Linear Approximation and Differentials

Practice 1: $\mathrm{f}(\mathrm{x})=\mathrm{x}^{1 / 2}$ so $\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{2 \sqrt{\mathrm{x}}}$. At the point $(16,4)$ on the graph of $\mathrm{f}$, the slope of the tangent line is $f^{\prime}(16)=\frac{1}{2 \sqrt{16}}=\frac{1}{8}$. The equation of the tangent line is

$y-4=\frac{1}{8}(x-16)$ or $y=\frac{1}{8} x+2: L(x)=\frac{1}{8} x+2$. Then

$\sqrt{16.1} \approx \mathrm{L}(16.1)=\frac{1}{8}(16.1)+2=\mathbf{4.0125}$ and $\sqrt{15.92} \approx \mathrm{L}(15.92)=\frac{1}{8}(15.92)+2=\mathbf{3.99}$

Practice 2: $\quad \mathrm{f}(\mathrm{x})=\mathrm{x}^{3}$ so $\mathrm{f}^{\prime}(\mathrm{x})=3 \mathrm{x}^{2}$. At $(1,1)$, the slope of the tangent line is $\mathrm{f}^{\prime}(1)=3$. The equation of the tangent line is $y-1=3(x-1)$ or $y=3 x-2: L(x)=3 x-2$. Then $(1.02)^{3} \approx \mathrm{L}(1.02)=3(1.02)-2=\mathbf{1.06}$ and $(0.97)^{3} \approx \mathrm{L}(0.97)=3(0.97)-2=\mathbf{0.91}$.

Practice 3: $f(x)=x^{4}$ so $f^{\prime}(x)=4 x^{3}. a=1$ and $\Delta x=0.06$.

$(1.06)^{4}=\mathrm{f}(1.06) \approx \mathrm{L}(1.06)=\mathrm{f}(1)+\mathrm{f}^{\prime}(1) \cdot(0.06)=1^{4}+4(1)^{3} \cdot(0.06)=1.24$

Practice 4: $\begin{aligned} &\mathrm{f}(1.1) \approx \mathrm{f}(1)+\mathrm{f}^{\prime}(1) \cdot(0.1)=0.7854+(0.5) \cdot(0.1)=\mathbf{0.8354}. \\ &\mathrm{f}(1.23) \approx \mathrm{f}(1.2)+\mathrm{f}^{\prime}(1.2) \cdot(0.03)=0.8761+(0.4098) \cdot(0.03)=\mathbf{0.888394}. \\ &\mathrm{f}(1.38) \approx \mathrm{f}(1.4)+\mathrm{f}^{\prime}(1.4) \cdot(-0.02)=0.9505+(0.3378) \cdot(-0.02)=\mathbf{0.943744}. \end{aligned}$

Practice 5: $\quad x=4 \mathrm{~cm}$ and $\Delta x=0.1 \mathrm{~cm}. f(x)=x^{3}$ so $f^{\prime}(x)=3 x^{2}$ and $f(4)=4^{3}=64 \mathrm{~cm}^{3}$. Then "error" $\Delta \mathrm{f} \approx \mathrm{f}^{\prime}(\mathrm{x}) \Delta \mathrm{x}=3 \mathrm{x}^{2} \cdot \Delta \mathrm{x}$. When $\mathrm{x}=4$ and $\Delta \mathrm{x}=0.1$, "error" $\Delta \mathrm{f} \approx 3(4)^{2}(0.1)=\mathbf{4. 8} \mathrm{cm}^{3}$.

Practice 6: $43 \pm 1^{\circ}$ is $0.75049 \pm 0.01745$ radians. $f(x)=3000 \cdot \tan (x)$ so $\mathrm{f}(0.75049)=3000 \cdot \tan (0.75049) \approx \mathbf{2 7 9 7. 5} \mathrm{m}. \quad \mathrm{f}^{\prime}(\mathrm{x})=3000 \sec ^{2}(\mathrm{x}) \mathrm{so}$ 

$\Delta f(x) \approx f^{\prime}(x) \cdot \Delta x=3000 \cdot \sec ^{2}(x) \cdot \Delta x=3000 \cdot \sec ^{2}(0.75049) \cdot(0.01745)=97.9 \mathrm{~m}$

The height of the rocket is $\mathbf{2797.5} \pm \mathbf{97.9} \mathbf{~ m}$.

