Corrected exercises: linearization and differentials (MATH 203)

a) $f(x) = x^{1/4}$, so by the power rule $f'(x) = \frac{1}{4}x^{-3/4}$. At the centre, the algebra first: $16^{-3/4} = \frac{1}{16^{3/4}} = \frac{1}{\left(\sqrt[4]{16}\right)^3} = \frac{1}{2^3} = \frac{1}{8}$. Take the root FIRST, then the power, then the reciprocal: $16^{3}$ is $4096$, a number nobody wants to root. A negative exponent means a reciprocal, never a negative number: writing $16^{-3/4} = -8$ or $-12$ makes the slope negative although the figure shows a rising curve. Hence $f(16) = 2$, $f'(16) = \frac{1}{4} \cdot \frac{1}{8} = \frac{1}{32}$, and $L(x) = 2 + \frac{1}{32}(x - 16)$. Check: $L(16) = 2 = f(16)$.

b) $\sqrt[4]{17} \approx L(17) = 2 + \frac{1}{32} = 2.03125$. For $15.5$ the increment is NEGATIVE: $x - a = 15.5 - 16 = -0.5$, so $\sqrt[4]{15.5} \approx 2 - \frac{0.5}{32} = 2 - \frac{1}{64} = 1.984375$. Losing the sign gives $2.015625$, a value above $2$ for a number below $16$, which a rising function cannot produce.

c) $f''(x) = \frac{1}{4} \cdot \left(-\frac{3}{4}\right)x^{-7/4} = -\frac{3}{16}x^{-7/4} < 0$ for $x > 0$: the curve is concave down, it lies below each of its tangents (the figure shows it on both sides of $(16, 2)$), so BOTH estimates are too large. The calculator agrees: $\sqrt[4]{17} = 2.030543$ and $\sqrt[4]{15.5} = 1.984188$ to six decimals. Actual errors: $2.03125 - 2.030543 = 0.000707$ and $1.984375 - 1.984188 = 0.000187$, both positive as predicted. As percentages of the true values, about $0.035\%$ and $0.009\%$: the tangent is excellent this close to $16$.

d) $g(x) = x^{-3/2}$ gives $g'(x) = -\frac{3}{2}x^{-5/2}$. At $4$: $4^{-3/2} = \frac{1}{(\sqrt 4)^3} = \frac{1}{8}$ and $4^{-5/2} = \frac{1}{(\sqrt 4)^5} = \frac{1}{32}$, so $g(4) = \frac{1}{8}$ and $g'(4) = -\frac{3}{2} \cdot \frac{1}{32} = -\frac{3}{64}$. Then $L_g(x) = \frac{1}{8} - \frac{3}{64}(x - 4)$ and $4.2^{-3/2} \approx 0.125 - \frac{3}{64}(0.2) = 0.125 - 0.009375 = 0.115625$. Side: $g''(x) = \frac{15}{4}x^{-7/2} > 0$, concave up, the tangent lies BELOW the curve, so the estimate is too small. Calculator: $4.2^{-3/2} = 0.116179$, larger than $0.115625$ as predicted, error $0.000554$. Two exponents with a minus sign, two reciprocals: the algebra of exponents is where this question is won or lost.

e) $L(16.5) = 2 + \frac{0.5}{32} = 2.015625$ and the calculator gives $\sqrt[4]{16.5} = 2.015445$, an error of $0.000180$. At $17$ the error was $0.000707$: the ratio is $\frac{0.000707}{0.000180}$, about $3.9$, so about $4$. The distance to the centre was halved, from $1$ to $0.5$, and the error was divided by about $2^2 = 4$, not by $2$. The error of a linearization behaves like a constant times $(x - a)^2$: this is why a centre CLOSE to the target matters more than anything else.

a) $f(\ln 2) = e^{\ln 2} = 2$ and $f'(x) = e^x$, so $f'(\ln 2) = 2$ as well. Hence $L(x) = 2 + 2(x - \ln 2)$. The calculator gives the increment, $0.7 - \ln 2 = 0.006853$, and $e^{0.7} \approx 2 + 2(0.006853) = 2.013706$. Side: $(e^x)'' = e^x > 0$, the curve is concave up and lies above its tangent, so the estimate is too small. Calculator: $e^{0.7} = 2.013753$, larger indeed, an error of $0.000047$. Keep $\ln 2$ exact inside $L$; round only the final number.

b) $44^\circ = 45^\circ - 1^\circ = \frac{\pi}{4} - \frac{\pi}{180}$, so $h = -\frac{\pi}{180}$ rad. For $f(x) = \tan x$: $f\left(\frac{\pi}{4}\right) = 1$ and $f'(x) = \sec^2 x$, $\sec^2\frac{\pi}{4} = \frac{1}{\cos^2(\pi/4)} = \frac{1}{1/2} = 2$. So $L(x) = 1 + 2\left(x - \frac{\pi}{4}\right)$ and $\tan 44^\circ \approx 1 + 2\left(-\frac{\pi}{180}\right) = 1 - \frac{\pi}{90} = 0.965093$. With $h = -1$, the degree left unconverted, one gets $L = -1$, a negative tangent for an acute angle. Side: $f''(x) = 2\sec^2 x \tan x > 0$ on $\left(0, \frac{\pi}{2}\right)$, the curve bends up away from its tangent on both sides (figure): too small. Calculator in degree mode: $\tan 44^\circ = 0.965689$, error $0.000595$.

