac{x_2x_3\De x_1+x_1x_3\De x_2+x_1x_2\De x_3}{x_1x_2x_3}\right|\\
&\hskip0.5in=\left|100\frac{\De x_1}{x_1}+100\frac{\De x_2}{x_2}
+100\frac{\De x_3}{x_3}\right|\\
&\hskip0.5in\le \veps_1+\veps_2+\veps_3
\end{align*}
More generally, if we take a product of \(n\text{,}\) rather than three, variables the percentage error in the product becomes at most (approximately) \(\
\smsum\limits_{i=1}^n\veps_i.
\ \) This is the basis of the experimentalist’s rule of thumb that when you take products, percentage errors add.
Still more generally, if we take a “product” \(\prod_{i=1}^n x_i^{m_i}\text{,}\) the percentage error in the “product” becomes at most (approximately) \(\
\smsum\limits_{i=1}^n|m_i|\veps_i.
\ \)
Subsection 2.6.1 Quadratic Approximation and Error Bounds
Recall that, in the CLP-1 text, we started with the constant approximation, then improved it to the linear approximation by adding in degree one terms, then improved that to the quadratic approximation by adding in degree two terms, and so on. We can do the same thing here. Once again, set
\begin{equation*}
g(t) = f\big(x_0+t\,\De x\,,\,y_0+t\,\De y\big)
\end{equation*}
and recall that
\begin{equation*}
g(0) = f\big(x_0\,,\,y_0\big)\qquad\text{and}\qquad
g(1) = f\big(x_0 + \De x\,,\,y_0+\De y\big)
\end{equation*}
We’ll now see what the quadratic approximation
\begin{equation*}
g(t_0+\De t) \approx g(t_0) +g'(t_0)\,\De t +\tfrac{1}{2} g''(t_0)\,\De t^2
\end{equation*}
and the corresponding exact formula (see (3.4.32) in the CLP-1 text)
\begin{equation*}
g(t_0+\De t) = g(t_0) +g'(t_0)\,\De t +\tfrac{1}{2} g''(t_0+c\De t)\,\De t^2
\qquad\text{for some } 0\le c\le 1
\end{equation*}
tells us about \(f\text{.}\) We have already found, using the chain rule, that
\begin{equation*}
g'(t) = \pdiff{f}{x}\big(x_0+t\,\De x\,,\,y_0+t\,\De y\big)\,\De x
+ \pdiff{f}{y}\big(x_0+t\,\De x\,,\,y_0+t\,\De y\big)\,\De y
\end{equation*}
We now need to evaluate \(g''(t)\text{.}\) Temporarily write \(f_1=\pdiff{f}{x}\) and \(f_2=\pdiff{f}{y}\) so that
\begin{equation*}
g'(t) = f_1\big(x_0+t\,\De x\,,\,y_0+t\,\De y\big)\,\De x
+ f_2\big(x_0+t\,\De x\,,\,y_0+t\,\De y\big)\,\De y
\end{equation*}
Then we have, again using the chain rule,
\begin{align*}
&\diff{}{t}\left[f_1\big(x_0+t\,\De x\,,\,y_0+t\,\De y\big)\right]\\
&=\frac{\partial f_1}{\partial x}\big(x_0+t\,\De x\,,\,y_0+t\,\De y\big)
\,\De x
+\frac{\partial f_1}{\partial y}
\big(x_0+t\,\De x\,,\,y_0+t\,\De y\big) \,\De y\\
&
=\frac{\partial^2 f}{\partial x^2}\big(x_0+t\,\De x\,,\,y_0+t\,\De y\big)
\,\De x
+\frac{\partial^2\ f}{\partial y\partial x}
\big(x_0+t\,\De x\,,\,y_0+t\,\De y\big) \,\De y
\tag{$*$}
\end{align*}
and
\begin{align*}
&\diff{}{t}\left[f_2\big(x_0+t\,\De x\,,\,y_0+t\,\De y\big)\right]\\
&=\frac{\partial f_2}{\partial x}\big(x_0+t\,\De x\,,\,y_0+t\,\De y\big)
\,\De x
+\frac{\partial f_2}{\partial y}
\big(x_0+t\,\De x\,,\,y_0+t\,\De y\big) \,\De y\\
&
=\frac{\partial^2\ f}{\partial x\partial y}
\big(x_0+t\,\De x\,,\,y_0+t\,\De y\big) \,\De x
+\frac{\partial^2 f}{\partial y^2}
\big(x_0+t\,\De x\,,\,y_0+t\,\De y\big) \,\De y
\tag{$**$}
\end{align*}
Adding \(\De x\) times \((*)\) to \(\De y\) times \((**)\) and recalling that \(\frac{\partial^2\ f}{\partial y\partial x}
=\frac{\partial^2\ f}{\partial x\partial y}\text{,}\) gives
\begin{align*}
g''(t) &=
\frac{\partial^2 f}{\partial x^2}\big(x_0+t\,\De x\,,\,y_0+t\,\De y\big)
\,\De x^2\\
&\hskip0.5in +2\frac{\partial^2\ f}{\partial x\partial y}
\big(x_0+t\,\De x\,,\,y_0+t\,\De y\big) \,\De x\De y\\
&\hskip1in+ \frac{\partial^2 f}{\partial y^2}\big(x_0+t\,\De x\,,\,y_0+t\,\De y\big)
\,\De y^2
\end{align*}
Now setting \(t_0=0\) and \(\De t=1\text{,}\) the quadratic approximation
\begin{align*}
f\big(x_0 + \De x\,,\,y_0+\De y\big)
&=g(1)\approx g(0) +g'(0) +\tfrac{1}{2} g''(0)
\end{align*}
is
Equation 2.6.12.
