KaTeX-DB demo

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Katex-DB uses a database approach to fast loading of beautiful math on the Web.

The math you see on the this demo page consists of pre-generated and saved KaTeX spans, so there is no javascript processing required when the page loads (except for a tiny script which aligns equal signs), resulting in extremely fast page load.

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As a comparison, this Khan Academy page has only 7 math expressions rendered by KaTeX, yet takes 2.6 s to load, has 115 requests (mostly .JS files), and the payload is 2.14 MB with a javascript blocking time of 1.1 s.

NOTE: The text on this demo page is obfuscated for copyright reasons.

On with the demo.

Gur tencu bs $y = 3 x \displaystyle{y}={3}{x}$ , jvgu gur nern haqre gur "pheir" orgjrra $x = 0 \displaystyle{x}={0}$ to $x = 1 \displaystyle{x}={1}$ funqrq.

Jura gur funqrq nern vf ebgngrq 360° nobhg gur $x \displaystyle{x}$ -nkvf, n ibyhzr vf trarengrq.

Gur erfhygvat fbyvq vf n pbar:

Gur ibyhzr bs n plyvaqre vf tvira ol:

$V = π r^{2} h \displaystyle{V}=\pi{r}^{2}{h}$

Orpnhfr $radius = r = y \displaystyle\text{radius}={r}={y}$ naq rnpu qvfx vf $d x \displaystyle{\left.{d}{x}\right.}$ uvtu, jr abgvpr gung gur ibyhzr bs rnpu fyvpr vf:

$V = π y^{2} d x \displaystyle{V}=\pi{y}^{2}\ {\left.{d}{x}\right.}$

Nqqvat gur ibyhzrf bs gur qvfxf (jvgu vasvavgryl fznyy $d x \displaystyle{\left.{d}{x}\right.}$ ), jr bognva gur sbezhyn:

$V = π \int_{a}^{b} y^{2} d x \displaystyle{V}=\pi{\int_{{a}}^{{b}}}{y}^{2}{\left.{d}{x}\right.}$ juvpu zrnaf $V = π \int_{a}^{b} {f (x)}^{2} d x \displaystyle{V}=\pi{\int_{{a}}^{{b}}}{\left\lbrace f{{\left({x}\right)}}\right\rbrace}^{2}{\left.{d}{x}\right.}$

jurer:

$y = f (x) \displaystyle{y}= f{{\left({x}\right)}}$ vf gur rdhngvba bs gur pheir jubfr nern vf orvat ebgngrq
$a \displaystyle{a}$ naq $b \displaystyle{b}$ ner gur yvzvgf bs gur nern orvat ebgngrq
$d x \displaystyle{\left.{d}{x}\right.}$ fubjf gung gur nern vf orvat ebgngrq nobhg gur $x \displaystyle{x}$ -nkvf

ABGR: Ba guvf cntr jr hfr gur qvfx zrgubq naq jnfure zrgubq (jurer jr phg gur funcr vagb pvephyne fyvprf) bayl, naq zrrg gur Furyy Zrgubq arkg).

Nccylvat gur sbezhyn $V = π \int_{a}^{b} y^{2} d x \displaystyle{V}=\pi{\int_{{a}}^{{b}}}{y}^{2}{\left.{d}{x}\right.}$ gb gur rneyvre rknzcyr, jr unir:

(3) $Vol = π \int_{a}^{b} y^{2} d x \displaystyle\text{Vol}=\pi{\int_{{a}}^{{b}}}{y}^{2}{\left.{d}{x}\right.}$

$= π \int_{0}^{1} {(3 x)}^{2} d x \displaystyle=\pi{\int_{{0}}^{{1}}}{\left({3}{x}\right)}^{2}{\left.{d}{x}\right.}$

$= π \int_{0}^{1} 9 x^{2} d x \displaystyle=\pi{\int_{{0}}^{{1}}}{9}{x}^{2}{\left.{d}{x}\right.}$

$= π {[3 x^{3}]}_{0}^{1} \displaystyle=\pi{{\left[{3}{x}^{3}\right]}_{{0}}^{{1}}}$

$= π [3] - π [0] \displaystyle=\pi{\left[{3}\right]}-\pi{\left[{0}\right]}$

$= 3 π {unit}^{3} \displaystyle={3}\pi\ \text{unit}^{3}$

PURPX: Qbrf gur zrgubq jbex? Jr pna svaq gur ibyhzr bs gur pbar hfvat

$Vol = \frac{π r^{2} h}{3} \displaystyle\text{Vol}=\frac{{\pi{r}^{2}{h}}}{{3}}$

$= \frac{π {(3)}^{2} (1)}{3} \displaystyle=\frac{{\pi{\left({3}\right)}^{2}{\left({1}\right)}}}{{3}}$

$= \frac{9 π}{3} \displaystyle=\frac{{{9}\pi}}{{3}}$

$= 3 π {unit}^{3} \displaystyle={3}\pi\ \text{unit}^{3}$ (Purpxf BX.)

Rknzcyr 2

Svaq gur ibyhzr vs gur nern obhaqrq ol gur pheir $y = x^{3} + 1 \displaystyle{y}={x}^{3}+{1}$ , gur $x \displaystyle{x}$ -nkvf naq gur yvzvgf bs $x = 0 \displaystyle{x}={0}$ naq $x = 3 \displaystyle{x}={3}$ vf ebgngrq nebhaq gur $x \displaystyle{x}$ -nkvf.

Nern haqre gur pheir $y = x^{3} + 1 \displaystyle{y}={x}^{3}+{1}$ sebz $x = 0 \displaystyle{x}={0}$ to $x = 3 \displaystyle{x}={3}$ ebgngrq nebhaq gur $x \displaystyle{x}$ -nkvf, fubjvat n glcvpny qvfx.

