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Subsection 3.4.8 Further Examples

In this subsection we give further examples of computation and use of Taylor approximations.

Estimate \(\tan 46^\circ\text{,}\) using the constant-, linear- and quadratic-approximations (equations 3.4.1, 3.4.3 and 3.4.6).

Solution Note that we need to be careful to translate angles measured in degrees to radians.

  • Set \(f(x)=\tan x\text{,}\) \(x=46\tfrac{\pi}{180}\) radians and \(a=45\tfrac{\pi}{180}=\tfrac{\pi}{4}\) radians. This is a good choice for \(a\) because

    • \(a=45^\circ\) is close to \(x=46^\circ\text{.}\) As noted above, it is generally the case that the closer \(x\) is to \(a\text{,}\) the better various approximations will be.
    • We know the values of all trig functions at \(45^\circ\text{.}\)
  • Now we need to compute \(f\) and its first two derivatives at \(x=a\text{.}\) It is a good time to recall the special \(1:1:\sqrt{2}\) triangle


    \begin{align*} f(x) &= \tan x & f(\pi/4) &= 1\\ f'(x) &= \sec^2 x = \frac{1}{\cos^2 x} & f'(\pi/4) &= \frac{1}{1/\sqrt{2}^2} = 2\\ f''(x) &= \frac{2\sin x}{\cos^3 x} & f''(\pi/4) &= \frac{2/\sqrt{2}}{1/\sqrt{2}^3} = 4 \end{align*}
  • As \(x-a=46\tfrac{\pi}{180}-45\tfrac{\pi}{180}=\tfrac{\pi}{180}\) radians, the three approximations are
    \begin{alignat*}{2} f(x)&\approx f(a) \\ \amp=1\\ f(x)&\approx f(a)+f'(a)(x-a) & &=1+2\tfrac{\pi}{180} \\ &=1.034907\\ f(x)&\approx f(a)+f'(a)(x\!-\!a)+\half f''(a)(x\!-\!a)^2& &=1+2\tfrac{\pi}{180}+\half 4\big(\tfrac{\pi}{180}\big)^2\\ & =1.035516 \end{alignat*}
    For comparison purposes, \(\tan 46^\circ\) really is \(1.035530\) to 6 decimal places.
Warning 3.4.23

All of our derivative formulae for trig functions were developed under the assumption that angles are measured in radians. Those derivatives appeared in the approximation formulae that we used in Example 3.4.22, so we were obliged to express \(x-a\) in radians.

Suppose that you are ten meters from a vertical pole. You were contracted to measure the height of the pole. You can't take it down or climb it. So you measure the angle subtended by the top of the pole. You measure \(\theta=30^\circ\text{,}\) which gives

\begin{gather*} h=10\tan 30^\circ=\tfrac{10}{\sqrt{3}}\approx 5.77\text{m}\qquad\qquad \end{gather*}

This is just standard trigonometry — if we know the angle exactly then we know the height exactly.

However, in the “real world” angles are hard to measure with such precision. If the contract requires you the measurement of the pole to be accurate within \(10\) cm, how accurate does your measurement of the angle \(\theta\) need to be?

Solution For simplicity  15 Mathematicians love assumptions that let us tame the real world. , we are going to assume that the pole is perfectly straight and perfectly vertical and that your distance from the pole was exactly 10 m.