\begin{align*}
&f\big(x_0 + \De x\,,\,y_0+\De y\big)\\
&\hskip0.25in\approx f\big(x_0\,,\,y_0\big)
+ \pdiff{f}{x}\big(x_0\,,\,y_0\big)\,\De x
+ \pdiff{f}{y}\big(x_0\,,\,y_0\big)\,\De y\\
&\hskip0.5in +\frac{1}{2}\left\{
\frac{\partial^2 f}{\partial x^2}\big(x_0,y_0\big)\,\De x^2
+2\frac{\partial^2\ f}{\partial x\partial y}
\big(x_0,y_0\big) \,\De x\De y
+ \frac{\partial^2 f}{\partial y^2}\big(x_0,y_0\big)\,\De y^2
\right\}
\end{align*}
and the corresponding exact formula
\begin{align*}
f\big(x_0 + \De x\,,\,y_0+\De y\big)
&=g(1) = g(0) +g'(0) +\tfrac{1}{2} g''(c)
\end{align*}
is
Equation 2.6.13.
\begin{align*}
&f\big(x_0 + \De x\,,\,y_0+\De y\big)\\
&\hskip0.25in= f\big(x_0\,,\,y_0\big)
+ \pdiff{f}{x}\big(x_0\,,\,y_0\big)\,\De x
+ \pdiff{f}{y}\big(x_0\,,\,y_0\big)\,\De y\\
&\hskip0.5in +\frac{1}{2}\left\{
\frac{\partial^2 f}{\partial x^2}\big(\vr(c)\big)\,\De x^2
+2\frac{\partial^2\ f}{\partial x\partial y}
\big(\vr(c)\big) \,\De x\De y
+ \frac{\partial^2 f}{\partial y^2}\big(\vr(c)\big)\,\De y^2
\right\}
\end{align*}
where \(\vr(c) = \big(x_0+c\,\De x\,,\,y_0+c\,\De y\big)\) and \(c\) is some (unknown) number satisfying \(0\le c\le 1\text{.}\)
Equation 2.6.14.
If we can bound the second derivatives
\begin{equation*}
\left|\frac{\partial^2 f}{\partial x^2}\big(\vr(c)\big)\right|\ ,\
\left|\frac{\partial^2\ f}{\partial x\partial y}
\big(\vr(c)\big)\right|\ ,\
\left|\frac{\partial^2 f}{\partial y^2}\big(\vr(c)\big)\right|\le M
\end{equation*}
we can massage 2.6.13 into the form
\begin{align*}
&\left|f\big(x_0 + \De x\,,\,y_0+\De y\big)
-\Big\{f\big(x_0\,,\,y_0\big)
+ \pdiff{f}{x}\big(x_0\,,\,y_0\big)\,\De x
+ \pdiff{f}{y}\big(x_0\,,\,y_0\big)\,\De y\Big\}\right|\\
&\hskip 3.2in
\le \frac{M}{2}\big(|\De x|^2 +2|\De x|\,|\De y| +|\De y|^2\big)
\end{align*}
Why might we want to do this? The left hand side of 2.6.14 is exactly the error in the linear approximation 2.6.1. So the right hand side is a rigorous bound on the error in the linear approximation.
Example 2.6.15. Example 2.6.7, continued.
