2.8: Linearization and Differentials

Equation of Tangent Line To calculate the slope of \(y = \sqrt x\) when \(x = 9,\) we differentiate \(\sqrt x = x^{1/2}\) using the Power Rule: \[\deriv{}{x} \par{x^{1/2}} = \tfrac{1}{2} x^{-1/2} = \frac{1}{2 \sqrt x} \pd\] Hence, the slope when \(x = 9\) is \[\frac{1}{2 \sqrt 9} = \frac{1}{6} \pd\] The point of tangency is \((9, 3),\) so an equation of the tangent line is \[y - 3 = \tfrac{1}{6} (x - 9) \or \boxed{y = 3 + \tfrac{1}{6} (x - 9)}\] (See Figure 2.)

Approximation The tangent line \(y = 3 + \tfrac{1}{6} (x - 9)\) is the best linear approximation to \(y = \sqrt x\) for \(x\) near \(9.\) We estimate \(\sqrt{10}\) as the output of this tangent line when \(x = 10.\) Substituting \(x = 10\) into the equation of the tangent line, we see \[\sqrt{10} \approx 3 + \tfrac{1}{6}(10 - 9) = \boxed{\tfrac{19}{6}} \approx 3.167 \pd\] The actual value of \(\sqrt{10}\) is approximately \(3.162.\) Our approximation of \(3.167\) is very accurate! Since \(x = 10\) is close to the center of the approximation, \(a = 9,\) the tangent line is an excellent estimate for the value of \(\sqrt x.\) (See Figure 3.) Using tangent lines allows you to estimate values of square roots in your head—a great trick.

Equation of Tangent Line The derivative of \(\sin x\) is \(\cos x,\) so the slope of the curve \(y = \sin x\) at \(x = 2 \pi/3\) is \[\cos \frac{2 \pi}{3} = -\frac{1}{2} \pd\] When \(x = 2 \pi/3,\) \[y = \sin \frac{2 \pi}{3} = \frac{\sqrt 3}{2} \pd\] Thus, an equation of the tangent line is \[y - \frac{\sqrt 3}{2} = -\frac{1}{2} \par{x - \frac{2 \pi}{3}} \or \boxed{y = -\frac{x}{2} + \frac{\pi}{3} + \frac{\sqrt 3}{2}} \]

Approximation We approximate \(\sin 3\) as the \(y\)-value the tangent line reaches when \(x = 3.\) Substituting \(x = 3\) into the equation of the tangent line shows that \[\sin 3 \approx -\frac{1}{2} (3) + \frac{\pi}{3} + \frac{\sqrt 3}{2} \approx \boxed{0.413}\] Unfortunately, the actual value of \(\sin 3\) turns out to be roughly \(0.141.\) Our estimate \(0.413\) is not very accurate because \(x = 3\) is relatively far away from the center of the linearization, \(x = 2 \pi/3.\) (See Figure 4.)

Accuracy to within \(0.1\) means that the difference between \(f\) and the linearization should be less than \(0.1 \col\) \[ \ba \abs{f(x) - \par{\tfrac{x}{5} + \tfrac{14}{5}}} &\lt 0.1 \nl \abs{\sqrt{2x + 3} - \par{\tfrac{x}{5} + \tfrac{14}{5}}} &\lt 0.1 \pd \ea \] Solving the inequality \[-0.1 \lt \sqrt{2x + 3} - \par{\tfrac{x}{5} + \tfrac{14}{5}} \lt 0.1\] gives the interval \(\boxed{6.5 \lt x \lt 16.5}.\)

1. The derivative of \(\sin x\) is \(\cos x,\) so we say \[\dd \par{\sin x} = \boxed{\cos x \di x}\]

---

2. By the Power Rule, the derivative of \(y^4\) is \(4y^3.\) So \[\dd \par{y^4} = \boxed{4y^3 \di y}\]

---

3. The Product Rule gives the derivative of \(te^t\) to be \(e^t + te^t.\) Thus, \[\dd \par{t e^t} = \boxed{\par{e^t + te^t} \di t}\]

From \(1\) to \(1.7\) We see \[\deriv{}{x} \par{\tfrac{1}{2} x^2 + 1} = x \pd\] Therefore, the differential \(\dd y\) is given by \begin{equation} \dd y = x \di x \pd \label{eq:dy-ex} \end{equation} Because \(x\) changes from \(1\) to \(1.7,\) we take \(\dd x = 0.7\) and so \(\eqrefer{eq:dy-ex}\) gives \[\dd y = (1)(0.7) = \boxed{0.7}\] Conversely, we see \[ \ba \Delta y &= \parbr{\tfrac{1}{2} (1.7)^2 + 1} - \parbr{\tfrac{1}{2} (1)^2 + 1} \nl &= \boxed{0.945} \ea \] (See Figure 6.)

From \(1\) to \(1.2\) Since \(x\) changes from \(1\) to \(1.2,\) we use \(\dd x = 0.2.\) By \(\eqref{eq:dy-ex},\) \[\dd y = (1)(0.2) = \boxed{0.2}\] In contrast, \[ \ba \Delta y &= \parbr{\tfrac{1}{2} (1.2)^2 + 1} - \parbr{\tfrac{1}{2} (1)^2 + 1} \nl &= \boxed{0.22} \ea \] Note that \(\dd y\) is much easier to compute than \(\Delta y.\) We see that \(\Delta y \approx \dd y\) increases in accuracy for small values of \(\dd x.\) For example, the difference between \(\dd y\) and \(\Delta y\) is even smaller if \(x\) changes from \(1\) to \(1.1.\) For small values of \(\dd x,\) the tangent line to \(y = \tfrac{1}{2} x^2 + 1\) at \(x = 1\) becomes nearly indistinguishable from the curve.

Maximum Error Let \(r\) be the sphere's radius and \(V\) be its volume. Then \(V = \tfrac{4}{3} \pi r^3.\) We take the maximum measurement error of \(r\) to be \(\Delta r = \dd r.\) Likewise, the error in \(V\) is \(\Delta V,\) as approximated by \[\dd V = 4 \pi r^2 \di r \pd\] For \(r = 14\) and \(\dd r = 0.07,\) we estimate the maximum error in the sphere's volume to be \[\dd V = 4 \pi(14)^2 (0.07) \approx \boxed{172.411 \un{cm}^3}\]

Percent Error The relative error is given by the error \(\Delta V\) divided by the measured volume \(V.\) The sphere's volume is measured to be \[V = \tfrac{4}{3} \pi r^3 = \tfrac{4}{3} \pi (14)^3 \approx 11494.040 \un{cm}^3 \pd\] Hence, the maximum relative error in the measurement is \[ \frac{\Delta V}{V} \approx \frac{\dd V}{V} = \frac{172.411}{11494.040} \approx 0.015 \pd \] This relative error is equivalent to a percent error of \(\boxed{1.500\%}.\)

Let \(V\) be the cube's volume and \(s\) be the length of one side. Then \(V = s^3.\) The length of one side changes by \(\dd s = -0.1,\) so we see \[\dd V = 3s^2 \di s = 3(2)^2(-0.1) = -1.2 \un{in}^3 \pd\] (This result is negative because the volume has decreased.) So the volume has shrunk by \(\boxed{1.2 \un{in}^3}.\) The actual amount by which the cube's volume changes is \[\Delta V = (1.9)^3 - (2)^3 \approx -1.141 \un{in}^3 \cma\] so our differential \(\dd V = -1.2\) is close. With calculators, using differentials seems pointless—but what if you needed to perform estimations in your head?