Linear Approximation & Error

$(a, b) = (2, 9)$: $9$ is a perfect square.

$f(2, 9) = 2 \cdot 3 = 6$.

$f_x = \sqrt{y} \Rightarrow f_x(2,9) = 3$. $\ f_y = \dfrac{x}{2\sqrt{y}} \Rightarrow f_y(2,9) = \dfrac{1}{3}$.

$L(x,y) = 6 + 3(x - 2) + \dfrac{1}{3}(y - 9)$.

$L(2.03, 8.97) = 6 + 3(0.03) + \dfrac{1}{3}(-0.03) = 6 + 0.09 - 0.01 = 6.08$.

$(a, b) = (1, 2)$, where the exponent is $0$.

$f(1, 2) = e^{0} = 1$.

$f_x = 2e^{2x - y} \Rightarrow 2$. $\ f_y = -e^{2x - y} \Rightarrow -1$.

$L(x,y) = 1 + 2(x - 1) - (y - 2)$.

$L(0.98, 2.03) = 1 + 2(-0.02) - 0.03 = 0.93$.

$(a, b, c) = (3, 4, 12)$: $9 + 16 + 144 = 169$ is a perfect square.

$f(3, 4, 12) = 13$.

$f_x = \dfrac{x}{\sqrt{x^2 + y^2 + z^2}}$, and likewise for $y$ and $z$: $\ \dfrac{3}{13},\ \dfrac{4}{13},\ \dfrac{12}{13}$.

$L = 13 + \dfrac{3}{13}(x - 3) + \dfrac{4}{13}(y - 4) + \dfrac{12}{13}(z - 12)$.

$L = 13 + \dfrac{3(-0.01) + 4(0.01) + 12(0.01)}{13} = 13 + \dfrac{0.13}{13} = 13.01$.

$(a, b) = (2, -1)$, given.

$f(2, -1) = 4$.

$f_x = 3$ and $f_y = -2$, given.

$L(x,y) = 4 + 3(x - 2) - 2(y + 1)$.

$L(2.1, -1.2) = 4 + 3(0.1) - 2(-0.2) = 4.7$.

$V = \tfrac{1}{3}\pi r^2h$, $\ \Delta r = 0.05$, $\ \Delta h = -0.1$.

$V_r = \tfrac{2}{3}\pi rh = 16\pi$, $\ V_h = \tfrac{1}{3}\pi r^2 = 3\pi$.

$\Delta V \approx 16\pi\,\Delta r + 3\pi\,\Delta h$.

$\Delta V \approx 16\pi(0.05) + 3\pi(-0.1) = 0.8\pi - 0.3\pi = 0.5\pi \approx 1.57\ \text{cm}^3$.

$z = x^2y^3$, $\ \Delta x = 0.05$, $\ \Delta y = -0.04$.

$z_x = 2xy^3 \Rightarrow 4$; $\ z_y = 3x^2y^2 \Rightarrow 12$.

$dz = 2xy^3\,dx + 3x^2y^2\,dy$; at $(2, 1)$ this is $dz = 4\,dx + 12\,dy$.

$\Delta z \approx 4(0.05) + 12(-0.04) = 0.2 - 0.48 = -0.28$.

$P = \dfrac{8T}{V}$, $\ \Delta T = 3$, $\ \Delta V = 1$.

$P_T = \dfrac{8}{V} = \dfrac{1}{5}$, $\ P_V = -\dfrac{8T}{V^2} = -\dfrac{2400}{1600} = -\dfrac{3}{2}$.

$\Delta P \approx \tfrac{1}{5}\,\Delta T - \tfrac{3}{2}\,\Delta V$.

$\Delta P \approx 0.6 - 1.5 = -0.9$.

$V = \pi r^2h$, $\ \Delta r = 0.1$, $\ \Delta h = 0.2$.

$V_r = 2\pi rh = 100\pi$, $\ V_h = \pi r^2 = 25\pi$.