Suppose that we approximate \(\frac{(0.998)^3}{1.003}\) as in Example 2.6.7 and we want a rigorous bound on the approximation. We can get such a rigorous bound by applying 2.6.13. Set
\begin{equation*}
f(x,y)=\frac{x^3}{y}
\end{equation*}
and
\begin{equation*}
x_0=1\qquad
\De x=-0.002\qquad
y_0=1\qquad
\De y=0.003
\end{equation*}
Then the exact answer is \(f\big(x_0 + \De x\,,\,y_0+\De y\big)\) and the approximate answer is \(f\big(x_0\,,\,y_0\big)
+ \pdiff{f}{x}\big(x_0\,,\,y_0\big)\,\De x
+ \pdiff{f}{y}\big(x_0\,,\,y_0\big)\,\De y\text{,}\) so that, by 2.6.13, the error in the approximation is exactly
\begin{gather*}
\frac{1}{2}\left|
\frac{\partial^2 f}{\partial x^2}\big(\vr(c)\big)\,\De x^2
+2\frac{\partial^2\ f}{\partial x\partial y}
\big(\vr(c)\big) \,\De x\De y
+ \frac{\partial^2 f}{\partial y^2}\big(\vr(c)\big)\,\De y^2
\right|
\end{gather*}
with \(\vr(c) = \big(1-0.002 c\,,\,1+0.003 c\big)\) for some, unknown, \(0\le c\le 1\text{.}\) For our function \(f\)
\begin{align*}
f(x,y)&=\frac{x^3}{y} &
\pdiff{f}{x}(x,y)&=\frac{3 x^2}{y} &
\pdiff{f}{y}(x,y)&=-\frac{x^3}{y^2}\\
\frac{\partial^2 f}{\partial x^2}(x,y)&=\frac{6 x}{y} &
\frac{\partial^2 f}{\partial x\partial y}(x,y)&=-\frac{3 x^2}{y^2} &
\frac{\partial^2 f}{\partial y^2}(x,y)&=\frac{2 x^3}{y^3}
\end{align*}
We don’t know what \(\vr(c)=\big(1-0.002 c\,,\,1+0.003 c\big)\) is. But we know that \(0\le c\le 1\text{,}\) so we definitely know that the \(x\) component of \(\vr(c)\) is smaller that \(1\) and the \(y\) component of \(\vr(c)\) is bigger than \(1\text{.}\) So
\begin{align*}
\left|\frac{\partial^2 f}{\partial x^2}\big(\vr(c)\big)\right|&\le 6 &
\left|\frac{\partial^2 f}{\partial x\partial y}\big(\vr(c)\big)\right|&\le 3 &
\left|\frac{\partial^2 f}{\partial y^2}\big(\vr(c)\big)\right|&\le 2
\end{align*}
and
\begin{align*}
\text{error}
&\le \frac{1}{2}\left[6\De x^2 +2\times 3 |\De x\,\De y| +2\De y^2\right]\\
&\le 3(0.002)^2 + 3(0.002)(0.003) +(0.003)^2\\
&= 0.000039
\end{align*}
By way of comparison, the exact error is 0.0000389, to seven decimal places.
Example 2.6.16.
In this example, we find the quadratic approximation of \(f(x,y)=\sqrt{1+4x^2+y^2}\) at \((x_0,y_0)=(1,2)\) and use it to compute approximately \(f(1.1\,,\,2.05)\text{.}\) We know that we will need all partial derivatives up to order 2, so we first compute them and evaluate them at \((x_0,y_0)=(1,2)\text{.}\)
\begin{align*}
f(x,y)&=\sqrt{1+4x^2+y^2} &
f(x_0,y_0)&=3\\
f_x(x,y)&=\frac{4x}{\sqrt{1+4x^2+y^2}} &
f_x(x_0,y_0)&=\frac{4}{3}\\
f_y(x,y)&=\frac{y}{\sqrt{1+4x^2+y^2}} &
f_y(x_0,y_0)&=\frac{2}{3}\\
f_{xx}(x,y)&=\frac{4}{\sqrt{1+4x^2+y^2}} -\frac{16x^2}{[1+4x^2+y^2]^{3/2}} &
f_{xx}(x_0,y_0)&=\frac{4}{3} -\frac{16}{27}\\
&&&=\frac{20}{27}\\
f_{xy}(x,y)&=-\frac{4xy}{[1+4x^2+y^2]^{3/2}} &
f_{xy}(x_0,y_0)&= -\frac{8}{27}\\
f_{yy}(x,y)&=\frac{1}{\sqrt{1+4x^2+y^2}} -\frac{y^2}{[1+4x^2+y^2]^{3/2}} &
f_{yy}(x_0,y_0)&=\frac{1}{3} -\frac{4}{27}\\
&&&=\frac{5}{27}
\end{align*}
We now just substitute them into 2.6.12 to get that the quadratic approximation to \(f\) about \((x_0,y_0)\) is
\begin{align*}
&f\big(x_0+\De x\,,\,y_0+\De y\big)\\
&\hskip0.5in \approx f(x_0, y_0)
+f_x(x_0, y_0)\De x+f_y(x_0, y_0)\De y \cr &\hskip1.0in
+\frac{1}{2}\bigg[f_{xx}(x_0, y_0)\De x^2
+2f_{xy}(x_0, y_0)\De x\De y
+f_{yy}(x_0, y_0)\De y^2\bigg] \cr
&\hskip0.5in= 3+\frac{4}{3} \De x+\frac{2}{3}\De y
+\frac{10}{27}\De x^2 -\frac{8}{27}\De x\De y+\frac{5}{54}\De y^2
\end{align*}
In particular, with \(\De x=0.1\) and \(\De y=0.05\text{,}\)
\begin{align*}
f(1.1\,,\,2.05)&\approx 3 \!+\!\frac{4}{3} (0.1)\!+\!\frac{2}{3}(0.05)
\!+\!\frac{10}{27}(0.01) \!-\!\frac{8}{27}(0.005)\!+\!\frac{5}{54}(0.0025)\\
&=3.1691
\end{align*}
The actual value, to four decimal places, is \(3.1690\text{.}\) The percentage error is about 0.004\%.