Nccylvat gur sbezhyn sbe gur fbyvq bs eribyhgvba, jr trg

$V = π \int_{a}^{b} y^{2} d x \displaystyle{V}=\pi{\int_{{a}}^{{b}}}{y}^{2}{\left.{d}{x}\right.}$

$= π \int_{0}^{3} {(x^{3} + 1)}^{2} d x \displaystyle=\pi{\int_{{0}}^{{3}}}{\left({x}^{3}+{1}\right)}^{2}{\left.{d}{x}\right.}$

$= π \int_{0}^{3} (x^{6} + 2 x^{3} + 1) d x \displaystyle=\pi{\int_{{0}}^{{3}}}{\left({x}^{6}+{2}{x}^{3}+{1}\right)}{\left.{d}{x}\right.}$

$= π {[\frac{x^{7}}{7} + \frac{x^{4}}{2} + x]}_{0}^{3} \displaystyle=\pi{{\left[\frac{{{x}^{7}}}{{{7}}}+\frac{{{x}^{4}}}{{{2}}}+{x}\right]}_{{0}}^{{3}}}$

$= π (∣ 355.93 ∣ - ∣ 0 ∣) \displaystyle=\pi{\left({\left|{355.93}\right|}-{\left|{0}\right|}\right)}$

$= 1118.2 {units}^{3} \displaystyle={1118.2}\ \text{units}^{3}$

Ibyhzr ol Ebgngvat gur Nern Rapybfrq Orgjrra 2 Pheirf

Vs jr unir 2 pheirf $y_{2} \displaystyle{y}_{{2}}$ naq $y_{1} \displaystyle{y}_{{1}}$ gung rapybfr fbzr nern naq jr ebgngr gung nern nebhaq gur $x \displaystyle{x}$ -nkvf, gura gur ibyhzr bs gur fbyvq sbezrq vf tvira ol:

$Volume = π \int_{a}^{b} [{(y_{2})}^{2} - {(y_{1})}^{2}] d x \displaystyle\text{Volume}=\pi{\int_{{a}}^{{b}}}{\left[{\left({y}_{{2}}\right)}^{2}-{\left({y}_{{1}}\right)}^{2}\right]}{\left.{d}{x}\right.}$

Va gur sbyybjvat trareny tencu, $y_{2} \displaystyle{y}_{{2}}$ vf nobir $y_{1} \displaystyle{y}_{{1}}$ . Gur ybjre naq hccre yvzvgf sbe gur ertvba gb or ebgngrq ner vaqvpngrq ol gur iregvpny yvarf ng $x = a \displaystyle{x}={a}$ naq $x = b \displaystyle{x}={b}$ .

Jura jr ebgngr fhpu n funcr nebhaq na nkvf, naq gnxr fyvprf, gur erfhyg vf n jnfure funcr (jvgu n ebhaq ubyr va gur zvqqyr).

Rknzcyr 3

N phc-yvxr bowrpg vf znqr ol ebgngvat gur nern orgjrra $y = 2 x^{2} \displaystyle{y}={2}{x}^{2}$ naq $y = x + 1 \displaystyle{y}={x}+{1}$ jvgu $x \geq 0 \displaystyle{x}\ge{0}$ nebhaq gur $x \displaystyle{x}$ -nkvf. Svaq gur ibyhzr bs gur zngrevny arrqrq gb znxr gur phc. Havgf ner $cm \displaystyle\text{cm}$ .

Nern obhaqrq ol $y = 2 x^{2} \displaystyle{y}={2}{x}^{2}$ (gur obggbz pheir), $y = x + 1 \displaystyle{y}={x}+{1}$ (gur yvar nobir), naq $x = 0 \displaystyle{x}={0}$ , fubjvat n glcvpny erpgnatyr.

Gur ybjre yvzvg bs vagrtengvba vf $x = 0 \displaystyle{x}={0}$ (fvapr gur dhrfgvba fnlf $x \geq 0 \displaystyle{x}\ge{0}$ ).

Arkg, jr arrq gb svaq jurer gur pheirf vagrefrpg fb jr xabj gur hccre yvzvg bs vagrtengvba.

Rdhngvat gur 2 rkcerffvbaf naq fbyivat:

$Volume = π \int_{0}^{1} [{(x + 1)}^{2} - {(2 x^{2})}^{2}] d x \displaystyle\text{Volume}=\pi{\int_{{0}}^{{1}}}{\left[{\left({x}+{1}\right)}^{2}-{\left({2}{x}^{2}\right)}^{2}\right]}{\left.{d}{x}\right.}$

$= π \int_{0}^{1} [(x^{2} + 2 x + 1) - (4 x^{4})] d x \displaystyle=\pi{\int_{{0}}^{{1}}}{\left[{\left({x}^{2}+{2}{x}+{1}\right)}-{\left({4}{x}^{4}\right)}\right]}{\left.{d}{x}\right.}$

$= π {[\frac{x^{3}}{3} + x^{2} + x - \frac{4 x^{5}}{5}]}_{0}^{1} \displaystyle=\pi{{\left[\frac{{{x}^{3}}}{{{3}}}+{x}^{2}+{x}-\frac{{{4}{x}^{5}}}{{{5}}}\right]}_{{0}}^{{1}}}$

$= π [(\frac{1}{3} + 1 + 1 - \frac{4}{5}) - (0)] \displaystyle=\pi{\left[{\left(\frac{1}{{3}}+{1}+{1}-\frac{4}{{5}}\right)}-{\left({0}\right)}\right]}$

$= π [\frac{5 + 30 - 12}{15}] \displaystyle=\pi{\left[\frac{{{5}+{30}-{12}}}{{{15}}}\right]}$

$= \frac{23 π}{15} \displaystyle=\frac{{{23}\pi}}{{{15}}}$

$= 4.817 {cm}^{3} \displaystyle={4.817}\ \text{cm}^{3}$

Sbe Z₁ (gur hccre cneg bs gur pvephvg), jr unir:

$X_{C} = \frac{1}{2 π (60) (1.20 \times 10^{- 6})} \displaystyle{X}_{{C}}=\frac{1}{{{2}\pi{\left({60}\right)}{\left({1.20}\times{10}^{ -{{6}}}\right)}}}$

$= 2210.485 Ω \displaystyle={2210.485}\ \Omega$

Jura gur funqrq nern vf ebgngrq 360° nobhg gur $y \displaystyle{y}$ -nkvf, gur ibyhzr gung vf trarengrq pna or sbhaq ol:

$V = π \int_{c}^{d} x^{2} d y \displaystyle{V}=\pi{\int_{{c}}^{{d}}}{x}^{2}{\left.{d}{y}\right.}$ juvpu zrnaf $V = π \int_{c}^{d} {f (y)}^{2} d y \displaystyle{V}=\pi{\int_{{c}}^{{d}}}{\left\lbrace f{{\left({y}\right)}}\right\rbrace}^{2}{\left.{d}{y}\right.}$

jurer:

$x = f (y) \displaystyle{x}= f{{\left({y}\right)}}$ vf gur rdhngvba bs gur pheir rkcerffrq va grezf bs $y \displaystyle{y}$
$c \displaystyle{c}$ naq $d \displaystyle{d}$ ner gur hccre naq ybjre l yvzvgf bs gur nern orvat ebgngrq

$d y \displaystyle{\left.{d}{y}\right.}$ fubjf gung gur nern vf orvat ebgngrq nobhg gur $y \displaystyle{y}$ -nkvf

Rknzcyr 4

Svaq gur ibyhzr bs gur fbyvq bs eribyhgvba trarengrq ol ebgngvat gur pheir $y = x^{3} \displaystyle{y}={x}^{3}$ orgjrra $y = 0 \displaystyle{y}={0}$ naq $y = 4 \displaystyle{y}={4}$ nobhg gur $y \displaystyle{y}$ -nkvf.

