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## problem statement
Let $k$ be the coefficient of similarity transformation with the center at the origin. Is it true that point $A$ belongs to the image of plane $a$?
$A(-1 ; 1 ; 1)$
$a: 3 x-y+2 z+4=0$
$k=\frac{1}{2}$
|
## Solution
When transforming similarity with the center at the origin of the plane
$a: A x+B y+C z+D=0_{\text{and coefficient }} k$ transitions to the plane
$a^{\prime}: A x+B y+C z+k \cdot D=0$. We find the image of the plane $a$:
$a^{\prime}: 3 x-y+2 z+2=0$
Substitute the coordinates of point $A$ into the equation $a^{\prime}$:
$3 \cdot(-1)-1+2 \cdot 1+2=0$
$-3-1+2+2=0$
$0=0$
Since $0=0$, point $A$ belongs to the image of the plane $a$.
## Problem Kuznetsov Analytical Geometry 12-3
|
0
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Based on the definition of the derivative, find $f^{\prime}(0)$:
$f(x)=\left\{\begin{array}{c}\frac{\ln \left(1+2 x^{2}+x^{3}\right)}{x}, x \neq 0 ; \\ 0, x=0\end{array}\right.$
|
## Solution
By definition, the derivative at the point $x=0$:
$f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}$
Based on the definition, we find:
$$
\begin{aligned}
& f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim _{\Delta x \rightarrow 0} \frac{\frac{\ln \left(1+2 \Delta x^{2}+\Delta x^{3}\right)}{\Delta x}-0}{\Delta x}= \\
& =\lim _{\Delta x \rightarrow 0} \frac{\ln \left(1+2 \Delta x^{2}+\Delta x^{3}\right)}{\Delta x^{2}}=
\end{aligned}
$$
Using the substitution of equivalent infinitesimals:
$$
\ln \left(1+2 \Delta x^{2}+\Delta x^{3}\right) \sim 2 \Delta x^{2}+\Delta x^{3}, \text { as } \Delta x \rightarrow 0\left(2 \Delta x^{2}+\Delta x^{3} \rightarrow 0\right)
$$
We get:
$$
=\lim _{\Delta x \rightarrow 0} \frac{2 \Delta x^{2}+\Delta x^{3}}{\Delta x^{2}}=\lim _{\Delta x \rightarrow 0}(2+\Delta x)=2+0=2
$$
Thus, $f^{\prime}(0)=2$
## Problem Kuznetsov Differentiation $2-29$
|
2
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{\sqrt{n+2}-\sqrt[3]{n^{3}+2}}{\sqrt[7]{n+2}-\sqrt[5]{n^{5}+2}}$
|
Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{\sqrt{n+2}-\sqrt[3]{n^{3}+2}}{\sqrt[7]{n+2}-\sqrt[5]{n^{5}+2}}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n}\left(\sqrt{n+2}-\sqrt[3]{n^{3}+2}\right)}{\frac{1}{n}\left(\sqrt[7]{n+2}-\sqrt[5]{n^{5}+2}\right)}= \\
& =\lim _{n \rightarrow \infty} \frac{\sqrt{\frac{1}{n}+\frac{2}{n^{2}}}-\sqrt[3]{1+\frac{2}{n^{3}}}}{\sqrt[7]{\frac{1}{n^{6}}+\frac{2}{n^{7}}}-\sqrt[5]{1+\frac{2}{n^{5}}}}=\frac{\sqrt{0+0}-\sqrt[3]{1+0}}{\sqrt[7]{0+0}-\sqrt[5]{1+0}}=\frac{-1}{-1}=1
\end{aligned}
$$
## Problem Kuznetsov Limits 4-25
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} n^{3}\left(\sqrt[3]{n^{2}\left(n^{6}+4\right)}-\sqrt[3]{n^{8}-1}\right)$
|
## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} n^{3}\left(\sqrt[3]{n^{2}\left(n^{6}+4\right)}-\sqrt[3]{n^{8}-1}\right)= \\
& =\lim _{n \rightarrow \infty} \frac{n^{3}\left(\sqrt[3]{n^{2}\left(n^{6}+4\right)}-\sqrt[3]{n^{8}-1}\right)\left(\left(\sqrt[3]{n^{2}\left(n^{6}+4\right)}\right)^{2}+\sqrt[3]{n^{2}\left(n^{6}+4\right)} \sqrt[3]{n^{8}-1}+\left(\sqrt[3]{n^{8}-1}\right)^{2}\right)}{\left(\sqrt[3]{n^{2}\left(n^{6}+4\right)}\right)^{2}+\sqrt[3]{n^{2}\left(n^{6}+4\right)} \sqrt[3]{n^{8}-1}+\left(\sqrt[3]{n^{8}-1}\right)^{2}}=
\end{aligned}
$$
$$
\begin{aligned}
& =\lim _{n \rightarrow \infty} \frac{n^{3}\left(n^{2}\left(n^{6}+4\right)-\left(n^{8}-1\right)\right)}{\sqrt[3]{n^{4}\left(n^{6}+4\right)^{2}}+\sqrt[3]{n^{2}\left(n^{6}+4\right)\left(n^{8}-1\right)}+\sqrt[3]{\left(n^{8}-1\right)^{2}}}= \\
& =\lim _{n \rightarrow \infty} \frac{n^{3}\left(n^{8}+4 n^{2}-n^{8}+1\right)}{\sqrt[3]{n^{4}\left(n^{6}+4\right)^{2}}+\sqrt[3]{n^{2}\left(n^{6}+4\right)\left(n^{8}-1\right)}+\sqrt[3]{\left(n^{8}-1\right)^{2}}}= \\
& =\lim _{n \rightarrow \infty} \frac{n^{3}\left(4 n^{2}+1\right)}{\sqrt[3]{n^{4}\left(n^{6}+4\right)^{2}}+\sqrt[3]{n^{2}\left(n^{6}+4\right)\left(n^{8}-1\right)}+\sqrt[3]{\left(n^{8}-1\right)^{2}}}= \\
& =\lim _{n \rightarrow \infty} \frac{n^{-\frac{1}{3}} n^{3}\left(4 n^{2}+1\right)}{n^{-\frac{16}{3}}\left(\sqrt[3]{n^{4}\left(n^{6}+4\right)^{2}}+\sqrt[3]{n^{2}\left(n^{6}+4\right)\left(n^{8}-1\right)}+\sqrt[3]{\left(n^{8}-1\right)^{2}}\right)}= \\
& =\lim _{n \rightarrow \infty} \frac{n^{-\frac{7}{3}}\left(4 n^{2}+1\right)}{\sqrt[3]{n^{-16} n^{4}\left(n^{6}+4\right)^{2}}+\sqrt[3]{n^{-16} n^{2}\left(n^{6}+4\right)\left(n^{8}-1\right)}+\sqrt[3]{n^{-16}\left(n^{8}-1\right)^{2}}}= \\
& =\lim _{n \rightarrow \infty} \frac{4 n^{-\frac{1}{3}}+n^{-\frac{5}{3}}}{\sqrt[3]{\left(1+\frac{4}{n^{6}}\right)}+\sqrt[3]{\left(1+\frac{4}{n^{6}}\right)\left(1-\frac{1}{n^{8}}\right)}+\sqrt[3]{\left(1-\frac{1}{n^{8}}\right)^{2}}}= \\
& =\frac{4 \cdot 0+0}{\sqrt[3]{(1+0)^{2}}+\sqrt[3]{(1+0)(1-0)}+\sqrt[3]{(1-0)^{2}}}=\frac{0}{3}=0
\end{aligned}
$$
## Problem Kuznetsov Limits $5-25$
|
0
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{\sqrt{1-2 x+3 x^{2}}-(1+x)}{\sqrt[3]{x}}$
|
## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{\sqrt{1-2 x+3 x^{2}}-(1+x)}{\sqrt[3]{x}}= \\
& =\lim _{x \rightarrow 0} \frac{\left(\sqrt{1-2 x+3 x^{2}}-(1+x)\right)\left(\sqrt{1-2 x+3 x^{2}}+(1+x)\right)}{\sqrt[3]{x}\left(\sqrt{1-2 x+3 x^{2}}+(1+x)\right)}= \\
& =\lim _{x \rightarrow 0} \frac{1-2 x+3 x^{2}-(1+x)^{2}}{\sqrt[3]{x}\left(\sqrt{1-2 x+3 x^{2}}+1+x\right)}=\lim _{x \rightarrow 0} \frac{1-2 x+3 x^{2}-\left(1+2 x+x^{2}\right)}{\sqrt[3]{x}\left(\sqrt{1-2 x+3 x^{2}}+1+x\right)}= \\
& =\lim _{x \rightarrow 0} \frac{1-2 x+3 x^{2}-1-2 x-x^{2}}{\sqrt[3]{x}\left(\sqrt{1-2 x+3 x^{2}}+1+x\right)}=\lim _{x \rightarrow 0} \frac{-4 x+2 x^{2}}{\sqrt[3]{x}\left(\sqrt{1-2 x+3 x^{2}}+1+x\right)}= \\
& =\lim _{x \rightarrow 0} \frac{x(2 x-4)}{\sqrt[3]{x}\left(\sqrt{1-2 x+3 x^{2}}+1+x\right)}=\lim _{x \rightarrow 0} \frac{\sqrt[3]{x^{2}}(2 x-4)}{\sqrt{1-2 x+3 x^{2}}+1+x}= \\
& =\frac{\sqrt[3]{0^{2}}(2 \cdot 0-4)}{\sqrt{1-2 \cdot 0+3 \cdot 0^{2}}+1+0}=\frac{0}{1+1}=0
\end{aligned}
$$
## Problem Kuznetsov Limits 11-25
|
0
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{\sin ^{2} x-\tan^{2} x}{x^{4}}$
|
## Solution
We will use the substitution of equivalent infinitesimals:
$$
\begin{aligned}
& \sin x \sim x, \text { as } x \rightarrow 0 \\
& \operatorname{tg} x \sim x, \text { as } x \rightarrow 0
\end{aligned}
$$
We get:
$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{\sin ^{2} x-\operatorname{tg}^{2} x}{x^{4}}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow 0} \frac{\frac{\sin ^{2} x \cdot \cos ^{2} x}{\cos ^{2} x}-\operatorname{tg}^{2} x}{x^{4}}= \\
& =\lim _{x \rightarrow 0} \frac{\operatorname{tg}^{2} x \cdot \cos ^{2} x-\operatorname{tg}^{2} x}{x^{4}}=\lim _{x \rightarrow 0} \frac{\operatorname{tg}^{2} x\left(\cos ^{2} x-1\right)}{x^{4}}=
\end{aligned}
$$
$$
=\lim _{x \rightarrow 0} \frac{x^{2}\left(-\sin ^{2} x\right)}{x^{4}}=\lim _{x \rightarrow 0} \frac{x^{2}\left(-x^{2}\right)}{x^{4}}=\lim _{x \rightarrow 0} \frac{-1}{1}=-1
$$
## Problem Kuznetsov Limits 12-25
|
-1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow 0}\left(e^{x}+x\right)^{\cos x^{4}}
$$
|
## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 0}\left(e^{x}+x\right)^{\cos x^{4}}=\left(e^{0}+0\right)^{\cos 0^{4}}= \\
& =(1+0)^{\cos 0}=1^{1}=1
\end{aligned}
$$
## Problem Kuznetsov Limits 18-25
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 2}(\cos \pi x)^{\tan(x-2)}$
|
## Solution
$\lim _{x \rightarrow 2}(\cos \pi x)^{\operatorname{tg}(x-2)}=(\cos (\pi \cdot 2))^{\operatorname{tg}(2-2)}=(\cos 2 \pi)^{\operatorname{tg} 0}=1^{0}=1$
## Problem Kuznetsov Limits 20-25
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Task Condition
Determine the work (in joules) performed when lifting a satellite from the Earth's surface to a height of $H$ km. The mass of the satellite is $m$ tons, the radius of the Earth $R_{3}=6380$ km. The acceleration due to gravity $g$ at the Earth's surface is taken to be $10 \mathrm{~m} / \mathrm{c}^{2}$.
$$
m=7.0 \mathrm{t}, H=200 \text { km. }
$$
|
## Solution
By definition, the elementary work $\Delta A=F(x) \Delta x$, where $F(x)=G \cdot \frac{m \cdot M}{\left(R_{3}+x\right)^{2}} ; G=6.67 \cdot 10^{-11} \mathrm{H}^{*} \mathrm{m}^{*} \mathrm{m} /($ kg*kg)
\[
\begin{aligned}
& F_{x}=G \cdot \frac{m \cdot M}{\left(R_{3}+x\right)^{2}}-\text { force of attraction at height } x \\
& F_{0}=G \cdot \frac{m \cdot M}{R_{3}^{2}}=m g-\text { force of attraction at the Earth's surface } \\
& d A=\frac{m \cdot g \cdot R_{3}^{2}}{\left(R_{3}+x\right)^{2}} d x \Rightarrow \\
& \Rightarrow A=\int_{0}^{H} \frac{m \cdot g \cdot R_{3}^{2}}{\left(R_{3}+x\right)^{2}} d\left(R_{3}+x\right)=-\left.m \cdot g \cdot R_{3}^{2} \cdot \frac{1}{R_{3}+x}\right|_{0} ^{H}=m \cdot g \cdot R_{3}^{2} \cdot\left(\frac{1}{R_{3}}-\frac{1}{R_{3}+H}\right)= \\
& =7 \cdot 10^{3}[\mathrm{~kg}] \cdot 10\left[\frac{M}{c^{2}}\right) \cdot\left(6380 \cdot 10^{3}[m]\right)^{2} \cdot\left(\frac{1}{6380 \cdot 10^{3}[m]}-\frac{1}{6380 \cdot 10^{3}[m]+200 \cdot 10^{3}[m]}\right)=13574468085 \\
& \text { J] }
\end{aligned}
\]
Source — "http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\�_\�\�\�\�\�\� \%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD\%D1\%82\%D0\%B5\%D0\%B3\%D1\%80\%D0\%B0\%D0\%BB $\%$ D1\%8B_22-11"
Categories: Kuznetsov Problem Book Integrals Problem $22 \mid$ Integrals
Ukrainian Banner Network
- Last modified on this page: 10:45, 31 May 2010.
- Content is available under CC-BY-SA 3.0.
Created by GeeTeatoo
## Problem Kuznetsov Integrals 22-12
## Material from PlusPi
|
13574468085
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Task Condition
Determine the work (in joules) performed when lifting a satellite from the Earth's surface to a height of $H$ km. The mass of the satellite is $m$ tons, the radius of the Earth $R_{3}=6380$ km. The acceleration due to gravity $g$ at the Earth's surface is taken to be $10 \mathrm{~m} / \mathbf{c}^{2}$.
$$
m=6.0 \mathrm{t}, H=300 \text { km. }
$$
|
## Solution
By definition, the elementary work $\Delta A=F(x) \Delta x$, where $F(x)=G \cdot \frac{m \cdot M}{\left(R_{3}+x\right)^{2}} ; G=6.67 \cdot 10^{-11} \mathrm{H}^{*} \mathrm{m}^{*} \mathrm{m} /($ kg*kg)
\[
\begin{aligned}
& F_{x}=G \cdot \frac{m \cdot M}{\left(R_{3}+x\right)^{2}}-\text { force of attraction at height } x \\
& F_{0}=G \cdot \frac{m \cdot M}{R_{3}^{2}}=m g-\text { force of attraction at the Earth's surface } \\
& d A=\frac{m \cdot g \cdot R_{3}^{2}}{\left(R_{3}+x\right)^{2}} d x \Rightarrow \\
& \Rightarrow A=\int_{0}^{H} \frac{m \cdot g \cdot R_{3}^{2}}{\left(R_{3}+x\right)^{2}} d\left(R_{3}+x\right)=-\left.m \cdot g \cdot R_{3}^{2} \cdot \frac{1}{R_{3}+x}\right|_{0} ^{H}=m \cdot g \cdot R_{3}^{2} \cdot\left(\frac{1}{R_{3}}-\frac{1}{R_{3}+H}\right)= \\
& =6 \cdot 10^{3}[\mathrm{~kg}] \cdot 10\left[\frac{M}{c^{2}}\right) \cdot\left(6380 \cdot 10^{3}[m]\right)^{2} \cdot\left(\frac{1}{6380 \cdot 10^{3}[m]}-\frac{1}{6380 \cdot 10^{3}[m]+300 \cdot 10^{3}[m]}\right)=17191616766 \\
& \text { J] }
\end{aligned}
\]
Source — "http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\�_\�\�\�\�\�\� \%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD\%D1\%82\%D0\%B5\%D0\%B3\%D1\%80\%D0\%B0\%D0\%BB $\%$ D1\%8B_22-13"
Categories: Kuznetsov Problem Book Integrals Problem $22 \mid$ Integrals
Ukrainian Banner Network
- Last modified on this page: 10:47, 31 May 2010.
- Content is available under CC-BY-SA 3.0.
Created by GeeTeatoo
## Problem Kuznetsov Integrals 22-14
## Material from PlusPi
|
17191616766
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Task Condition
Determine the work (in joules) performed when lifting a satellite from the Earth's surface to a height of $H$ km. The mass of the satellite is $m$ tons, the radius of the Earth $R_{3}=6380$ km. The acceleration due to gravity $g$ at the Earth's surface is taken to be $10 \mathrm{~m} / \mathbf{c}^{2}$.
$$
m=6.0 \mathrm{t}, H=350 \text { km. }
$$
|
## Solution
By definition, the elementary work $\Delta A=F(x) \Delta x$, where $F(x)=G \cdot \frac{m \cdot M}{\left(R_{3}+x\right)^{2}} ; G=6.67 \cdot 10^{-11} \mathrm{H}^{*} \mathrm{m}^{*} \mathrm{m} /($ kg*kg)
\[
\begin{aligned}
& F_{x}=G \cdot \frac{m \cdot M}{\left(R_{3}+x\right)^{2}}-\text { force of attraction at height } x \\
& F_{0}=G \cdot \frac{m \cdot M}{R_{3}^{2}}=m g-\text { force of attraction at the Earth's surface } \\
& d A=\frac{m \cdot g \cdot R_{3}^{2}}{\left(R_{3}+x\right)^{2}} d x \Rightarrow \\
& \Rightarrow A=\int_{0}^{H} \frac{m \cdot g \cdot R_{3}^{2}}{\left(R_{3}+x\right)^{2}} d\left(R_{3}+x\right)=-\left.m \cdot g \cdot R_{3}^{2} \cdot \frac{1}{R_{3}+x}\right|_{0} ^{H}=m \cdot g \cdot R_{3}^{2} \cdot\left(\frac{1}{R_{3}}-\frac{1}{R_{3}+H}\right)= \\
& =6 \cdot 10^{3}[\mathrm{~kg}] \cdot 10\left[\frac{M}{c^{2}}\right] \cdot\left(6380 \cdot 10^{3}[m]\right)^{2} \cdot\left(\frac{1}{6380 \cdot 10^{3}[m]}-\frac{1}{6380 \cdot 10^{3}[m]+350 \cdot 10^{3}[m]}\right)=19907875186 \\
& \text { J] }
\end{aligned}
\]
Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\�_\�\�\�\�\�\� \%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD\%D1\%82\%D0\%B5\%D0\%B3\%D1\%80\%D0\%B0\%D0\%BB $\%$ D1\%8B_22-14)
Categories: Kuznetsov Problem Book Integrals Problem $22 \mid$ Integrals
Ukrainian Banner Network
- Last modified on this page: 10:48, 31 May 2010.
- Content is available under CC-BY-SA 3.0.
