Work - 4-masala yechimi | MathModel.uz

6.6-bo'lim : Ish

4. Suv idishining uzunligi 15 fut bo'lib, tomonlari 2 fut uzunlikdagi (uchburchak nuqtasi to'g'ridan-to'g'ri pastga qaragan) teng qirrali uchburchak shaklida kesmaga ega. Tank 9 dyuym chuqurlikda suv bilan to'ldirilgan. Barcha suvni tankning yuqori qismiga quyish uchun zarur bo'lgan ish hajmini aniqlang. Suvning zichligi 62 lb/fut deb faraz qilaylik³.

Yechim

Xo'sh, keling, boshlaylik va $x = 0$ ni tankning pastki nuqtasi va idishdagi suv balandligini $x = \frac{9}{{12}} = \frac{3}{4}$ fut deb belgilaymiz (chunki boshqa barcha miqdorlar futda bo'lgani uchun biz buni ham futga aylantirdik). Bu shuni anglatadiki, biz ushbu muammo uchun $\left[ {0,\frac{3}{4}} \right]$ oralig'ida ishlaymiz.

Keyinchalik $\left[ {0,\frac{3}{4}} \right]$ oralig'ini $\Delta x$ kengligidagi $n$ subintervallariga ajratamiz va $x_i^*$ $i$ ning istalgan nuqtasi bo'lishiga ruxsat beramiz.^th subinterval bu erda $n = 1,2, \ldots ,n$. Har bir subinterval uchun biz ushbu subintervalga mos keladigan idishdagi suvni uzunligi 15 fut, kengligi $s_i^*$ va balandligi $\Delta x$ bo'lgan quti sifatida taxmin qilishimiz mumkin.

Mana tankning qisqacha eskizi. Qizil chiziq $i$-dagi idishdagi suvni taxminiy hisoblash uchun foydalanadigan qutini anglatadi.^th subinterval.

$A three dimensional sketch of the tank. It is a long trough 15 feet in length whose ends are triangles pointing straight down, and the water in the bottom is shaded in. A red strip running the length of the tank across the surface region represents the box being used to approximate the water in the $i$th subinterval.$

Tankning eskizi juda yaxshi va u tankni tasavvur qilishimizga yordam bersa-da, bizga tankning to'g'ridan-to'g'ri old tomonidagi eskizi kerak bo'ladi (i.e. tank uchun odatiy vertikal teng qirrali uchburchak kesim). Mana o'sha eskiz.

$A sketch of the tank seen from directly in front, so the cross-section is an equilateral triangle pointing straight down. The top of the triangle is marked as 2 ft across and the height of the triangle is marked as $\sqrt 3 $ ft. The water fills the bottom of the triangle to a depth of 9 inches and a red strip across the water represents the box being used to approximate the water in the $i$th subinterval, with its width labeled ${s_i}$ and its position labeled $x_i^*$.$

Qizil chiziq yana $i$-dagi suvni taxmin qilish uchun foydalanadigan qutini anglatadi.^th subinterval. Muammo bayonida ta'kidlanganidek, kesma teng tomonli uchburchak bo'lib, tomonlari 2 fut uzunlikda bo'ladi.

Biz yuqoridagi eskizga balandlikni kiritdik va bu to'g'ri burchakli uchburchakning asosiy triglaridan foydalanish oson. Mana, kesmaning yana bir eskizi.

$A sketch of the equilateral triangular cross-section split down the middle by its height. Each of the two upper interior angles is labeled $\frac{\pi }{3}$, the two halves of the top side are each labeled 1, the two slanted sides are each labeled 2, the two angles at the bottom point are each labeled $\frac{\pi }{6}$, and the vertical line down the middle is labeled $h$.$

Uchburchak teng qirrali uchburchak bo'lgani uchun biz uchburchakning har bir ichki burchagi $\frac{\pi }{3}$ bo'lishi kerakligini bilamiz va bizga har bir tomonning uzunligi 2 ga teng deb aytiladi. Uchburchakning balandligi ko'rsatilgandek uchburchakni ikkiga bo'luvchi chiziqdir. Uchburchakning har bir yarmi bir xil to'g'ri burchakli uchburchak bo'ladi va trig funktsiyalaridan foydalanib, biz uchburchakning balandligini tezda aniqlashimiz mumkin. Biz bu erda kosinusdan foydalanamiz.

\[\cos \left( {\frac{\pi }{6}} \right) = \frac{h}{2}\hspace{0.25in} \Rightarrow \hspace{0.25in}h = 2\cos \left( {\frac{\pi }{6}} \right) = \sqrt 3 \]

2-qadam

Biz $i$ dagi suv hajmini taxmin qilish uchun biz foydalanadigan suv qutisi hajmiga kerak bo'ladi.^th subinterval (1-bosqichdagi dastlabki ikkita rasmda qizil chiziq bilan ifodalangan).

