Asosiy tarkibga o'tish

4.1-bo'lim: O'zgarish stavkalari

Ushbu bo'limning maqsadi bizga hosilalarning eng muhim ilovalaridan birini eslatishdir. Ya'ni, \(f'\left( x \right)\) \(f\left( x \right)\) ning o'zgarish tezligini ifodalaydi. Bu avvalgi bobda biz qayta-qayta ko'rgan dastur. Oldingi bobning deyarli har bir qismida hosilalarni qo'llash bilan bog'liq kamida bitta muammo bor edi. Ushbu ilova vaqti-vaqti bilan ushbu bobda paydo bo'lsa-da, biz ushbu bobdagi boshqa ilovalarga ko'proq e'tibor qaratamiz.

Shunday qilib, biz ushbu dastur haqida unutmasligimiz uchun bu erda hosilalarning o'zgarish tezligiga e'tibor qaratadigan qisqacha misollar to'plami. E'tibor bering, bu misollarning maqsadi oldingi bobda ko'rib chiqilgan materialni eslatish va bunday muammolarni qanday qilishni o'rgatish emas. Agar siz bunday misollarni qanday qilishni eslamasangiz, oldingi bobga qaytib, ko'rib chiqishingiz kerak bo'ladi.

1-misol Quyidagi funktsiya o'zgarmaydigan barcha nuqtalarni aniqlang. \[g\left( x \right) = 5 - 6x - 10\cos \left( {2x} \right)\]
Yechim

Birinchidan, biz funktsiyaning hosilasini olishimiz kerak.

\[g'\left( x \right) = - 6 + 20\sin \left( {2x} \right)\]

Endi, agar o'zgarish tezligi nolga teng bo'lsa, funktsiya o'zgarmaydi va shuning uchun bu savolga javob berish uchun hosila nolga teng ekanligini aniqlashimiz kerak. Shunday qilib, keling, buni nolga tenglashtiramiz va hal qilamiz.

\[ - 6 + 20\sin \left( {2x} \right) = 0\hspace{0.5in} \Rightarrow \hspace{0.5in}\sin \left( {2x} \right) = \frac{6}{{20}} = 0.3\]

Buning yechimi shundan iboratki,

\[\begin{alignat}{4}2x = & 0.3047 + 2\pi n & & \hspace{0.5in}\,\,\,\,{\mbox{OR}}\hspace{0.5in}\,\,\,\, & 2x = & 2.8369 + 2\pi n & & \hspace{0.25in}n = 0, \pm 1, \pm 2, \ldots \\ x = & 0.1524 + \pi n & & \hspace{0.5in}\,\,\,\,{\mbox{OR}}\hspace{0.5in}\,\,\, & x = & 1.4185 + \pi n & & \hspace{0.25in}n = 0, \pm 1, \pm 2, \ldots \end{alignat}\]

Agar trig tenglamalarni qanday yechish kerakligini eslamasangiz, tekshiring Trig tenglamalarni yechish Ko'rib chiqish bo'limidagi bo'limlar.

2-misol Quyidagi funksiya qayerda ortib, kamayib borayotganini aniqlang. \[A\left( t \right) = 27{t^5} - 45{t^4} - 130{t^3} + 150\]
Yechim

Birinchi misoldagi kabi, avval funksiyaning hosilasini olishimiz kerak.

\[A'\left( t \right) = 135{t^4} - 180{t^3} - 390{t^2} = 15{t^2}\left( {9{t^2} - 12t - 26} \right)\]

Keyinchalik, funktsiya qayerda o'zgarmasligini aniqlashimiz kerak. Bu yerda,

\[\begin{align*}t & = 0\\ t & = \frac{{12 \pm \sqrt {144 - 4\left( 9 \right)\left( { - 26} \right)} }}{{18}} = \frac{{12 \pm \sqrt {1080} }}{{18}} = \frac{{12 \pm 6\sqrt {30} }}{{18}} = \frac{{2 \pm \sqrt {30} }}{3} = - 1.159,\,\,\,\,2.492\end{align*}\]

So, the function is not changing at three values of \(t\). Va nihoyat, funktsiya qayerda ortib yoki kamayib borayotganini aniqlash uchun hosila ijobiy yoki salbiy qaerda ekanligini aniqlashimiz kerak. Eslatib o'tamiz, agar hosila musbat bo'lsa, funktsiya ortib borishi kerak, agar hosila manfiy bo'lsa, funktsiya kamayishi kerak. Quyidagi raqam qatori bu ma'lumotni beradi.