Created by GeeTeatoo
## Problem Kuznetsov Integrals $22-15$
## Material from PlusPi
|
19907875186
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Task Condition
Determine the work (in joules) performed when lifting a satellite from the Earth's surface to a height of $H$ km. The mass of the satellite is $m$ tons, the radius of the Earth $R_{3}=6380$ km. The acceleration due to gravity $g$ at the Earth's surface is taken to be $10 \mathrm{~m} / \mathbf{c}^{2}$.
$$
m=5,0 \mathrm{t}, H=400 \text { km. }
$$
|
## Solution
By definition, the elementary work $\Delta A=F(x) \Delta x$, where $F(x)=G \cdot \frac{m \cdot M}{\left(R_{3}+x\right)^{2}} ; G=6.67 \cdot 10^{-11} \mathrm{H}^{*} \mathrm{m}^{*} \mathrm{m} /($ kg*kg)
\[
\begin{aligned}
& F_{x}=G \cdot \frac{m \cdot M}{\left(R_{3}+x\right)^{2}}-\text { force of attraction at height } x \\
& F_{0}=G \cdot \frac{m \cdot M}{R_{3}^{2}}=m g-\text { force of attraction at the Earth's surface } \\
& d A=\frac{m \cdot g \cdot R_{3}^{2}}{\left(R_{3}+x\right)^{2}} d x \Rightarrow \\
& \Rightarrow A=\int_{0}^{H} \frac{m \cdot g \cdot R_{3}^{2}}{\left(R_{3}+x\right)^{2}} d\left(R_{3}+x\right)=-\left.m \cdot g \cdot R_{3}^{2} \cdot \frac{1}{R_{3}+x}\right|_{0} ^{H}=m \cdot g \cdot R_{3}^{2} \cdot\left(\frac{1}{R_{3}}-\frac{1}{R_{3}+H}\right)= \\
& =5 \cdot 10^{3}[\mathrm{~kg}] \cdot 10\left[\frac{m}{s^{2}}\right] \cdot\left(6380 \cdot 10^{3}[m]\right)^{2} \cdot\left(\frac{1}{6380 \cdot 10^{3}[m]}-\frac{1}{6380 \cdot 10^{3}[m]+400 \cdot 10^{3}[m]}\right)=18820058997 \\
& \text { J] }
\end{aligned}
\]
Source — "http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\�_\�\�\�\�\�\� \%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD\%D1\%82\%D0\%B5\%D0\%B3\%D1\%80\%D0\%B0\%D0\%BB $\%$ D1\%8B_22-15\%
Categories: Kuznetsov Problem Book Integrals Problem $22 \mid$ Integrals
Ukrainian Banner Network
- Last modified on this page: 10:50, 31 May 2010.
- Content is available under CC-BY-SA 3.0.
Created by GeeTeatoo
## Problem Kuznetsov Integrals 22-16
## Material from PlusPi
|
18820058997
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Task Condition
Determine the work (in joules) performed when lifting a satellite from the Earth's surface to a height of $H$ km. The mass of the satellite is $m$ tons, the radius of the Earth $R_{3}=6380$ km. The acceleration due to gravity $g$ at the Earth's surface is taken to be $10 \mathrm{~m} / \mathrm{c}^{2}$.
$$
m=5,0 \mathrm{t}, H=450 \text { km. }
$$
|
## Solution
By definition, the elementary work $\Delta A=F(x) \Delta x$, where $F(x)=G \cdot \frac{m \cdot M}{\left(R_{3}+x\right)^{2}} ; G=6.67 \cdot 10^{-11} \mathrm{H}^{*} \mathrm{m}^{*} \mathrm{m} /($ kg*kg)
$$
\begin{aligned}
& F_{x}=G \cdot \frac{m \cdot M}{\left(R_{3}+x\right)^{2}}-\text { force of attraction at height } x \\
& F_{0}=G \cdot \frac{m \cdot M}{R_{3}^{2}}=m g-\text { force of attraction at the Earth's surface } \\
& d A=\frac{m \cdot g \cdot R_{3}^{2}}{\left(R_{3}+x\right)^{2}} d x \Rightarrow \\
& \Rightarrow A=\int_{0}^{H} \frac{m \cdot g \cdot R_{3}^{2}}{\left(R_{3}+x\right)^{2}} d\left(R_{3}+x\right)=-\left.m \cdot g \cdot R_{3}^{2} \cdot \frac{1}{R_{3}+x}\right|_{0} ^{H}=m \cdot g \cdot R_{3}^{2} \cdot\left(\frac{1}{R_{3}}-\frac{1}{R_{3}+H}\right)= \\
& =5 \cdot 10^{3}[\mathrm{~kg}] \cdot 10\left[\frac{M}{c^{2}}\right) \cdot\left(6380 \cdot 10^{3}[m]\right)^{2} \cdot\left(\frac{1}{6380 \cdot 10^{3}[m]}-\frac{1}{6380 \cdot 10^{3}[m]+450 \cdot 10^{3}[m]}\right)=21017569546 \\
& \text { J] }
\end{aligned}
$$
Source — "http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\�_\�\�\�\�\�\� \%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD\%D1\%82\%D0\%B5\%D0\%B3\%D1\%80\%D0\%B0\%D0\%BB $\%$ D1\%8B_22-16"
Categories: Kuznetsov Problem Book Integrals Problem 22 | Integrals
Ukrainian Banner Network
- Last modified on this page: 10:38, 31 May 2010.
- Content is available under CC-BY-SA 3.0.
Created by GeeTeatoo
## Problem Kuznetsov Integrals $22-17$
## Material from PlusPi
|
21017569546
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Task Condition
Determine the work (in joules) performed when lifting a satellite from the Earth's surface to a height of $H$ km. The mass of the satellite is $m$ tons, the radius of the Earth $R_{3}=6380$ km. The acceleration due to gravity $g$ at the Earth's surface is taken to be $10 \mathrm{~m} / \mathbf{c}^{2}$.
$$
m=4.0 \mathrm{t}, H=500 \text { km. }
$$
|
## Solution
By definition, the elementary work $\Delta A=F(x) \Delta x$, where $F(x)=G \cdot \frac{m \cdot M}{\left(R_{3}+x\right)^{2}} ; G=6.67 \cdot 10^{-11} \mathrm{H}^{*} \mathrm{m}^{*} \mathrm{m} /($ kg*kg)
\[
\begin{aligned}
& F_{x}=G \cdot \frac{m \cdot M}{\left(R_{3}+x\right)^{2}}-\text { force of attraction at height } x \\
& F_{0}=G \cdot \frac{m \cdot M}{R_{3}^{2}}=m g-\text { force of attraction at the Earth's surface } \\
& d A=\frac{m \cdot g \cdot R_{3}^{2}}{\left(R_{3}+x\right)^{2}} d x \Rightarrow \\
& \Rightarrow A=\int_{0}^{H} \frac{m \cdot g \cdot R_{3}^{2}}{\left(R_{3}+x\right)^{2}} d\left(R_{3}+x\right)=-\left.m \cdot g \cdot R_{3}^{2} \cdot \frac{1}{R_{3}+x}\right|_{0} ^{H}=m \cdot g \cdot R_{3}^{2} \cdot\left(\frac{1}{R_{3}}-\frac{1}{R_{3}+H}\right)= \\
& =4 \cdot 10^{3}[\mathrm{~kg}] \cdot 10\left[\frac{m}{s^{2}}\right] \cdot\left(6380 \cdot 10^{3}[m]\right)^{2} \cdot\left(\frac{1}{6380 \cdot 10^{3}[m]}-\frac{1}{6380 \cdot 10^{3}[m]+500 \cdot 10^{3}[m]}\right)=18546511628 \\
& \text { J] }
\end{aligned}
\]
Source — "http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\�_\�\�\�\�\�\� \%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD\%D1\%82\%D0\%B5\%D0\%B3\%D1\%80\%D0\%B0\%D0\%BB $\%$ D1\%8B_22-17"
Categories: Kuznetsov Problem Book Integrals Problem $22 \mid$ Integrals
Ukrainian Banner Network
- Last modified on this page: 10:53, 31 May 2010.
- Content is available under CC-BY-SA 3.0.
Created by GeeTeatoo
## Problem Kuznetsov Integrals $22-18$
## Material from PlusPi
|
18546511628
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Task Condition
Determine the work (in joules) performed when lifting a satellite from the Earth's surface to a height of $H$ km. The mass of the satellite is $m$ tons, the radius of the Earth $R_{3}=6380$ km. The acceleration due to gravity $g$ at the Earth's surface is taken to be $10 \mathrm{~m} / \mathbf{c}^{2}$.
$$
m=4.0 \mathrm{t}, H=550 \text { km. }
$$
|
## Solution
By definition, the elementary work $\Delta A=F(x) \Delta x$, where $F(x)=G \cdot \frac{m \cdot M}{\left(R_{3}+x\right)^{2}} ; G=6.67 \cdot 10^{-11} \mathrm{H}^{*} \mathrm{m}^{*} \mathrm{m} /($ kg*kg)
\[
\begin{aligned}
& F_{x}=G \cdot \frac{m \cdot M}{\left(R_{3}+x\right)^{2}}-\text { force of attraction at height } x \\
& F_{0}=G \cdot \frac{m \cdot M}{R_{3}^{2}}=m g-\text { force of attraction at the Earth's surface } \\
& d A=\frac{m \cdot g \cdot R_{3}^{2}}{\left(R_{3}+x\right)^{2}} d x \Rightarrow \\
& \Rightarrow A=\int_{0}^{H} \frac{m \cdot g \cdot R_{3}^{2}}{\left(R_{3}+x\right)^{2}} d\left(R_{3}+x\right)=-\left.m \cdot g \cdot R_{3}^{2} \cdot \frac{1}{R_{3}+x}\right|_{0} ^{H}=m \cdot g \cdot R_{3}^{2} \cdot\left(\frac{1}{R_{3}}-\frac{1}{R_{3}+H}\right)= \\
& =4 \cdot 10^{3}[\mathrm{~kg}] \cdot 10\left[\frac{m}{s^{2}}\right] \cdot\left(6380 \cdot 10^{3}[m]\right)^{2} \cdot\left(\frac{1}{6380 \cdot 10^{3}[m]}-\frac{1}{6380 \cdot 10^{3}[m]+550 \cdot 10^{3}[\mathrm{~m}]}\right)=20253968254[ \\
& \text { J }]
\end{aligned}
\]
Source — "http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\�_\�\�\�\�\�\� \%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD\%D1\%82\%D0\%B5\%D0\%B3\%D1\%80\%D0\%B0\%D0\%BB $\%$ D1\%8B_22-18"
Categories: Kuznetsov Problem Book Integrals Problem $22 \mid$ Integrals
Ukrainian Banner Network
- Last modified on this page: 10:54, 31 May 2010.
- Content is available under CC-BY-SA 3.0.
Created by GeeTeatoo
## Problem Kuznetsov Integrals 22-19
## Material from PlusPi
|
20253968254
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Task Condition
Determine the work (in joules) performed when lifting a satellite from the Earth's surface to a height of $H$ km. The mass of the satellite is $m$ tons, the radius of the Earth $R_{3}=6380$ km. The acceleration due to gravity $g$ at the Earth's surface is taken to be $10 \mathrm{~m} / \mathbf{c}^{2}$.
$$
m=3.0 \mathrm{t}, H=600 \text { km. }
$$
|
## Solution
By definition, the elementary work $\Delta A=F(x) \Delta x$, where $F(x)=G \cdot \frac{m \cdot M}{\left(R_{3}+x\right)^{2}} ; G=6.67 \cdot 10^{-11} \mathrm{H}^{*} \mathrm{m}^{*} \mathrm{m} /($ kg*kg)
\[
\begin{aligned}
& F_{x}=G \cdot \frac{m \cdot M}{\left(R_{3}+x\right)^{2}}-\text { force of attraction at height } x \\
& F_{0}=G \cdot \frac{m \cdot M}{R_{3}^{2}}=m g-\text { force of attraction at the Earth's surface } \\
& d A=\frac{m \cdot g \cdot R_{3}^{2}}{\left(R_{3}+x\right)^{2}} d x \Rightarrow \\
& \Rightarrow A=\int_{0}^{H} \frac{m \cdot g \cdot R_{3}^{2}}{\left(R_{3}+x\right)^{2}} d\left(R_{3}+x\right)=-\left.m \cdot g \cdot R_{3}^{2} \cdot \frac{1}{R_{3}+x}\right|_{0} ^{H}=m \cdot g \cdot R_{3}^{2} \cdot\left(\frac{1}{R_{3}}-\frac{1}{R_{3}+H}\right)= \\
& =3 \cdot 10^{3}[\mathrm{~kg}] \cdot 10\left[\frac{m}{s^{2}}\right] \cdot\left(6380 \cdot 10^{3}[m]\right)^{2} \cdot\left(\frac{1}{6380 \cdot 10^{3}[m]}-\frac{1}{6380 \cdot 10^{3}[m]+600 \cdot 10^{3}[m]}\right)=16452722063 \\
& \text { J] }
\end{aligned}
\]
Source — "http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\�_\�\�\�\�\�\� \%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD\%D1\%82\%D0\%B5\%D0\%B3\%D1\%80\%D0\%B0\%D0\%BB $\%$ D1\%8B_22-19"
Categories: Kuznetsov Problem Book Integrals Problem $22 \mid$ Integrals
Ukrainian Banner Network
- Last modified on this page: 10:56, 31 May 2010.
- Content is available under CC-BY-SA 3.0.
Created by GeeTeatoo
## Problem Kuznetsov Integrals 22-20
## Material from PlusPi
|
16452722063
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Task Condition
Determine the work (in joules) performed when lifting a satellite from the Earth's surface to a height of $H$ km. The mass of the satellite is $m$ tons, the radius of the Earth $R_{3}=6380$ km. The acceleration due to gravity $g$ at the Earth's surface is taken to be $10 \mathrm{~m} / \mathbf{c}^{2}$.
$$
m=3.0 \mathrm{t}, H=650 \text { km. }
$$
|
## Solution
By definition, the elementary work $\Delta A=F(x) \Delta x$, where $F(x)=G \cdot \frac{m \cdot M}{\left(R_{3}+x\right)^{2}} ; G=6.67 \cdot 10^{-11} \mathrm{H}^{*} \mathrm{m}^{*} \mathrm{m} /($ kg*kg)
\[
\begin{aligned}
& F_{x}=G \cdot \frac{m \cdot M}{\left(R_{3}+x\right)^{2}}-\text { force of attraction at height } x \\
& F_{0}=G \cdot \frac{m \cdot M}{R_{3}^{2}}=m g-\text { force of attraction at the Earth's surface } \\
& d A=\frac{m \cdot g \cdot R_{3}^{2}}{\left(R_{3}+x\right)^{2}} d x \Rightarrow \\
& \Rightarrow A=\int_{0}^{H} \frac{m \cdot g \cdot R_{3}^{2}}{\left(R_{3}+x\right)^{2}} d\left(R_{3}+x\right)=-\left.m \cdot g \cdot R_{3}^{2} \cdot \frac{1}{R_{3}+x}\right|_{0} ^{H}=m \cdot g \cdot R_{3}^{2} \cdot\left(\frac{1}{R_{3}}-\frac{1}{R_{3}+H}\right)= \\
& =3 \cdot 10^{3}[\mathrm{~kg}] \cdot 10\left[\frac{m}{s^{2}}\right] \cdot\left(6380 \cdot 10^{3}[m]\right)^{2} \cdot\left(\frac{1}{6380 \cdot 10^{3}[m]}-\frac{1}{6380 \cdot 10^{3}[m]+650 \cdot 10^{3}[m]}\right)=17697012802 \\
& \text { J] }
\end{aligned}
\]
Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\�_\�\�\�\�\�\� \%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD\%D1\%82\%D0\%B5\%D0\%B3\%D1\%80\%D0\%B0\%D0\%BB $\%$ D1\%8B_22-20»
Categories: Kuznetsov Problem Book Integrals Problem $22 \mid$ Integrals
Ukrainian Banner Network
- Last edited on this page: 10:57, 31 May 2010.
- Content is available under CC-BY-SA 3.0.
Created by GeeTeatoo
## Problem Kuznetsov Integrals 22-21
## Material from PlusPi
|
17697012802
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
A cylinder is filled with gas at atmospheric pressure (103.3 kPa). Assuming the gas is ideal, determine the work (in joules) during the isothermal compression of the gas by a piston moving inside the cylinder by $h$ meters (see figure).
Hint: The equation of state for the gas
$\rho V = \text{const}$, where $\rho$ is the pressure, and $V$ is the volume.
$$
H = 0.4 \text{ m, } h = 0.35 \text{ m, } R = 0.1 \text{ m. }
$$
|
## Solution
Piston area: $S=\pi R^{2}$
Volume of gas during compression: $V(x)=S \cdot(H-x) ; 0 \leq x \leq h$
Pressure of gas during compression: $p(x)=\frac{p_{0} \cdot S \cdot H}{V(x)}$
Force of pressure on the piston: $F(x)=p(x) \cdot S$
By definition, the elementary work $\Delta A=F(x) \Delta x \Rightarrow$
$$
\begin{aligned}
& \Rightarrow d A=\frac{p_{0} \cdot S \cdot H}{S \cdot(H-x)} \cdot S d x=\frac{p_{0} \cdot S \cdot H}{H-x} d x \Rightarrow \\
& \Rightarrow A=\int_{0}^{h} \frac{p_{0} \cdot S \cdot H}{H-x} d x=-\int_{0}^{h} \frac{p_{0} \cdot S \cdot H}{H-x} d(H-x)=-\left.p_{0} \cdot S \cdot H \cdot \ln (H-x)\right|_{0} ^{h}= \\
& =p_{0} \cdot S \cdot H \cdot \ln \left(\frac{H}{H-h}\right)=p_{0} \cdot \pi \cdot R^{2} \cdot H \cdot \ln \left(\frac{H}{H-h}\right) \\
& =\pi \cdot(0.1[m])^{2} \cdot 103.3[k P a] \cdot 0.4[m] \cdot \ln \left(\frac{0.4[m]}{0.4[m]-0.35[m]}\right)=2699 \approx 2700[ \\
& \text { kJ] }
\end{aligned}
$$
Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\� \%D0\%9A\%D1\%83\%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+22-21$ » Categories: Kuznetsov Integral Problems 22 | Integrals
- Last edited on this page: 12:46, 31 May 2010.
- Content is available under CC-BY-SA 3.0.
Created by GeeTeatoo
## Problem Kuznetsov Integrals 22-22
## Material from PlusPi
|
2700
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Task Condition
A cylinder is filled with gas at atmospheric pressure (103.3 kPa). Assuming the gas is ideal, determine the work (in joules) during the isothermal compression of the gas by a piston moving inside the cylinder by $h$ meters (see figure).
Hint: The equation of state of the gas
$\rho V=$ const, where $\rho$ - pressure, $V$ - volume.
$$
H=0.4 \mathrm{m}, h=0.3 \mathrm{m}, R=0.1 \mathrm{m}
$$
|
## Solution
Let the piston be at a distance $x, \quad 0 \leq x \leq h$
The force with which the gas presses on the walls is: $F(x)=p(x) \cdot S$
where: $S=\pi R^{2}-$ area of the piston, $p=p(x)-$ pressure of the gas.