Practice 7: $\mathrm{f}(\theta)=2000 \cdot \tan (\theta)$ so $\mathrm{f}^{\prime}(\theta)=2000 \cdot \sec ^{2}(\theta).$ We know $4000=2000 \cdot \tan (\theta)$ so

$\tan (\theta)=2$ and $\theta \approx 1.10715$ (radians). $\mathrm{f}^{\prime}(\theta)=2000 \cdot \sec ^{2}(\theta)$ so $\mathrm{f}^{\prime}(1.10715)=2000 \cdot \sec ^{2}(1.10715) \approx 10,000$.

Finally, "error" $\Delta \mathrm{f} \approx \mathrm{f}^{\prime}(\theta) \cdot \Delta \theta$ so $10 \approx 10,000 \cdot \Delta \theta$ and $\Delta \theta \approx 10 / 10,000=\mathbf{0. 0 0 1}($ radians $) \approx 0.057^{\circ}$.

Practice 8: $\mathrm{A}(\mathrm{r})=6 \mathrm{r}^{2}$ so $\mathrm{A}^{\prime}(\mathrm{r})=12 \mathrm{r}$ and $\Delta \mathrm{A} \approx \mathrm{A}^{\prime}(\mathrm{r}) \cdot \Delta \mathrm{r}=12 \mathrm{r} \cdot \Delta \mathrm{r}.$ We are also told that $\Delta \mathrm{r} / \mathrm{r} < 0.03$ Percentage error is $\frac{\Delta \mathrm{A}}{\mathrm{A}} \cdot 100=\frac{12 \mathrm{r} \cdot \Delta \mathrm{r}}{6 \mathrm{r}^{2}} \cdot 100=\frac{2 \cdot \Delta \mathrm{r}}{\mathrm{r}} \cdot 100 < 200 \cdot(0.03)=\mathbf{6}$.

Practice 9: $\begin{aligned} &\mathrm{f}(\mathrm{x})=\ln (\mathrm{x}) \qquad \mathrm{df}=\mathrm{f}^{\prime}(\mathrm{x}) \cdot \mathrm{dx}=\frac{1}{\mathbf{x}} \mathrm{d} \mathrm{x} \\ &\mathrm{u}=\sqrt{1-3 \mathrm{x}} \qquad \mathrm{du}=\frac{\mathrm{du}}{\mathrm{dx}} \cdot \mathrm{dx}=\frac{-3}{2 \sqrt{1-3 \mathrm{x}}} \cdot \mathrm{dx} \\ &\mathrm{r}=3 \cos (\theta) \mathrm{dr}=\frac{\mathrm{dr}}{\mathrm{d} \theta} \cdot \mathrm{d} \theta=-\mathbf{3} \sin (\boldsymbol{\theta}) \cdot \mathrm{d} \theta \end{aligned}$

Practice 10: $f(x)=x^{3}, f^{\prime}(x)=3 x^{2}$, and $L(4+\Delta x)=f(4)+f^{\prime}(4) \Delta x=4^{3}+3(4)^{2} \Delta x=64+48 \cdot \Delta x$.

$\begin{array}{l|l|l|l} { }_­ \Delta \mathrm{x} & \mathrm{f}(4+\Delta \mathrm{x}) & \mathrm{L}(4+\Delta \mathrm{x}) & \mathrm{f}(4+\Delta \mathrm{x})-\mathrm{L}(4+\Delta \mathrm{x}) \\ \hline 0.1 & 68.921 & 68.8 & 0.121 \\ 0.05 & 66.430125 & 66.4 & 0.030125 \\ 0.01 & 64.481201 & 64.48 & 0.001201 \\ 0.001 & 64.048012 & 64.048 & 0.000012 \end{array}$

$\mathrm{f}(4+\Delta \mathrm{x})$ is the actual volume of the cube with side length $4+\Delta \mathrm{x}$.

$\mathrm{L}(4+\Delta \mathrm{x})$ is the volume of the cube with side length $4(\mathrm{v}=64)$ plus the volume of the 3 "slabs" $\left(\mathrm{v}=3 \cdot 4^{2} \cdot \Delta \mathrm{x}\right)$

$|f(4+\Delta x)-L(4+\Delta x)|$ is the volume of the "leftover" pieces from $L$ : the 3 "rods" $\left(v=3 \cdot 4 \cdot(\Delta x)^{2}\right)$ and the tiny cube $\left(\mathrm{v}=(\Delta \mathrm{x})^{3}\right)$.