Corrected exercises: linear approximation and differentials (MATH 140)

a) $f(x) = x^{1/2}$, so $f'(x) = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt x}$ by the power rule. At $a = 4$: $f(4) = 2$ and $f'(4) = \frac{1}{2 \cdot 2} = \frac{1}{4}$. Hence $L(x) = 2 + \frac{1}{4}(x - 4)$. Check against the figure: the orange line passes through $(4, 2)$ and rises by $1$ when $x$ goes from $4$ to $8$, a slope of $\frac{1}{4}$; it also meets the vertical axis at $L(0) = 1$, as drawn. Write the factor $(x - 4)$, not $x$: the line $2 + \frac{1}{4}x$ has the right slope but passes through $(4, 3)$, and it is the most frequent lost mark of the chapter.

b) $\sqrt{4.1} \approx L(4.1) = 2 + \frac{1}{4}(0.1) = 2 + 0.025 = 2.025$. And $\sqrt{3.96} \approx L(3.96) = 2 + \frac{1}{4}(-0.04) = 2 - 0.01 = 1.99$. The only arithmetic is the product $f'(a)\,(x - a)$, which is why the centre must make $f'(a)$ a simple number. For $x < a$ the increment $x - a$ is negative, and forgetting its sign gives $2.01$, a value on the wrong side of $2$ although $3.96 < 4$.

c) $f''(x) = -\frac{1}{4}x^{-3/2} < 0$ for every $x > 0$: $f$ is concave down, so its graph lies below each of its tangent lines, as the figure shows on both sides of $(4, 2)$. Therefore $L(x) \ge \sqrt x$, and both estimates are too large. Confirmation by squaring, which is exact arithmetic: $2.025^2 = 4.100625 > 4.1$, so $2.025 > \sqrt{4.1}$; and $1.99^2 = 3.9601 > 3.96$, so $1.99 > \sqrt{3.96}$. Squaring the estimate is the cheapest check of the chapter, and it must agree with the side predicted by $f''$.

d) The centre must be a point NEAR the target where $f(a)$ and $f'(a)$ are known exactly: $4$ is the perfect square next to $4.1$, and $f'(4) = \frac{1}{4}$ is a clean fraction. At $a = 0$, $f'(0) = \frac{1}{2\sqrt 0}$ does not exist: the tangent at the origin is vertical and there is no linearization. At $a = 1$, $L_1(x) = 1 + \frac{1}{2}(x - 1)$ gives $L_1(4.1) = 1 + 1.55 = 2.55$, while $\sqrt{4.1}$ is barely above $2$: an error of more than $0.5$, because $4.1$ is far from $1$ and the curve has bent a lot in between.

e) $L(5) = 2 + \frac{1}{4} = 2.25$. Now $2.23^2 = 4.9729 < 5$ and $2.24^2 = 5.0176 > 5$, so $2.23 < \sqrt 5 < 2.24$ and the error $2.25 - \sqrt 5$ lies between $0.01$ and $0.02$. For $\sqrt{4.1}$: $2.024^2 = 4.096576 < 4.1$, so $2.024 < \sqrt{4.1} < 2.025$ and the error is below $0.001$. The distance to the centre went from $0.1$ to $1$, ten times more, and the error grew by a factor of more than ten, about a hundred. The tangent is a LOCAL tool: its quality collapses as soon as $x$ moves away from $a$.

a) $f(x) = x^{1/3}$, so $f'(x) = \frac{1}{3}x^{-2/3} = \frac{1}{3\sqrt[3]{x^2}}$. At $a = 8$, the perfect cube next to both targets: $f(8) = 2$ and $f'(8) = \frac{1}{3 \cdot 4} = \frac{1}{12}$. So $L(x) = 2 + \frac{1}{12}(x - 8)$. Then $\sqrt[3]{7.9} \approx 2 + \frac{1}{12}(-0.1) = 2 - \frac{1}{120} = \frac{239}{120}$, and $\sqrt[3]{8.24} \approx 2 + \frac{0.24}{12} = 2 + 0.02 = 2.02$. The value $f'(8) = \frac{1}{12}$ is the whole reason for choosing $8$: $\sqrt[3]{8^2} = 4$ is exact.

b) $g(x) = x^{-1}$, $g'(x) = -x^{-2}$, so $g(1) = 1$ and $g'(1) = -1$: $L(x) = 1 - (x - 1) = 2 - x$, the orange line of the figure. Hence $\frac{1}{0.98} \approx 2 - 0.98 = 1.02$ and $\frac{1}{1.05} \approx 2 - 1.05 = 0.95$. The rule a student remembers, $\frac{1}{1 + h} \approx 1 - h$, is exactly this linearization with $x = 1 + h$.