Bizning taxminiy hajmimiz qutining hajmi va shuning uchun biz $i$ uchun hajm ekanligini bilamiz.^th subinterval bo'ladi,

\[{V_i} = \left( \mbox{length} \right)\left( \mbox{width} \right)\left( \mbox{height} \right) = \left( {15} \right)\left( {{s_i}} \right)\left( {\Delta x} \right) = 15\,{s_i}\,\Delta x\]

Oxir-oqibat bizga $x_i^*$ bo'yicha ovoz balandligi kerak bo'ladi va baxtimizga buni qilish juda oson.

1-bosqichdagi qizil chiziqli kesma eskizdan biz ikkita o'xshash uchburchaklar mavjudligini ko'ramiz (yaxshi, bizda uchtasi bor, lekin ulardan faqat ikkitasi kerak). Bizga kerak bo'lgan ikkitasi - kengligi 2 va balandligi $\sqrt 3 $ va kengligi ${s_i}$ bo'lgan uchburchak (i.e. uchburchak, uning tepasi qizil chiziq) va balandligi $x_i^*$. Bu ikki uchburchak bir-biriga o'xshash bo'lgani uchun, biz endi quyidagi ikkita nisbat teng bo'lishi kerak.

\[\frac{{{s_i}}}{{x_i^*}} = \frac{2}{{\sqrt 3 }}\hspace{0.25in} \Rightarrow \hspace{0.25in} {s_i} = \frac{2}{{\sqrt 3 }}x_i^*\]

Buni yuqoridagi hajm formulasiga ulab, biz quyidagilarni olamiz:

\[{V_i} = \frac{{30}}{{\sqrt 3 }}x_i^*\,\Delta x\]

3-qadam

Keyinchalik $i$ suvga ta'sir qiladigan tortishish kuchini engish uchun qancha kuch kerakligini bilishimiz kerak.^th subinterval. Bu taxminan 2-bosqichda topilgan suv hajmiga ta'sir qiluvchi tortishish kuchini engish uchun zarur bo'lgan majburiy bo'ladi. Chunki biz Britaniya tizimi bilan ishlayotganimiz uchun bu erda kuch,

\[{F_i} = {\mbox{density}} \times {V_i} \approx \left( {62} \right)\left( {\frac{{30}}{{\sqrt 3 }}x_i^*\,\Delta x} \right) = \frac{{1860}}{{\sqrt 3 }}x_i^*\,\Delta x\]

4-qadam

$i$-dagi suv hajmini oshirish uchun bizga zarur bo'lgan ish hajmi kerak bo'ladi^th tankning yuqori qismiga subinterval, i.e. uni $\sqrt 3 - x_i^*$ masofasiga ko'taring. Bu taxminan,

\[{W_i} \approx {F_i}\left( {\sqrt 3 - x_i^*} \right) = \frac{{1860}}{{\sqrt 3 }}x_i^*\left( {\sqrt 3 - x_i^*} \right)\Delta x\]

5-qadam

Barcha suvni idishning yuqori qismiga ko'tarish bo'yicha ishlarning umumiy miqdori $i = 1,2, \ldots n$ uchun barcha ${W_i}$larning taxminan yig'indisiga teng yoki,

\[W \approx \sum\limits_{i = 1}^n {\frac{{1860}}{{\sqrt 3 }}x_i^*\left( {\sqrt 3 - x_i^*} \right)\Delta x} \]

Kerakli aniq ish keyin $n \to \infty $ ruxsat berish orqali topiladi yoki,

\[W = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {\frac{{1860}}{{\sqrt 3 }}x_i^*\left( {\sqrt 3 - x_i^*} \right)\Delta x} \]

Biroq, bu faqat quyidagi aniq integralning ta'rifi,

\[W = \int_{0}^{{\frac{3}{4}}}{{\frac{{1860}}{{\sqrt 3 }}x\left( {\sqrt 3 - x} \right)\,dx}}\]

Keyin barcha suvni tankning yuqori qismiga quyish uchun zarur bo'lgan ish,

\[W = \int_{0}^{{\frac{3}{4}}}{{\frac{{1860}}{{\sqrt 3 }}x\left( {\sqrt 3 - x} \right)\,dx}} = \left. {\frac{{1860}}{{\sqrt 3 }}\left( {\frac{{\sqrt 3 }}{2}\,{x^2} - \frac{1}{3}{x^3}} \right)} \right|_0^{\frac{3}{4}} = \require{bbox} \bbox[2pt,border:1px solid black]{{{\mbox{372}}{\mbox{.112 ft - lbs}}}}\]