21b. Differentials · Newton's Method — Practice

Drill sheet for 21. Differentials · Newton's Method. Book-style solved ideas as guided checks and regenerable worksheets.

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1. Linear approximations

(a) $\sqrt[3]{124}$: $x=125$, $\Delta x=-1$, $f'(125)=1/75$

$$\sqrt[3]{124}\sim 5-\frac1{75}=\frac{374}{75}\approx 4.9867$$

(true $\approx 4.9866$).

(b) $\sin 61^\circ$: use radians — $x=\pi/3$, $\Delta x=\pi/180$, $f'=\cos(\pi/3)=1/2$

$$\sin 61^\circ\sim \frac{\sqrt3}{2}+\frac{\pi}{360}\approx 0.8747$$

(true $\approx 0.8746$).

Radians

Trig derivatives in calculus assume the argument is in radians. Convert $1^\circ=\pi/180$.

Linear approximation

Roots, sine, percent change

(1)Approximate $\sqrt[3]{124}$ using $f(x)=x^{1/3}$, $x=125$, $\Delta x=-1$.

$f(125)=5$, $f'(x)=\frac13 x^{-2/3}$, $f'(125)=1/75$.

$\sqrt[3]{124}\approx 5+(1/75)(-1)=374/75\approx 4.9867$.

$\approx 4.9867$ (true $\approx 4.9866$).

(2)Approximate $\sqrt[4]{17}$ from $x=16$, $\Delta x=1$.

$f=x^{1/4}$, $f(16)=2$, $f'(16)=1/(4\cdot 8)=1/32$.

$2+(1/32)=2.03125$.

$2.03125$.

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2. Cube volume, $1\%$ side increase

$V=x^{3}$, $\Delta x=0.01x$:

$$\Delta V\sim 3x^{2}\cdot(0.01x)=0.03x^{3}.$$

Volume grows by roughly $3\%$.

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3. Find $dy$

$y$	$dy$
$x^{3}+4x^{2}-5x+6$	$(3x^{2}+8x-5)\,dx$
$(2x^{3}+5)^{3/2}$	$9x^{2}\sqrt{2x^{3}+5}\,dx$
$\dfrac{x^{3}+2x+1}{x^{2}+3}$	$\dfrac{x^{4}+7x^{2}-2x+6}{(x^{2}+3)^{2}}\,dx$
$\cos^{2}2x+\sin 3x$	$(-2\sin 4x+3\cos 3x)\,dx$

Compute dy

Polynomial · chain · quotient · trig

(1)Find $dy$ if $y=x^{3}+4x^{2}-5x+6$.

$dy=(3x^{2}+8x-5)\,dx$.

$(3x^{2}+8x-5)\,dx$.

(2)Find $dy$ if $y=(2x^{3}+5)^{3/2}$.

$dy=\tfrac32(2x^{3}+5)^{1/2}\cdot 6x^{2}\,dx=9x^{2}(2x^{3}+5)^{1/2}\,dx$.

$9x^{2}\sqrt{2x^{3}+5}\,dx$.

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4. $dy/dx$ via differentials

(a) $xy+x-2y=5$ ⇒ $\dfrac{dy}{dx}=-\dfrac{y+1}{x-2}$.

(b) $\dfrac{2x}{y}-\dfrac{3y}{x}=8$ ⇒ $\dfrac{dy}{dx}=\dfrac{y}{x}$ (after canceling $2x^{2}+3y^{2}\neq 0$).

(c) Parametric $x=3\cos\theta-\cos 3\theta$, $y=3\sin\theta-\sin 3\theta$:

$$\frac{dy}{dx}=\frac{\cos\theta-\cos 3\theta}{-\sin\theta+\sin 3\theta}.$$

dy/dx from differentials

Implicit and parametric

(1)Parametric: $x=3\cos\theta-\cos 3\theta$, $y=3\sin\theta-\sin 3\theta$. Find $dy/dx$.

$dx=(-3\sin\theta+3\sin 3\theta)\,d\theta$, $dy=(3\cos\theta-3\cos 3\theta)\,d\theta$.

$dy/dx=(\cos\theta-\cos 3\theta)/(-\sin\theta+\sin 3\theta)$.

