02. 선형화와 미분Linearization and Differentials

Solution. Since $f(x)=\sqrt{1+x}$, we have $f'(x) = \dfrac{1}{2\sqrt{1+x}}$.

(1) At $x=0$, $f(0)=1$ and $f'(0)=\dfrac{1}{2}$, so the linearization is $$L(x) = f(0)+f'(0)(x-0) = 1+\frac{1}{2}x.$$

(2) At $x=3$, $f(3)=\sqrt{4}=2$ and $f'(3)=\dfrac{1}{2\sqrt{4}}=\dfrac{1}{4}$, so the linearization is $$L(x) = f(3)+f'(3)(x-3) = 2+\frac{1}{4}(x-3).$$

Solution.

(1) Since $f(x)=\cos x$, $f'(x) = -\sin x$. At $x=\pi/2$, $f(\pi/2)=0$ and $f'(\pi/2)=-1$, so $$L(x) = 0 + (-1)\left(x-\frac{\pi}{2}\right) = \frac{\pi}{2} - x.$$

(2) Since $f(x)=(1+x)^k$, $f'(x)=k(1+x)^{k-1}$. At $x=0$, $f(0)=1$ and $f'(0)=k$, so $$L(x) = 1+kx.$$

(3) Since $f(x)=x+\dfrac1x$, $f'(x)=1-\dfrac{1}{x^2}$. At $x=1$, $f(1)=2$ and $f'(1)=0$, so $$L(x) = 2 + 0\cdot(x-1) = 2.$$ That is, near $x=1$, $f$ is linearly approximated by the constant function $2$.

Solution.

(1) Since $y=x^5+37x$, $dy = (5x^4+37)\,dx$. At $x=1$, $dx=0.2$, $$dy = (5\cdot1^4+37)(0.2) = 42\times0.2 = 8.4.$$

(2) For $\tan 2x$, $d(\tan2x) = \sec^2(2x)\cdot 2\,dx = 2\sec^2(2x)\,dx$. For $\dfrac{x}{x+1}$, the quotient rule gives $\left(\dfrac{x}{x+1}\right)' = \dfrac{(x+1)-x}{(x+1)^2} = \dfrac{1}{(x+1)^2}$, so $$d\left(\frac{x}{x+1}\right) = \frac{1}{(x+1)^2}\,dx.$$

Solution.

(1) Differentiating both sides of $2y^{3/2}+xy-x=0$, $$2\cdot\frac32 y^{1/2}\,dy + (x\,dy+y\,dx) - dx = 0$$ $$3\sqrt{y}\,dy + x\,dy + y\,dx - dx = 0$$ $$(3\sqrt{y}+x)\,dy = (1-y)\,dx,$$ so $$dy = \frac{1-y}{3\sqrt{y}+x}\,dx.$$

(2) Let $y=3\csc(1-2\sqrt{x})$. Set $u=1-2\sqrt x$, so $\dfrac{du}{dx} = -\dfrac{1}{\sqrt x}$, and $(\csc u)' = -\csc u\cot u\cdot u'$ gives $$\frac{dy}{dx} = 3\left(-\csc u\cot u\right)\left(-\frac{1}{\sqrt x}\right) = \frac{3\csc(1-2\sqrt x)\cot(1-2\sqrt x)}{\sqrt x}.$$ Hence $$dy = \frac{3\csc(1-2\sqrt x)\cot(1-2\sqrt x)}{\sqrt x}\,dx.$$

Solution. Since the area of a circle is $A=\pi r^2$, $dA = 2\pi r\,dr$.

With $r=10$ and $dr=10.1-10=0.1$, $$dA = 2\pi(10)(0.1) = 2\pi \approx 6.28\ (\text{m}^2).$$ That is, as the radius changes from 10 m to 10.1 m, the area increases by about $2\pi\ \text{m}^2$ (about $6.28\ \text{m}^2$).

Solution. The volume of the cylinder is $V = \pi r^2 h_0$ with $h_0$ fixed, so $$dV = 2\pi r_0 h_0\,dr.$$ That is, as the radius changes from $r_0$ to $r_0+dr$, the change in volume is, by linear approximation, $2\pi r_0 h_0\,dr$.

Solution. By the quotient rule, the derivative of $C(t) = \dfrac{4t}{1+t^3}+0.06t$ is $$C'(t) = \frac{4(1+t^3)-4t(3t^2)}{(1+t^3)^2}+0.06 = \frac{4-8t^3}{(1+t^3)^2}+0.06.$$

20 minutes is $t=\dfrac{1}{3}$ hour, and the change from 20 to 30 minutes is $dt = \dfrac{1}{6}$ hour. At $t=\dfrac13$, $t^3=\dfrac{1}{27}$, so $$C'\left(\frac13\right) = \frac{4-\frac{8}{27}}{\left(\frac{28}{27}\right)^2}+0.06 = \frac{\frac{100}{27}}{\frac{784}{729}}+0.06 = \frac{675}{196}+0.06 \approx 3.444+0.06 = 3.504.$$

Hence, by linear approximation, $$dC = C'\left(\frac13\right)dt \approx 3.504\times\frac16 \approx 0.584\ (\text{mg/ml}).$$ That is, the blood concentration increases by about $0.584\ \text{mg/ml}$ from 20 to 30 minutes.