Example 2.6.17.
In this example, we find the quadratic approximation of \(f(x,y)=e^{2x}\sin(3y)\) about \((x_0,y_0)=(0,0)\) in two different ways.
The first way uses the canned formula 2.6.12. We compute all partial derivatives up to order 2 at \((x_0,y_0)\text{.}\)
\begin{align*}
f(x,y)&= e^{2x}\sin(3y) &
f(x_0,y_0)&=0\\
f_x(x,y)&= 2e^{2x}\sin(3y) &
f_x(x_0,y_0)&=0\\
f_y(x,y)&= 3e^{2x}\cos(3y) &
f_y(x_0,y_0)&=3\\
f_{xx}(x,y)&=4e^{2x}\sin(3y) &
f_{xx}(x_0,y_0)&=0\\
f_{xy}(x,y)&=6e^{2x}\cos(3y) &
f_{xy}(x_0,y_0)&= 6\\
f_{yy}(x,y)&=-9e^{2x}\sin(3y) &\qquad
f_{yy}(x_0,y_0)&=0
\end{align*}
So the quadratic approximation to \(f\) about \((0,0)\) is
\begin{align*}
f\big(x\,,\,y\big)
& \approx f(x, y)
+f_x(x, y) x+f_y(0, 0)y\\
&\hskip1in+\frac{1}{2}\bigg[f_{xx}(0, 0)x^2
+2f_{xy}(0, 0) x y
+f_{yy}(0, 0) y^2\bigg]\\
&=3y+6xy
\end{align*}
That’s pretty simple — just compute a bunch of partial derivatives and substitute into the formula 2.6.12.
But there is also a sneakier, and often computationally more efficient, method to get the same result. It exploits the single variable Taylor expansions
\begin{align*}
e^{x}&=1+x+\frac{1}{2!}x^2+\cdots\\
\sin y &=y - \frac{1}{3!}y^3+\cdots
\end{align*}
Replacing \(x\) by \(2x\) in the first and \(y\) by \(3y\) in the second and multiplying the two together, keeping track only of terms of degree at most two, gives
\begin{align*}
f(x,y)&= e^{2x}\sin(3y)\cr
&= \Big[1+(2x)+\frac{1}{2!}(2x)^2+\cdots\Big]
\Big[(3y) - \frac{1}{3!}(3y)^3+\cdots\Big]\\
&= \Big[1+2x+2x^2+\cdots\Big]
\Big[3y - \frac{9}{2}y^3+\cdots\Big]\\
&= 3y+6xy+6x^2y+\cdots
- \frac{9}{2}y^3- 9xy^3- 9x^2y^3+\cdots\\
&=3y + 6xy + \cdots
\end{align*}
just as in the first computation.