Since the process is isothermal, then $p V=$ Const $\Rightarrow p_{0} \cdot \pi \cdot R^{2} \cdot H=p V$
The pressure of the gas during compression: $p(x)=\frac{p_{0} \cdot S \cdot H}{V(x)}$
The volume of the gas during compression: $V(x)=S \cdot(H-x)$
By definition, the elementary work $\Delta A=F(x) \Delta x \Rightarrow$
$$
\begin{aligned}
& \Rightarrow d A=\frac{p_{0} \cdot S \cdot H}{S \cdot(H-x)} \cdot S d x=\frac{p_{0} \cdot S \cdot H}{H-x} d x \Rightarrow \\
& \Rightarrow A=\int_{0}^{h} \frac{p_{0} \cdot S \cdot H}{H-x} d x=-\int_{0}^{h} \frac{p_{0} \cdot S \cdot H}{H-x} d(H-x)=-\left.p_{0} \cdot S \cdot H \cdot \ln (H-x)\right|_{0} ^{h}= \\
& =p_{0} \cdot S \cdot H \cdot \ln \left(\frac{H}{H-h}\right)=p_{0} \cdot \pi \cdot R^{2} \cdot H \cdot \ln \left(\frac{H}{H-h}\right) \\
& =\pi \cdot(0.1[m])^{2} \cdot 103.3[k P a] \cdot 0.4[m] \cdot \ln \left(\frac{0.4[m]}{0.4[m]-0.3[m]}\right)=1800[\text { kJ }]
\end{aligned}
$$
Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\� \%D0\%9A\%D1\%83\%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+22-22 »
Categories: Kuznetsov Integral Problems 22 | Integrals
Ukrainian Banner Network
- Last modified: 13:22, 31 May 2010.
- Content is available under CC-BY-SA 3.0.
|
1800
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
A cylinder is filled with gas at atmospheric pressure (103.3 kPa). Assuming the gas is ideal, determine the work (in joules) done during the isothermal compression of the gas by a piston moving inward by $h$ meters (see figure).
Hint: The equation of state for the gas is
$\rho V = \text{const}$, where $\rho$ is the pressure and $V$ is the volume.
$$
H = 0.4 \text{m}, h = 0.2 \text{m}, R = 0.1 \text{m}
$$
|
## Solution
Piston area: $S=\pi R^{2}$
Volume of gas during compression: $V(x)=S \cdot(H-x) ; 0 \leq x \leq h$
Pressure of gas during compression: $p(x)=\frac{p_{0} \cdot S \cdot H}{V(x)}$
Force of pressure on the piston: $F(x)=p(x) \cdot S$
By definition, the elementary work $\Delta A=F(x) \Delta x \Rightarrow$
$$
\begin{aligned}
& \Rightarrow d A=\frac{p_{0} \cdot S \cdot H}{S \cdot(H-x)} \cdot S d x=\frac{p_{0} \cdot S \cdot H}{H-x} d x \Rightarrow \\
& \Rightarrow A=\int_{0}^{h} \frac{p_{0} \cdot S \cdot H}{H-x} d x=-\int_{0}^{h} \frac{p_{0} \cdot S \cdot H}{H-x} d(H-x)=-\left.p_{0} \cdot S \cdot H \cdot \ln (H-x)\right|_{0} ^{h}= \\
& =p_{0} \cdot S \cdot H \cdot \ln \left(\frac{H}{H-h}\right)=p_{0} \cdot \pi \cdot R^{2} \cdot H \cdot \ln \left(\frac{H}{H-h}\right) \\
& =\pi \cdot(0.1[m])^{2} \cdot 103.3[k P a] \cdot 0.4[m] \cdot \ln \left(\frac{0.4[m]}{0.4[m]-0.2[m]}\right)=900[\text { kJ }]
\end{aligned}
$$
Source — "http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\� \%D0\%9A\%D1\%83\%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+22-23$ "
Categories: Kuznetsov Integral Problems 22 | Integrals
- Last modified: 12:49, 31 May 2010.
- Content is available under CC-BY-SA 3.0.
Created by GeeTeatoo
## Problem Kuznetsov Integrals 22-24
## Material from PlusPi
|
900
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
A cylinder is filled with gas at atmospheric pressure (103.3 kPa). Assuming the gas is ideal, determine the work (in joules) during the isothermal compression of the gas by a piston moving inside the cylinder by $h$ meters (see figure).
Hint: The equation of state for the gas
$\rho V=$ const, where $\rho$ - pressure, $V$ - volume.
$$
H=0.8 \mathrm{m}, h=0.6 \mathrm{m}, R=0.2 \mathrm{m} .
$$
|
## Solution
Piston area: $S=\pi R^{2}$
Volume of gas during compression: $V(x)=S \cdot(H-x) ; 0 \leq x \leq h$
Pressure of gas during compression: $p(x)=\frac{p_{0} \cdot S \cdot H}{V(x)}$
Force of pressure on the piston: $F(x)=p(x) \cdot S$
By definition, the elementary work $\Delta A=F(x) \Delta x \Rightarrow$
$$
\begin{aligned}
& \Rightarrow d A=\frac{p_{0} \cdot S \cdot H}{S \cdot(H-x)} \cdot S d x=\frac{p_{0} \cdot S \cdot H}{H-x} d x \Rightarrow \\
& \Rightarrow A=\int_{0}^{h} \frac{p_{0} \cdot S \cdot H}{H-x} d x=-\int_{0}^{h} \frac{p_{0} \cdot S \cdot H}{H-x} d(H-x)=-\left.p_{0} \cdot S \cdot H \cdot \ln (H-x)\right|_{0} ^{h}= \\
& =p_{0} \cdot S \cdot H \cdot \ln \left(\frac{H}{H-h}\right)=p_{0} \cdot \pi \cdot R^{2} \cdot H \cdot \ln \left(\frac{H}{H-h}\right) \\
& =\pi \cdot(0.2[m])^{2} \cdot 103.3[k P a] \cdot 0.8[m] \cdot \ln \left(\frac{0.8[m]}{0.8[m]-0.6[m]}\right)=14396 \approx 14400[ \\
& \text { kJ] }
\end{aligned}
$$
Source — "http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\� \%D0\%9A\%D1\%83\%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+22-25$ "
Categories: Kuznetsov Integral Problems 22-26 | Integrals
- Last edited: 12:53, 31 May 2010.
- Content is available under CC-BY-SA 3.0.
|
14400
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Task Condition
A cylinder is filled with gas at atmospheric pressure (103.3 kPa). Assuming the gas is ideal, determine the work (in joules) during the isothermal compression of the gas by a piston moving inside the cylinder by $h$ meters (see figure).
Hint: The equation of state of the gas
$\rho V=$ const, where $\rho$ - pressure, $V$ - volume.
$$
H=1.6 \mathrm{m}, h=1.4 \mathrm{m}, R=0.3 \mathrm{m}
$$
|
## Solution
Piston area: $S=\pi R^{2}$
Volume of gas during compression: $V(x)=S \cdot(H-x) ; 0 \leq x \leq h$
Pressure of gas during compression: $p(x)=\frac{p_{0} \cdot S \cdot H}{V(x)}$
Force of pressure on the piston: $F(x)=p(x) \cdot S$
By definition, the elementary work $\Delta A=F(x) \Delta x \Rightarrow$
$$
\begin{aligned}
& \Rightarrow d A=\frac{p_{0} \cdot S \cdot H}{S \cdot(H-x)} \cdot S d x=\frac{p_{0} \cdot S \cdot H}{H-x} d x \Rightarrow \\
& \Rightarrow A=\int_{0}^{h} \frac{p_{0} \cdot S \cdot H}{H-x} d x=-\int_{0}^{h} \frac{p_{0} \cdot S \cdot H}{H-x} d(H-x)=-\left.p_{0} \cdot S \cdot H \cdot \ln (H-x)\right|_{0} ^{h}= \\
& =p_{0} \cdot S \cdot H \cdot \ln \left(\frac{H}{H-h}\right)=p_{0} \cdot \pi \cdot R^{2} \cdot H \cdot \ln \left(\frac{H}{H-h}\right) \\
& =\pi \cdot(0.3[m])^{2} \cdot 103.3[k P a] \cdot 1.6[m] \cdot \ln \left(\frac{1.6[m]}{1.6[m]-1.4[m]}\right)=97176 \approx 97200[ \\
& \text { kJ] }
\end{aligned}
$$
Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\� \%D0\%9A\%D1\%83\%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+22-27$ »
Categories: Kuznetsov Integral Problems 22 | Integrals
- Last edited on this page: 12:55, 31 May 2010.
- Content is available under CC-BY-SA 3.0.
|
97200
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Task Condition
A cylinder is filled with gas at atmospheric pressure (103.3 kPa). Assuming the gas is ideal, determine the work (in joules) during the isothermal compression of the gas by a piston moving inside the cylinder by $h$ meters (see figure).
Hint: The equation of state of the gas
$\rho V=$ const, where $\rho$ - pressure, $V$ - volume.
$$
H=1.6 \text{ m}, h=1.2 \text{ m}, R=0.3 \text{ m.}
$$
|
## Solution
Piston area: $S=\pi R^{2}$
Volume of gas during compression: $V(x)=S \cdot(H-x) ; 0 \leq x \leq h$
Pressure of gas during compression: $p(x)=\frac{p_{0} \cdot S \cdot H}{V(x)}$
Force of pressure on the piston: $F(x)=p(x) \cdot S$
By definition, the elementary work $\Delta A=F(x) \Delta x \Rightarrow$
$$
\begin{aligned}
& \Rightarrow d A=\frac{p_{0} \cdot S \cdot H}{S \cdot(H-x)} \cdot S d x=\frac{p_{0} \cdot S \cdot H}{H-x} d x \Rightarrow \\
& \Rightarrow A=\int_{0}^{h} \frac{p_{0} \cdot S \cdot H}{H-x} d x=-\int_{0}^{h} \frac{p_{0} \cdot S \cdot H}{H-x} d(H-x)=-\left.p_{0} \cdot S \cdot H \cdot \ln (H-x)\right|_{0} ^{h}= \\
& =p_{0} \cdot S \cdot H \cdot \ln \left(\frac{H}{H-h}\right)=p_{0} \cdot \pi \cdot R^{2} \cdot H \cdot \ln \left(\frac{H}{H-h}\right) \\
& =\pi \cdot(0.3[m])^{2} \cdot 103.3[k P a] \cdot 1.6[m] \cdot \ln \left(\frac{1.6[m]}{1.6[m]-1.2[m]}\right)=64784 \approx 64800[ \\
& \text { kJ] }
\end{aligned}
$$
Source — "http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\� \%D0\%9A\%D1\%83\%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+22-28$ "
Categories: Kuznetsov Integral Problems 22-29 | Integrals
- Last edited on this page: 12:56, 31 May 2010.
- Content is available under CC-BY-SA 3.0.
|
64800
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Task Condition
A cylinder is filled with gas at atmospheric pressure (103.3 kPa). Assuming the gas is ideal, determine the work (in joules) during the isothermal compression of the gas by a piston moving inside the cylinder by $h$ meters (see figure).
Hint: The equation of state of the gas
$\rho V=$ const, where $\rho$ - pressure, $V$ - volume.
$$
H=1.6 \mathrm{m}, h=0.8 \mathrm{m}, R=0.3 \mathrm{m}
$$
|
## Solution
Piston area: $S=\pi R^{2}$
Volume of gas during compression: $V(x)=S \cdot(H-x) ; 0 \leq x \leq h$
Pressure of gas during compression: $p(x)=\frac{p_{0} \cdot S \cdot H}{V(x)}$
Force of pressure on the piston: $F(x)=p(x) \cdot S$
By definition, the elementary work $\Delta A=F(x) \Delta x \Rightarrow$
$$
\begin{aligned}
& \Rightarrow d A=\frac{p_{0} \cdot S \cdot H}{S \cdot(H-x)} \cdot S d x=\frac{p_{0} \cdot S \cdot H}{H-x} d x \Rightarrow \\
& \Rightarrow A=\int_{0}^{h} \frac{p_{0} \cdot S \cdot H}{H-x} d x=-\int_{0}^{h} \frac{p_{0} \cdot S \cdot H}{H-x} d(H-x)=-\left.p_{0} \cdot S \cdot H \cdot \ln (H-x)\right|_{0} ^{h}= \\
& =p_{0} \cdot S \cdot H \cdot \ln \left(\frac{H}{H-h}\right)=p_{0} \cdot \pi \cdot R^{2} \cdot H \cdot \ln \left(\frac{H}{H-h}\right) \\
& =\pi \cdot(0.3[m])^{2} \cdot 103.3[k P a] \cdot 1.6[m] \cdot \ln \left(\frac{1.6[m]}{1.6[m]-0.8[m]}\right)=32392 \approx 32400[ \\
& \text { kJ] }
\end{aligned}
$$
Source — "http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\� \%D0\%9A\%D1\%83\%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+22-29$ "
Categories: Kuznetsov Integral Problems 22 | Integrals
- Last edited on this page: 12:57, 31 May 2010.
- Content is available under CC-BY-SA 3.0.
Created by GeeTeatoo
## Problem Kuznetsov Integrals 22-30
## Material from PlusPi
|
32400
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Task Condition
A cylinder is filled with gas at atmospheric pressure (103.3 kPa). Assuming the gas is ideal, determine the work (in joules) during the isothermal compression of the gas by a piston moving inside the cylinder by $h$ meters (see figure).
Hint: The equation of state of the gas $\rho V=$ const, where $\rho$ - pressure, $V$ - volume.
$$
H=2.0 \mathrm{m}, h=1.5 \mathrm{m}, R=0.4 \mathrm{m}
$$
|
## Solution
Piston area: $S=\pi R^{2}$
Volume of gas during compression: $V(x)=S \cdot(H-x) ; 0 \leq x \leq h$
Pressure of gas during compression: $p(x)=\frac{p_{0} \cdot S \cdot H}{V(x)}$
Force of pressure on the piston: $F(x)=p(x) \cdot S$
By definition, the elementary work $\Delta A=F(x) \Delta x \Rightarrow$
$$
\begin{aligned}
& \Rightarrow d A=\frac{p_{0} \cdot S \cdot H}{S \cdot(H-x)} \cdot S d x=\frac{p_{0} \cdot S \cdot H}{H-x} d x \Rightarrow \\
& \Rightarrow A=\int_{0}^{h} \frac{p_{0} \cdot S \cdot H}{H-x} d x=-\int_{0}^{h} \frac{p_{0} \cdot S \cdot H}{H-x} d(H-x)=-\left.p_{0} \cdot S \cdot H \cdot \ln (H-x)\right|_{0} ^{h}= \\
& =p_{0} \cdot S \cdot H \cdot \ln \left(\frac{H}{H-h}\right)=p_{0} \cdot \pi \cdot R^{2} \cdot H \cdot \ln \left(\frac{H}{H-h}\right) \\
& =\pi \cdot(0.4[m])^{2} \cdot 103.3[k P a] \cdot 2.0[m] \cdot \ln \left(\frac{2.0[m]}{2.0[m]-1.5[m]}\right)=143965 \approx 144000[ \\
& \text { kJ }]
\end{aligned}
$$
Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\� \%D0\%9A\%D1\%83\%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+22-30$ » Categories: Kuznetsov's Problem Book Integrals Problem 22 | Integrals
- Last edited on this page: 12:59, 31 May 2010.
- Content is available under CC-BY-SA 3.0.
Created by GeeTeatoo
## Problem Kuznetsov Integrals 22-31
## Material from PlusPi
|
144000
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
A cylinder is filled with gas at atmospheric pressure (103.3 kPa). Assuming the gas is ideal, determine the work (in joules) during the isothermal compression of the gas by a piston moving inside the cylinder by $h$ meters (see figure).
Hint: The equation of state for the gas
$\rho V=$ const, where $\rho$ - pressure, $V$ - volume.
$$
H=2.0 \mathrm{~m}, h=1.0 \mathrm{~m}, R=0.4 \mathrm{~m}
$$
|
## Solution
Area of the piston: $S=\pi R^{2}$
Volume of the gas during compression: $V(x)=S \cdot(H-x) ; 0 \leq x \leq h$
Pressure of the gas during compression: $p(x)=\frac{p_{0} \cdot S \cdot H}{V(x)}$
Force of pressure on the piston: $F(x)=p(x) \cdot S$
By definition, the elementary work $\Delta A=F(x) \Delta x \Rightarrow$
$$
\begin{aligned}
& \Rightarrow d A=\frac{p_{0} \cdot S \cdot H}{S \cdot(H-x)} \cdot S d x=\frac{p_{0} \cdot S \cdot H}{H-x} d x \Rightarrow \\
& \Rightarrow A=\int_{0}^{h} \frac{p_{0} \cdot S \cdot H}{H-x} d x=-\int_{0}^{h} \frac{p_{0} \cdot S \cdot H}{H-x} d(H-x)=-\left.p_{0} \cdot S \cdot H \cdot \ln (H-x)\right|_{0} ^{h}= \\
& =p_{0} \cdot S \cdot H \cdot \ln \left(\frac{H}{H-h}\right)=p_{0} \cdot \pi \cdot R^{2} \cdot H \cdot \ln \left(\frac{H}{H-h}\right) \\
& =\pi \cdot(0.4[m])^{2} \cdot 103.3[k P a] \cdot 2.0[m] \cdot \ln \left(\frac{2.0[m]}{2.0[m]-1.0[m]}\right)=71982 \approx 72000[ \\
& \text { kJ] }
\end{aligned}
$$
Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\� \%D0\%9A\%D1\%83\%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+22-31$ » Categories: Kuznetsov Problem Book Integrals Problem 22 | Integrals
- Last edited: 13:00, 31 May 2010.
- Content is available under CC-BY-SA 3.0.