  • Write \(\theta=\theta_0+\De\theta\) where \(\theta\) is the exact angle, \(\theta_0\) is the measured angle and \(\De \theta\) is the error.
  • Similarly write \(h=h_0+\De h\text{,}\) where \(h\) is the exact height and \(h_0=\tfrac{10}{\sqrt{3}}\) is the computed height. Their difference, \(\De h\text{,}\) is the error.
  • Then
    \begin{align*} h_0&=10\tan\theta_0 & h_0+\De h&=10\tan(\theta_0+\De\theta)\\ \De h &= 10\tan(\theta_0+\De\theta) - 10\tan\theta_0 \end{align*}
    We could attempt to solve this equation for \(\De\theta\) in terms of \(\De h\) — but it is far simpler to approximate \(\De h\) using the linear approximation in equation 3.4.20.
  • To use equation 3.4.20, replace \(y\) with \(h\text{,}\) \(x\) with \(\theta\) and \(a\) with \(\theta_0\text{.}\) Our function \(f(\theta) = 10 \tan\theta\) and \(\theta_0 = 30^\circ = \pi/6\) radians. Then
    \begin{align*} \De y &\approx f'(a) \De x & \text{ becomes }&& \De h &\approx f'(\theta_0) \De \theta \end{align*}
    Since \(f(\theta)=10 \tan \theta\text{,}\) \(f'(\theta) = 10\sec^2\theta\) and
    \begin{gather*} f'(\theta_0) = 10\sec^2(\pi/6) = 10 \cdot \left(\frac{2}{\sqrt{3}} \right)^2 = \frac{40}{3} \end{gather*}
  • Putting things together gives
    \begin{align*} \De h &\approx f'(\theta_0) \De \theta & \text{ becomes }&& \De h & \approx \frac{40}{3} \De \theta \end{align*}
    We can then solve this equation for \(\De\theta\) in terms of \(\De h\text{:}\)
    \begin{align*} \De \theta & \approx \frac{3}{40} \De h \end{align*}
  • We are told that we must have \(|\De h| \lt 0.1\text{,}\) so we must have
    \begin{align*} |\De \theta| &\leq \frac{3}{400} \end{align*}
    This is measured in radians, so converting back to degrees
    \begin{align*} \frac{3}{400} \cdot \frac{180}{\pi} &= 0.43^\circ \end{align*}
Definition 3.4.25

Suppose that you measure, approximately, some quantity. Suppose that the exact value of that quantity is \(Q_0\) and that your measurement yielded \(Q_0+\De Q\text{.}\) Then \(|\De Q|\) is called the absolute error of the measurement and \(100\frac{|\De Q|}{Q_0}\) is called the percentage error of the measurement. As an example, if the exact value is \(4\) and the measured value is \(5\text{,}\) then the absolute error is \(|5-4|=1\) and the percentage error is \(100\frac{|5-4|}{4}=25\text{.}\) That is, the error, \(1\text{,}\) was \(25\%\) of the exact value, \(4\text{.}\)

Suppose that the radius of a sphere has been measured with a percentage error of at most \(\varepsilon\)%. Find the corresponding approximate percentage errors in the surface area and volume of the sphere.

Solution We need to be careful in this problem to convert between absolute and percentage errors correctly.

  • Suppose that the exact radius is \(r_0\) and that the measured radius is \(r_0+\De r\text{.}\)
  • Then the absolute error in the measurement is \(|\De r|\) and, by definition, the percentage error is \(100\tfrac{|\De r|}{r_0}\text{.}\) We are told that \(100\tfrac{|\De r|}{r_0}\le\varepsilon\text{.}\)
  • The surface area  16 We do not expect you to remember the surface areas of solids for this course. of a sphere of radius \(r\) is \(A(r)=4\pi r^2\text{.}\) The error in the surface area computed with the measured radius is
    \begin{align*} \De A &=A(r_0+\De r)-A(r_0)\approx A'(r_0)\De r\\ &= 8\pi r_0 \Delta r \end{align*}
    where we have made use of the linear approximation, equation 3.4.20.
  • The corresponding percentage error is then
    \begin{gather*} 100\frac{|\De A|}{A(r_0)} \approx 100\frac{|A'(r_0)\De r|}{A(r_0)} = 100\frac{8\pi r_0|\De r|}{4\pi r_0^2} = 2\times 100\frac{|\De r|}{r_0} \le 2\varepsilon \end{gather*}
  • The volume of a sphere  17 We do expect you to remember the formula for the volume of a sphere. of radius \(r\) is \(V(r)=\frac{4}{3}\pi r^3\text{.}\) The error in the volume computed with the measured radius is
    \begin{align*} \De V &=V(r_0+\De r)-V(r_0)\approx V'(r_0)\De r\\ &= 4\pi r_0^2 \Delta r \end{align*}
    where we have again made use of the linear approximation, equation 3.4.20.
  • The corresponding percentage error is
    \begin{gather*} 100\frac{|\De V|}{V(r_0)} \approx 100\frac{|V'(r_0)\De r|}{V(r_0)} = 100\frac{4\pi r_0^2|\De r|}{4\pi r_0^3/3} = 3\times 100\frac{|\De r|}{r_0} \le 3\varepsilon \end{gather*}