Gur tencu bs $y = x \displaystyle{y}={x}$ , jvgu gur nern haqre gur "pheir" orgjrra $x = 0 \displaystyle{x}={0}$ to $x = 2 \displaystyle{x}={2}$ funqrq.

Urapr, gur ibyhzr trarengrq pna or sbhaq hfvat gur sbezhyn sbe ibyhzr bs fbyvq bs eribyhgvba:

$Vol = π \int_{a}^{b} y^{2} d x \displaystyle\text{Vol}=\pi{\int_{{a}}^{{b}}}{y}^{2}{\left.{d}{x}\right.}$

$= π \int_{0}^{2} {(x)}^{2} d x \displaystyle=\pi{\int_{{0}}^{{2}}}{\left({x}\right)}^{2}{\left.{d}{x}\right.}$

$= π {[\frac{x^{3}}{3}]}_{0}^{2} \displaystyle=\pi{{\left[\frac{{{x}^{3}}}{{{3}}}\right]}_{{0}}^{{2}}}$

$= π [\frac{8}{3}] - π [0] \displaystyle=\pi{\left[\frac{{{8}}}{{{3}}}\right]}-\pi{\left[{0}\right]}$

$= \frac{8}{3} π {units}^{3} \displaystyle=\frac{8}{{3}}\pi\ \text{units}^{3}$

$\approx 8.378 {units}^{3} \displaystyle\approx{8.378}\ \text{units}^{3}$

Gur tencu bs $y = 2 x - x^{2} \displaystyle{y}={2}{x}-{x}^{2}$ , jvgu gur nern haqre gur pheir orgjrra $x = 0 \displaystyle{x}={0}$ to $x = 2 \displaystyle{x}={2}$ funqrq.

Gur yvar $y = 0 \displaystyle{y}={0}$ fvzcyl zrnaf gur $x \displaystyle{x}$ -nkvf.

Gur ibyhzr trarengrq vf:

$Vol = π \int_{a}^{b} y^{2} d x \displaystyle\text{Vol}=\pi{\int_{{a}}^{{b}}}{y}^{2}{\left.{d}{x}\right.}$

$= π \int_{0}^{2} {(2 x - x^{2})}^{2} d x \displaystyle=\pi{\int_{{0}}^{{2}}}{\left({2}{x}-{x}^{2}\right)}^{2}{\left.{d}{x}\right.}$

$= π \int_{0}^{2} (4 x^{2} - 4 x^{3} + x^{4}) d x \displaystyle=\pi{\int_{{0}}^{{2}}}{\left({4}{x}^{2}-{4}{x}^{3}+{x}^{4}\right)}{\left.{d}{x}\right.}$

$= π {[\frac{4 x^{3}}{3} - \frac{4 x^{4}}{4} + \frac{x^{5}}{5}]}_{0}^{2} \displaystyle=\pi{{\left[\frac{{{4}{x}^{3}}}{{{3}}}-\frac{{{4}{x}^{4}}}{{{4}}}+\frac{{{x}^{5}}}{{{5}}}\right]}_{{0}}^{{2}}}$

$= π [\frac{32}{3} - 16 + \frac{32}{5}] - π [0] \displaystyle=\pi{\left[\frac{32}{{3}}-{16}+\frac{32}{{5}}\right]}-\pi{\left[{0}\right]}$

$= \frac{16}{15} π {units}^{3} \displaystyle=\frac{{{16}}}{{{15}}}\pi\ \text{units}^{3}$

$\approx 3.351 {units}^{3} \displaystyle\approx{3.351}\ \text{units}^{3}$

Nccyvpngvbaf

1. Ibyhzr bs n jvar pnfx

N jvar pnfx unf n enqvhf ng gur gbc bs 30 pz naq n enqvhf ng gur zvqqyr bs 40 pz. Gur urvtug bs gur pnfx vf 1 z. Jung vf gur ibyhzr bs gur pnfx (va Y), nffhzvat gung gur funcr bs gur fvqrf vf cnenobyvp?

Jr arrq gb svaq gur rdhngvba bs n cnenobyn jvgu iregrk ng $(0, 40) \displaystyle{\left({0},{40}\right)}$ naq cnffvat guebhtu $(50, 30) \displaystyle{\left({50},{30}\right)}$ .

Jr hfr gur sbezhyn:

Fb gur rdhngvba bs gur fvqr bs gur oneery vf

$y = - \frac{x^{2}}{250} + 40 \displaystyle{y}=-\frac{{{x}^{2}}}{{{250}}}+{40}$

Jr arrq gb svaq gur ibyhzr bs gur pnfx juvpu vf trarengrq jura jr ebgngr guvf cnenobyn orgjrra x = -50 naq x = 50 nebhaq gur x-nkvf.

2. Ibyhzr bs n jngrezryba

N jngrezryba unf na ryyvcfbvqny funcr jvgu znwbe nkvf 28 pz naq zvabe nkvf 25 pz. Svaq vgf ibyhzr.

Uvfgbevpny Nccebnpu: Orsber pnyphyhf, bar jnl bs nccebkvzngvat gur ibyhzr jbhyq or gb fyvpr gur jngrezryba (fnl va 2 pz guvpx fyvprf) naq nqq hc gur ibyhzrf bs rnpu $s l i c e \displaystyle{s}{l}{i}{c}{e}$ hfvat $V = π r^{2} h . \displaystyle{V}=\pi{r}^{2}{h}.$

Vagrerfgvatyl, Nepuvzrqrf (gur bar jub snzbhfyl whzcrq bhg bs uvf ongu naq ena qbja gur fgerrg fubhgvat "Rherxn! V'ir tbg vg") hfrq guvf nccebnpu gb svaq ibyhzrf bs fcurerf nebhaq 200 OP. Gur grpuavdhr jnf nyzbfg sbetbggra hagvy gur rneyl 1700f jura pnyphyhf jnf qrirybcrq ol Arjgba naq Yrvoavm.