$\dfrac{\cos\theta-\cos 3\theta}{-\sin\theta+\sin 3\theta}$.

(2)From $xy=\sin(x-y)$, find $dy/dx$.

$x\,dy+y\,dx=\cos(x-y)(dx-dy)$.

$dy/dx=\dfrac{\cos(x-y)-y}{\cos(x-y)+x}$.

$\dfrac{\cos(x-y)-y}{\cos(x-y)+x}$.

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5. Newton: $x^{3}+2x-5=0$

One real root between $1$ and $2$. With $x_0=1$:

$$x_{n+1}=\frac{2x_n^{3}+5}{3x_n^{2}+2}.$$

$x_1=1.4$; stabilizes at $\approx 1.328268856$.

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6. Newton: $2\cos x-x^{2}=0$

Even $f$ ⇒ roots $\pm r$. From $x_0=1$:

$$x_{n+1}=\frac{x_n^{2}+2(x_n\sin x_n+\cos x_n)}{2(x_n+\sin x_n)},$$

converging to $r\approx 1.02168995$ (and $-r$).

Newton roots

Cubic and trig–polynomial

(1)Newton for $2\cos x-x^{2}=0$ with $x_{0}=1$. Approximate the positive root.

$f'=-2(x+\sin x)$; even $f$ ⇒ roots $\pm r$.

Converges to $r\approx 1.02168995$.

$\approx\pm 1.02169$.

(2)Newton $x-\cos x=0$ to four decimals.

Fixed-point of $\cos$; $\approx 0.7391$.

$0.7391$.

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Mixed review

Differentials & Newton mixed

Approx · errors · dy · implicit · roots · failures

(1)Piecewise $f=\pm\sqrt{|x-2|}$ with $x_{0}=3$: what goes wrong?

Cusp / infinite slope at the root $x=2$; tangent from $x_{0}=3$ does not settle.

Not a smooth simple root; Newton leaves the basin.

(2)Circular plate: radius $5\to 5.06$. Estimate $\Delta A$ for $A=\pi r^{2}$.

$dA=2\pi r\,dr=2\pi\cdot 5\cdot 0.06=0.6\pi\approx 1.88$ in².

$0.6\pi\approx 1.88$ in².

(3)Approximate $\sin 61^\circ$ from $\sin(\pi/3)$ with $\Delta x=\pi/180$.

$f=\sin$, $f(\pi/3)=\sqrt{3}/2$, $f'=\cos(\pi/3)=1/2$.

$\sin 61^\circ\approx \sqrt{3}/2+(\pi/360)\approx 0.8747$.

$\approx 0.8747$ (true $\approx 0.8746$).

(4)Piecewise $f=\pm\sqrt{|x-2|}$ with $x_{0}=3$: what goes wrong?

Cusp / infinite slope at the root $x=2$; tangent from $x_{0}=3$ does not settle.

Not a smooth simple root; Newton leaves the basin.

(5)From $xy=\sin(x-y)$, find $dy/dx$.

$x\,dy+y\,dx=\cos(x-y)(dx-dy)$.

$dy/dx=\dfrac{\cos(x-y)-y}{\cos(x-y)+x}$.

$\dfrac{\cos(x-y)-y}{\cos(x-y)+x}$.

(6)Approximate $\Delta(1/x)$ as $x$ goes $1\to 0.98$.

$d(1/x)=-dx/x^{2}=-( -0.02)/1=0.02$.

$0.02$ (value increases).

Supplementary Problems

7–10. More linear approximations and $\Delta y$

#	Setup	Ans
7	$\sqrt[4]{17}$; $\sqrt[5]{1020}$; $\cos 59^\circ$; $\tan 44^\circ$	$2.03125$; $3.99688$; $0.5151$; $0.9651$
8	$\Delta(x^{3})$ for $5\to 5.01$; $\Delta(1/x)$ for $1\to 0.98$	$0.75$; $0.02$
9	Plate $r:5\to 5.06$	$\Delta A\approx 0.6\pi\approx 1.88$ in²
10	Ice ball $r:10\to 9.8$	$\Delta V\approx -80\pi$; $\Delta S\approx -16\pi$