Created by GeeTeatoo
[^0]: - Last edited: 12:42, 31 May 2010.
|
72000
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Based on the definition of the derivative, find $f^{\prime}(0)$:
$$
f(x)=\left\{\begin{array}{c}
\ln \left(1-\sin \left(x^{3} \sin \frac{1}{x}\right)\right), x \neq 0 \\
0, x=0
\end{array}\right.
$$
|
## Solution
By definition, the derivative at the point $x=0$:
$f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}$
Based on the definition, we find:
$$
\begin{aligned}
& f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim _{\Delta x \rightarrow 0} \frac{\ln \left(1-\sin \left(\Delta x^{3} \sin \frac{1}{\Delta x}\right)\right)-0}{\Delta x}= \\
& =\lim _{\Delta x \rightarrow 0} \frac{\ln \left(1-\sin \left(\Delta x^{3} \sin \frac{1}{\Delta x}\right)\right)}{\Delta x}=
\end{aligned}
$$
Using the substitution of equivalent infinitesimals:
$$
\begin{aligned}
& \ln \left(1-\sin \left(\Delta x^{3} \sin \frac{1}{\Delta x}\right)\right) \sim-\sin \left(\Delta x^{3} \sin \frac{1}{\Delta x}\right), \text { as } \\
& \Delta x \rightarrow 0\left(-\sin \left(\Delta x^{3} \sin \frac{1}{\Delta x}\right) \rightarrow 0\right)
\end{aligned}
$$
We get:
$=\lim _{\Delta x \rightarrow 0} \frac{-\sin \left(\Delta x^{3} \sin \frac{1}{\Delta x}\right)}{\Delta x}=$
Using the substitution of equivalent infinitesimals:
We get:
$$
=\lim _{\Delta x \rightarrow 0} \frac{-\Delta x^{3} \sin \frac{1}{\Delta x}}{\Delta x}=\lim _{\Delta x \rightarrow 0}-\Delta x^{2} \sin \frac{1}{\Delta x}=
$$
Since $\sin \frac{1}{\Delta x}$ is bounded, then
$\Delta x^{2} \sin \frac{1}{\Delta x} \rightarrow{ }_{, \text {as }} \Delta x \rightarrow 0\left(\Delta^{2} x \rightarrow 0\right)$
Thus,
$=-0=0$
Therefore, $f^{\prime}(0)=0$
## Problem Kuznetsov Differentiation 2-4
|
0
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Find the derivative $y_{x}^{\prime}$.
$$
\left\{\begin{array}{l}
x=\arcsin (\sin t) \\
y=\arccos (\cos t)
\end{array}\right.
$$
|
## Solution
$x_{t}^{\prime}=(\arcsin (\sin t))^{\prime}=t^{\prime}=1$
$y_{t}^{\prime}=(\arccos (\cos t))^{\prime}=t^{\prime}=1$
We obtain:
$y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{t}^{\prime}}=\frac{1}{1}=1$
## Kuznetsov Differentiation 16-4
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the lengths of the arcs of the curves given by the equations in the Cartesian coordinate system.
$$
y=2+\arcsin \sqrt{x}+\sqrt{x-x^{2}}, \frac{1}{4} \leq x \leq 1
$$
|
## Solution
The length of the arc of a curve defined by the equation $y=f(x) ; a \leq x \leq b$, is determined by the formula
$$
L=\int_{a}^{b} \sqrt{1+\left(f^{\prime}(x)\right)^{2}} d x
$$
Let's find the derivative of the given function:
$$
\begin{aligned}
f^{\prime}(x)=\left(2+\arcsin \sqrt{x}+\sqrt{x-x^{2}}\right)^{\prime} & =\frac{1}{\sqrt{1-x}} \cdot \frac{1}{2 \sqrt{x}}+\frac{1}{2 \sqrt{x} \sqrt{1-x}} \cdot(1-2 x)= \\
& =\frac{1}{2 \sqrt{x} \sqrt{1-x}}+\frac{1}{2 \sqrt{x} \sqrt{1-x}}-\frac{2 x}{2 \sqrt{x} \sqrt{1-x}}= \\
& =\frac{1}{\sqrt{x} \sqrt{1-x}}-\frac{x}{\sqrt{x} \sqrt{1-x}}= \\
& =\frac{1-x}{\sqrt{x} \sqrt{1-x}}=\sqrt{\frac{1-x}{x}}
\end{aligned}
$$
Then, using the formula above, we get:
$$
\begin{aligned}
L & =\int_{1 / 4}^{1} \sqrt{1+\left(\sqrt{\frac{1-x}{x}}\right)^{2}} d x= \\
& =\int_{1 / 4}^{1} \sqrt{1+\frac{1-x}{x}} d x= \\
& =\int_{1 / 4}^{1} \sqrt{\frac{x+1-x}{x}} d x= \\
& =\int_{1 / 4}^{1} \frac{1}{\sqrt{x}} d x=\left.2 \sqrt{x}\right|_{1 / 4} ^{1}=2-\frac{2}{\sqrt{4}}=1
\end{aligned}
$$
Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\�_\�\�
\%D1\%83\%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD\%D1\%82\%D0\%B5 $\% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+17-7$ »
Categories: Kuznetsov's Problem Book Integrals Problem 17 | Integrals
- Last modified: 07:28, 26 May 2010.
- Content is available under CC-BY-SA 3.0.
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the lengths of the arcs of the curves given by equations in a rectangular coordinate system.
$$
y=-\arccos \sqrt{x}+\sqrt{x-x^{2}}, 0 \leq x \leq \frac{1}{4}
$$
|
## Solution
The length of the arc of a curve defined by the equation $y=f(x) ; a \leq x \leq b$, is determined by the formula
$$
L=\int_{a}^{b} \sqrt{1+\left(f^{\prime}(x)\right)^{2}} d x
$$
Let's find the derivative of the given function:
$$
\begin{aligned}
f^{\prime}(x)=\left(-\arccos \sqrt{x}+\sqrt{x-x^{2}}\right)^{\prime} & =\frac{1}{\sqrt{1-x}} \cdot \frac{1}{2 \sqrt{x}}+\frac{1}{2 \sqrt{x} \sqrt{1-x}} \cdot(1-2 x)= \\
& =\frac{1}{2 \sqrt{x} \sqrt{1-x}}+\frac{1}{2 \sqrt{x} \sqrt{1-x}}-\frac{2 x}{2 \sqrt{x} \sqrt{1-x}}= \\
& =\frac{1}{\sqrt{x} \sqrt{1-x}}-\frac{x}{\sqrt{x} \sqrt{1-x}}= \\
& =\frac{1-x}{\sqrt{x} \sqrt{1-x}}=\sqrt{\frac{1-x}{x}}
\end{aligned}
$$
Then, using the above formula, we get:
$$
\begin{aligned}
L & =\int_{1 / 4}^{1} \sqrt{1+\left(\sqrt{\frac{1-x}{x}}\right)^{2}} d x= \\
& =\int_{1 / 4}^{1} \sqrt{1+\frac{1-x}{x}} d x= \\
& =\int_{1 / 4}^{1} \sqrt{\frac{x+1-x}{x}} d x= \\
& =\int_{1 / 4}^{1} \frac{1}{\sqrt{x}} d x=\left.2 \sqrt{x}\right|_{1 / 4} ^{1}=2-\frac{2}{\sqrt{4}}=1
\end{aligned}
$$
Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\� $\% \mathrm{D} 0 \% 9 \mathrm{~A} \% \mathrm{D} 1 \% 83 \% \mathrm{D} 0 \% \mathrm{~B} 7 \% \mathrm{D} 0 \% \mathrm{BD} \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 1 \% 86 \% \mathrm{D} 0 \% \mathrm{BE} \% \mathrm{D} 0 \% \mathrm{~B} 2 \mathrm{O} 0 \% 98 \% \mathrm{D} 0 \% \mathrm{BD} \% \mathrm{D1} \% 82$ $\% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+17-14$ " Categories: Kuznetsov's Problem Book Integrals Problem 17| Integrals | Problems for checking
Ukrainian Banner Network
- Last modified on this page: 07:46, 26 May 2010.
- Content is available under CC-BY-SA 3.0.
Created by GeeTeatoo
## Problem Kuznetsov Integrals 17-15
## Material from Plusi
## Contents
- 1 Problem Statement
- 2 Solution
- 3 Then
- 4 Using the substitution
- 5 Therefore
- 6 Using the substitution we get
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Based on the definition of the derivative, find $f^{\prime}(0)$:
$$
f(x)=\left\{\begin{array}{c}
x+\arcsin \left(x^{2} \sin \frac{6}{x}\right), x \neq 0 \\
0, x=0
\end{array}\right.
$$
|
## Solution
By definition, the derivative at the point $x=0$:
$f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}$
Based on the definition, we find:
$$
\begin{aligned}
& f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim _{\Delta x \rightarrow 0} \frac{\Delta x+\arcsin \left(\Delta x^{2} \sin \frac{6}{\Delta x}\right)-0}{\Delta x}= \\
& =\lim _{\Delta x \rightarrow 0} \frac{\Delta x}{\Delta x}+\lim _{\Delta x \rightarrow 0} \frac{\arcsin \left(\Delta x^{2} \sin \frac{6}{\Delta x}\right)}{\Delta x}=
\end{aligned}
$$
Using the substitution of equivalent infinitesimals:
$\arcsin \left(\Delta x^{2} \sin \frac{6}{\Delta x}\right) \sim \Delta x^{2} \sin \frac{6}{\Delta x} \text{ as } \Delta x \rightarrow 0 \left(\Delta x^{2} \sin \frac{6}{\Delta x} \rightarrow 0\right)$
We get:
$=\lim _{\Delta x \rightarrow 0} 1+\lim _{\Delta x \rightarrow 0} \frac{\Delta x^{2} \sin \frac{6}{\Delta x}}{\Delta x}=1+\lim _{\Delta x \rightarrow 0} \Delta x \sin \frac{6}{\Delta x}=$
Since $\sin \frac{6}{\Delta x}$ is bounded, then
$\Delta x \sin \frac{6}{\Delta x} \rightarrow 0$, as $\Delta x \rightarrow 0$
Thus:
$=1+0=1$
Therefore, $f^{\prime}(0)=1$
## Problem Kuznetsov Differentiation 2-11
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Condition of the problem
To derive the equation of the normal to the given curve at the point with abscissa $x_{0}$.
$y=\sqrt{x}-3 \sqrt[3]{x}, x_{0}=64$
|
## Solution
Let's find $y^{\prime}:$
$$
y^{\prime}=(\sqrt{x}-3 \sqrt[3]{x})^{\prime}=\left(\sqrt{x}-3 \cdot x^{\frac{1}{3}}\right)^{\prime}=\frac{1}{2 \sqrt{x}}-3 \cdot \frac{1}{3} \cdot x^{-\frac{2}{3}}=\frac{1}{2 \sqrt{x}}-\frac{1}{\sqrt[3]{x^{2}}}
$$
Then:
$y_{0}^{\prime}=y^{\prime}\left(x_{0}\right)=\frac{1}{2 \sqrt{x_{0}}}-\frac{1}{\sqrt[3]{x_{0}^{2}}}=\frac{1}{2 \sqrt{64}}-\frac{1}{\sqrt[3]{64^{2}}}=\frac{1}{16}-\frac{1}{16}=0$
Since $y^{\prime}\left(x_{0}\right)=0$, the equation of the normal line is:
$x=x_{0}$
$x=64$
Thus, the equation of the normal line is:
$x=64$
## Problem Kuznetsov Differentiation 3-11
|
64
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Based on the definition of the derivative, find $f^{\prime}(0)$:
$$
f(x)=\left\{\begin{array}{c}
\frac{\cos x-\cos 3 x}{x}, x \neq 0 \\
0, x=0
\end{array}\right.
$$
|
## Solution
By definition, the derivative at the point $x=0$:
$f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}$
Based on the definition, we find:
$$
\begin{aligned}
& f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim _{\Delta x \rightarrow 0} \frac{\frac{\cos \Delta x-\cos (3 \Delta x)}{\Delta x}-0}{\Delta x}= \\
& =\lim _{\Delta x \rightarrow 0} \frac{\cos \Delta x-\cos (3 \Delta x)}{\Delta x^{2}}=\lim _{\Delta x \rightarrow 0} \frac{1-2 \sin ^{2} \frac{\Delta x}{2}-\left(1-2 \sin ^{2}\left(\frac{3 \Delta x}{2}\right)\right)}{\Delta x^{2}}= \\
& =\lim _{\Delta x \rightarrow 0} \frac{-2 \sin ^{2} \frac{\Delta x}{2}+2 \sin ^{2}\left(\frac{3 \Delta x}{2}\right)}{\Delta x^{2}}=\lim _{\Delta x \rightarrow 0} \frac{-2 \sin ^{2} \frac{\Delta x}{2}}{\Delta x^{2}}+\lim _{\Delta x \rightarrow 0} \frac{2 \sin ^{2}\left(\frac{3 \Delta x}{2}\right)}{\Delta x^{2}}=
\end{aligned}
$$
Using the substitution of equivalent infinitesimals:
$\sin \frac{\Delta x}{2} \sim \frac{\Delta x}{2}$, as $\Delta x \rightarrow 0\left(\frac{\Delta x}{2} \rightarrow 0\right)$
$\sin \left(\frac{3 \Delta x}{2}\right) \sim \frac{3 \Delta x}{2}$, as $\Delta x \rightarrow 0\left(\frac{3 \Delta x}{2} \rightarrow 0\right)$
We get:
$$
=\lim _{\Delta x \rightarrow 0} \frac{-2\left(\frac{\Delta x}{2}\right)^{2}}{\Delta x^{2}}+\lim _{\Delta x \rightarrow 0} \frac{2\left(\frac{3 \Delta x}{2}\right)^{2}}{\Delta x^{2}}=\lim _{\Delta x \rightarrow 0} \frac{-1}{2}+\lim _{\Delta x \rightarrow 0} \frac{9}{2}=-\frac{1}{2}+\frac{9}{2}=4
$$
Thus, $f^{\prime}(0)=4$
## Problem Kuznetsov Differentiation 2-30
|
4
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Find the second-order derivative $y_{x x}^{\prime \prime}$ of the function given parametrically.
$\left\{\begin{array}{l}x=\cos t+\sin t \\ y=\sin 2 t\end{array}\right.$
|
## Solution
$x_{t}^{\prime}=(\cos t+\sin t)^{\prime}=-\sin t+\cos t$
$y_{t}^{\prime}=(\sin 2 t)^{\prime}=2 \cos 2 t$
We obtain:
$$
\begin{aligned}
& y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{t}^{\prime}}=\frac{2 \cos 2 t}{-\sin t+\cos t}=2 \cdot \frac{\cos ^{2} t-\sin ^{2} t}{\cos t-\sin t}=2(\sin t+\cos t) \\
& \left(y_{x}^{\prime}\right)_{t}^{\prime}=(2(\sin t+\cos t))^{\prime}=2(\cos t-\sin t)
\end{aligned}
$$
Then:
$y_{x x}^{\prime \prime}=\frac{\left(y_{x}^{\prime}\right)_{t}^{\prime}}{x_{t}^{\prime}}=\frac{2(\cos t-\sin t)}{-\sin t+\cos t}=2$
## Problem Kuznetsov Differentiation 20-30
|
2
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the numerical sequence:
$$
\lim _{n \rightarrow \infty} \frac{(n+1)^{3}+(n-1)^{3}}{n^{3}-3 n}
$$
|
## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{(n+1)^{3}+(n-1)^{3}}{n^{3}-3 n}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n^{3}}\left((n+1)^{3}+(n-1)^{3}\right)}{\frac{1}{n^{3}}\left(n^{3}-3 n\right)}= \\
& =\lim _{n \rightarrow \infty} \frac{\left(1+\frac{1}{n}\right)^{3}+\left(1-\frac{1}{n}\right)^{3}}{1-\frac{3}{n^{2}}}=\frac{1^{3}+1^{3}}{1-0}=2
\end{aligned}
$$
## Problem Kuznetsov Limits 3-28
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the numerical sequence:
$$
\lim _{n \rightarrow \infty} \frac{n!+(n+2)!}{(n-1)!+(n+2)!}
$$
|
## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{n!+(n+2)!}{(n-1)!+(n+2)!}=\lim _{n \rightarrow \infty} \frac{(n-1)!(n+n(n+1)(n+2))}{(n-1)!(1+n(n+1)(n+2))}= \\
& =\lim _{n \rightarrow \infty} \frac{n+n(n+1)(n+2)}{1+n(n+1)(n+2)}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n^{3}}(n+n(n+1)(n+2))}{\frac{1}{n^{3}}(1+n(n+1)(n+2))}= \\
& =\lim _{n \rightarrow \infty} \frac{\frac{1}{n^{2}}+\left(1+\frac{1}{n}\right)\left(1+\frac{2}{n}\right)}{\frac{1}{n^{3}}+\left(1+\frac{1}{n}\right)\left(1+\frac{2}{n}\right)}=\frac{0+(1+0)(1+0)}{0+(1+0)(1+0)}=1
\end{aligned}
$$
## Problem Kuznetsov Limits 6-28
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Condition of the problem
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{(1+x)^{3}-(1+3 x)}{x^{2}+x^{5}}$
|
## Solution
$\lim _{x \rightarrow 0} \frac{(1+x)^{3}-(1+3 x)}{x^{2}+x^{5}}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow 0} \frac{1^{3}+3 \cdot 1^{2} \cdot x+3 \cdot 1 \cdot x^{2}+x^{3}-1-3 x}{x^{2}\left(1+x^{3}\right)}=$
$=\lim _{x \rightarrow 0} \frac{1+3 x+3 x^{2}+x^{3}-1-3 x}{x^{2}\left(1+x^{3}\right)}=\lim _{x \rightarrow 0} \frac{3 x^{2}+x^{3}}{x^{2}\left(1+x^{3}\right)}=$
$=\lim _{x \rightarrow 0} \frac{x^{2}(3+x)}{x^{2}\left(1+x^{3}\right)}=\lim _{x \rightarrow 0} \frac{3+x}{1+x^{3}}=\frac{3+0}{1+0^{3}}=\frac{3}{1}=3$
## Problem Kuznetsov Limits 10-28
|
3
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow-2} \frac{\sqrt[3]{x-6}+2}{\sqrt[3]{x^{3}+8}}
$$
|
## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow-2} \frac{\sqrt[3]{x-6}+2}{\sqrt[3]{x^{3}+8}}=\lim _{x \rightarrow-2} \frac{(\sqrt[3]{x-6}+2)\left(\sqrt[3]{(x-6)^{2}}-2 \sqrt[3]{x-6}+4\right)}{\sqrt[3]{x^{3}+8}\left(\sqrt[3]{(x-6)^{2}}-2 \sqrt[3]{x-6}+4\right)}= \\
& =\lim _{x \rightarrow-2} \frac{(\sqrt[3]{x-6}+2)\left(\sqrt[3]{(x-6)^{2}}-2 \sqrt[3]{x-6}+4\right)}{\sqrt[3]{x^{3}+8}\left(\sqrt[3]{(x-6)^{2}}-2 \sqrt[3]{x-6}+4\right)}= \\
& =\lim _{x \rightarrow-2} \frac{x-6+8}{\sqrt[3]{x^{3}+8}\left(\sqrt[3]{(x-6)^{2}}-2 \sqrt[3]{x-6}+4\right)}= \\
& =\lim _{x \rightarrow-2} \frac{x+2}{\sqrt[3]{(x+2)\left(x^{2}+2 x+4\right)}\left(\sqrt[3]{(x-6)^{2}}-2 \sqrt[3]{x-6}+4\right)}= \\