Jr frr ubj gb qb gur ceboyrz hfvat obgu nccebnpurf.

Ibyhzr hfvat uvfgbevpny zrgubq:

Orpnhfr gur zryba vf flzzrgevpny, jr pna jbex bhg gur ibyhzr bs bar unys bs gur zryba, naq gura qbhoyr bhe nafjre.

Gur enqvv sbe gur fyvprf sbe bar unys bs n cnegvphyne jngrezryba ner sbhaq sebz zrnfherzrag gb or:

$0, 6.4, 8.7, \displaystyle{0},{6.4},{8.7},$ $10.3, 11.3, \displaystyle{10.3},{11.3},$ $12.0, 12.4, 12.5. \displaystyle{12.0},{12.4},{12.5}.$

Gur nccebkvzngr ibyhzr sbe bar unys bs gur zryba hfvat fyvprf 2 pz guvpx jbhyq or:

$V_{half} = π \times [{6.4}^{2} + {8.7}^{2} + {10.3}^{2} + {11.3}^{2} \displaystyle\text{V}_{{\text{half}}}=\pi\times{\left[{6.4}^{2}+{8.7}^{2}+{10.3}^{2}+{11.3}^{2}\right.}$ $+ {12.0}^{2} \displaystyle+{12.0}^{2}$ $+ {12.4}^{2} \displaystyle+{12.4}^{2}$ $+ {12.5}^{2}] \times 2 \displaystyle{\left.+{12.5}^{2}\right]}\times{2}$

$= π \times 8040.44 \times 2 \displaystyle=\pi\times{8040.44}\times{2}$

$= 5054.4 \displaystyle={5054.4}$

Fb gur ibyhzr sbe gur jubyr jngrezryba vf nobhg

$5054.4 \times 2 = 10109 {cm}^{3} = 10.1 L \displaystyle{5054.4}\times{2}={10109}\ \text{cm}^{3}={10.1}\ \text{L}$ .

Va gur sbyybjvat dhrfgvba, jr frr ubj gb svaq gur "rknpg" inyhr hfvat gur ibyhzr bs fbyvq bs eribyhgvba sbezhyn.

Jr ner gbyq gur zryba vf na ryyvcfbvq. Jr arrq gb svaq gur rdhngvba bs gur pebff-frpgvbany ryyvcfr jvgu znwbe nkvf 28 pz naq zvabe nkvf 25 pz.

Jr hfr gur sbezhyn (sebz gur frpgvba ba ryyvcfrf):

$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 \displaystyle\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}={1}$

jurer a vf unys gur yratgu bs gur znwbe nkvf naq b vf unys gur yratgu bs gur zvabe nkvf.

Sbe gur ibyhzr sbezhyn, jr jvyy arrq gur rkcerffvba sbe y² naq vg vf rnfvre gb fbyir sbe guvf abj (orsber fhofgvghgvat bhe a naq b).

$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 \displaystyle\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}={1}$

$b^{2} x^{2} + a^{2} y^{2} = a^{2} b^{2} \displaystyle{b}^{2}{x}^{2}+{a}^{2}{y}^{2}={a}^{2}{b}^{2}$

$a^{2} y^{2} = a^{2} b^{2} - b^{2} x^{2} = b^{2} (a^{2} - x^{2}) \displaystyle{a}^{2}{y}^{2}={a}^{2}{b}^{2}-{b}^{2}{x}^{2}={b}^{2}{\left({a}^{2}-{x}^{2}\right)}$

$y^{2} = \frac{b^{2}}{a^{2}} (a^{2} - x^{2}) \displaystyle{y}^{2}=\frac{{{b}^{2}}}{{{a}^{2}}}{\left({a}^{2}-{x}^{2}\right)}$

Fvapr $a = 14 \displaystyle{a}={14}$ naq $b = 12.5 \displaystyle{b}={12.5}$ , jr unir:

$y^{2} = \frac{{12.5}^{2}}{14^{2}} (14^{2} - x^{2}) \displaystyle{y}^{2}=\frac{{{12.5}^{2}}}{{{14}^{2}}}{\left({14}^{2}-{x}^{2}\right)}$

$= 0.797 (196 - x^{2}) \displaystyle={0.797}{\left({196}-{x}^{2}\right)}$

ABGR: Gur a naq b gung jr ner hfvat sbe gur ryyvcfr sbezhyn ner abg gur fnzr a naq b jr hfr va gur vagrtengvba fgrc. Gurl ner pbzcyrgryl qvssrerag cnegf bs gur ceboyrz.

Hfvat guvf, jr pna abj svaq gur ibyhzr hfvat vagrtengvba. (Bapr ntnva jr svaq gur ibyhzr sbe unys naq gura qbhoyr vg ng gur raq).

$V_{half} = π \int_{0}^{14} y^{2} d x \displaystyle\text{V}_{{\text{half}}}=\pi{\int_{{0}}^{{14}}}{y}^{2}{\left.{d}{x}\right.}$

$= π \int_{0}^{14} 0.797 (196 - x^{2}) d x \displaystyle=\pi{\int_{{0}}^{{14}}}{0.797}{\left({196}-{x}^{2}\right)}{\left.{d}{x}\right.}$

$= 0.797 π \int_{0}^{14} (196 - x^{2}) d x \displaystyle={0.797}\pi{\int_{{0}}^{{14}}}{\left({196}-{x}^{2}\right)}{\left.{d}{x}\right.}$

$= 2.504 {[196 x - \frac{x^{3}}{3}]}_{0}^{14} \displaystyle={2.504}{{\left[{196}{x}-\frac{{{x}^{3}}}{{{3}}}\right]}_{{0}}^{{14}}}$

$= 2.504 [196 (14) - \frac{14^{3}}{3}] \displaystyle={2.504}{\left[{196}{\left({14}\right)}-\frac{{{14}^{3}}}{{{3}}}\right]}$

$= 2.504 \times 1829.33 \displaystyle={2.504}\times{1829.33}$

$= 4580.65 {cm}^{3} \displaystyle={4580.65}\ \text{cm}^{3}$

Fb gur jngrezryba'f gbgny ibyhzr vf $2 \times 4580.65 = 9161 {cm}^{3} \displaystyle{2}\times{4580.65}={9161}\ \text{cm}^{3}$ or $9.161 L \displaystyle{9.161}\ \text{L}$ . Guvf vf nobhg gur fnzr nf jung jr tbg ol fyvpvat gur jngrezryba naq nqqvat gur ibyhzr bs gur fyvprf.