Subsection 2.6.2 Optional — Taylor Polynomials
We have just found linear and quadratic approximations to the function \(f(x,y)\text{,}\) for \((x,y)\) near the point \((x_0,y_0)\text{.}\) In CLP-1, we found not only linear and quadratic approximations, but in fact a whole hierarchy of approximations. For each integer \(n\ge 0\text{,}\) the \(n^\mathrm{th}\) order Taylor polynomial for \(f(x)\) about \(x=a\) was defined, in Definition 3.4.11 of the CLP-1 text, to be
\begin{gather*}
\sum_{k=0}^n \frac{1}{k!} f^{(k)}(a) \cdot (x-a)^k
\end{gather*}
We’ll now define, and find, the Taylor polynomial of order \(n\) for the function \(f(x,y)\) about \((x,y)=(x_0,y_0)\text{.}\) It is going to be a polynomial of degree at most \(n\) in \(\De x\) and \(\De y\text{.}\) The most general such polynomial is
\begin{equation*}
T_n(\De x,\De y)
=\sum_{\Atop{\ell,m\ge 0}{\ell+m\le n}} a_{\ell,m}\ (\De x)^\ell (\De y)^m
\end{equation*}
with all of the coefficients \(a_{\ell,m}\) being constants. The specific coefficients for the Taylor polynomial are determined by the requirement that all partial derivatives of \(T_n(\De x,\De y)\) at \(\De x=\De y=0\) are the same as the corresponding partial derivatives of \(f\big(x_0 + \De x\,,\,y_0+\De y\big)\) at \(\De x=\De y=0\text{.}\)
By way of preparation for our computation of the derivatives of \(T_n(\De x,\De y)\text{,}\) consider
\begin{align*}
\diff{}{t}t^4&=4t^3 &
\difftwo{}{t}t^4&=(4)(3)t^2 &
\frac{\dee{}^3}{\dee{t}^3}t^4&=(4)(3)(2)t\\
\frac{\dee{}^4}{\dee{t}^4}t^4&=(4)(3)(2)(1)=4! &
\frac{\dee{}^5}{\dee{t}^5}t^4&=0 &
\frac{\dee{}^6}{\dee{t}^6}t^4&=0
\end{align*}
and
\begin{align*}
\left.\diff{}{t}t^4\right|_{t=0}&=0 &
\left.\difftwo{}{t}t^4\right|_{t=0}&=0 &
\left.\frac{\dee{}^3}{\dee{t}^3}t^4\right|_{t=0}&=0\\
\left.\frac{\dee{}^4}{\dee{t}^4}t^4\right|_{t=0}&=4! &
\left.\frac{\dee{}^5}{\dee{t}^5}t^4\right|_{t=0}&=0 &
\left.\frac{\dee{}^6}{\dee{t}^6}t^4\right|_{t=0}&=0
\end{align*}
More generally, for any natural numbers \(p\text{,}\) \(m\text{,}\)
\begin{equation*}
\frac{\dee{}^p}{\dee{t}^p} t^m =
\begin{cases}
m(m-1)\cdots(m-p+1) t^{m-p} &\text {if } p\le m \\
0 &\text{if }p \gt m
\end{cases}
\end{equation*}
so that
\begin{equation*}
\left.\frac{\dee{}^p}{\dee{t}^p} t^m\right|_{t=0} =
\begin{cases}
m! &\text {if } p = m \\
0 &\text{if } p\ne m
\end{cases}
\end{equation*}
Consequently
\begin{equation*}
\left.\frac{\partial^p}{\partial (\De x)^p}\frac{\partial^q}{\partial (\De y)^q}
(\De x)^\ell (\De y)^m\right|_{\De x=\De y=0}
=\begin{cases}
\ell!\,m! &\text {if $p = \ell$ and $q=m$} \\
0 &\text{if $p\ne\ell$ or $q\ne m$}
\end{cases}
\end{equation*}
and
\begin{align*}
\frac{\partial^{p+q}\ \ T_n}{\partial (\De x)^p\, \partial (\De y)^q}(0,0)
&=\sum_{\Atop{\ell,m\ge 0}{\ell+m\le n}} a_{\ell,m}\
\left.\frac{\partial^p}{\partial (\De x)^p}\frac{\partial^q}{\partial (\De y)^q}
(\De x)^\ell (\De y)^m\right|_{\De x=\De y=0}\\
&=\begin{cases}
p!\,q!\,a_{p,q} &\text {if } p+q\le n \\
0 &\text{if } p+q \gt n
\end{cases}
\end{align*}
Our requirement that the derivatives of \(f\) and \(T_n\) match is the requirement that, for all \(p+q\le n\text{,}\)
\begin{align*}
\frac{\partial^{p+q}\ \ T_n}{\partial (\De x)^p\, \partial (\De y)^q}(0,0)
&=\frac{\partial^{p+q}\ }{\partial (\De x)^p\, \partial (\De y)^q}
f\big(x_0 + \De x\,,\,y_0+\De y\big)\Big|_{\De x=\De y=0}\\
&=\frac{\partial^{p+q}\ f}{\partial x^p\, \partial y^q}(x_0,y_0)
\end{align*}
This requirement gives
\begin{equation*}
p!\,q!\,a_{p,q} = \frac{\partial^{p+q}\ f}{\partial x^p\, \partial y^q}(x_0,y_0)
\end{equation*}
So the Taylor polynomial of order \(n\) for the function \(f(x,y)\) about \((x,y)=(x_0,y_0)\) is the right hand side of
Equation 2.6.18.
\begin{align*}
f\big(x_0 + \De x\,,\,y_0+\De y\big)
&\approx
\sum_{\Atop{\ell,m\ge 0}{\ell+m\le n}} \frac{1}{\ell!\ m!}\
\frac{\partial^{\ell+m}\ f}{\partial x^\ell\,\partial y^m}(x_0,y_0)\
(\De x)^\ell (\De y)^m
\end{align*}
This is for functions, \(f(x,y)\text{,}\) of two variables. There are natural extensions of this for functions of any (finite) number of variables. For example, the Taylor polynomial of order \(n\) for a function, \(f(x,y,z)\text{,}\) of three variables is the right hand side of
\begin{align*}
&f\big(x_0 + \De x\,,\,y_0+\De y\,,\,z_0+\De z\big)\\
&\hskip1in\approx
\sum_{\Atop{k,\ell,m\ge 0}{k+\ell+m\le n}} \frac{1}{k!\ \ell!\ m!}\
\frac{\partial^{k+\ell+m}\ f}
{\partial x^k\,\partial y^\ell\,\partial z^m}(x_0,y_0,z_0)\
(\De x)^k (\De y)^\ell (\De z)^m
\end{align*}
Exercises 2.6.3 Exercises
Exercise Group.