c) $f''(x) = -\frac{2}{9}x^{-5/3} < 0$ for $x > 0$: the cube root is concave down there, its tangent lies above, and both estimates are too large. Check: $\left(\frac{239}{120}\right)^3 = \frac{13\,651\,919}{1\,728\,000}$ while $7.9 = \frac{13\,651\,200}{1\,728\,000}$, so $\frac{239}{120} > \sqrt[3]{7.9}$; and $2.02^3 = 8.242408 > 8.24$. For the reciprocal, $g''(x) = \frac{2}{x^3} > 0$ for $x > 0$: concave up, tangent below, both estimates too small, which is what the figure shows on both sides of $(1, 1)$. Check: $1.02 \times 0.98 = 0.9996 < 1$, so $1.02 < \frac{1}{0.98}$; and $0.95 \times 1.05 = 0.9975 < 1$, so $0.95 < \frac{1}{1.05}$. The side does not depend on whether $x$ is left or right of $a$, as long as the concavity keeps its sign between them.

d) $\frac{50}{49} - 1.02 = \frac{50}{49} - \frac{51}{50} = \frac{2500 - 2499}{2450} = \frac{1}{2450}$, positive as predicted in c), and about $0.0004$. An error of that size for a distance $0.02$ to the centre is typical: it is roughly the square of the distance, which is why a linear approximation is excellent very close to $a$ and nowhere else.

e) On the figure the orange line reaches the horizontal axis at $x = 2$: $L(2) = 0$, while $g(2) = \frac{1}{2}$. The estimate is not slightly wrong, it is meaningless, since a positive number is approximated by $0$. The tangent follows the curve only in a small neighbourhood of the point of contact; one unit away, the curve has bent up so far that the line says nothing about it. Before using $L$, look at the distance $|x - a|$ compared with the scale on which $f$ changes.

a) Each time $L(x) = f(0) + f'(0)\,x$. For $e^x$: $f(0) = 1$, $f'(0) = e^0 = 1$, so $L = 1 + x$. For $\ln(1 + x)$: $f(0) = 0$, $f'(x) = \frac{1}{1 + x}$ by the chain rule with inner function $1 + x$, $f'(0) = 1$, so $L = x$. For $\sin x$: $f(0) = 0$, $f'(0) = \cos 0 = 1$, so $L = x$. For $\tan x$: $f(0) = 0$, $f'(0) = \sec^2 0 = 1$, so $L = x$. For $(1 + x)^k$: $f(0) = 1$, $f'(x) = k(1 + x)^{k-1}$, $f'(0) = k$, so $L = 1 + kx$. The derivatives of $\sin$ and $\tan$ are $\cos$ and $\sec^2$ only in radians; in degrees every one of these rules is false.

b) $e^{-0.03} \approx 1 - 0.03 = 0.97$; $(e^x)'' = e^x > 0$, the curve lies above its tangent, so $0.97$ is too small. $\ln 1.02 = \ln(1 + 0.02) \approx 0.02$; $(\ln(1 + x))'' = -\frac{1}{(1 + x)^2} < 0$, the curve lies below its tangent, as the figure shows on both sides of $0$, so $0.02$ is too large. $\sin 0.01 \approx 0.01$; $(\sin x)'' = -\sin x < 0$ on $(0, \pi)$, so on the right of $0$ the curve lies below the tangent and $0.01$ is too large. The three values look alike and their errors do not have the same sign: the side is part of the answer.

c) $\sqrt{1.06} = (1 + 0.06)^{1/2} \approx 1 + \frac{1}{2}(0.06) = 1.03$. The second derivative of $(1 + x)^{1/2}$ is $-\frac{1}{4}(1 + x)^{-3/2} < 0$, so $1.03$ is too large; indeed $1.03^2 = 1.0609 > 1.06$. Next, $\frac{1}{1.01^3} = (1 + 0.01)^{-3} \approx 1 - 3(0.01) = 0.97$. The second derivative of $(1 + x)^{-3}$ is $12(1 + x)^{-5} > 0$ near $0$, so $0.97$ is too small; indeed $0.97 \times 1.01^3 = 0.97 \times 1.030301 = 0.99939197 < 1$. The exponent $k$ goes in front of $x$ WITH its sign: $1 + 3(0.01)$ would put the estimate above $1$, although $1.01^3 > 1$ makes the quotient smaller than $1$.