$100\pi(0.1) + 25\pi(0.2) = 10\pi + 5\pi = 15\pi$.

Maximum error $\approx 15\pi \approx 47.1\ \text{cm}^3$.

$\rho = \dfrac{m}{V} = 2.5$, $\ \Delta m = 0.5$, $\ \Delta V = 0.4$.

$\rho_m = \dfrac{1}{V} = 0.05$, $\ \rho_V = -\dfrac{m}{V^2} = -0.125$.

$0.05(0.5) + 0.125(0.4) = 0.025 + 0.05 = 0.075$.

$100 \times \dfrac{0.075}{2.5} = 3\%$.

$A = \tfrac{1}{2}(10)(8)\sin 30^\circ = 20$; $\ \Delta a = \Delta b = 0.1$, $\ \Delta C = 1^\circ = \tfrac{\pi}{180}$.

$A_a = \tfrac{1}{2}b\sin C = 2$, $\ A_b = \tfrac{1}{2}a\sin C = 2.5$, $\ A_C = \tfrac{1}{2}ab\cos C = 20\sqrt{3}$.

$2(0.1) + 2.5(0.1) + 20\sqrt{3} \cdot \tfrac{\pi}{180} = 0.45 + \tfrac{\sqrt{3}\,\pi}{9}$.

Maximum error $\approx 0.45 + 0.60 = 1.05$.

$F = k\,\dfrac{r^4}{L}$, $\ \Delta r = \Delta L = 0.01$.

$F_r = \dfrac{4kr^3}{L} = \dfrac{k}{40}$, $\ F_L = -\dfrac{kr^4}{L^2} = -\dfrac{k}{6400}$.

Compare the terms: $\dfrac{k}{40}(0.01)$ against $\dfrac{k}{6400}(0.01)$, which is $160$ times smaller.

The error in $r$ matters far more.

$z_0 = 1$, given. Check: $1 + 3 - 4 = 0$.

In $x$: $3z^2 z_x + z + x\,z_x = 0$. $\$ In $y$: $3z^2 z_y + x\,z_y - 2y = 0$.

At $(3, 2, 1)$: $6z_x + 1 = 0 \Rightarrow z_x = -\tfrac{1}{6}$; $\ 6z_y - 4 = 0 \Rightarrow z_y = \tfrac{2}{3}$.

$L = 1 - \tfrac{1}{6}(x - 3) + \tfrac{2}{3}(y - 2)$, so $L(2.94, 2.03) = 1 + 0.01 + 0.02 = 1.03$.

$z^2 + z - 2 = 0 \Rightarrow (z + 2)(z - 1) = 0$. Since $z > 0$, $z_0 = 1$.

In $x$: $2z\,z_x + z + x\,z_x = 0$. $\$ In $y$: $2z\,z_y + x\,z_y = 1$.

At $(1, 2, 1)$: $3z_x + 1 = 0 \Rightarrow z_x = -\tfrac{1}{3}$; $\ 3z_y = 1 \Rightarrow z_y = \tfrac{1}{3}$.

$L = 1 - \tfrac{1}{3}(x - 1) + \tfrac{1}{3}(y - 2)$, so $L(1.06, 1.97) = 1 - 0.02 - 0.01 = 0.97$.

$x_0 = 2$, given. Check: $8 + 0 + 1 = 9$.

In $y$: $3x^2 x_y + x_y\,y + x = 0$. $\$ In $z$: $3x^2 x_z + x_z\,y + e^{z} = 0$.

At $(2, 0, 0)$: $12x_y + 2 = 0 \Rightarrow x_y = -\tfrac{1}{6}$; $\ 12x_z + 1 = 0 \Rightarrow x_z = -\tfrac{1}{12}$.

$L = 2 - \tfrac{1}{6}y - \tfrac{1}{12}z$, so $x \approx 2 - 0.02 + 0.01 = 1.99$.