Exercises — Stage 1
1.
Let \(x_0\) and \(y_0\) be constants and let \(m\) and \(n\) be integers. If \(m \lt 0\) assume that \(x_0\ne 0\text{,}\) and if \(n \lt 0\) assume that \(y_0\ne 0\text{.}\) Define \(P(x,y) = x^m y^n\text{.}\)
- Find the linear approximation to \(P(x_0+\De x,y_0+\De y)\text{.}\)
- Denote by\begin{align*} P_\% &= 100\left|\frac{P(x_0+\De x,y_0+\De y)-P(x_0,y_0)}{P(x_0,y_0)}\right|\\ x_\% &= 100\left|\frac{\De x}{x_0}\right|\\ y_\% &= 100\left|\frac{\De y}{y_0}\right| \end{align*}the percentage errors in \(P\text{,}\) \(x\) and \(y\) respectively. Use the linear approximation to find an (approximate) upper bound on \(P_\%\) in terms of \(m\text{,}\) \(n\text{,}\) \(x_\%\) and \(y_\%\text{.}\)
2.
Consider the following work.
We compute, approximately, the \(y\)-coordinate of the point whose polar coordinates are \(r=0.9\) and \(\theta=2^\circ\text{.}\) In general, the \(y\)-coordinate of the point whose polar coordinates are \(r\) and \(\theta\) is \(Y(r,\theta) = r\sin\theta\text{.}\) The partial derivatives
\begin{equation*}
Y_r(r,\theta)=\sin\theta\qquad Y_\theta(r,\theta) = r\cos\theta
\end{equation*}
So the linear approximation to \(Y(r_0+\De r,\theta_0+\De\theta)\) with \(r_0=1\) and \(\theta_0=0\) is
\begin{align*}
Y(1+\De r,0+\De\theta)
&\approx Y(1,0) + Y_r(1,0)\,\De r + Y_\theta(1,0)\,\De\theta\\
&= 0\ +\ (0)\,\De r\ +\ (1)\De\theta
\end{align*}
Applying this with \(\De r=-0.1\) and \(\De\theta=2\) gives the (approximate) \(y\)-coordinate
\begin{gather*}
Y(0.9,2) = Y(1-0.1\,,\, 0+2)\approx 0\ +\ (0)\,(-0.1)\ +\ (1)(2)
=2
\end{gather*}
This conclusion is ridiculous. We’re saying that the \(y\)-coordinate is more than twice the distance from the point to the origin. What was the mistake?
Exercise Group.
Exercises — Stage 2
3.
Find an approximate value for \(f(x,y)=\sin(\pi xy+\ln y)\) at \((0.01,1.05)\) without using a calculator or computer.
4. (✳).
Let \(\displaystyle f(x,y)=\frac{x^2y}{x^4+2y^2}\text{.}\) Find an approximate value for \(f (-0.9\,,\, 1.1)\) without using a calculator or computer.
5.
Four numbers, each at least zero and each at most 50, are rounded to the first decimal place and then multiplied together. Estimate the maximum possible error in the computed product.
6. (✳).
One side of a right triangle is measured to be \(3\) with a maximum possible error of \(\pm 0.1\text{,}\) and the other side is measured to be \(4\) with a maximum possible error of \(\pm 0.2\text{.}\) Use the linear approximation to estimate the maximum possible error in calculating the length of the hypotenuse of the right triangle.
7. (✳).
If two resistors of resistance \(R_1\) and \(R_2\) are wired in parallel, then the resulting resistance R satisfies the equation \(\frac{1}{R} =\frac{1}{R_1}+\frac{1}{R_2}\text{.}\) Use the linear approximation to estimate the change in \(R\) if \(R_1\) decreases from \(2\) to \(1.9\) ohms and \(R_2\) increases from 8 to 8.1 ohms.
8.
The total resistance \(R\) of three resistors, \(R_1\text{,}\) \(R_2\text{,}\) \(R_3\text{,}\) connected in parallel is determined by
\begin{equation*}
\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}
\end{equation*}
If the resistances, measured in Ohms, are \(R_1=25\Om\text{,}\) \(R_2=40\Om\) and \(R_3=50\Om\text{,}\) with a possible error of 0.5\% in each case, estimate the maximum error in the calculated value of \(R\text{.}\)
9.