11–16. Measurement error and surfaces

• 11. $v=\sqrt{64.4 h}$, $h=100\pm 0.5$ ⇒ $|\Delta v|\approx 0.2$ ft/s.
• 12. Flight $2$ mi above equator: extra $\approx 2\pi\cdot 2\approx 12.6$ mi.
• 13. $|dr|\le 0.001$, need $|dA|\le 0.1$ ⇒ $r\le 16$ in.
• 14. $pV=20$, $p=5\pm 0.02$ ⇒ $V=4\pm 0.016$.
• 15. $F=1/r^{2}$, $F=4\pm 0.05$ ⇒ $r=0.5\pm 0.003$.
• 16. Cone total surface $S=\pi r\sqrt{r^{2}+h^{2}}+\pi r^{2}$:
  • (a) fixed $r$: $dS=\dfrac{\pi r h}{\sqrt{r^{2}+h^{2}}}\,\Delta h$
  • (b) fixed $h$: $dS=\pi\!\left(\dfrac{h^{2}+2r^{2}}{\sqrt{r^{2}+h^{2}}}+2r\right)\Delta r$

Error & small-change estimates

Plates · spheres · gas law · equator

(1)Area to $\pm 0.1$ in²; $|dr|\le 0.001$. Max $r$ for $A=\pi r^{2}$?

$|dA|=2\pi r|dr|\le 0.1$ ⇒ $r\le 0.1/(2\pi\cdot 0.001)\approx 16$.

$r\le 16$ in.

(2)Area to $\pm 0.1$ in²; $|dr|\le 0.001$. Max $r$ for $A=\pi r^{2}$?

$|dA|=2\pi r|dr|\le 0.1$ ⇒ $r\le 0.1/(2\pi\cdot 0.001)\approx 16$.

$r\le 16$ in.

(3)$pV=20$, $p=5\pm 0.02$. Estimate $V$.

$V=4$; $dV=-20\,dp/p^{2}=-20(0.02)/25=-0.016$.

$V=4\pm 0.016$.

17–18. More $dy$ and implicit $dy/dx$

(17) $y=(5-x)^{3}$, $\sin x/x$, $\arccos(2x)$, $\cos(bx^{2})$ — answers in the worksheet pool.

(18) $2xy^{3}+3x^{2}y=1$ and $xy=\sin(x-y)$ — see $dy/dx$ worksheet.

19–22, 25–26. Newton (calculator)

Equation / target	≈ answer
$x^{3}+3x+1=0$	$-0.3222$
$x-\cos x=0$	$0.7391$
$x^{3}+2x^{2}-4=0$	$1.1304$
$\sqrt[4]{3}$; $\sqrt[5]{247}$	$1.3161$; $3.0098$
Heron $\sqrt{r}$: $x\leftarrow\tfrac12(x+r/x)$; $\sqrt{5}$	$2.2361$
$x^{3}+x^{2}-3=0$ in $(1,2)$	$1.1746$
$\cos x+1=0$ (→ $\pi$)	$3.141592654$ (slow: $f'(\pi)=0$)
$\cos x=x/2$ (positive)	$1.029866529$

23–24. Failure cases

• 23. $x^{1/3}=0$, $x_0=1$: update $x\mapsto -2x$ diverges.
• 24. (a) $f'(1)=0$; (b) bad basin / horizontal issues; (c) cusp at root $x=2$.

Newton failures

Why the iteration breaks

(1)Piecewise $f=\pm\sqrt{|x-2|}$ with $x_{0}=3$: what goes wrong?

Cusp / infinite slope at the root $x=2$; tangent from $x_{0}=3$ does not settle.

Not a smooth simple root; Newton leaves the basin.

(2)For $x^{3}-3x^{2}+3x+2=0$ with $x_{0}=1$, why does Newton fail?

$f'(x)=3(x-1)^{2}$; $f'(1)=0$ — division by zero / horizontal tangent.

$f'(x_{0})=0$.

Self-check

Quiz

Differentials & Newton review

1 / 7

∛124 ≈ 5 − 1/75 uses f′(125) = …

Pick an answer to see if you are right.

Further Reading

• 21. Differentials · Newton's Method — theory and √3 demo
• 20b. Related Rates — Practice — prior applied rates
• 17. Differentiation of Trig Functions — $\sin/\cos$ in radians
• Computational physics: Newton–Raphson