c) $f(x) = \arcsin x$, $f\left(\frac{1}{2}\right) = \frac{\pi}{6}$, and $f'(x) = \frac{1}{\sqrt{1 - x^2}}$. The algebra: $1 - \frac{1}{4} = \frac{3}{4}$, $\sqrt{\frac{3}{4}} = \frac{\sqrt 3}{2}$, and its reciprocal is $\frac{2}{\sqrt 3}$, not $\frac{\sqrt 3}{2}$: the fraction is flipped by the $\frac{1}{\ldots}$. So $L(x) = \frac{\pi}{6} + \frac{2}{\sqrt 3}\left(x - \frac{1}{2}\right)$ and $\arcsin 0.52 \approx \frac{\pi}{6} + \frac{2}{\sqrt 3}(0.02) = \frac{\pi}{6} + \frac{0.04}{\sqrt 3} = 0.523599 + 0.023094 = 0.546693$. Side: $f''(x) = x(1 - x^2)^{-3/2} > 0$ for $0 < x < 1$: too small. Calculator (radian mode): $\arcsin 0.52 = 0.546851$, error $0.000158$.

d) $\ln e = 1$ and $(\ln x)' = \frac{1}{x}$ gives $\frac{1}{e}$: $L(x) = 1 + \frac{1}{e}(x - e)$. At $2.7$: $1 + \frac{2.7 - e}{e} = 1 + \frac{2.7}{e} - \frac{e}{e} = \frac{2.7}{e}$. Distributing the $\frac{1}{e}$ over $(x - e)$ makes the $1$ disappear, a simplification worth doing BEFORE the calculator. Value: $\frac{2.7}{e} = 0.993274$. Side: $(\ln x)'' = -\frac{1}{x^2} < 0$, concave down, too large. Calculator: $\ln 2.7 = 0.993252$, error $0.000023$.

e) Centred at $0$: $e^{0.7} \approx 1 + 0.7 = 1.7$, and the percentage error is $\frac{2.013753 - 1.7}{2.013753} \times 100$, about $15.58\%$, against $0.002\%$ in a). The target $0.7$ is at distance $0.7$ from $0$ but only $0.0069$ from $\ln 2$. The centre is chosen for two properties together: $f(a)$ and $f'(a)$ exact (here both equal $2$), and $|x - a|$ small. Whether $a$ itself looks simple does not matter, since only $f(a)$ and $f'(a)$ enter the computation.

a) $f(x) = (1 + x)^k$ gives $f(0) = 1$ and, by the chain rule with inner function $1 + x$, $f'(x) = k(1 + x)^{k-1}$, so $f'(0) = k$. Hence $L(x) = 1 + kx$. For $\frac{1}{1 - x} = (1 + (-x))^{-1}$, replace $x$ by $-x$ and $k$ by $-1$: $1 + (-1)(-x) = 1 + x$. The two minus signs multiply to a plus, and dropping one of them gives $1 - x$, the classic slip. For $\sqrt{1 + x} = (1 + x)^{1/2}$: $1 + \frac{x}{2}$.

b) $(1.002)^{25} = (1 + 0.002)^{25} \approx 1 + 25(0.002) = 1.05$. Side: $f''(x) = k(k - 1)(1 + x)^{k-2} = 600(1 + x)^{23} > 0$, concave up, the tangent is below: too small. Calculator: $(1.002)^{25} = 1.051219$, and the percentage error is $\frac{1.051219 - 1.05}{1.051219} \times 100$, about $0.12\%$.

c) The rule needs $1 +$ something small, and $9.18$ is not of that form. Factor: $\sqrt{9.18} = \sqrt{9 \times 1.02} = \sqrt 9\,\sqrt{1.02} = 3(1 + 0.02)^{1/2} \approx 3\left(1 + \frac{0.02}{2}\right) = 3(1.01) = 3.03$. Applying the rule to $9.18 = 1 + 8.18$ instead gives $1 + \frac{8.18}{2} = 5.09$, a tangent at $1$ used eight units away. Side: $(1 + x)^{1/2}$ is concave down ($k(k - 1) = -\frac{1}{4} < 0$), so $3.03$ is too large. Calculator: $\sqrt{9.18} = 3.029851$; and $3.03^2 = 9.1809 > 9.18$ confirms it.

d) $(2 + x)^{-2} = \left(2\left(1 + \frac{x}{2}\right)\right)^{-2} = 2^{-2}\left(1 + \frac{x}{2}\right)^{-2} = \frac{1}{4}\left(1 + \frac{x}{2}\right)^{-2} \approx \frac{1}{4}\left(1 - 2 \cdot \frac{x}{2}\right) = \frac{1}{4}(1 - x) = \frac{1}{4} - \frac{x}{4}$. The factored constant is raised to the power $-2$ too: $2^{-2} = \frac{1}{4}$. At $x = 0.06$: $\frac{1}{4}(0.94) = 0.235$. Calculator: $\frac{1}{2.06^2} = 0.235649$ (the estimate is too small, $(1 + u)^{-2}$ being concave up). The student applied $(1 + x)^k$ to $2 + x$ as if the $2$ were a $1$: he forgot to factor out $2^{-2}$ and to halve $x$. At $x = 0$ his formula gives $1$ while $(2 + 0)^{-2} = \frac{1}{4}$: any linearization must be exact at its centre, and this one is not.