A number line from -3 to 4 with dashed vertical lines at approximately \(t = -1.159\), \(t = 0\) and \(t = 2.492\) dividing it into four regions. In the region to the left of -1.159 it is noted that \(A'\left( { - 2} \right) = 2040\) and so the derivative is positive there. In the region between -1.159 and 0 a box with an arrow pointing into the region notes that \(A'\left( { - 1} \right) = - 75\) and so the derivative is negative there. In the region between 0 and 2.492 it is noted that \(A'\left( 1 \right) = - 435\) and so the derivative is negative there. In the region to the right of 2.492 it is noted that \(A'\left( 3 \right) = 2565\) and so the derivative is positive there.

So, from this number line we can see that we have the following increasing and decreasing information.

\[{\mbox{Increasing :}}\,\, - \infty < t < - 1.159,\,\,\,2.492 < t < \infty \,\,\,\,\,\,\,\,{\mbox{Decreasing :}} - 1.159 < t < 0,\,\,\,0 < t < 2.492\]

Agar siz polinom va ratsional tengsizliklarni qanday yechish kerakligini eslay olmasangiz, "Ko'rib chiqish" bo'limidagi tegishli bo'limlarni ko'rib chiqing.

Nihoyat, biz unuta olmaymiz Tegishli tariflar muammolar.

3-misol Ikki mashina bir-biridan 500 milya masofada harakatlana boshlaydi. A avtomobili B avtomashinasidan g'arbda va sharqqa qarab harakatlana boshlaydi (i.e. towards Car B) at 35 mph and at the same time Car B starts driving south at 50 mph. After 3 hours of driving at what rate is the distance between the two cars changing? U ortib bormoqdami yoki kamaymoqdami?
Yechim

Bu erda qilish kerak bo'lgan birinchi narsa - vaziyatni ko'rsatadigan rasmning eskizini olish.

A sketch of the positions of the two cars. A horizontal line across the top is labeled 500 miles and runs from Car A’s initial position on the left to Car B’s initial position on the right. Car A’s current position is marked part of the way along this line, with the distance from its initial position labeled “Distance Driven By Car A” and the remaining distance to Car B’s initial position labeled \(x\). Car B is shown directly below its initial position with that vertical distance labeled \(y\). A diagonal line labeled \(z\) connects Car A’s current position to Car B, forming a right triangle with legs \(x\) and \(y\) and hypotenuse \(z\).

Bu rasmda \(y\) B avtomobili bosib o'tgan masofani, \(x\) esa A avtomobilini B avtomobilining dastlabki holatidan ajratib turadigan masofani, \(z\) esa ikkita mashinani ajratuvchi masofani ifodalaydi. After 3 hours driving time with have the following values of \(x\) and \(y\).

\[x = 500 - 35\left( 3 \right) = 395\hspace{0.5in}\hspace{0.25in}y = 50\left( 3 \right) = 150\]

We can use the Pythagorean theorem to find \(z\) at this time as follows,

\[{z^2} = {395^2} + {150^2} = 178525\hspace{0.5in} \Rightarrow \hspace{0.5in}z = \sqrt {178525} = 422.5222\]

Now, to answer this question we will need to determine \(z'\) given that \(x' = - 35\) and \(y' = 50\). Do you agree with the signs on the two given rates? Remember that a rate is negative if the quantity is decreasing and positive if the quantity is increasing.

Bu erda yana Pifagor teoremasidan foydalanishimiz mumkin. Birinchidan, uni yozib oling va \(x\), \(y\) va \(z\) vaqt o'tishi bilan o'zgarib borayotganini yodda tuting va shuning uchun tenglamadan foydalanib farqlang. Yashirin farqlash.

\[{z^2} = {x^2} + {y^2}\hspace{0.5in} \Rightarrow \hspace{0.5in}2zz' = 2xx' + 2yy'\]

Finally, all we need to do is cancel a two from everything, plug in for the known quantities and solve for \(z'\).

\[z'\left( {422.5222} \right) = \left( {395} \right)\left( { - 35} \right) + \left( {150} \right)\left( {50} \right)\hspace{0.25in} \Rightarrow \hspace{0.5in}z' = \frac{{ - 6325}}{{422.5222}} = - 14.9696\]

So, after three hours the distance between them is decreasing at a rate of 14.9696 mph.

Shunday qilib, biz ushbu bo'limda funktsiyaning hosilasi funktsiyaning o'zgarish tezligini beradi degan fikrdan foydalangan holda uchta "standart" masalani ko'rib chiqdik. Yuqorida aytib o'tilganidek, ushbu bobda o'zgarish tezligi g'oyasidan ko'ra ko'proq boshqa ilovalarga e'tibor qaratiladi, ammo biz bu ilovani unuta olmaymiz, chunki u juda muhim.