$v = \frac{d s}{d t} \displaystyle{v}=\frac{{{d}{s}}}{{{\left.{d}{t}\right.}}}$

naq gur nppryrengvba sebz gur irybpvgl shapgvba (be qvfcynprzrag shapgvba), hfvat:

$a = \frac{d v}{d t} = \frac{d^{2} s}{{d t}^{2}} \displaystyle{a}=\frac{{{d}{v}}}{{\left.{d}{t}\right.}}=\frac{{{d}^{2}{s}}}{{{\left.{d}{t}\right.}^{2}}}$

Gurfr sbezhynr ner bayl nccebcevngr sbe erpgvyvarne zbgvba (v.r. irybpvgl naq nppryrengvba va n fgenvtug yvar). Guvf vf vanqrdhngr sbe zbfg erny fvghngvbaf, fb jr vagebqhpr urer gur pbaprcg bs pheivyvarne zbgvba, jurer na bowrpg vf zbivat va n cynar nybat n fcrpvsvrq pheirq cngu.

Jr trarenyyl rkcerff gur x naq y pbzcbaragf bs gur zbgvba nf shapgvbaf bs gvzr. Guvf sbez vf pnyyrq cnenzrgevp sbez.

Jr frr gung jr unir sbezrq n pvepyr, prager $(0, 0) \displaystyle{\left({0},{0}\right)}$ , enqvhf $1 \displaystyle{1}$ havg.

Abgvpr gung gur inevnoyr t qbrf abg nccrne va gur nkrf bs guvf tencu, whfg gur inevnoyrf x naq y.

Ubevmbagny naq Iregvpny Pbzcbaragf bs Irybpvgl

Gur ubevmbagny pbzcbarag bs gur irybpvgl vf jevggra:

$v_{x} = \frac{d x}{d t} \displaystyle{v}_{{x}}=\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}$

naq gur iregvpny pbzcbarag vf jevggra:

$v_{y} = \frac{d y}{d t} \displaystyle{v}_{{y}}=\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}$

Jr jnag gb svaq gur zntavghqr bs gur erfhygnag irybpvgl v bapr jr xabj gur ubevmbagny naq iregvpny pbzcbaragf. Jr hfr:

$v = \sqrt{{(v_{x})}^{2} + {(v_{y})}^{2}} \displaystyle{v}=\sqrt{{{\left({v}_{{x}}\right)}^{2}+{\left({v}_{{y}}\right)}^{2}}}$

Gur qverpgvba θ gung gur bowrpg vf zbivat va, vf sbhaq hfvat:

$\tan θ_{v} = \frac{v_{y}}{v_{x}} \displaystyle \tan{\ }\theta_{{v}}=\frac{{{v}_{{y}}}}{{{v}_{{x}}}}$

Rknzcyr 2

$\frac{d x}{d t} = 15 t^{2} \displaystyle\frac{{\left.{d}{x}\right.}}{{\left.{d}{t}\right.}}={15}{t}^{2}$

$\frac{d x}{d t} = v_{x} = 15 {(10)}^{2} = 1500 {ms}^{- 1} \displaystyle\frac{{\left.{d}{x}\right.}}{{\left.{d}{t}\right.}}={v}_{{x}}={15}{\left({10}\right)}^{2}={1500}\ \text{ms}^{ -{{1}}}$

$\frac{d y}{d t} = 8 t \displaystyle\frac{{\left.{d}{y}\right.}}{{\left.{d}{t}\right.}}={8}{t}$

$\frac{d y}{d t} = v_{y} = 8 (10) = 80 {ms}^{- 1} \displaystyle\frac{{\left.{d}{y}\right.}}{{\left.{d}{t}\right.}}={v}_{{y}}={8}{\left({10}\right)}={80}\ \text{ms}^{ -{{1}}}$

Fb gur zntavghqr bs gur irybpvgl jvyy or:

$v = \sqrt{{(v_{x})}^{2} + {(v_{y})}^{2}} \displaystyle{v}=\sqrt{{{\left({v}_{{x}}\right)}^{2}+{\left({v}_{{y}}\right)}^{2}}}$

$= \sqrt{1500^{2} + 80^{2}} \displaystyle=\sqrt{{{1500}^{2}+{80}^{2}}}$

$= 1502.1 {ms}^{- 1} \displaystyle={1502.1}\ \text{ms}^{ -{{1}}}$

Abj sbe gur qverpgvba bs gur irybpvgl (vg vf na natyr, eryngvir gb gur cbfvgvir x-nkvf):

$\tan θ_{v} = \frac{v_{y}}{v_{x}} = \frac{80}{1500} \displaystyle \tan{\ }\theta_{{v}}=\frac{{v}_{{y}}}{{v}_{{x}}}=\frac{80}{{1500}}$

Fb $θ_{v} = 0.053 \displaystyle\theta_{{v}}={0.053}$ enqvnaf $= {3.05}^{\circ} \displaystyle={3.05}^{\circ}$ .

Rknzcyr 3

$x = \frac{20 t}{2 t + 1} \displaystyle{x}=\frac{{{20}{t}}}{{{2}{t}+{1}}}$

naq

$y = 0.1 (t^{2} + t) \displaystyle{y}={0.1}{\left({t}^{2}+{t}\right)}$

Nppryrengvba bs n Obql va Pheivyvarne Zbgvba

Gur rkcerffvbaf sbe nppryrengvba ner irel fvzvyne gb gubfr sbe irybpvgl:

Ubevmbagny pbzcbarag bs nppryrengvba:

$a_{x} = \frac{d v_{x}}{d t} \displaystyle{a}_{{x}}=\frac{{{d}{v}_{{x}}}}{{{\left.{d}{t}\right.}}}$

Iregvpny pbzcbarag bs nppryrengvba:

$a_{y} = \frac{d v_{y}}{d t} \displaystyle{a}_{{y}}=\frac{{{d}{v}_{{y}}}}{{{\left.{d}{t}\right.}}}$

Zntavghqr bs $p (x) = 0 \displaystyle{p}{\left({x}\right)}={0}$ nppryrengvba:

$a = \sqrt{{(a_{x})}^{2} + {(a_{y})}^{2}} \displaystyle{a}=\sqrt{{{\left({a}_{{x}}\right)}^{2}+{\left({a}_{{y}}\right)}^{2}}}$

Qverpgvba bs nppryrengvba:

$\tan θ_{a} = \frac{a_{y}}{a_{x}} \displaystyle \tan{\ }\theta_{{a}}=\frac{{{a}_{{y}}}}{{{a}_{{x}}}}$

Rknzcyr 4

(vv) Svaq gur nppryrengvba bs gur pne ng $t = 3.0 \displaystyle{t}={3.0}$ frpbaqf.