& =\lim _{x \rightarrow-2} \frac{\sqrt[3]{(x+2)^{2}}}{\sqrt[3]{x^{2}+2 x+4}\left(\sqrt[3]{(x-6)^{2}}-2 \sqrt[3]{x-6}+4\right)}= \\
& =\frac{\sqrt[3]{(-2+2)^{2}}}{\sqrt[3]{(-2)^{2}+2 \cdot(-2)+4}\left(\sqrt[3]{(-2-6)^{2}}-2 \sqrt[3]{-2-6}+4\right)}= \\
& =\frac{0}{\sqrt[3]{4-4+4}\left(\sqrt[3]{8^{2}}-2 \sqrt[3]{-8}+4\right)}=\frac{0}{\sqrt[3]{4}\left(2^{2}+2 \cdot 2+4\right)}=0
\end{aligned}
$$
|
0
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{\ln \left(x^{2}+1\right)}{1-\sqrt{x^{2}+1}}$
|
## Solution
We will use the substitution of equivalent infinitesimals:
$\ln \left(1+x^{2}\right) \sim x^{2}$, as $x \rightarrow 0\left(x^{2} \rightarrow 0\right)$
We get:
$\lim _{x \rightarrow 0} \frac{\ln \left(x^{2}+1\right)}{1-\sqrt{x^{2}+1}}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow 0} \frac{x^{2}}{1-\sqrt{x^{2}+1}}=$
$=\lim _{x \rightarrow 0} \frac{x^{2}\left(1+\sqrt{x^{2}+1}\right)}{\left(1-\sqrt{x^{2}+1}\right)\left(1+\sqrt{x^{2}+1}\right)}=$
$=\lim _{x \rightarrow 0} \frac{x^{2}\left(1+\sqrt{x^{2}+1}\right)}{1-\left(x^{2}+1\right)}=\lim _{x \rightarrow 0} \frac{x^{2}\left(1+\sqrt{x^{2}+1}\right)}{x^{2}}=$
$=\lim _{x \rightarrow 0}\left(1+\sqrt{x^{2}+1}\right)=\left(1+\sqrt{0^{2}+1}\right)=2$
## Problem Kuznetsov Limits 12-28
|
2
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{e^{2 x}-e^{x}}{\sin 2 x-\sin x}$
|
## Solution
$\lim _{x \rightarrow 0} \frac{e^{2 x}-e^{x}}{\sin 2 x-\sin x}=\lim _{x \rightarrow 0} \frac{\left(e^{2 x}-1\right)-\left(e^{x}-1\right)}{\sin 2 x-\sin x}=$
$=\lim _{x \rightarrow 0} \frac{\frac{1}{x}\left(\left(e^{2 x}-1\right)-\left(e^{x}-1\right)\right)}{\frac{1}{x}(\sin 2 x-\sin x)}=$
$=\frac{\lim _{x \rightarrow 0} \frac{1}{x}\left(\left(e^{2 x}-1\right)-\left(e^{x}-1\right)\right)}{\lim _{x \rightarrow 0} \frac{1}{x}(\sin 2 x-\sin x)}=$
$=\left(\lim _{x \rightarrow 0} \frac{e^{2 x}-1}{x}-\lim _{x \rightarrow 0} \frac{e^{x}-1}{x}\right) /\left(\lim _{x \rightarrow 0} \frac{\sin 2 x}{x}-\lim _{x \rightarrow 0} \frac{\sin x}{x}\right)=$
Using the substitution of equivalent infinitesimals:
$e^{2 x}-1 \sim 2 x$, as $x \rightarrow 0(2 x \rightarrow 0)$
$e^{x}-1 \sim x$, as $x \rightarrow 0$
$\sin 2 x \sim 2 x$, as $x \rightarrow 0(2 x \rightarrow 0)$
$\sin x \sim x$, as $x \rightarrow 0$
We get:
$$
=\frac{\lim _{x \rightarrow 0} \frac{2 x}{x}-\lim _{x \rightarrow 0} \frac{x}{x}}{\lim _{x \rightarrow 0} \frac{2 x}{x}-\lim _{x \rightarrow 0} \frac{x}{x}}=\frac{\lim _{x \rightarrow 0} 2-\lim _{x \rightarrow 0} 1}{\lim _{x \rightarrow 0} 2-\lim _{x \rightarrow 0} 1}=\frac{2-1}{2-1}=1
$$
## Problem Kuznetsov Limits 15-28
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Condition of the problem
Calculate the limit of the function:
$$
\lim _{x \rightarrow 0}\left(6-\frac{5}{\cos x}\right)^{\operatorname{tg}^{2} x}
$$
|
## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 0}\left(6-\frac{5}{\cos x}\right)^{\operatorname{tg}^{2} x}=\left(6-\frac{5}{\cos 0}\right)^{\operatorname{tg}^{2} 0}= \\
& =\left(6-\frac{5}{1}\right)^{0^{2}}=1^{0}=1
\end{aligned}
$$
## Problem Kuznetsov Limits $18-28$
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow \frac{\pi}{2}}(\sin x)^{\frac{18 \sin x}{\operatorname{ctg} x}}$
|
## Solution
Substitution:
$$
\begin{aligned}
& x=y+\frac{\pi}{2} \Rightarrow y=x-\frac{\pi}{2} \\
& x \rightarrow \frac{\pi}{2} \Rightarrow y \rightarrow 0
\end{aligned}
$$
We obtain:
$$
\begin{aligned}
& \lim _{x \rightarrow \frac{\pi}{2}}(\sin x)^{\frac{18 \sin x}{\operatorname{ctg} x}}=\lim _{y \rightarrow 0}\left(\sin \left(y+\frac{\pi}{2}\right)\right)^{\frac{18 \sin \left(y+\frac{\pi}{2}\right)}{\operatorname{ctg}\left(y+\frac{\pi}{2}\right)}}= \\
& =\lim _{y \rightarrow 0}(\cos y)^{\frac{18 \cos y}{-\operatorname{tg} y}}=\lim _{y \rightarrow 0}\left(e^{\ln (\cos y)}\right)^{-\frac{18 \cos y}{\operatorname{tg} y}}= \\
& =\lim _{y \rightarrow 0} e^{-\frac{18 \cos y}{\operatorname{tg} y} \cdot \ln (\cos y)}=\exp \left\{\lim _{y \rightarrow 0}-\frac{18 \cos y}{\operatorname{tg} y} \cdot \ln (\cos y)\right\}= \\
& =\exp \left\{\lim _{y \rightarrow 0}-\frac{18 \cos y}{\operatorname{tg} y} \cdot \ln \left(1-2 \sin ^{2} \frac{y}{2}\right)\right\}=
\end{aligned}
$$
Using the substitution of equivalent infinitesimals:
$\ln \left(1-2 \sin ^{2} \frac{y}{2}\right) \sim-2 \sin ^{2} \frac{y}{2}$, as $\quad y \rightarrow 0\left(-2 \sin ^{2} \frac{y}{2} \rightarrow 0\right)$ $\operatorname{tg} y \sim y_{\text {, as }} y \rightarrow 0$
We get:
$=\exp \left\{\lim _{y \rightarrow 0}-\frac{18 \cos y}{y} \cdot\left(-2 \sin ^{2} \frac{y}{2}\right)\right\}=$
Using the substitution of equivalent infinitesimals:
$$
\sin \frac{y}{2} \sim \frac{y}{2}_{\text {as }} y \rightarrow 0\left(\frac{y}{2} \rightarrow 0\right)
$$
We get:
$$
\begin{aligned}
& =\exp \left\{\lim _{y \rightarrow 0} \frac{36 \cos y}{y} \cdot\left(\frac{y}{2}\right)^{2}\right\}=\exp \left\{\lim _{y \rightarrow 0} 9 \cdot y \cdot \cos y\right\}= \\
& =\exp \{9 \cdot 0 \cdot \cos 0\}=\exp \{0\}=e^{0}=1
\end{aligned}
$$
## Problem Kuznetsov Limits 19-28
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow 1}(\sqrt[3]{x}+x-1)^{\sin \left(\frac{\pi x}{4}\right)}
$$
|
## Solution
$\lim _{x \rightarrow 1}(\sqrt[3]{x}+x-1)^{\sin \left(\frac{\pi x}{4}\right)}=(\sqrt[3]{1}+1-1)^{\sin \left(\frac{\pi \cdot 1}{4}\right)}=(1)^{\frac{\sqrt{2}}{2}}=1$
## Problem Kuznetsov Limits 20-28
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{1 / \sqrt{2}} \frac{d x}{\left(1-x^{2}\right) \sqrt{1-x^{2}}}
$$
|
## Solution
$$
\int_{0}^{1 / \sqrt{2}} \frac{d x}{\left(1-x^{2}\right) \sqrt{1-x^{2}}}=
$$
Substitution:
$$
\begin{aligned}
& x=\sin t ; d x=\cos t d t \\
& x=0 \Rightarrow t=\arcsin 0=0 \\
& x=\frac{1}{\sqrt{2}} \Rightarrow t=\arcsin \frac{1}{\sqrt{2}}=\frac{\pi}{4}
\end{aligned}
$$
We get:
$$
\begin{aligned}
& =\int_{0}^{\pi / 4} \frac{\cos t d t}{\sqrt{\left(1-\sin ^{2} t\right)^{3}}}=\int_{0}^{\pi / 4} \frac{\cos t d t}{\sqrt{\left(1-\sin ^{2} t\right)^{3}}}=\int_{0}^{\pi / 4} \frac{\cos t d t}{\cos ^{3} t}=\int_{0}^{\pi / 4} \frac{d t}{\cos ^{2} t}= \\
& =\left.\operatorname{tg} t\right|_{0} ^{\pi / 4}=\operatorname{tg} \frac{\pi}{4}-\operatorname{tg} 0=1
\end{aligned}
$$
Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\� \%D0\%9A\%D1\%83\%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+12-29$ » Categories: Kuznetsov's Problem Book Integrals Problem 12 | Integrals
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Created by Geeteatoo
## Problem Kuznetsov Integrals 12-30
## Material from PlusPi
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Task Condition
Calculate the area of the parallelogram constructed on vectors $a_{\text {and }} b$.
$$
\begin{aligned}
& a=5 p-q \\
& b=p+q \\
& |p|=5 \\
& |q|=3 \\
& (\widehat{p, q})=\frac{5 \pi}{6}
\end{aligned}
$$
|
## Solution
The area of the parallelogram constructed on vectors $a$ and $b$ is numerically equal to the modulus of their vector product:
$S=|a \times b|$
We compute $a \times b$ using the properties of the vector product:
$a \times b=(5 p-q) \times(p+q)=5 \cdot p \times p+5 \cdot p \times q-q \times p-q \times q=$
$=5 \cdot p \times q-q \times p=5 \cdot p \times q+p \times q=(5+1) \cdot p \times q=6 \cdot p \times q$
We compute the area:
$S=|a \times b|=|6 \cdot p \times q|=6 \cdot|p \times q|=6 \cdot|p| \cdot|q| \cdot \sin (\widehat{p, q})=$
$=6 \cdot 5 \cdot 3 \cdot \sin \frac{5 \pi}{6}=90 \cdot \sin \frac{5 \pi}{6}=90 \cdot \frac{1}{2}=45$
Thus, the area of the parallelogram constructed on vectors $a$ and $b$ is 45.
## Problem Kuznetsov Analytical Geometry 5-26
|
45
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Task Condition
Are the vectors $a, b$ and $c$ coplanar?
$a=\{1 ;-1 ; 4\}$
$b=\{1 ; 0 ; 3\}$
$c=\{1 ;-3 ; 8\}$
|
## Solution
For three vectors to be coplanar (lie in the same plane or parallel planes), it is necessary and sufficient that their scalar triple product $(a, b, c)$ be equal to zero.
$(a, b, c)=\left|\begin{array}{ccc}1 & -1 & 4 \\ 1 & 0 & 3 \\ 1 & -3 & 8\end{array}\right|=$
$=1 \cdot\left|\begin{array}{cc}0 & 3 \\ -3 & 8\end{array}\right|-(-1) \cdot\left|\begin{array}{ll}1 & 3 \\ 1 & 8\end{array}\right|+4 \cdot\left|\begin{array}{cc}1 & 0 \\ 1 & -3\end{array}\right|=$
$=1 \cdot 9+1 \cdot 5+4 \cdot(-3)=9+5-12=2$
Since $(a, b, c)=2 \neq 0$, the vectors $a, b$ and $c$ are not coplanar.
## Problem Kuznetsov Analytic Geometry 6-26
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the numerical sequence:
$$
\lim _{n \rightarrow \infty} \frac{\sqrt{n+3}-\sqrt[3]{8 n^{3}+3}}{\sqrt[4]{n+4}-\sqrt[5]{n^{5}+5}}
$$
|
## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{\sqrt{n+3}-\sqrt[3]{8 n^{3}+3}}{\sqrt[4]{n+4}-\sqrt[5]{n^{5}+5}}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n}\left(\sqrt{n+3}-\sqrt[3]{8 n^{3}+3}\right)}{\frac{1}{n}\left(\sqrt[4]{n+4}-\sqrt[5]{n^{5}+5}\right)}= \\
& =\lim _{n \rightarrow \infty} \frac{\sqrt{\frac{1}{n}+\frac{3}{n^{2}}}-\sqrt[3]{8+\frac{3}{n^{3}}}}{\sqrt[4]{\frac{1}{n^{3}}+\frac{4}{n^{4}}}-\sqrt[5]{1+\frac{5}{n^{5}}}}=\frac{\sqrt{0+0}-\sqrt[3]{8+0}}{\sqrt[4]{0+0}-\sqrt[5]{1+0}}=\frac{-2}{-1}=2
\end{aligned}
$$
## Problem Kuznetsov Limits 4-20
|
2
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{(2 n+1)!+(2 n+2)!}{(2 n+3)!-(2 n+2)!}$
|
## Solution
$\lim _{n \rightarrow \infty} \frac{(2 n+1)!+(2 n+2)!}{(2 n+3)!-(2 n+2)!}=\lim _{n \rightarrow \infty} \frac{(2 n+1)!+(2 n+2)!}{(2 n+3) \cdot(2 n+2)!-(2 n+2)!}=$
$$
=\lim _{n \rightarrow \infty} \frac{(2 n+1)!+(2 n+2)!}{(2 n+2)!((2 n+3)-1)}=\lim _{n \rightarrow \infty} \frac{(2 n+1)!+(2 n+2)!}{(2 n+2)!\cdot(2 n+2)}=
$$
$$
\begin{aligned}
& =\lim _{n \rightarrow \infty}\left(\frac{(2 n+1)!}{(2 n+2)!\cdot(2 n+2)}+\frac{(2 n+2)!}{(2 n+2)!\cdot(2 n+2)}\right)= \\
& =\lim _{n \rightarrow \infty}\left(\frac{1}{(2 n+2) \cdot(2 n+2)}+\frac{1}{2 n+2}\right)=0+0=0
\end{aligned}
$$
|
0
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 2} \frac{x^{3}-3 x-2}{x-2}$
|
## Solution
$\lim _{x \rightarrow 2} \frac{x^{3}-3 x-2}{x-2}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow 2} \frac{(x-2)\left(x^{2}+2 x+1\right)}{x-2}=$
$=\lim _{x \rightarrow 2}\left(x^{2}+2 x+1\right)=2^{2}+2 \cdot 2+1=4+4+1=9$
## Problem Kuznetsov Limits 10-20
|
9
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{e^{2 x}-e^{-5 x}}{2 \sin x-\tan x}$
|
## Solution
$\lim _{x \rightarrow 0} \frac{e^{2 x}-e^{-5 x}}{2 \sin x-\tan x}=\lim _{x \rightarrow 0} \frac{\left(e^{2 x}-1\right)-\left(e^{-5 x}-1\right)}{2 \sin x-\tan x}=$
$=\lim _{x \rightarrow 0} \frac{\frac{1}{x}\left(\left(e^{2 x}-1\right)-\left(e^{-5 x}-1\right)\right)}{\frac{1}{x}(2 \sin x-\tan x)}=$
$=\frac{\lim _{x \rightarrow 0} \frac{1}{x}\left(\left(e^{2 x}-1\right)-\left(e^{-5 x}-1\right)\right)}{\lim _{x \rightarrow 0} \frac{1}{x}(2 \sin x-\tan x)}=$
$=\left(\lim _{x \rightarrow 0} \frac{e^{2 x}-1}{x}-\lim _{x \rightarrow 0} \frac{e^{-5 x}-1}{x}\right) /\left(\lim _{x \rightarrow 0} \frac{2 \sin x}{x}-\lim _{x \rightarrow 0} \frac{\tan x}{x}\right)=$
Using the substitution of equivalent infinitesimals:
$e^{2 x}-1 \sim 2 x$, as $x \rightarrow 0(2 x \rightarrow 0)$
$e^{-5 x}-1 \sim -5 x$, as $x \rightarrow 0(-5 x \rightarrow 0)$
$\sin x \sim x$, as $x \rightarrow 0$
$\tan x \sim x$, as $x \rightarrow 0$
We get:
$=\frac{\lim _{x \rightarrow 0} \frac{2 x}{x}-\lim _{x \rightarrow 0} \frac{-5 x}{x}}{\lim _{x \rightarrow 0} \frac{2 x}{x}-\lim _{x \rightarrow 0} \frac{x}{x}}=\frac{\lim _{x \rightarrow 0} 2-\lim _{x \rightarrow 0}-5}{\lim _{x \rightarrow 0} 2-\lim _{x \rightarrow 0} 1}=\frac{2+5}{2-1}=7$
## Problem Kuznetsov Limits 15-20
|
7
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{2+\ln \left(e+x \sin \left(\frac{1}{x}\right)\right)}{\cos x+\sin x}$
|
## Solution
Since $\sin \left(\frac{1}{x}\right)_{\text {- is bounded as }} x \rightarrow 0$, then
$x \sin \left(\frac{1}{x}\right) \rightarrow 0 \quad$ as $x \rightarrow 0$
Then:
$\lim _{x \rightarrow 0} \frac{2+\ln \left(e+x \sin \left(\frac{1}{x}\right)\right)}{\cos x+\sin x}=\frac{2+\ln (e+0)}{\cos 0+\sin 0}=\frac{2+\ln e}{1+0}=\frac{2+1}{1+0}=3$
|
3
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Task Condition
Find the angle between the planes
$x-y \sqrt{2}+z-1=0$
$x+y \sqrt{2}-z+3=0$
|
## Solution
The dihedral angle between planes is equal to the angle between their normal vectors. The normal vectors of the given planes are:
$\overrightarrow{n_{1}}=\{1 ;-\sqrt{2} ; 1\}$
$\overrightarrow{n_{2}}=\{1 ; \sqrt{2} ;-1\}$
The angle $\phi$ between the planes is determined by the formula:
$$
\begin{aligned}
& \cos \phi=\frac{\left(\overrightarrow{n_{1}}, \overrightarrow{n_{2}}\right)}{\left|\overrightarrow{n_{1}}\right| \cdot\left|\overrightarrow{n_{2}}\right|}=\frac{1 \cdot 1+(-\sqrt{2}) \cdot \sqrt{2}+1 \cdot(-1)}{\sqrt{1^{2}+(-\sqrt{2})^{2}+1^{2}} \cdot \sqrt{1^{2}+(\sqrt{2})^{2}+(-1)^{2}}}= \\
& =\frac{1-2-1}{\sqrt{1+2+1} \cdot \sqrt{1+2+1}}=\frac{-2}{\sqrt{4} \cdot \sqrt{4}}=-\frac{1}{2} \\
& \phi=\arccos -\frac{1}{2}=120^{\circ}
\end{aligned}
$$
## Problem Kuznetsov Analytic Geometry 10-6
|
120
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Task Condition
Let $k$ be the coefficient of similarity transformation with the center at the origin. Is it true that point $A$ belongs to the image of plane $a$?
$A\left(\frac{1}{2} ; \frac{1}{3} ; 1\right)$
$a: 2 x-3 y+3 z-2=0$
$k=1.5$
|
## Solution
When transforming similarity with the center at the origin of the plane
$a: A x+B y+C z+D=0_{\text{and coefficient }} k$ transitions to the plane
$a^{\prime}: A x+B y+C z+k \cdot D=0$. We find the image of the plane $a$:
$a^{\prime}: 2 x-3 y+3 z-3=0$
Substitute the coordinates of point $A$ into the equation $a^{\prime}$:
$2 \cdot \frac{1}{2}-3 \cdot \frac{1}{3}+3 \cdot 1-3=0$
$1-1+3-3=0$
$0=0$
Since $0=0$, point $A$ belongs to the image of the plane $a$.