We have just computed an approximation to \(\Delta V\text{.}\) This problem is actually sufficiently simple that we can compute \(\Delta V\) exactly:

\begin{align*} \Delta V &= V(r_0 + \Delta r) - V(r_0) = \tfrac{4}{3} \pi (r_0 + \Delta r)^3 - \tfrac{4}{3} \pi r_0^3 \end{align*}
  • Applying \((a+b)^3=a^3+3a^2b+3ab^2+b^3\) with \(a=r_0\) and \(b=\De r\text{,}\) gives
    \begin{align*} V(r_0+\De r)-V(r_0)&=\tfrac{4}{3}\pi \left[r_0^3+3r_0^2\De r+3r_0\,(\De r)^2+(\De r)^3\right] - \tfrac{4}{3}\pi r_0^3\\ &=\tfrac{4}{3}\pi[3r_0^2\De r+3r_0\,(\De r)^2+(\De r)^3] \end{align*}
  • Thus the difference between the exact error and the linear approximation to the error is obtained by retaining only the last two terms in the square brackets. This has magnitude
    \begin{gather*} \tfrac{4}{3}\pi\big|3r_0\,(\De r)^2+(\De r)^3\big| =\tfrac{4}{3}\pi\big|3r_0+\De r\big|(\De r)^2 \end{gather*}
    or in percentage terms
    \begin{align*} 100\cdot \dfrac{1}{\tfrac{4}{3}\pi r_0^3} \cdot \tfrac{4}{3}\pi \big|3r_0\,(\De r)^2+(\De r)^3\big| &=100\left|3\frac{\De r^2}{r_0^2}+\frac{\De r^3}{r_0^3}\right|\\ &=\left(100 \frac{3\De r}{r_0}\right) \cdot \left(\frac{\De r}{r_0}\right) \left|1 +\frac{\De r}{3r_0}\right|\\ & \le 3\varepsilon \left(\frac{\varepsilon}{100}\right)\cdot \left(1+\frac{\varepsilon}{300}\right) \end{align*}
    Since \(\varepsilon\) is small, we can assume that \(1 + \frac{\varepsilon}{300} \approx 1\text{.}\) Hence the difference between the exact error and the linear approximation of the error is roughly a factor of \(\tfrac{\varepsilon}{100}\) smaller than the linear approximation \(3\varepsilon\text{.}\)
  • As an aside, notice that if we argue that \(\De r\) is very small and so we can ignore terms involving \((\De r)^2\) and \((\De r)^3\) as being really really small, then we obtain
    \begin{align*} V(r_0+\De r)-V(r_0) &=\tfrac{4}{3}\pi[3r_0^2\De r \underbrace{+3r_0\,(\De r)^2+(\De r)^3}_\text{really really small}]\\ &\approx \tfrac{4}{3}\pi \cdot 3r_0^2\De r = 4 \pi r_0^2 \De r \end{align*}
    which is precisely the result of our linear approximation above.

To compute the height \(h\) of a lamp post, the length \(s\) of the shadow of a two meter pole is measured. The pole is 6 m from the lamp post. If the length of the shadow was measured to be 4 m, with an error of at most one cm, find the height of the lamp post and estimate the percentage error in the height.

Solution We should first draw a picture  18 We get to reuse that nice lamp post picture from Example 3.2.4.

  • By similar triangles we see that
    \begin{align*} \frac{2}{s} &= \frac{h}{6+s} \end{align*}
    from which we can isolate \(h\) as a function of \(s\text{:}\)
    \begin{align*} h &= \frac{2(6+s)}{s} = \frac{12}{s} + 2 \end{align*}
  • The length of the shadow was measured to be \(s_0=4\) m. The corresponding height of the lamp post is
    \begin{align*} h_0 &= \frac{12}{4} + 2 = 5m \end{align*}
  • If the error in the measurement of the length of the shadow was \(\De s\text{,}\) then the exact shadow length was \(s=s_0+\De s\) and the exact lamp post height is \(h=f(s_0+\De s)\text{,}\) where \(f(s)=\tfrac{12}{s}+2\text{.}\) The error in the computed lamp post height is
    \begin{gather*} \De h=h-h_0=f(s_0+\De s)-f(s_0) \end{gather*}
  • We can then make a linear approximation of this error using equation 3.4.20:
    \begin{align*} \De h &\approx f'(s_0)\De s =-\frac{12}{s_0^2}\De s =-\frac{12}{4^2}\De s \end{align*}
  • We are told that \(|\De s|\le\frac{1}{100}\) m. Consequently, approximately,
    \begin{gather*} |\De h|\le \frac{12}{4^2}\frac{1}{100}=\frac{3}{400} \end{gather*}
    The percentage error is then approximately
    \begin{align*} 100\frac{|\De h|}{h_0} & \le 100\frac{3}{400\times 5}=0.15\% \end{align*}