Jr svaq gur erny (ubevmbagny) naq vzntvanel (iregvpny) pbzcbaragf va grezf bs r (gur yratgu bs gur irpgbe) naq θ (gur natyr znqr jvgu gur erny nkvf):

Sebz Clguntbenf, jr unir: $r^{2} = x^{2} + y^{2} \displaystyle{r}^{2}={x}^{2}+{y}^{2}$ naq onfvp gevtbabzrgel tvirf hf:

$\tan θ = \frac{y}{x} \displaystyle \tan{\ }\theta=\frac{y}{{x}}$ $x = r \cos θ \displaystyle{x}={r}\ \cos{\theta}$ $y = r \sin θ \displaystyle{y}={r}\ \sin{\theta}$

Zhygvcylvat gur ynfg rkcerffvba guebhtubhg ol $j \displaystyle{j}$ tvirf hf:

$y j = j r \sin θ \displaystyle{y}{j}={j}{r}\ \sin{\theta}$

Fb jr pna jevgr gur cbyne sbez bs n pbzcyrk ahzore nf:

$x + y j = r (\cos θ + j \sin θ) \displaystyle{x}+{y}{j}={r}{\left( \cos{\theta}+{j}\ \sin{\theta}\right)}$

r vf gur nofbyhgr inyhr (be zbqhyhf) bs gur pbzcyrk ahzore

θ vf gur nethzrag bs gur pbzcyrk ahzore.

Gurer ner gjb bgure jnlf bs jevgvat gur cbyne sbez bs n pbzcyrk ahzore:

$r cis θ \displaystyle{r}\ \text{cis}\ \theta$ [Guvf vf whfg n fubegunaq sbe $r (\cos θ + j \sin θ) \displaystyle{r}{\left( \cos{\theta}+{j}\ \sin{\theta}\right)}$ ]

$r ∠ θ \displaystyle{r}\ \angle\ \theta$ [zrnaf bapr ntnva, $r (\cos θ + j \sin θ) \displaystyle{r}{\left( \cos{\theta}+{j}\ \sin{\theta}\right)}$ ]

ABGR: Jura jevgvat n pbzcyrk ahzore va cbyne sbez, gur natyr θ pna or va QRTERRF be ENQVNAF.

Jr arrq gb svaq r naq θ.

$r = \sqrt{x^{2} + y^{2}} \displaystyle{r}=\sqrt{{{x}^{2}+{y}^{2}}}$

$= \sqrt{7^{2} + {(- 5)}^{2}} \displaystyle=\sqrt{{{7}^{2}+{\left(-{5}\right)}^{2}}}$

$= \sqrt{49 + 25} \displaystyle=\sqrt{{{49}+{25}}}$

$= \sqrt{74} \approx 8.6 \displaystyle=\sqrt{{{74}}}\approx{8.6}$

$α = \tan^{- 1} (\frac{y}{x}) \displaystyle\alpha={{\tan}^{{-{1}}}{\left(\frac{y}{{x}}\right)}}$

$= \tan^{- 1} (\frac{5}{7}) \displaystyle={{\tan}^{{-{1}}}{\left(\frac{5}{{7}}\right)}}$

$\approx {35.54}^{o} \displaystyle\approx{35.54}^\text{o}$

Abj, $7 - 5 j \displaystyle{7}-{5}{j}$ vf va gur sbhegu dhnqenag, fb

$θ = 360^{\circ} - {35.54}^{\circ} = {324.46}^{\circ} \displaystyle\theta={360}^{\circ}-{35.54}^{\circ}={324.46}^{\circ}$

Fb, rkcerffvat $7 - 5 j \displaystyle{7}-{5}{j}$ in cbyne sbez, jr unir:

$7 - 5 j \displaystyle{7}-{5}{j}$ $= 8.6 (\cos {324.5}^{\circ} + j \sin {324.5}^{\circ}) \displaystyle={8.6}{\left({ \cos{{324.5}}^{\circ}+}{j}\ \sin{{324.5}}^{\circ}\right)}$

Jr pbhyq nyfb jevgr guvf nafjre nf $7 - 5 j = 8.6 cis {324.5}^{\circ} \displaystyle{7}-{5}{j}={8.6}\ \text{cis}\ {324.5}^{\circ}$ .

Nyfb jr pbhyq jevgr: $7 - 5 j = 8.6 ∠ {324.5}^{\circ} \displaystyle{7}-{5}{j}={8.6}\angle{324.5}^{\circ}$

Rkcerff $3 (\cos 232^{\circ} + j \sin 232^{\circ}) \displaystyle{3}{\left({ \cos{{232}}^{\circ}+}{j} \sin{{232}}^{\circ}\right)}$ va erpgnathyne sbez.

Guvf vf ubj gur pbzcyrk ahzore ybbxf ba na Netnaq qvntenz. Gur qvfgnapr sebz gur bevtva vf $3 \displaystyle{3}$ naq gur natyr sebz gur cbfvgvir $R \displaystyle{R}$ nkvf vf $232^{\circ} \displaystyle{232}^{\circ}$ .

Gb trg gur erdhverq nafjre, jr fvzcyl zhygvcyl bhg gur rkcerffvba:

(v) $3 (\cos 232^{\circ} + j \sin 232^{\circ}) \displaystyle{3}{\left({ \cos{{232}}^{\circ}+}{j}\ \sin{{232}}^{\circ}\right)}$ $= 3 \cos 232^{\circ} + j (3 \sin 232^{\circ}) \displaystyle={3}\ { \cos{{232}}^{\circ}+}{j}{\left({3}\ \sin{{232}}^{\circ}\right)}$

$= - 1.85 - 2.36 j \displaystyle=-{1.85}-{2.36}{j}$

N arj ovg bs $\int_{- π}^{π} \sin^{2} (x) d x \displaystyle{\int_{{-\pi}}^{\pi}}{{\sin}^{2}{\left({x}\right)}}{\left.{d}{x}\right.}$ pbagrag.