d) $1.02$ is not small: $\arctan x \approx x$ would give $1.02$, a tangent at $0$ used at distance $1.02$. The right centre is $a = 1$, where $\arctan 1 = \frac{\pi}{4}$ is exact and $(\arctan x)' = \frac{1}{1 + x^2}$ gives $\frac{1}{2}$. So $L(x) = \frac{\pi}{4} + \frac{1}{2}(x - 1)$ and $\arctan 1.02 \approx \frac{\pi}{4} + 0.01$, about $0.795$ with $\pi \approx 3.14$, far from $1.02$. The second derivative is $-\frac{2x}{(1 + x^2)^2} < 0$ for $x > 0$: the estimate is too large. The answer stays exact, $\frac{\pi}{4} + 0.01$; the decimal is only an order of magnitude.

e) On the figure the line $y = x$ and the curve $y = \ln(1 + x)$ separate quickly: at $x = 1$ the line is at $1$ and the curve at $\ln 2 \approx 0.69$, an error of about $0.31$, a third of the value. At $x = -0.5$, the line gives $-0.5$ and the curve $\ln 0.5 = -\ln 2 \approx -0.69$. The rule $\ln(1 + x) \approx x$ is a statement about SMALL $x$, like $0.02$ in b), where the error is about $0.0002$. A small-$x$ rule used at $x = 1$ is a tangent used far from its point.

a) $\frac{dy}{dx} = 3x^2$, so $dy = 3x^2\,dx$. At $x = 2$ with $dx = 0.5$: $dy = 12 \times 0.5 = 6$. And $\Delta y = 2.5^3 - 2^3 = 15.625 - 8 = 7.625$. On the figure, $R(2.5, 8)$ is level with $P$, the tangent at $P$ has equation $y = 8 + 12(x - 2)$ and reaches $T(2.5, 14)$, and the curve reaches $Q(2.5, 15.625)$. So $dy$ is the segment $RT$, of length $6$, and $\Delta y$ is the segment $RQ$, of length $7.625$. The gap $TQ = 1.625$ is the error of the linear approximation, large here only because $dx$ was chosen large to be visible.

b) For $dx = 0.1$: $dy = 12 \times 0.1 = 1.2$ and $\Delta y = 2.1^3 - 8 = 9.261 - 8 = 1.261$, a gap of $0.061$. For $dx = 0.01$: $dy = 0.12$ and $\Delta y = 2.01^3 - 8 = 8.120601 - 8 = 0.120601$, a gap of $0.000601$. When $dx$ is divided by $10$, $dy$ is divided by $10$, but the gap is divided by about $100$: relative to $dy$, the error of the tangent shrinks. This is the whole content of the approximation $\Delta y \approx dy$.

c) $(x + dx)^3 = x^3 + 3x^2\,dx + 3x(dx)^2 + (dx)^3$, so $\Delta y = 3x^2\,dx + 3x(dx)^2 + (dx)^3$ and $\Delta y - dy = 3x(dx)^2 + (dx)^3$. At $x = 2$: $6(0.1)^2 + (0.1)^3 = 0.06 + 0.001 = 0.061$, as found in b). Dividing by $dx$: $\frac{\Delta y - dy}{dx} = 3x\,dx + (dx)^2 \to 0$ as $dx \to 0$. The error is not only small, it is small COMPARED WITH $dx$; that is what makes the tangent the best line through $P$, since any other line through $P$ leaves an error of the same order as $dx$.

d) $\frac{dy}{dx} = \frac{x}{\sqrt{x^2 + 9}}$ by the chain rule with inner function $x^2 + 9$. At $x = 4$: $\sqrt{25} = 5$, so $dy = \frac{4}{5}(-0.2) = -0.16$. The exact change is $\Delta y = \sqrt{3.8^2 + 9} - 5 = \sqrt{23.44} - 5$. Since $4.84^2 = 23.4256 < 23.44$, $\sqrt{23.44} > 4.84$ and $\Delta y > 4.84 - 5 = -0.16 = dy$; and $4.842^2 = 23.444964 > 23.44$ gives $\Delta y < -0.158$. The function decreases less than the tangent predicts: the curve is concave up, the tangent lies below it, and with $dx < 0$ that means $dy$ is the more negative of the two. A negative $dx$ gives a negative $dy$ here; dropping the sign of $dx$ is the common slip.

e) $A = s^2$ gives $dA = 2s\,ds$, while $\Delta A = (s + ds)^2 - s^2 = 2s\,ds + (ds)^2$. Geometrically, enlarge the square by $ds$ to the right and upwards: the added region is two strips of size $s \times ds$ along two sides, total $2s\,ds$, plus a small corner square $ds \times ds$. The differential keeps the two strips and drops the corner, whose area $(ds)^2$ is negligible compared with $ds$ when $ds$ is small. The same picture, with a cube and its three slabs, explains $dV = 3s^2\,ds$.