e) $(1.08)^5 \approx 1 + 5(0.08) = 1.4$, while the calculator gives $1.469328$: a percentage error of $\frac{1.469328 - 1.4}{1.469328} \times 100$, about $4.72\%$, forty times worse than in b). Yet $x = 0.08$ looks small. What enters the approximation is the product $kx$: $25 \times 0.002 = 0.05$ in b), $5 \times 0.08 = 0.4$ here. On the figure the tangent $y = 1 + 5x$ and the curve are almost together up to about $x = 0.02$, then separate: at $x = 0.08$ the gap between the two marked points is about $0.07$. The rule is safe when $kx$ is small, not merely when $x$ is.

a) Product rule, then the chain rule on $\sqrt{1 - x^2}$ with inner function $1 - x^2$: $\frac{dy}{dx} = \sqrt{1 - x^2} + x \cdot \frac{-2x}{2\sqrt{1 - x^2}} = \sqrt{1 - x^2} - \frac{x^2}{\sqrt{1 - x^2}}$. Common denominator, the step that costs the marks: $\sqrt{1 - x^2} = \frac{1 - x^2}{\sqrt{1 - x^2}}$, so $\frac{dy}{dx} = \frac{1 - x^2 - x^2}{\sqrt{1 - x^2}} = \frac{1 - 2x^2}{\sqrt{1 - x^2}}$ and $dy = \frac{1 - 2x^2}{\sqrt{1 - x^2}}\,dx$. At $x = 0.6$: $1 - 2(0.36) = 0.28$ and $\sqrt{1 - 0.36} = 0.8$, so $dy = \frac{0.28}{0.8}(0.02) = 0.35 \times 0.02 = 0.007$.

b) Quotient rule: $\frac{dy}{dx} = \frac{(x + 3) - (x - 1)}{(x + 3)^2}$. The numerator needs its brackets: $(x + 3) - (x - 1) = x + 3 - x + 1 = 4$; without them one gets $x + 3 - x - 1 = 2$, half the answer. So $dy = \frac{4}{(x + 3)^2}\,dx$. At $x = 1$, $dx = 0.2$: $dy = \frac{4}{16}(0.2) = 0.05$. Exact change: $y(1.2) - y(1) = \frac{0.2}{4.2} - 0 = 0.047619$. The differential is close, and slightly too large, the function being concave down there ($y'' = -\frac{8}{(x + 3)^3} < 0$).

c) Product rule and chain rule on $e^{-x}$ (inner function $-x$): $\frac{dy}{dx} = 2xe^{-x} + x^2(-e^{-x}) = 2xe^{-x} - x^2e^{-x}$. Factor out the common $xe^{-x}$: $\frac{dy}{dx} = x(2 - x)e^{-x}$, so $dy = x(2 - x)e^{-x}\,dx$. The factored form shows at once where $dy = 0$ ($x = 0$ and $x = 2$) and makes the evaluation clean: at $x = 1$, $dy = 1 \cdot 1 \cdot e^{-1}(0.1) = \frac{0.1}{e} = 0.036788$.

d) Take the differential of each term: $d(xy^2) = y^2\,dx + x \cdot 2y\,dy$ (product rule in differential form) and $d(y) = dy$, while $d(6) = 0$. So $y^2\,dx + 2xy\,dy + dy = 0$, that is $(2xy + 1)\,dy = -y^2\,dx$ and $dy = -\frac{y^2}{2xy + 1}\,dx$. Check the point: $1 \cdot 4 + 2 = 6$. At $(1, 2)$: $dy = -\frac{4}{5}\,dx = -0.8\,dx$. With $dx = 0.05$: $dy = -0.04$, so $y \approx 2 - 0.04 = 1.96$. Exactly, $1.05y^2 + y - 6 = 0$ gives $y = \frac{-1 + \sqrt{1 + 4(1.05)(6)}}{2(1.05)} = \frac{-1 + \sqrt{26.2}}{2.1} = 1.961235$ (the positive root, the one near $2$). The estimate is off by about $0.0012$.

e) $d(uv) = u\,dv + v\,du = 3(-0.2) + 5(0.1) = -0.6 + 0.5 = -0.1$: the product decreases slightly although $u$ increased. And $d\left(\frac{u}{v}\right) = \frac{v\,du - u\,dv}{v^2} = \frac{5(0.1) - 3(-0.2)}{25} = \frac{0.5 + 0.6}{25} = \frac{1.1}{25} = 0.044$. The minus sign of the quotient rule meets the minus sign of $dv$: two negatives, a plus. These two forms are exactly what propagates measurement errors through a product or a quotient.