Problem Kuznetsov Analytical Geometry 12-6
|
0
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{\sqrt{n+6}-\sqrt{n^{2}-5}}{\sqrt[3]{n^{3}+3}+\sqrt[4]{n^{3}+1}}$
|
## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{\sqrt{n+6}-\sqrt{n^{2}-5}}{\sqrt[3]{n^{3}+3}+\sqrt[4]{n^{3}+1}}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n}\left(\sqrt{n+6}-\sqrt{n^{2}-5}\right)}{\frac{1}{n}\left(\sqrt[3]{n^{3}+3}+\sqrt[4]{n^{3}+1}\right)}= \\
& =\lim _{n \rightarrow \infty} \frac{\sqrt{\frac{1}{n}+\frac{6}{n^{2}}}-\sqrt{1-\frac{5}{n^{2}}}}{\sqrt[3]{1+\frac{3}{n^{3}}}+\sqrt[4]{\frac{1}{n}+\frac{1}{n^{4}}}}=\frac{\sqrt{0+0}-\sqrt{1-0}}{\sqrt[3]{1+0}+\sqrt[4]{0+0}}=\frac{-1}{1}=-1
\end{aligned}
$$
## Problem Kuznetsov Limits 4-27
|
-1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{2^{n}+7^{n}}{2^{n}-7^{n-1}}$
|
## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{2^{n}+7^{n}}{2^{n}-7^{n-1}}=\lim _{n \rightarrow \infty} \frac{\frac{1}{7^{n}}\left(2^{n}+7^{n}\right)}{\frac{1}{7^{n}}\left(2^{n}-7^{n-1}\right)}= \\
& =\lim _{n \rightarrow \infty} \frac{\left(\frac{2}{7}\right)^{n}+1}{\left(\frac{2}{7}\right)^{n}-\frac{1}{7}}=\frac{0+1}{0-\frac{1}{7}}=-7
\end{aligned}
$$
## Problem Kuznetsov Limits 6-27
|
-7
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow 16} \frac{\sqrt[4]{x}-2}{\sqrt[3]{(\sqrt{x}-4)^{2}}}
$$
|
## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 16} \frac{\sqrt[4]{x}-2}{\sqrt[3]{(\sqrt{x}-4)^{2}}}=\lim _{x \rightarrow 16} \frac{\sqrt[4]{x}-2}{\sqrt[3]{(\sqrt[4]{x}-2)^{2}(\sqrt[4]{x}+2)^{2}}}= \\
& =\lim _{x \rightarrow 16} \frac{\sqrt[4]{x}-2}{(\sqrt[4]{x}-2)^{\frac{2}{3}} \sqrt[3]{(\sqrt[4]{x}+2)^{2}}}=\lim _{x \rightarrow 16} \frac{(\sqrt[4]{x}-2)^{\frac{1}{3}}}{\sqrt[3]{(\sqrt[4]{x}+2)^{2}}}= \\
& =\frac{(\sqrt[4]{16}-2)^{\frac{1}{3}}}{\sqrt[3]{(\sqrt[4]{16}+2)^{2}}}=\frac{(2-2)^{\frac{1}{3}}}{\sqrt[3]{(2+2)^{2}}}=\frac{0}{\sqrt[3]{4^{2}}}=0
\end{aligned}
$$
## Problem Kuznetsov Limits 11-27
|
0
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0}\left(\tan\left(\frac{\pi}{4}-x\right)\right)^{\left(e^{x}-1\right) / x}$
|
## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 0}\left(\tan\left(\frac{\pi}{4}-x\right)\right)^{\left(e^{x}-1\right) / x}=\left(\lim _{x \rightarrow 0} \tan\left(\frac{\pi}{4}-x\right)\right)^{\lim _{x \rightarrow 0}\left(e^{x}-1\right) / x}= \\
& =\left(\tan\left(\frac{\pi}{4}-0\right)\right)^{\lim _{x \rightarrow 0}\left(e^{x}-1\right) / x}=(1)^{\lim _{x \rightarrow 0}\left(e^{x}-1\right) / x}=
\end{aligned}
$$
Using the substitution of equivalent infinitesimals:
$e^{x}-1 \sim x$, as $x \rightarrow 0$
We get:
$=1^{\lim _{x \rightarrow 0} \frac{x}{x}}=1^{\lim _{x \rightarrow 0} 1}=1^{1}=1$
## Problem Kuznetsov Limits 18-27
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow \frac{\pi}{2}}(\cos x+1)^{\sin x}$
|
## Solution
$\lim _{x \rightarrow \frac{\pi}{2}}(\cos x+1)^{\sin x}=\left(\cos \frac{\pi}{2}+1\right)^{\sin \frac{\pi}{2}}=(0+1)^{1}=1^{1}=1$
## Problem Kuznetsov Limits 20-27
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Based on the definition of the derivative, find $f^{\prime}(0)$:
$$
f(x)=\left\{\begin{array}{c}
2 x^{2}+x^{2} \cos \frac{1}{9 x}, x \neq 0 \\
0, x=0
\end{array}\right.
$$
|
## Solution
By definition, the derivative at the point $x=0$:
$f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}$
Based on the definition, we find:
$$
\begin{aligned}
& f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim _{\Delta x \rightarrow 0} \frac{2 \Delta x^{2}+\Delta x^{2} \cos \frac{1}{9 \Delta x}-0}{\Delta x}= \\
& =\lim _{\Delta x \rightarrow 0} \frac{2 \Delta x^{2}+\Delta x^{2} \cos \frac{1}{9 \Delta x}}{\Delta x}=\lim _{\Delta x \rightarrow 0}\left(2 \Delta x+\Delta x \cos \frac{1}{9 \Delta x}\right)=
\end{aligned}
$$
Since $\cos \frac{1}{9 \Delta x}$ is bounded, then
$\Delta x \cdot \cos \frac{1}{9 \Delta x} \rightarrow 0$, as $\Delta x \rightarrow 0$
Then:
$=2 \cdot 0+0=0$
Therefore, $f^{\prime}(0)=0$
## Problem Kuznetsov Differentiation 2-14
|
0
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Based on the definition of the derivative, find $f^{\prime}(0):$
$f(x)=\left\{\begin{array}{c}\tan\left(x^{3}+x^{2} \sin \left(\frac{2}{x}\right)\right), x \neq 0 \\ 0, x=0\end{array}\right.$
|
## Solution
By definition, the derivative at the point $x=0$:
$f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}$
Based on the definition, we find:
$$
\begin{aligned}
& f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim _{\Delta x \rightarrow 0} \frac{\tan\left(\Delta x^{3}+\Delta x^{2} \sin \left(\frac{2}{\Delta x}\right)\right)-0}{\Delta x}= \\
& =\lim _{\Delta x \rightarrow 0} \frac{\tan\left(\Delta x^{3}+\Delta x^{2} \sin \left(\frac{2}{\Delta x}\right)\right)}{\Delta x}=
\end{aligned}
$$
Using the substitution of equivalent infinitesimals:
$$
\begin{aligned}
& \tan\left(\Delta x^{3}+\Delta x^{2} \sin \left(\frac{2}{\Delta x}\right)\right) \sim \Delta x^{3}+\Delta x^{2} \sin \left(\frac{2}{\Delta x}\right), \text { as } \\
& \Delta x \rightarrow 0\left(\Delta x^{3}+\Delta x^{2} \sin \left(\frac{2}{\Delta x}\right) \rightarrow 0\right)
\end{aligned}
$$
We get:
$=\lim _{\Delta x \rightarrow 0} \frac{\Delta x^{3}+\Delta x^{2} \sin \left(\frac{2}{\Delta x}\right)}{\Delta x}=\lim _{\Delta x \rightarrow 0} \Delta x^{2}+\Delta x \sin \left(\frac{2}{\Delta x}\right)=$
Since $\sin \left(\frac{2}{\Delta x}\right)$ is bounded, then
$$
\Delta x \cdot \sin \left(\frac{2}{\Delta x}\right) \rightarrow 0 \quad \text { as } \Delta x \rightarrow 0
$$
Thus:
$=0^{2}+0=0$
Therefore, $f^{\prime}(0)=0$
## Problem Kuznetsov Differentiation 2-1
|
0
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Condition of the problem
To derive the equation of the normal to the given curve at the point with abscissa $x_{0}$.
$y=\frac{4 x-x^{2}}{4}, x_{0}=2$
|
## Solution
Let's find $y^{\prime}:$
$$
y^{\prime}=\left(\frac{4 x-x^{2}}{4}\right)^{\prime}=\frac{4-2 x}{4}=\frac{2-x}{2}
$$
Then:
$y_{0}^{\prime}=y^{\prime}\left(x_{0}\right)=\frac{2-x_{0}}{2}=\frac{2-2}{2}=0$
Since $y^{\prime}\left(x_{0}\right)=0$, the equation of the normal line is:
$x=x_{0}$
$x=2$
Thus, the equation of the normal line is:
$x=2$
## Problem Kuznetsov Differentiation 3-1
|
2
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Based on the definition of the derivative, find $f^{\prime}(0)$:
$$
f(x)=\left\{\begin{array}{c}
3^{x^{2} \sin \frac{2}{x}}-1+2 x, x \neq 0 \\
0, x=0
\end{array}\right.
$$
|
## Solution
By definition, the derivative at the point $x=0$:
$f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}$
Based on the definition, we find:
$$
\begin{aligned}
& f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim _{\Delta x \rightarrow 0}\left(3^{\Delta x^{2} \sin \frac{2}{\Delta x}}-1+2 x-0\right) / \Delta x= \\
& =\lim _{\Delta x \rightarrow 0}\left(\left(e^{\ln 3}\right)^{\Delta x^{2} \sin \frac{2}{\Delta x}}-1-2 \Delta x\right) / \Delta x= \\
& =\lim _{\Delta x \rightarrow 0}\left(e^{\ln 3 \cdot \Delta x^{2} \sin \frac{2}{\Delta x}}-1\right) / \Delta x-\lim _{\Delta x \rightarrow 0} \frac{2 \Delta x}{\Delta x}=
\end{aligned}
$$
Using the substitution of equivalent infinitesimals:
$$
e^{\ln 3 \cdot \Delta x^{2} \sin \frac{2}{\Delta x}}-1 \sim \Delta \ln 3 \cdot \Delta x^{2} \sin \frac{2}{\Delta x}_{, \text {as }} \Delta x \rightarrow 0\left(\ln 3 \cdot \Delta x^{2} \sin \frac{2}{\Delta x} \rightarrow 0\right)
$$
We get:
$$
\begin{aligned}
& =\lim _{\Delta x \rightarrow 0}\left(\ln 3 \cdot \Delta x^{2} \sin \frac{2}{\Delta x}\right) / \Delta x-\lim _{\Delta x \rightarrow 0} 2= \\
& =\lim _{\Delta x \rightarrow 0} \ln 3 \cdot \Delta x \sin \frac{2}{\Delta x}-2=
\end{aligned}
$$
Since $\sin \frac{2}{\Delta x}$ is bounded, then
$$
\Delta x \cdot \sin \frac{2}{\Delta x} \rightarrow 0, \text { as } \Delta x \rightarrow 0
$$
Thus
$=\ln 3 \cdot 0-2=-2$
Therefore, $f^{\prime}(0)=-2$
## Problem Kuznetsov Differentiation $2-21$
|
-2
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{(n+1)^{3}-(n-1)^{3}}{(n+1)^{2}+(n-1)^{2}}$
|
## Solution
$\lim _{n \rightarrow \infty} \frac{(n+1)^{3}-(n-1)^{3}}{(n+1)^{2}+(n-1)^{2}}=\lim _{n \rightarrow \infty} \frac{n^{3}+3 n^{2}+3 n+1-n^{3}+3 n^{2}-3 n+1}{n^{2}+2 n+1+n^{2}-2 n+1}=$
$=\lim _{n \rightarrow \infty} \frac{6 n^{2}+2}{2 n^{2}+2}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n^{2}}\left(3 n^{2}+1\right)}{\frac{1}{n^{2}}\left(n^{2}+1\right)}=\lim _{n \rightarrow \infty} \frac{3+\frac{1}{n^{2}}}{1+\frac{1}{n^{2}}}=\frac{3+0}{1+0}=3$
## Problem Kuznetsov Limits 3-26
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{n \sqrt{71 n}-\sqrt[3]{64 n^{6}+9}}{(n-\sqrt[3]{n}) \sqrt{11+n^{2}}}$
|
## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{n \sqrt{71 n}-\sqrt[3]{64 n^{6}+9}}{(n-\sqrt[3]{n}) \sqrt{11+n^{2}}}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n^{2}}\left(n \sqrt{71 n}-\sqrt[3]{64 n^{6}+9}\right)}{\frac{1}{n^{2}}(n-\sqrt[3]{n}) \sqrt{11+n^{2}}}= \\
& =\lim _{n \rightarrow \infty} \frac{\sqrt{\frac{71}{n}}-\sqrt[3]{64+\frac{9}{n^{6}}}}{\left(\frac{1}{n}(n-\sqrt[3]{n})\right)\left(\frac{1}{n} \sqrt{11+n^{2}}\right)}=\lim _{n \rightarrow \infty} \frac{\sqrt{\frac{71}{n}}-\sqrt[3]{64+\frac{9}{n^{6}}}}{\left(1-\sqrt[3]{\frac{1}{n^{2}}}\right) \sqrt{\frac{11}{n^{2}}+1}}= \\
& =\frac{\sqrt{0}-\sqrt[3]{64+0}}{(1-\sqrt[3]{0}) \sqrt{0+1}}=\frac{-4}{1}=-4
\end{aligned}
$$
## Problem Kuznetsov Limits 4-26
|
-4
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the numerical sequence:
$$
\lim _{n \rightarrow \infty} \frac{1-2+3-4+\ldots+(2 n-1)-2 n}{\sqrt[3]{n^{3}+2 n+2}}
$$
|
## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{1-2+3-4+\ldots+(2 n-1)-2 n}{\sqrt[3]{n^{3}+2 n+2}}= \\
& =\{1-2=3-4=\ldots=(2 n-1)-2 n=-1\}= \\
& =\lim _{n \rightarrow \infty} \frac{-1 \cdot n}{\sqrt[3]{n^{3}+2 n+2}}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n} \cdot(-1) \cdot n}{\frac{1}{n} \sqrt[3]{n^{3}+2 n+2}}= \\
& =\lim _{n \rightarrow \infty} \frac{-1}{\sqrt[3]{1+\frac{2}{n^{2}}+\frac{2}{n^{3}}}}=\frac{-1}{\sqrt[3]{1+0+0}}=\frac{-1}{1}=-1
\end{aligned}
$$
## Problem Kuznetsov Limits 6-26
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{e^{x}-e^{-2 x}}{x+\sin x^{2}}$
|
## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{e^{x}-e^{-2 x}}{x+\sin x^{2}}=\lim _{x \rightarrow 0} \frac{\left(e^{x}-1\right)-\left(e^{-2 x}-1\right)}{x+\sin x^{2}}= \\
& =\lim _{x \rightarrow 0} \frac{\frac{1}{x}\left(\left(e^{x}-1\right)-\left(e^{-2 x}-1\right)\right)}{\frac{1}{x}\left(x+\sin x^{2}\right)}=
\end{aligned}
$$
$$
\begin{aligned}
& =\frac{\lim _{x \rightarrow 0} \frac{1}{x}\left(\left(e^{x}-1\right)-\left(e^{-2 x}-1\right)\right)}{\lim _{x \rightarrow 0} \frac{1}{x}\left(x+\sin x^{2}\right)}= \\
& =\left(\lim _{x \rightarrow 0} \frac{e^{x}-1}{x}-\lim _{x \rightarrow 0} \frac{e^{-2 x}-1}{x}\right) /\left(\lim _{x \rightarrow 0} \frac{x}{x}+\lim _{x \rightarrow 0} \frac{\sin x^{2}}{x}\right)=
\end{aligned}
$$
Using the substitution of equivalent infinitesimals:
$e^{x}-1 \sim x$, as $x \rightarrow 0$
$e^{-2 x}-1 \sim-2 x$, as $x \rightarrow 0(-2 x \rightarrow 0)$
$\sin x^{2} \sim x^{2}$, as $x \rightarrow 0\left(x^{2} \rightarrow 0\right)$
We get:
$=\frac{\lim _{x \rightarrow 0} \frac{x}{x}-\lim _{x \rightarrow 0} \frac{-2 x}{x}}{\lim _{x \rightarrow 0} \frac{x}{x}+\lim _{x \rightarrow 0} \frac{x^{2}}{x}}=\frac{\lim _{x \rightarrow 0} 1-\lim _{x \rightarrow 0}-2}{\lim _{x \rightarrow 0} 1+\lim _{x \rightarrow 0} x}=\frac{1+2}{1+0}=3$
## Problem Kuznetsov Limits 15-26
|
3
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0}\left(\frac{\sin 5 x^{2}}{\sin x}\right)^{\frac{1}{x+6}}$
|
## Solution
$\lim _{x \rightarrow 0}\left(\frac{\sin 5 x^{2}}{\sin x}\right)^{\frac{1}{x+6}}=\left(\lim _{x \rightarrow 0} \frac{\sin 5 x^{2}}{\sin x}\right)^{\lim _{x \rightarrow 0} \frac{1}{x+6}}=$
$=\left(\lim _{x \rightarrow 0} \frac{\sin 5 x^{2}}{\sin x}\right)^{\frac{1}{0+6}}=\left(\lim _{x \rightarrow 0} \frac{\sin 5 x^{2}}{\sin x}\right)^{\frac{1}{6}}=$
Using the substitution of equivalent infinitesimals:
$\sin 5 x^{2} \sim 5 x^{2}$, as $x \rightarrow 0\left(5 x^{2} \rightarrow 0\right)$
$\sin x \sim x$, as $x \rightarrow 0(x \rightarrow 0)$
We get:
$=\left(\lim _{x \rightarrow 0} \frac{5 x^{2}}{x}\right)^{\frac{1}{6}}=\left(\lim _{x \rightarrow 0} 5 x\right)^{\frac{1}{6}}=0^{\frac{1}{6}}=0$
## Problem Kuznetsov Limits 18-26
|
0
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## problem statement
Let $k$ be the coefficient of similarity transformation with the center at the origin. Is it true that point $A$ belongs to the image of plane $a$?
$A(-2; -1; 1)$
$a: x-2y+6z-10=0$
$k=\frac{3}{5}$
|
## Solution
When transforming similarity with the center at the origin of the plane
$a: A x+B y+C z+D=0_{\text{and coefficient }} k$ transitions to the plane
$a^{\prime}: A x+B y+C z+k \cdot D=0$. We find the image of the plane $a$:
$a^{\prime}: x-2 y+6 z-6=0$
Substitute the coordinates of point $A$ into the equation of $a^{\prime}$:
$-2-2 \cdot(-1)+6 \cdot 1-6=0$
$-2+2+6-6=0$
$0=0$
Since $0=0$, point $A$ belongs to the image of the plane $a$.