The specific gravity \(S\) of an object is given by \(\ S=\frac{A}{A-W}\ \) where \(A\) is the weight of the object in air and \(W\) is the weight of the object in water. If \(\ A=20\pm .01\ \) and \(\ W=12\pm.02\ \) find the approximate percentage error in calculating \(S\) from the given measurements.
10. (✳).
The pressure in a solid is given by
\begin{equation*}
P(s,r) = sr(4s^2 - r^2 - 2)
\end{equation*}
where \(s\) is the specific heat and \(r\) is the density. We expect to measure \((s,r)\) to be approximately \((2,2)\) and would like to have the most accurate value for \(P\text{.}\) There are two different ways to measure \(s\) and \(r\text{.}\) Method \(1\) has an error in \(s\) of \(\pm 0.01\) and an error in \(r\) of \(\pm 0.1\text{,}\) while method 2 has an error of \(\pm 0.02\) for both \(s\) and \(r\text{.}\)
Should we use method 1 or method 2? Explain your reasoning carefully.
11.
A rectangular beam that is supported at its two ends and is subjected to a uniform load sags by an amount
\begin{equation*}
S=C\frac{p\ell^4}{w h^3}
\end{equation*}
where \(p={\rm load}\text{,}\) \(\ell={\rm length}\text{,}\) \(h={\rm height}\text{,}\) \(w={\rm width}\) and \(C\) is a constant. Suppose \(p\approx 100\text{,}\) \(\ell\approx 4\text{,}\) \(w\approx .1\) and \(h\approx.2\text{.}\) Will the sag of the beam be more sensitive to changes in the height of the beam or to changes in the width of the beam.
12. (✳).
Let \(z=f(x,y)=\frac{2y}{x^2+y^2}\text{.}\) Find an approximate value for \(f(-0.8,2.1)\text{.}\)
13. (✳).
Suppose that a function \(z = f (x, y)\) is implicitly defined by an equation:
\begin{equation*}
xyz + x + y^2 + z^3 = 0
\end{equation*}
- Find \(\pdiff{z}{x}\text{.}\)
- If \(f(-1, 1) \lt 0\text{,}\) find the linear approximation of the function \(z = f (x, y)\) at \((-1, 1)\text{.}\)
- If \(f(-1, 1) \lt 0\text{,}\) use the linear approximation in (b) to approximate \(f(-1.02, 0.97)\text{.}\)
14. (✳).
Let \(z = f(x,y)\) be given implicitly by
\begin{equation*}
e^z + yz = x + y.
\end{equation*}
- Find the differential \(\dee{z}\text{.}\)
- Use linear approximation at the point \((1,0)\) to approximate \(f(0.99,0.01)\text{.}\)
15. (✳).
Two sides and the enclosed angle of a triangle are measured to be \(3\pm.1\)m, \(4\pm.1\)m and \(90\pm 1^\circ\) respectively. The length of the third side is then computed using the cosine law \(C^2=A^2+B^2-2AB\cos\theta\text{.}\) What is the approximate maximum error in the computed value of \(C\text{?}\)
16. (✳).
Use differentials to find a reasonable approximation to the value of \(f(x,y)=xy\sqrt{x^2+y^2}\) at \(x=3.02\text{,}\) \(y=3.96\text{.}\) Note that \(3.02\approx 3\) and \(3.96\approx 4\text{.}\)
17. (✳).
Use differentials to estimate the volume of metal in a closed metal can with diameter 8cm and height 12cm if the metal is 0.04cm thick.
18. (✳).
Let \(z\) be a function of \(x\text{,}\) \(y\) such that
\begin{equation*}
z^3 - z + 2xy - y^2 = 0,\qquad z(2, 4) = 1.
\end{equation*}
- Find the linear approximation to \(z\) at the point \((2, 4)\text{.}\)
- Use your answer in (a) to estimate the value of \(z\) at \((2.02, 3.96)\text{.}\)
Exercise Group.
Exercises — Stage 3
19. (✳).
Consider the surface given by:
\begin{equation*}
z^3 - xyz^2 - 4x = 0.
\end{equation*}
- Find expressions for \(\pdiff{z}{x}\text{,}\) \(\pdiff{z}{y}\) as functions of \(x\text{,}\) \(y\text{,}\) \(z\text{.}\)
- Evaluate \(\pdiff{z}{x}\text{,}\) \(\pdiff{z}{y}\) at \((1, 1, 2)\text{.}\)
- Measurements are made with errors, so that \(x = 1 \pm 0.03\) and \(y = 1 \pm 0.02\text{.}\) Find the corresponding maximum error in measuring \(z\text{.}\)
- A particle moves over the surface along the path whose projection in the \(xy\)--plane is given in terms of the angle \(\theta\) as\begin{equation*} x(\theta) = 1 + \cos\theta,\ y(\theta) = \sin\theta \end{equation*}from the point \(A : x = 2,\ y = 0\) to the point \(B : x = 1,\ y = 1\text{.}\) Find \(\diff{z}{\theta}\) at points \(A\) and \(B\text{.}\)
20. (✳).