$X_{C} = \frac{1}{2 π (60) (2.40 \times 10^{- 6})} \displaystyle{X}_{{C}}=\frac{1}{{{2}\pi{\left({60}\right)}{\left({2.40}\times{10}^{ -{{6}}}\right)}}}$

$= 1105.243 Ω \displaystyle={1105.243}\ \Omega$

Gura

$Z_{2} = 1110.7 ∠ - {84.32}^{\circ} Ω \displaystyle{Z}_{{2}}={1110.7}\angle-{84.32}^{\circ}\ \Omega$

Fb gur gbgny vzcrqnapr, Z_T, vf tvira ol:

$Z_{T} = \frac{Z_{1} Z_{2}}{Z_{1} + Z_{2}} \displaystyle{Z}_{{T}}=\frac{{{Z}_{{1}}{Z}_{{2}}}}{{{Z}_{{1}}+{Z}_{{2}}}}$

$= \frac{2449326.75 ∠ - {171.72}^{o}}{210 - 3308.188 j} \displaystyle=\frac{{{2449326.75}\angle-{171.72}^\text{o}}}{{{210}-{3308.188}{j}}}$

$= \frac{2449326.75 ∠ - {171.72}^{o}}{3314.85 ∠ - {86.37}^{o}} \displaystyle=\frac{{{2449326.75}\angle-{171.72}^\text{o}}}{{{3314.85}\angle-{86.37}^\text{o}}}$

$= 738.9 ∠ - {85.35}^{o} \displaystyle={738.9}\angle-{85.35}^\text{o}$

Guvf ynfg yvar va erpgnathyne sbez vf

Abj:

$I_{T} = \frac{V_{T}}{Z_{T}} \displaystyle{I}_{{T}}=\frac{{{V}_{{T}}}}{{{Z}_{{T}}}}$

$= \frac{150 ∠ 0^{o}}{738.9 ∠ - {85.35}^{o}} \displaystyle=\frac{{{150}\angle{0}^\text{o}}}{{{738.9}\angle-{85.35}^\text{o}}}$

$= 0.203 ∠ {85.35}^{o} \displaystyle={0.203}\angle{85.35}^\text{o}$

Fb gur gbgny pheerag gnxra sebz gur fhccyl vf $203 mA \displaystyle{203}\ \text{mA}$ naq gur cunfr natyr bs gur pheerag vf $\approx 85^{\circ} \displaystyle\approx{85}^{\circ}$ .

Rkrepvfrf

1. Ercerfrag $1 + j \sqrt{3} \displaystyle{1}+{j}\sqrt{{3}}$ tencuvpnyyl naq jevgr vg va cbyne sbez.

Erpnyy Buz'f ynj sbe cher erfvfgnaprf:

$V = I R \displaystyle{V}={I}{R}$

Va gur pnfr bs NP pvephvgf, jr ercerfrag gur vzcrqnapr (rssrpgvir erfvfgnapr) nf n pbzcyrk ahzore, Z. Gur havgf ner buzf ( $Ω \displaystyle\Omega$ ).

Va guvf pnfr, Buz'f Ynj orpbzrf:

Erpnyy nyfb, vs jr unir frireny erfvfgbef (R₁, R₂, R₃, R₄, ...) pbaarpgrq va cnenyyry, gura gur gbgny erfvfgnapr R_T, vf tvira ol:

$\frac{1}{R_{T}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}} + \dots \displaystyle\frac{1}{{{R}_{{T}}}}=\frac{1}{{R}_{{1}}}+\frac{1}{{R}_{{2}}}+\frac{1}{{R}_{{3}}}+\ldots$

Va gur pnfr bs NP pvephvgf, guvf orpbzrf:

$\frac{1}{Z_{T}} = \frac{1}{Z_{1}} + \frac{1}{Z_{2}} + \frac{1}{Z_{3}} + \dots \displaystyle\frac{1}{{{Z}_{{T}}}}=\frac{1}{{Z}_{{1}}}+\frac{1}{{Z}_{{2}}}+\frac{1}{{Z}_{{3}}}+\ldots$

Fvzcyr pnfr:

Vs jr unir 2 vzcrqnaprf Z₁ naq Z₂, pbaarpgrq va cnenyyry, gura gur gbgny erfvfgnapr Z_T, vf tvira ol

$\frac{1}{Z_{T}} = \frac{1}{Z_{1}} + \frac{1}{Z_{2}} \displaystyle\frac{1}{{{Z}_{{T}}}}=\frac{1}{{Z}_{{1}}}+\frac{1}{{Z}_{{2}}}$

Jr pna jevgr guvf nf:

$\frac{1}{Z_{T}} = \frac{Z_{2} + Z_{1}}{Z_{1} Z_{2}} \displaystyle\frac{1}{{{Z}_{{T}}}}=\frac{{{Z}_{{2}}+{Z}_{{1}}}}{{{Z}_{{1}}{Z}_{{2}}}}$

Svaqvat gur erpvcebpny bs obgu fvqrf tvirf hf:

$Z_{T} = \frac{Z_{1} Z_{2}}{Z_{1} + Z_{2}} \displaystyle{Z}_{{T}}=\frac{{{Z}_{{1}}{Z}_{{2}}}}{{{Z}_{{1}}+{Z}_{{2}}}}$

Rknzcyr 1

Svaq gur pbzovarq vzcrqnapr bs gur sbyybjvat pvephvg:

Pnyy gur vzcrqnapr tvira ol gur gbc cneg bs gur pvephvg Z₁ naq gur vzcrqnapr tvira ol gur obggbz cneg Z₂.

$Z_{T} = \frac{Z_{1} Z_{2}}{Z_{1} + Z_{2}} \displaystyle{Z}_{{T}}=\frac{{{Z}_{{1}}{Z}_{{2}}}}{{{Z}_{{1}}+{Z}_{{2}}}}$

$= \frac{(70 + 60 j) (40 - 25 j)}{(70 + 60 j) + (40 - 25 j)} \displaystyle=\frac{{{\left({70}+{60}{j}\right)}{\left({40}-{25}{j}\right)}}}{{{\left({70}+{60}{j}\right)}+{\left({40}-{25}{j}\right)}}}$

$= \frac{(70 + 60 j) (40 - 25 j)}{110 + 35 j} \displaystyle=\frac{{{\left({70}+{60}{j}\right)}{\left({40}-{25}{j}\right)}}}{{{110}+{35}{j}}}$

(Nqqvat pbzcyrk ahzoref fubhyq or qbar va erpgnathyne sbez.