## Problem Kuznetsov Analytical Geometry 12-15
|
0
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the numerical sequence:
$$
\lim _{n \rightarrow \infty} \frac{n \sqrt[5]{n}-\sqrt[3]{27 n^{6}+n^{2}}}{(n+\sqrt[4]{n}) \sqrt{9+n^{2}}}
$$
|
## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{n \sqrt[5]{n}-\sqrt[3]{27 n^{6}+n^{2}}}{(n+\sqrt[4]{n}) \sqrt{9+n^{2}}}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n^{2}}\left(n \sqrt[5]{n}-\sqrt[3]{27 n^{6}+n^{2}}\right)}{\frac{1}{n^{2}}(n+\sqrt[4]{n}) \sqrt{9+n^{2}}}= \\
& =\lim _{n \rightarrow \infty} \frac{\sqrt[5]{\frac{1}{n^{4}}}-\sqrt[3]{27+\frac{1}{n^{4}}}}{\left(\frac{1}{n}(n+\sqrt[4]{n})\right)\left(\frac{1}{n} \sqrt{9+n^{2}}\right)}=\lim _{n \rightarrow \infty} \frac{\sqrt[5]{\frac{1}{n^{4}}}-\sqrt[3]{27+\frac{1}{n^{4}}}}{\left(1+\sqrt[4]{\frac{1}{n^{3}}}\right) \sqrt{\frac{9}{n^{2}}+1}}= \\
& =\frac{\sqrt[5]{0}-\sqrt[3]{27+0}}{(1+\sqrt[4]{0}) \sqrt{0+1}}=\frac{-3}{1}=-3
\end{aligned}
$$
## Problem Kuznetsov Limits 4-6
|
-3
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the numerical sequence:
$$
\lim _{n \rightarrow \infty} \frac{1+3+5+\ldots+(2 n-1)}{1+2+3+\ldots+n}
$$
|
## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{1+3+5+\ldots+(2 n-1)}{1+2+3+\ldots+n}= \\
& =\lim _{n \rightarrow \infty} \frac{1}{1+2+3+\ldots+n} \cdot \frac{(1+(2 n-1)) n}{2}= \\
& =\lim _{n \rightarrow \infty} \frac{1}{1+2+3+\ldots+n} \cdot n^{2}=\lim _{n \rightarrow \infty} \frac{1}{\frac{(1+n) n}{2}} \cdot n^{2}= \\
& =\lim _{n \rightarrow \infty} \frac{2 n}{1+n}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n} 2 n}{\frac{1}{n}(1+n)}=\lim _{n \rightarrow \infty} \frac{2}{\frac{1}{n}+1}= \\
& =\frac{2}{0+1}=2
\end{aligned}
$$
## Problem Kuznetsov Limits 6-6
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Condition of the problem
Calculate the limit of the function:
$\lim _{x \rightarrow-1} \frac{\left(x^{3}-2 x-1\right)^{2}}{x^{4}+2 x+1}$
|
Solution
$\lim _{x \rightarrow-1} \frac{\left(x^{3}-2 x-1\right)^{2}}{x^{4}+2 x+1}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow-1} \frac{\left(x^{2}-x-1\right)^{2}(x+1)^{2}}{\left(x^{3}-x^{2}+x+1\right)(x+1)}=$
$=\lim _{x \rightarrow-1} \frac{\left(x^{2}-x-1\right)^{2}(x+1)}{x^{3}-x^{2}+x+1}=\frac{\left((-1)^{2}-(-1)-1\right)^{2}(-1+1)}{(-1)^{3}-(-1)^{2}+(-1)+1}=$
$=\frac{(1+1-1)^{2} \cdot 0}{-1-1-1+1}=\frac{0}{-2}=0$
## Problem Kuznetsov Limits 10-6
|
0
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{e^{2 x}-e^{3 x}}{\operatorname{arctg} x-x^{2}}$
|
## Solution
$\lim _{x \rightarrow 0} \frac{e^{2 x}-e^{3 x}}{\operatorname{arctg} x-x^{2}}=\lim _{x \rightarrow 0} \frac{\left(e^{2 x}-1\right)-\left(e^{3 x}-1\right)}{\operatorname{arctg} x-x^{2}}=$
$=\lim _{x \rightarrow 0} \frac{\frac{1}{x}\left(\left(e^{2 x}-1\right)-\left(e^{3 x}-1\right)\right)}{\frac{1}{x}\left(\operatorname{arctg} x-x^{2}\right)}=$
$=\frac{\lim _{x \rightarrow 0} \frac{1}{x}\left(\left(e^{2 x}-1\right)-\left(e^{3 x}-1\right)\right)}{\lim _{x \rightarrow 0} \frac{1}{x}\left(\operatorname{arctg} x-x^{2}\right)}=$
$=\left(\lim _{x \rightarrow 0} \frac{e^{2 x}-1}{x}-\lim _{x \rightarrow 0} \frac{e^{3 x}-1}{x}\right) /\left(\lim _{x \rightarrow 0} \frac{\operatorname{arctg} x}{x}-\lim _{x \rightarrow 0} \frac{x^{2}}{x}\right)=$
Using the substitution of equivalent infinitesimals:
$e^{2 x}-1 \sim 2 x$, as $x \rightarrow 0(2 x \rightarrow 0)$
$e^{3 x}-1 \sim 3 x$, as $x \rightarrow 0(3 x \rightarrow 0)$
$\operatorname{arctg} x \sim x$, as $x \rightarrow 0$
We get:
$$
=\frac{\lim _{x \rightarrow 0} \frac{2 x}{x}-\lim _{x \rightarrow 0} \frac{3 x}{x}}{\lim _{x \rightarrow 0} \frac{x}{x}-\lim _{x \rightarrow 0} \frac{x^{2}}{x}}=\frac{\lim _{x \rightarrow 0} 2-\lim _{x \rightarrow 0} 3}{\lim _{x \rightarrow 0} 1-\lim _{x \rightarrow 0} x}=\frac{2-3}{1-0}=-1
$$
## Problem Kuznetsov Limits 15-6
|
-1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{e^{\alpha x}-e^{\beta x}}{\sin \alpha x-\sin \beta x}$
|
## Solution
$\lim _{x \rightarrow 0} \frac{e^{\alpha x}-e^{\beta x}}{\sin \alpha x-\sin \beta x}=\lim _{x \rightarrow 0} \frac{\left(e^{\alpha x}-1\right)-\left(e^{\beta x}-1\right)}{2 \sin \frac{x(\alpha-\beta)}{2} \cos \frac{x(\alpha+\beta)}{2}}=$
$=\lim _{x \rightarrow 0} \frac{e^{\alpha x}-1}{2 \sin \frac{x(\alpha-\beta)}{2} \cos \frac{x(\alpha+\beta)}{2}}-\lim _{x \rightarrow 0} \frac{e^{\beta x}-1}{2 \sin \frac{x(\alpha-\beta)}{2} \cos \frac{x(\alpha+\beta)}{2}}=$
Using the substitution of equivalent infinitesimals:
$e^{\alpha x}-1 \sim \alpha x$, as $x \rightarrow 0(\alpha x \rightarrow 0)$
$e^{\beta x}-1 \sim \beta x$, as $x \rightarrow 0(\beta x \rightarrow 0)$
$\sin \frac{x(\alpha-\beta)}{2} \sim \frac{x(\alpha-\beta)}{2}, \quad x \rightarrow 0\left(\frac{x(\alpha-\beta)}{2} \rightarrow 0\right)$
We get:
$$
\begin{aligned}
& =\lim _{x \rightarrow 0} \frac{\alpha x}{2 \frac{x(\alpha-\beta)}{2} \cos \frac{x(\alpha+\beta)}{2}}-\lim _{x \rightarrow 0} \frac{\beta x}{2 \frac{x(\alpha-\beta)}{2} \cos \frac{x(\alpha+\beta)}{2}}= \\
& =\lim _{x \rightarrow 0} \frac{\alpha}{(\alpha-\beta) \cos \frac{x(\alpha+\beta)}{2}}-\lim _{x \rightarrow 0} \frac{\beta}{(\alpha-\beta) \cos \frac{x(\alpha+\beta)}{2}}= \\
& =\frac{\alpha}{(\alpha-\beta) \cos \frac{0 \cdot(\alpha+\beta)}{2}}-\frac{\beta}{(\alpha-\beta) \cos \frac{0 \cdot(\alpha+\beta)}{2}}= \\
& =\frac{\alpha}{(\alpha-\beta) \cos 0}-\frac{\alpha}{(\alpha-\beta) \cos 0}=\frac{\alpha}{(\alpha-\beta) \cdot 1}-\frac{\beta}{(\alpha-\beta) \cdot 1}= \\
& =\frac{\alpha-\beta}{\alpha-\beta}=1
\end{aligned}
$$
## Problem Kuznetsov Limits 16-6
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0}\left(\frac{x^{2}+4}{x+2}\right)^{x^{2}+3}$
|
## Solution
$\lim _{x \rightarrow 0}\left(\frac{x^{2}+4}{x+2}\right)^{x^{2}+3}=\left(\frac{0^{2}+4}{0+2}\right)^{0^{2}+3}=$
$=\left(\frac{4}{2}\right)^{3}=2^{3}=8$
## Problem Kuznetsov Limits 18-6
|
8
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{(2 n+1)^{3}-(2 n+3)^{3}}{(2 n+1)^{2}+(2 n+3)^{2}}$
|
## Solution
$\lim _{n \rightarrow \infty} \frac{(2 n+1)^{3}-(2 n+3)^{3}}{(2 n+1)^{2}+(2 n+3)^{2}}=\lim _{n \rightarrow \infty} \frac{8 n^{3}+3 \cdot 4 n^{2}+3 \cdot 2 n+1-8 n^{3}-3 \cdot 3 \cdot 4 n^{2}-3 \cdot 3^{2} \cdot 2 n-3^{3}}{(2 n+1)^{2}+(2 n+3)^{2}}=$
$=\lim _{n \rightarrow \infty} \frac{\frac{1}{n^{2}}\left(3 \cdot 4 n^{2}(1-3)+3 \cdot 2 n\left(1-3^{2}\right)+\left(1-3^{3}\right)\right)}{\frac{1}{n^{2}}\left((2 n+1)^{2}+(2 n+3)^{2}\right)}=\lim _{n \rightarrow \infty} \frac{-24-\frac{48}{n}-\frac{26}{n^{2}}}{\left(2+\frac{1}{n}\right)^{2}+\left(2+\frac{3}{n}\right)^{2}}=$
$=\frac{-24-0-0}{2^{2}+2^{2}}=-\frac{24}{8}=-3$
## Problem Kuznetsov Limits 3-21
|
-3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{n \sqrt[4]{11 n}+\sqrt{25 n^{4}-81}}{(n-7 \sqrt{n}) \sqrt{n^{2}-n+1}}$
|
## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{n \sqrt[4]{11 n}+\sqrt{25 n^{4}-81}}{(n-7 \sqrt{n}) \sqrt{n^{2}-n+1}}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n^{2}}\left(n \sqrt[4]{11 n}+\sqrt{25 n^{4}-81}\right)}{\frac{1}{n^{2}}(n-7 \sqrt{n}) \sqrt{n^{2}-n+1}}= \\
& =\lim _{n \rightarrow \infty} \frac{\sqrt[4]{\frac{11}{n^{3}}}+\sqrt{25-\frac{81}{n^{4}}}}{\left(\frac{1}{n}(n-7 \sqrt{n})\right)\left(\frac{1}{n} \sqrt{n^{2}-n+1}\right)}=\lim _{n \rightarrow \infty} \frac{\sqrt[4]{\frac{11}{n^{3}}}+\sqrt{25-\frac{81}{n^{4}}}}{\left(1-7 \sqrt{\frac{1}{n}}\right) \sqrt{1-\frac{1}{n}+\frac{1}{n^{2}}}}= \\
& =\frac{\sqrt[4]{0}+\sqrt{25-0}}{(1-7 \sqrt{0}) \sqrt{1-0+0}}=\frac{5}{1}=5
\end{aligned}
$$
## Problem Kuznetsov Limits 4-21
|
5
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the numerical sequence:
$$
\lim _{n \rightarrow \infty}\left(\sqrt{\left(n^{2}+1\right)\left(n^{2}+2\right)}-\sqrt{\left(n^{2}-1\right)\left(n^{2}-2\right)}\right)
$$
|
## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty}\left(\sqrt{\left(n^{2}+1\right)\left(n^{2}+2\right)}-\sqrt{\left(n^{2}-1\right)\left(n^{2}-2\right)}\right)= \\
& =\lim _{n \rightarrow \infty} \frac{\left(\sqrt{\left(n^{2}+1\right)\left(n^{2}+2\right)}-\sqrt{\left(n^{2}-1\right)\left(n^{2}-2\right)}\right)\left(\sqrt{\left(n^{2}+1\right)\left(n^{2}+2\right)}+\sqrt{\left(n^{2}-1\right)\left(n^{2}-2\right)}\right)}{\sqrt{\left(n^{2}+1\right)\left(n^{2}+2\right)}+\sqrt{\left(n^{2}-1\right)\left(n^{2}-2\right)}}= \\
& =\lim _{n \rightarrow \infty} \frac{\left(n^{2}+1\right)\left(n^{2}+2\right)-\left(n^{2}-1\right)\left(n^{2}-2\right)}{\sqrt{\left(n^{2}+1\right)\left(n^{2}+2\right)}+\sqrt{\left(n^{2}-1\right)\left(n^{2}-2\right)}}= \\
& =\lim _{n \rightarrow \infty} \frac{n^{4}+n^{2}+2 n^{2}+2-n^{4}+n^{2}+2 n^{2}-2}{\sqrt{\left(n^{2}+1\right)\left(n^{2}+2\right)}+\sqrt{\left(n^{2}-1\right)\left(n^{2}-2\right)}}= \\
& =\lim _{n \rightarrow \infty} \frac{6 n^{2}}{\sqrt{\left(n^{2}+1\right)\left(n^{2}+2\right)}+\sqrt{\left(n^{2}-1\right)\left(n^{2}-2\right)}}= \\
& =\lim _{n \rightarrow \infty} \frac{\frac{1}{n^{2}} 6 n^{2}}{\frac{1}{n^{2}}\left(\sqrt{\left(n^{2}+1\right)\left(n^{2}+2\right)}+\sqrt{\left(n^{2}-1\right)\left(n^{2}-2\right)}\right)}= \\
& =\lim _{n \rightarrow \infty} \frac{6}{\sqrt{\left(1+\frac{1}{n^{2}}\right)\left(1+\frac{2}{n^{2}}\right)}+\sqrt{\left(1-\frac{1}{n^{2}}\right)\left(1-\frac{2}{n^{2}}\right)}}= \\
& =\frac{6}{\sqrt{(1+0)(1+0)}+\sqrt{(1-0)(1-0)}}=\frac{6}{2}=3
\end{aligned}
$$
## Problem Kuznetsov Limits 5-21
|
3
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty}\left(\frac{3 n^{2}-5 n}{3 n^{2}-5 n+7}\right)^{n+1}$
|
## Solution
$\lim _{n \rightarrow \infty}\left(\frac{3 n^{2}-5 n}{3 n^{2}-5 n+7}\right)^{n+1}=\lim _{n \rightarrow \infty}\left(\frac{3 n^{2}-5 n+7}{3 n^{2}-5 n}\right)^{-n-1}=$
$=\lim _{n \rightarrow \infty}\left(1+\frac{7}{3 n^{2}-5 n}\right)^{-n-1}=\lim _{n \rightarrow \infty}\left(1+\frac{1}{\left(\frac{3 n^{2}-5 n}{7}\right)}\right)^{-n-1}=$
$=\lim _{n \rightarrow \infty}\left(1+\frac{1}{\left(\frac{3 n^{2}-5 n}{7}\right)}\right)^{\left(\frac{3 n^{2}-5 n}{7}\right)\left(\frac{7}{3 n^{2}-5 n}\right)(-n-1)}=$
$=\{$ Using the second remarkable limit $\}=$
$=e^{\lim _{n \rightarrow \infty}\left(\frac{7}{3 n^{2}-5 n}\right)(-n-1)}=e^{\lim _{n \rightarrow \infty} \frac{7(-n-1)}{3 n^{2}-5 n}}=$
$=e^{\lim _{n \rightarrow \infty} \frac{\frac{1}{n^{2}} 7(-n-1)}{n^{2}\left(3 n^{2}-5 n\right)}}=e^{\lim _{n \rightarrow \infty} \frac{7\left(-\frac{1}{n}-\frac{1}{n^{2}}\right)}{3-\frac{5}{n^{2}}}}=e^{\frac{7(-0-0)}{3-0}}=e^{0}=1$
## Problem Kuznetsov Limits 7-21
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Condition of the problem
Calculate the limit of the function:
$$
\lim _{x \rightarrow-1} \frac{x^{3}-3 x-2}{x^{2}+2 x+1}
$$
|
## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow-1} \frac{x^{3}-3 x-2}{x^{2}+2 x+1}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow-1} \frac{(x+1)\left(x^{2}-x-2\right)}{(x+1)^{2}}= \\
& =\lim _{x \rightarrow-1} \frac{x^{2}-x-2}{x+1}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow-1} \frac{(x+1)(x-2)}{x+1}= \\
& =\lim _{x \rightarrow-1}(x-2)=-1-2=-3
\end{aligned}
$$
## Problem Kuznetsov Limits 10-21
|
-3
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the function:
$\lim _{h \rightarrow 0} \frac{a^{x+h}+a^{x-h}-2 a^{x}}{h}$
|
Solution
$\lim _{h \rightarrow 0} \frac{a^{x+h}+a^{x-h}-2 a^{x}}{h}=\lim _{h \rightarrow 0} \frac{a^{x+h}-a^{x}+a^{x-h}-a^{x}}{h}=$
$=\lim _{h \rightarrow 0} \frac{a^{x+h}-a^{x}}{h}+\lim _{h \rightarrow 0} \frac{a^{x-h}-a^{x}}{h}=$
$=\lim _{h \rightarrow 0} \frac{a^{x}\left(a^{h}-1\right)}{h}+\lim _{h \rightarrow 0} \frac{a^{x}\left(a^{-h}-1\right)}{h}=$
$=\lim _{h \rightarrow 0} \frac{a^{x}\left(\left(e^{\ln a}\right)^{h}-1\right)}{h}+\lim _{h \rightarrow 0} \frac{a^{x}\left(\left(e^{\ln a}\right)^{-h}-1\right)}{h}=$
$=\lim _{h \rightarrow 0} \frac{a^{x}\left(e^{h \ln a}-1\right)}{h}+\lim _{h \rightarrow 0} \frac{a^{x}\left(e^{-h \ln a}-1\right)}{h}=$
Using the substitution of equivalent infinitesimals:
$e^{h \ln a}-1 \sim h \ln a$, as $h \rightarrow 0(h \ln a \rightarrow 0)$
$e^{-h \ln a}-1 \sim -h \ln a$, as $h \rightarrow 0(-h \ln a \rightarrow 0)$
We get:
$=\lim _{h \rightarrow 0} \frac{a^{x} \cdot h \ln a}{h}+\lim _{h \rightarrow 0} \frac{a^{x} \cdot(-h) \ln a}{h}=$
$=\lim _{h \rightarrow 0} a^{x} \ln a+\lim _{h \rightarrow 0}-a^{x} \ln a=a^{x} \ln a-a^{x} \ln a=0$
## Problem Kuznetsov Limits 16-21
|
0
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0}\left(\frac{\ln \left(1+x^{2}\right)}{x^{2}}\right)^{\frac{3}{x+8}}$
|
## Solution
$\lim _{x \rightarrow 0}\left(\frac{\ln \left(1+x^{2}\right)}{x^{2}}\right)^{\frac{3}{x+8}}=\left(\lim _{x \rightarrow 0} \frac{\ln \left(1+x^{2}\right)}{x^{2}}\right)^{\lim _{x \rightarrow 0} \frac{3}{x+8}}=$
$=\left(\lim _{x \rightarrow 0} \frac{\ln \left(1+x^{2}\right)}{x^{2}}\right)^{\frac{3}{0+8}}=\left(\lim _{x \rightarrow 0} \frac{\ln \left(1+x^{2}\right)}{x^{2}}\right)^{\frac{3}{8}}=$
Using the substitution of equivalent infinitesimals:
$\ln \left(1+x^{2}\right) \sim x^{2}$, as $x \rightarrow 0\left(x^{2} \rightarrow 0\right)$
We get:
$=\left(\lim _{x \rightarrow 0} \frac{x^{2}}{x^{2}}\right)^{\frac{3}{8}}=\left(\lim _{x \rightarrow 0} 1\right)^{\frac{3}{8}}=1^{\frac{3}{8}}=1$
## Problem Kuznetsov Limits 18-21
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow 1}\left(\ln ^{2} e x\right)^{\frac{1}{x^{2}+1}}
$$
|
## Solution
$$
\lim _{x \rightarrow 1}\left(\ln ^{2} e x\right)^{\frac{1}{x^{2}+1}}=\left(\ln ^{2}(e \cdot 1)\right)^{\frac{1}{1^{2}+1}}=\left(1^{2}\right)^{\frac{1}{2}}=1
$$
## Problem Kuznetsov Limits 20-21
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Task Condition
Calculate the area of the parallelogram constructed on vectors $a$ and $b$.