Proof. Let $f(u) = \dfrac{1}{\sqrt{1-u}} = (1-u)^{-1/2}$. Then $f$ is differentiable near $u=0$, with $$f(0)=1, \qquad f'(u)=\frac12(1-u)^{-3/2}, \qquad f'(0)=\frac12,$$ so the linearization of $f$ at $u=0$ is $$L(u) = f(0)+f'(0)u = 1+\frac u2.$$ For $u$ near $0$, $f(u)\approx L(u)$, that is, $$\frac{1}{\sqrt{1-u}}\approx 1+\frac u2.$$ Substituting $u=x^2$ gives, for $x$ near $0$, $$\frac{1}{\sqrt{1-x^2}}\approx 1+\frac12 x^2.$$

Now let $x=v/c$ (which is near $0$ since $v$ is much smaller than $c$). By the approximation above, $$m = \frac{m_0}{\sqrt{1-v^2/c^2}} = m_0\cdot\frac{1}{\sqrt{1-x^2}} \approx m_0\left(1+\frac12 x^2\right) = m_0+\frac{m_0v^2}{2c^2}.$$ Hence $$\Delta m = m-m_0 \approx \frac{m_0v^2}{2c^2}.$$

Solution.

(1) Let $f(x)=x^{1/3}$, so $f'(x)=\dfrac13x^{-2/3}$. Since $8^{1/3}=2$, we linearize at $x=8$. We have $f(8)=2$, $f'(8)=\dfrac13\cdot 8^{-2/3} = \dfrac13\cdot\dfrac14=\dfrac1{12}$, and $dx = 7.97-8=-0.03$, so $$7.97^{1/3} \approx f(8)+f'(8)\,dx = 2+\frac1{12}(-0.03) = 2-0.0025 = 1.9975.$$

(2) Let $f(x)=\sin x$, so $f'(x)=\cos x$. At $x=\pi/6$, $f(\pi/6)=\dfrac12$, $f'(\pi/6)=\dfrac{\sqrt3}{2}$, and $dx=0.01$, so $$\sin\left(\frac{\pi}{6}+0.01\right) \approx f\left(\frac\pi6\right)+f'\left(\frac\pi6\right)(0.01) = \frac12+\frac{\sqrt3}{2}(0.01) \approx 0.5+0.00866 = 0.50866.$$

Solution. Since $s=4.9t^2$, $ds = 9.8t\,dt$.

With a time-measurement error of $dt=0.1$ seconds, the error in the computed distance is, by linear approximation, $$ds = 9.8t(0.1) = 0.98t\ (\text{m}).$$ That is, at fall time $t$ seconds, a $0.1$-second error in the time measurement produces an error of about $0.98t$ m in the computed distance (the size of the error grows in proportion to the fall time $t$).

Solution. Let $d$ be the diameter of the sphere. Then the volume is $V = \dfrac43\pi r^3 = \dfrac43\pi\left(\dfrac d2\right)^3 = \dfrac{\pi}{6}d^3$.

Since $dV = \dfrac{\pi}{2}d^2\,dd$, $$\frac{dV}{V} = \frac{\frac\pi2 d^2\,dd}{\frac\pi6 d^3} = 3\cdot\frac{dd}{d}.$$

For the volume error to be at most 3%, we need $\left|\dfrac{dV}{V}\right| \leq 0.03$, so $$3\left|\frac{dd}{d}\right| \leq 0.03 \quad\Longrightarrow\quad \left|\frac{dd}{d}\right| \leq 0.01.$$ That is, the diameter must be measured with an error of at most $1\%$.

Solution. Let $h$ be the height of the building and $\theta$ the angle, so $h = 9\tan\theta$ (with $\theta$ in radians, where $75° = \dfrac{5\pi}{12}$ radians).

Since $dh = 9\sec^2\theta\,d\theta$, $$\frac{dh}{h} = \frac{9\sec^2\theta\,d\theta}{9\tan\theta} = \frac{d\theta}{\sin\theta\cos\theta} = \frac{2\,d\theta}{\sin2\theta}.$$

At $\theta=75°$, $2\theta = 150°$ and $\sin150° = \dfrac12$, so $$\frac{dh}{h} = \frac{2\,d\theta}{1/2} = 4\,d\theta.$$

For the height error to be at most 4%, we need $\left|\dfrac{dh}{h}\right| \leq 0.04$, so $$4|d\theta| \leq 0.04 \quad\Longrightarrow\quad |d\theta| \leq 0.01\ (\text{radians}).$$ That is, the angle must be measured with an error of at most $0.01$ radian (about $0.57°$).