Abj, jr pbaireg rirelguvat gb cbyne sbez naq gura zhygvcyl naq qvivqr nf sbyybjf):

$= \frac{(70 + 60 j) (40 - 25 j)}{110 + 35 j} \displaystyle=\frac{{{\left({70}+{60}{j}\right)}{\left({40}-{25}{j}\right)}}}{{{110}+{35}{j}}}$

$= \frac{(92.20 ∠ {40.60}^{o}) (47.17 ∠ - {32.01}^{o})}{115.4 ∠ {17.65}^{o}} \displaystyle=\frac{{{\left({92.20}\angle{40.60}^\text{o}\right)}{\left({47.17}\angle-{32.01}^\text{o}\right)}}}{{{115.4}\angle{17.65}^\text{o}}}$

(Jr qb gur cebqhpg ba gur gbc svefg.)

$= \frac{(92.20 \times 47.17) ∠ ({40.60}^{o} - {32.01}^{o})}{115.4 ∠ {17.65}^{o}} \displaystyle=\frac{{{\left({92.20}\times{47.17}\right)}\angle{\left({40.60}^\text{o}-{32.01}^\text{o}\right)}}}{{{115.4}\angle{17.65}^\text{o}}}$

$= \frac{4349.074 ∠ {8.59}^{o}}{115.4 ∠ {17.65}^{o}} \displaystyle=\frac{{{4349.074}\angle{8.59}^\text{o}}}{{{115.4}\angle{17.65}^\text{o}}}$

(Abj jr qb gur qvivfvba.)

$= \frac{4349.074}{115.4} ∠ ({8.59}^{o} - {17.65}^{o}) \displaystyle=\frac{{{4349.074}}}{{115.4}}\angle{\left({8.59}^\text{o}-{17.65}^\text{o}\right)}$

$= 37.69 ∠ - {9.06}^{o} \displaystyle={37.69}\angle-{9.06}^\text{o}$

(Jr pbaireg onpx gb erpgnathyne sbez.)

$= 37.22 - 5.93 j \displaystyle={37.22}-{5.93}{j}$

(Jura zhygvcylvat pbzcyrk ahzoref va cbyne sbez, jr zhygvcyl gur r grezf (gur ahzoref bhg gur sebag) naq nqq gur natyrf. Jura qvivqvat pbzcyrk ahzoref va cbyne sbez, jr qvivqr gur r grezf naq fhogenpg gur natyrf. Frr gur Cebqhpgf naq Dhbgvragf frpgvba sbe zber vasbezngvba.)

Fb jr pbapyhqr gung gur pbzovarq vzcrqnapr vf

$Z_{T} = 37 - 5.9 j Ω \displaystyle{Z}_{{T}}={37}-{5.9}{j}\ \Omega$

Rknzcyr 2

Svaq

n) gur gbgny vzcrqnapr

o) gur cunfr natyr

p) gur gbgny yvar pheerag

$Z_{T} = \frac{Z_{1} Z_{2}}{Z_{1} + Z_{2}} \displaystyle{Z}_{{T}}={\frac{{{Z}_{{1}}{Z}_{{2}}}}{{{Z}_{{1}}+{Z}_{{2}}}}}$

$= \frac{(200 - 40 j) (60 + 130 j)}{(200 - 40 j) + (60 + 130 j)} \displaystyle={\frac{{{\left({200}-{40}{j}\right)}{\left({60}+{130}{j}\right)}}}{{{\left({200}-{40}{j}\right)}+{\left({60}+{130}{j}\right)}}}}$

$= \frac{(200 - 40 j) (60 + 130 j)}{260 + 90 j} \displaystyle={\frac{{{\left({200}-{40}{j}\right)}{\left({60}+{130}{j}\right)}}}{{{260}+{90}{j}}}}$

$= \frac{(204.0 ∠ - {11.31}^{\circ}) (143.2 ∠ {65.22}^{\circ})}{(275.1 ∠ {19.09}^{\circ})} \displaystyle={\frac{{{\left({204.0}\angle-{11.31}^{\circ}\right)}{\left({143.2}\angle{65.22}^{\circ}\right)}}}{{{\left({275.1}\angle{19.09}^{\circ}\right)}}}}$

$= \frac{204.0 \times 143.2}{275.1} ∠ (- {11.31}^{\circ} + \displaystyle={\frac{{{204.0}\times{143.2}}}{{{275.1}}}}\angle{\left(-{11.31}^{\circ}+\right.}$ ${65.22}^{\circ} - \displaystyle{65.22}^{\circ}-$ ${19.09}^{\circ}) \displaystyle{\left.{19.09}^{\circ}\right)}$

$= 106.2 ∠ {34.82}^{\circ} \displaystyle={106.2}\angle{34.82}^{\circ}$

$= 87.18 + 60.64 j \displaystyle={87.18}+{60.64}{j}$

Fb jr pbapyhqr gung gur gbgny vzcrqnapr vf

$Z_{T} = 87.2 + 60.6 j Ω \displaystyle{Z}_{{T}}={87.2}+{60.6}{j}\ \Omega$

o) Jr frr sebz gur frpbaq ynfg yvar bs bhe ynfg nafjre gung gur cunfr natyr vf $\approx 35^{\circ} \displaystyle\approx{35}^{\circ}$ .

p) Gbgny yvar pheerag:

Jr hfr

gur snpg gung gur vzcrqnapr vf $106.2 ∠ {34.82}^{\circ} \displaystyle{106.2}\angle{34.82}^{\circ}$ naq
gur snpg gung gur ibygntr fhccyvrq vf $12 V = 12 ∠ 0^{\circ} V . \displaystyle{12}{V}={12}\angle{0}^{\circ}\ {V}.$

$I = \frac{V}{Z} \displaystyle{I}=\frac{V}{{Z}}$

$= \frac{12 ∠ 0^{\circ}}{106.2 ∠ {34.82}^{\circ}} \displaystyle={\frac{{{12}\angle{0}^{\circ}}}{{{106.2}\angle{34.82}^{\circ}}}}$

$= 0.113 ∠ - {34.82}^{\circ} A \displaystyle={0.113}\angle-{34.82}^{\circ}\text{A}$

Fbzr grkg $a = 0 \displaystyle{a}={0}$ .

Fbzr zber grkg $b = 0 \displaystyle{b}={0}$ .

Pbagrag $r = \sqrt{3 a^{2} + b^{2}} \displaystyle{r}=\sqrt{{{3}{a}^{2}+{b}^{2}}}$

Fbzr zber $acos 0 = \frac{π}{2} \displaystyle\text{acos }\ {0}=\frac{\pi}{{2}}$ pbagrag

Svavfu enaqbz $π \neq 3 \displaystyle\pi\ne{3}$ pbagrag.