Consider the function \(f\) that maps each point \((x, y)\) in \(\bbbr^2\) to \(ye^{-x}\text{.}\)
- Suppose that \(x = 1\) and \(y = e\text{,}\) but errors of size \(0.1\) are made in measuring each of \(x\) and \(y\text{.}\) Estimate the maximum error that this could cause in \(f(x,y)\text{.}\)
- The graph of the function \(f\) sits in \(\bbbr^3\) , and the point \((1, e, 1)\) lies on that graph. Find a nonzero vector that is perpendicular to that graph at that point.
21. (✳).
A surface is defined implicitly by \(z^4 - xy^2 z^2 + y = 0\text{.}\)
- Compute \(\pdiff{z}{x}\text{,}\) \(\pdiff{z}{y}\) in terms of \(x\text{,}\) \(y\text{,}\) \(z\text{.}\)
- Evaluate \(\pdiff{z}{x}\) and \(\pdiff{z}{y}\) at \((x, y, z) = (2, -1/2, 1)\text{.}\)
- If \(x\) decreases from \(2\) to \(1.94\text{,}\) and \(y\) increases from \(-0.5\) to \(-0.4\text{,}\) find the approximate change in \(z\) from \(1\text{.}\)
- Find the equation of the tangent plane to the surface at the point \((2, -1/2, 1)\text{.}\)
22. (✳).
A surface \(z = f (x, y)\) has derivatives \(\pdiff{f}{x}=3\) and \(\pdiff{f}{y}=-2\) at \((x, y, z) = (1, 3, 1)\text{.}\)
- If \(x\) increases from \(1\) to \(1.2\text{,}\) and \(y\) decreases from \(3\) to \(2.6\text{,}\) find the change in \(z\) using a linear approximation.
- Find the equation of the tangent plane to the surface at the point \((1, 3, 1)\text{.}\)
23. (✳).
According to van der Waal’s equation, a gas satisfies the equation
\begin{equation*}
(pV^2 + 16)(V - 1) = T V^2 ,
\end{equation*}
where \(p\text{,}\) \(V\) and \(T\) denote pressure, volume and temperature respectively. Suppose the gas is now at pressure \(1\text{,}\) volume \(2\) and temperature \(5\text{.}\) Find the approximate change in its volume if \(p\) is increased by \(0.2\) and \(T\) is increased by \(0.3\text{.}\)
24. (✳).
Consider the function \(f(x, y) = e^{-x^2 +4y^2}\text{.}\)
- Find the equation of the tangent plane to the graph \(z = f (x,y)\) at the point where \((x, y) = (2, 1)\text{.}\)
- Find the tangent plane approximation to the value of \(f(1.99, 1.01)\) using the tangent plane from part (a).
25. (✳).
Let \(z = f (x, y) = \ln(4x^2 + y^2)\text{.}\)
- Use a linear approximation of the function \(z = f (x, y)\) at \((0, 1)\) to estimate \(f(0.1, 1.2)\text{.}\)
- Find a point \(P (a, b, c)\) on the graph of \(z = f(x, y)\) such that the tangent plane to the graph of \(z = f (x, y)\) at the point \(P\) is parallel to the plane \(2x + 2y - z = 3\text{.}\)
26. (✳).
- Find the equation of the tangent plane to the surface \(x^2 z^3 + y \sin(\pi x) = -y^2\) at the point \(P = (1,1,-1)\text{.}\)
- Let \(z\) be defined implicitly by \(x^2 z^3 + y \sin(\pi x) = -y^2\text{.}\) Find \(\pdiff{z}{x}\) at the point \(P = (1,1,-1)\text{.}\)
- Let \(z\) be the same implicit function as in part (ii), defined by the equation \(x^2 z^3 + y \sin(\pi x) = -y^2\text{.}\) Let \(x = 0.97\text{,}\) and \(y = 1\text{.}\) Find the approximate value of \(z\text{.}\)
27. (✳).
The surface \(x^4+y^4+z^4+xyz=17\) passes through \((0,1,2)\text{,}\) and near this point the surface determines \(x\) as a function, \(x=F(y,z)\text{,}\) of \(y\) and \(z\text{.}\)
- Find \(F_y\) and \(F_z\) at \((x,y,z)=(0,1,2)\text{.}\)
- Use the tangent plane approximation (also known as linear, first order or differential approximation) to find the approximate value of \(x\) (near \(0\)) such that \((x,1.01, 1.98)\) lies on the surface.
