Linear Approximation and Differentials

$f(9)=3$ and $f'(x)=\frac{1}{2\sqrt{x}}$, so $f'(9)=\frac16$ and

$L(x)=3+\tfrac16(x-9),\qquad L(9.1)=3+\tfrac{0.1}{6}=\tfrac{181}{60}\approx3.016667.$

$f''(x)=-\frac14 x^{-3/2}<0$ on $(0,\infty)$, so $f$ is concave down and the tangent lies above the graph: the estimate is too large.

Check by squaring: $\left(3+\frac{1}{60}\right)^{2}=9.100278>9.1$. (True value $3.0166206$.)

(a) $f'(x)=3x^{2}-2$, so $dy=(3x^{2}-2)\,dx$.

(b) At $x=2$ with $dx=0.1$: $dy=(12-2)(0.1)=1$. Along the curve, $f(2.1)=9.261-4.2=5.061$ and $f(2)=4$, so $\Delta y=1.061$.

(c) For a general step $h$, $\Delta y=10h+6h^{2}+h^{3}$ while $dy=10h$, so

$\Delta y-dy=6h^{2}+h^{3}.$

At $h=0.1$ that is $0.061$; at $h=0.01$ it is $0.000601$. A tenth of the step gives about a hundredth of the gap — the error is quadratic.

Take $f(x)=\cos x$, $a=\frac{\pi}{6}$, $dx=\frac{29\pi}{180}-\frac{\pi}{6}=-\frac{\pi}{180}$. Then $f(a)=\frac{\sqrt3}{2}$ and $f'(a)=-\sin\frac{\pi}{6}=-\frac12$, so

$\cos 29^{\circ}\approx\frac{\sqrt3}{2}-\frac12\left(-\frac{\pi}{180}\right)=\frac{\sqrt3}{2}+\frac{\pi}{360}\approx0.874752.$

$f''(x)=-\cos x<0$ across $[29^{\circ},30^{\circ}]$, so $f$ is concave down and this over-estimates.

Since $\left|f''(t)\right|=\left|\cos t\right|\le1$, take $M=1$: $\left|\text{error}\right|\le\frac12\left(\frac{\pi}{180}\right)^{2}\approx1.5\times10^{-4}$. (True value $0.874620$.)

$V=\frac43\pi r^{3}$, so $dV=4\pi r^{2}\,dr$. With $r=21$ and $\left|dr\right|\le0.05$:

$\left|dV\right|\le4\pi(441)(0.05)=88.2\pi\approx277.1\ \text{cm}^{3}.$

For the relative error divide first: $\frac{dV}{V}=\frac{4\pi r^{2}\,dr}{\frac43\pi r^{3}}=3\,\frac{dr}{r}$, hence

$\frac{\left|dV\right|}{V}\le3\cdot\frac{0.05}{21}=\frac{1}{140}\approx0.00714,$

i.e. about $0.71\%$ — three times the $0.238\%$ error in $r$, as the exponent $3$ predicts.

$A=\pi r^{2}$ gives $dA=2\pi r\,dr$, so $\frac{dA}{A}=2\,\frac{dr}{r}$ — the power rule with $n=2$.

Impose the tolerance $\left|\frac{dA}{A}\right|\le0.005$:

$2\left|\frac{dr}{r}\right|\le0.005\ \Longrightarrow\ \left|\frac{dr}{r}\right|\le0.0025.$

So the radius needs $0.25\%$ accuracy — half the tolerance on the area, because the exponent is $2$. In centimetres, $\left|dr\right|\le0.0025(24)=0.06$ cm.

(a) $L(x)=5-3(x-2)$, so $L(1.9)=5+0.3=5.3$ and $L(2.2)=5-0.6=4.4$.

(b) $f''>0$ everywhere means $f$ is concave up, so its graph lies above every tangent line on both sides of $2$: $f(x)>L(x)$ for $x\ne2$. Both are under-estimates. That $f$ is decreasing at $2$ changes nothing.

(c) Taylor's inequality with $M=4$ gives

$\left|f(2.2)-4.4\right|\le\frac{4}{2}(0.2)^{2}=0.08,$

so $4.32\le f(2.2)\le4.48$. Combining with (b): $4.4<f(2.2)\le4.48$.