$a=10 p+q$
$b=3 p-2 q$
$|p|=4$
$|q|=1$
$(\widehat{p, q})=\frac{\pi}{6}$
|
## Solution
The area of the parallelogram constructed on vectors $a$ and $b$ is numerically equal to the modulus of their vector product:
$S=|a \times b|$
We compute $a \times b$ using the properties of the vector product:
$a \times b=(10 p+q) \times(3 p-2 q)=10 \cdot 3 \cdot p \times p+10 \cdot(-2) \cdot p \times q+3 \cdot q \times p-2 q \times q=$
$=-20 \cdot p \times q+3 \cdot q \times p=-20 \cdot p \times q-3 \cdot p \times q=(-20-3) \cdot p \times q=-23 \cdot p \times q$
We compute the area:
$$
\begin{aligned}
& S=|a \times b|=|-23 \cdot p \times q|=23 \cdot|p \times q|=23 \cdot|p| \cdot|q| \cdot \sin (\widehat{p, q})= \\
& =23 \cdot 4 \cdot 1 \cdot \sin \frac{\pi}{6}=92 \cdot \sin \frac{\pi}{6}=92 \cdot \frac{1}{2}=46
\end{aligned}
$$
Thus, the area of the parallelogram constructed on vectors $a$ and $b$ is 46.
## Problem Kuznetsov Analytical Geometry 5-20
|
46
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## problem statement
Based on the definition of the derivative, find $f^{\prime}(0)$ :
$$
f(x)=\left\{\begin{array}{c}
\sqrt[3]{1-2 x^{3} \sin \frac{5}{x}}-1+x, x \neq 0 \\
0, x=0
\end{array}\right.
$$
|
## Solution
By definition, the derivative at the point $x=0$:
$$
f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}
$$
Based on the definition, we find:
$$
\begin{aligned}
& f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim _{\Delta x \rightarrow 0} \frac{\sqrt[3]{1-2 \Delta x^{3} \sin \frac{5}{\Delta x}}-1+\Delta x-0}{\Delta x}= \\
& =\lim _{\Delta x \rightarrow 0} \frac{\sqrt[3]{1-2 \Delta x^{3} \sin \frac{5}{\Delta x}}-1}{\Delta x}+\lim _{\Delta x \rightarrow 0} \frac{\Delta x}{\Delta x}= \\
& =\lim _{\Delta x \rightarrow 0} \frac{\left(\sqrt[3]{1-2 \Delta x^{3} \sin \frac{5}{\Delta x}}-1\right)\left(\sqrt[3]{\left(1-2 \Delta x^{3} \sin \frac{5}{\Delta x}\right)^{2}}+\sqrt[3]{1-2 \Delta x^{3} \sin \frac{5}{\Delta x}}+1\right)}{\Delta x\left(\sqrt[3]{\left(1-2 \Delta x^{3} \sin \frac{5}{\Delta x}\right)^{2}}+\sqrt[3]{1-2 \Delta x^{3} \sin \frac{5}{\Delta x}}+1\right)}+\lim _{\Delta x \rightarrow 0} 1= \\
& =\lim _{\Delta x \rightarrow 0} \frac{1-2 \Delta x^{3} \sin \frac{5}{\Delta x}-1}{\Delta x\left(\sqrt[3]{\left(1-2 \Delta x^{3} \sin \frac{5}{\Delta x}\right)^{2}}+\sqrt[3]{1-2 \Delta x^{3} \sin \frac{5}{\Delta x}}+1\right)}+1= \\
& =\lim _{\Delta x \rightarrow 0} \frac{-2 \Delta x^{3} \sin \frac{5}{\Delta x}}{\Delta x\left(\sqrt[3]{\left(1-2 \Delta x^{3} \sin \frac{5}{\Delta x}\right)^{2}}+\sqrt[3]{1-2 \Delta x^{3} \sin \frac{5}{\Delta x}}+1\right)}+1= \\
& =\lim _{\Delta x \rightarrow 0} \frac{-2 \Delta x^{2} \sin \frac{5}{\Delta x}}{\sqrt[3]{\left(1-2 \Delta x^{3} \sin \frac{5}{\Delta x}\right)^{2}}+\sqrt[3]{1-2 \Delta x^{3} \sin \frac{5}{\Delta x}}+1}+1= \\
& \text { Since } \sin \frac{5}{\Delta x} \text { is bounded, then }
\end{aligned}
$$
$\Delta x \cdot \sin \frac{5}{\Delta x} \rightarrow 0$, as $\Delta x \rightarrow 0$
Then:
$=\frac{-2 \cdot 0}{\sqrt[3]{(1-2 \cdot 0)^{2}}+\sqrt[3]{1-2 \cdot 0}+1}+1=\frac{0}{\sqrt[3]{1}+\sqrt[3]{1}+1}+1=1$
Therefore, $f^{\prime}(0)=1$
## Problem Kuznetsov Differentiation $2-27$
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Task Condition
Calculate the area of the parallelogram constructed on vectors $a$ and $b$.
$a=7 p-2 q$
$b=p+3 q$
$|p|=\frac{1}{2}$
$|q|=2$
$(\widehat{p, q})=\frac{\pi}{2}$
|
## Solution
The area of the parallelogram constructed on vectors $a$ and $b$ is numerically equal to the modulus of their vector product:
$S=|a \times b|$
We compute $a \times b$ using the properties of the vector product:
$a \times b=(7 p-2 q) \times(p+3 q)=7 \cdot p \times p+7 \cdot 3 \cdot p \times q-2 \cdot q \times p-2 \cdot 3 \cdot q \times q=$ $=21 \cdot p \times q-2 \cdot q \times p=21 \cdot p \times q+2 \cdot p \times q=(21+2) \cdot p \times q=23 \cdot p \times q$
We compute the area:
$$
\begin{aligned}
& S=|a \times b|=|23 \cdot p \times q|=23 \cdot|p \times q|=23 \cdot|p| \cdot|q| \cdot \sin (\widehat{p, q})= \\
& =23 \cdot \frac{1}{2} \cdot 2 \cdot \sin \frac{\pi}{2}=23 \cdot \sin \frac{\pi}{2}=23 \cdot 1=23
\end{aligned}
$$
Thus, the area of the parallelogram constructed on vectors $a$ and $b$ is 23.
## Problem Kuznetsov Analytical Geometry 5-18
|
23
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Find the distance from point $M_{0}$ to the plane passing through three points $M_{1}, M_{2}, M_{3}$.
$M_{1}(7 ; 2 ; 4)$
$M_{2}(7 ;-1 ;-2)$
$M_{3}(-5 ;-2 ;-1)$
$M_{0}(10 ; 1 ; 8)$
|
## Solution
Find the equation of the plane passing through three points $M_{1}, M_{2}, M_{3}$:
$$
\left|\begin{array}{ccc}
x-7 & y-2 & z-4 \\
7-7 & -1-2 & -2-4 \\
-5-7 & -2-2 & -1-4
\end{array}\right|=0
$$
Perform transformations:
$$
\begin{aligned}
& \left|\begin{array}{ccc}
x-7 & y-2 & z-4 \\
0 & -3 & -6 \\
-12 & -4 & -5
\end{array}\right|=0 \\
& (x-7) \cdot\left|\begin{array}{ll}
-3 & -6 \\
-4 & -5
\end{array}\right|-(y-2) \cdot\left|\begin{array}{cc}
0 & -6 \\
-12 & -5
\end{array}\right|+(z-4) \cdot\left|\begin{array}{cc}
0 & -3 \\
-12 & -4
\end{array}\right|=0 \\
& (x-7) \cdot(-9)-(y-2) \cdot(-72)+(z-4) \cdot(-36)=0 \\
& -9 x+63+72 y-144-36 z+144=0 \\
& -9 x+72 y-36 z+63=0 \\
& -x+8 y-4 z+7=0
\end{aligned}
$$
The distance $d$ from a point $M_{0}\left(x_{0} ; y_{0} ; z_{0}\right)$ to the plane $A x+B y+C z+D=0$:
$d=\frac{\left|A x_{0}+B y_{0}+C z_{0}+D\right|}{\sqrt{A^{2}+B^{2}+C^{2}}}$
Find:
$$
d=\frac{|-10+8 \cdot 1-4 \cdot 8+7|}{\sqrt{(-1)^{2}+8^{2}+(-4)^{2}}}=\frac{|-10+8-32+7|}{\sqrt{1+64+16}}=\frac{27}{\sqrt{81}}=3
$$
## Problem Kuznetsov Analytic Geometry 8-18
|
3
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Let $k$ be the coefficient of similarity transformation with the center at the origin. Is it true that point $A$ belongs to the image of plane $a$?
$A\left(\frac{1}{3} ; 1 ; 1\right)$
$a: 3 x-y+5 z-6=0$
$k=\frac{5}{6}$
|
## Solution
When transforming similarity with the center at the origin of the plane
$a: A x+B y+C z+D=0_{\text{and coefficient }} k$ transitions to the plane
$a^{\prime}: A x+B y+C z+k \cdot D=0$. We find the image of the plane $a$:
$a^{\prime}: 3 x-y+5 z-5=0$
Substitute the coordinates of point $A$ into the equation $a^{\prime}$:
$3 \cdot \frac{1}{3}-1+5 \cdot 1-5=0$
$1-1+5-5=0$
$0=0$
Since $0=0$, point $A$ belongs to the image of the plane $a$.
Problem Kuznetsov Analytical Geometry 12-18
|
0
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Based on the definition of the derivative, find $f^{\prime}(x)$:
$f(x)=\left\{\begin{array}{c}\sin \left(e^{x^{2} \sin \frac{5}{x}}-1\right)+x, x \neq 0 \\ 0, x=0\end{array}\right.$
|
## Solution
By definition, the derivative at the point $x=0$:
$$
f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}
$$
Based on the definition, we find:
$$
\begin{aligned}
& f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim _{\Delta x \rightarrow 0}\left(\sin \left(e^{\Delta x^{2} \sin \frac{s}{\Delta x}}-1\right)+\Delta x-0\right) / \Delta x= \\
& =\lim _{\Delta x \rightarrow 0}\left(\sin \left(e^{\Delta x^{2} \sin \frac{5}{\Delta x}}-1\right)\right) / \Delta x+\lim _{\Delta x \rightarrow 0} \frac{\Delta x}{\Delta x}=
\end{aligned}
$$
Using the substitution of equivalent infinitesimals:
$$
e^{\Delta x^{2} \sin \frac{5}{\Delta x}}-1 \sim \Delta x^{2} \sin \frac{5}{\Delta x}, \text { as } \Delta x \rightarrow 0\left(\Delta x^{2} \sin \frac{5}{\Delta x} \rightarrow 0\right)
$$
We get:
$$
=\lim _{\Delta x \rightarrow 0}\left(\sin \left(\Delta x^{2} \sin \frac{5}{\Delta x}\right)\right) / \Delta x+\lim _{\Delta x \rightarrow 0} 1=
$$
Using the substitution of equivalent infinitesimals:
We get:
$=\lim _{\Delta x \rightarrow 0} \frac{\Delta x^{2} \sin \frac{5}{\Delta x}}{\Delta x}+1=\lim _{\Delta x \rightarrow 0} \Delta x \sin \frac{5}{\Delta x}+1=$
Since $\sin \left(\frac{5}{\Delta x}\right)$ is bounded,
$\Delta x \cdot \sin \left(\frac{5}{\Delta x}\right) \rightarrow 0 \quad$, as $\Delta x \rightarrow 0$
Then:
$=0+1=1$
Thus, $f^{\prime}(0)=1$
## Problem Kuznetsov Differentiation $2-7$
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the volumes of the bodies bounded by the surfaces.
$$
\frac{x^{2}}{9}+y^{2}=1, z=y, z=0(y \geq 0)
$$
|
## Solution
The base of the considered area is a semi-ellipse, in which
$$
\begin{aligned}
& x=0 \text { when } y=1 \\
& y=0 \text { when } x=3
\end{aligned}
$$
That is, $x \in[-3,3], y \in[0,1]$
Consider the surface $z=y:$
$$
V_{z}=\int_{0}^{1} z d y=\int_{0}^{1} y d y=\left.\frac{y^{2}}{2}\right|_{0} ^{1}=\frac{1}{2}
$$
Now consider the area of the base and find the volume of the given body:
$$
\begin{aligned}
& \frac{x^{2}}{9}+y^{2}=1 \\
& y^{2}=1-\frac{x^{2}}{9} \\
& V=\frac{1}{2} \int_{-3}^{3}\left(1-\frac{x^{2}}{9}\right) d x=\left.\frac{1}{2}\left(x-\frac{x^{3}}{9 \cdot 3}\right)\right|_{-3} ^{3} d x=\frac{1}{2}\left(3-\frac{3^{3}}{27}\right)-\frac{1}{2}\left(-3-\frac{-3^{3}}{27}\right) \\
& =\frac{1}{2}(3-1)-\frac{1}{2}(-3+1)=1+1=2
\end{aligned}
$$
Answer: $V=2$
Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\� \%D0\%9A\%D1\%83\%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B} \_20-1$ »
Categories: Kuznetsov's Problem Book Integrals Problem 20 | Integrals | Problems for Checking
Ukrainian Banner Network
- Last modified on this page: 08:06, 24 June 2010.
- Content is available under CC-BY-SA 3.0.
|
2
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the volumes of the bodies bounded by the surfaces.
$$
\frac{x^{2}}{3}+\frac{y^{2}}{4}=1, z=y \sqrt{3}, z=0(y \geq 0)
$$
|
## Solution
The base of the considered area is a semi-ellipse, in which
$$
\begin{aligned}
& x=0 \text { when } y=2 \\
& y=0 \text { when } x=\sqrt{3}
\end{aligned}
$$
That is, $x \in[-\sqrt{3}, \sqrt{3}], y \in[0,2]$
Consider the surface $z=y \sqrt{3}$:
$$
V_{z}=\int_{0}^{2} z d y=\int_{0}^{2} y \sqrt{3} d y=\left.\sqrt{3} \cdot \frac{y^{2}}{2}\right|_{0} ^{2}=2 \sqrt{3}
$$
Now consider the area of the base and find the volume of the given body:
$$
\begin{aligned}
& \frac{x^{2}}{3}+\frac{y^{2}}{4}=1 \\
& y^{2}=4\left(1-\frac{x^{2}}{3}\right) \\
& V=2 \sqrt{3} \int_{-\sqrt{3}}^{\sqrt{3}} 4\left(1-\frac{x^{2}}{3}\right) d x=\left.8 \sqrt{3}\left(x-\frac{x^{3}}{3 \cdot 3}\right)\right|_{-\sqrt{3}} ^{\sqrt{3}} d x=
\end{aligned}
$$
$$
\begin{aligned}
& =8 \sqrt{3}\left(\sqrt{3}-\frac{\sqrt{3}^{3}}{9}\right)-8 \sqrt{3}\left(-\sqrt{3}-\frac{-\sqrt{3}^{3}}{9}\right)= \\
& =8\left(3-\frac{3}{3}\right)+8\left(3-\frac{3}{3}\right)=16+16=32
\end{aligned}
$$
Answer: $V=32$
Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\� \%D0\%9A\%D1\%83\%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+20-11$ »
Categories: Kuznetsov's Problem Book Integrals Problem 20 | Integrals | Problems for Checking
- Last modified: 08:29, 24 June 2010.
- Content is available under CC-BY-SA 3.0.
|
32
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the volumes of the bodies bounded by the surfaces.
$$
\frac{x^{2}}{3}+\frac{y^{2}}{16}=1, z=y \sqrt{3}, z=0(y \geq 0)
$$
|
## Solution
The base of the considered area is a semi-ellipse, in which
$$
\begin{aligned}
& x=0 \text { when } y=4 \\
& y=0 \text { when } x=\sqrt{3}
\end{aligned}
$$
That is, $x$ belongs to the interval $[-\sqrt{3}, \sqrt{3}]$, and $y \in [0,4]$
Consider the surface $z=y \sqrt{3}$:
$$
V_{z}=\int_{0}^{4} z d y=\int_{0}^{4} y \sqrt{3} d y=\sqrt{3} \int_{0}^{4} y d y=\left.\sqrt{3} \cdot \frac{y^{2}}{2}\right|_{0} ^{4}=\sqrt{3} \cdot \frac{16}{2}=8 \sqrt{3}
$$
Now consider the area of the base and find the volume of the given body:
$$
\begin{aligned}
& \frac{x^{2}}{3}+\frac{y^{2}}{16}=1 \\
& \frac{y^{2}}{16}=1-\frac{x^{2}}{3} \\
& V=8 \sqrt{3} \int_{-\sqrt{3}}^{\sqrt{3}}\left(1-\frac{x^{2}}{3}\right) d x=\left.8 \sqrt{3}\left(x-\frac{x^{3}}{9}\right)\right|_{-\sqrt{3}} ^{\sqrt{3}} d x=16 \sqrt{3}\left(\sqrt{3}-\frac{\sqrt{3}^{2}}{9}\right)=32
\end{aligned}
$$
Answer: $V=32$
Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\� \%D0\%9A\%D1\%83\%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+20-16 \%$
Categories: Kuznetsov's Problem Book Integrals Problem $20 \mid$ Integrals
Kuznetsov Integrals 20-16 — PlusPi
Created by GeeTeatoo
Creste
View $\qquad$
2
## Problem Kuznetsov Integrals 20-17
## Material from Plusi
|
32
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the volumes of the bodies bounded by the surfaces.
$$
\frac{x^{2}}{27}+\frac{y^{2}}{25}=1, z=\frac{y}{\sqrt{3}}, z=0(y \geq 0)
$$
|
## Solution
The base of the considered area is a semi-ellipse, in which
$$
\begin{aligned}
& x=0 \text { when } y=5 \\
& y=0 \text { when } x=\sqrt{27}=3 \sqrt{3}
\end{aligned}
$$
That is,
$x \in[-3 \sqrt{3}, 3 \sqrt{3}], y \in[0,5]$
Consider the surface $z=\frac{y}{\sqrt{3}}:$
$$
V_{z}=\int_{0}^{5} z d y=\int_{0}^{5} \frac{y}{\sqrt{3}} d y=\left.\frac{1}{\sqrt{3}} \cdot \frac{y^{2}}{2}\right|_{0} ^{5}=\frac{25}{2 \sqrt{3}}
$$
Now consider the area of the base and find the volume of the given body:
$$
\begin{aligned}
& \frac{x^{2}}{27}+\frac{y^{2}}{25}=1 \\
& y^{2}=25\left(1-\frac{x^{2}}{27}\right) \\
& V=\frac{25}{2 \sqrt{3}} \int_{-3 \sqrt{3}}^{3 \sqrt{3}} 25\left(1-\frac{x^{2}}{27}\right) d x=\left.\frac{625}{2 \sqrt{3}}\left(x-\frac{x^{3}}{3 \cdot 27}\right)\right|_{-3 \sqrt{3}} ^{3 \sqrt{3}} d x=
\end{aligned}
$$
$$
\begin{aligned}
& =\frac{625}{2 \sqrt{3}}\left(3 \sqrt{3}-\frac{(3 \sqrt{3})^{3}}{81}\right)-\frac{625}{2 \sqrt{3}}\left(-3 \sqrt{3}-\frac{(-3 \sqrt{3})^{3}}{81}\right)= \\
& =\frac{625}{2 \sqrt{3}}\left(3 \sqrt{3}-\frac{81 \sqrt{3}}{81}\right)+\frac{625}{2 \sqrt{3}}\left(3 \sqrt{3}-\frac{81 \sqrt{3}}{81}\right)= \\
& =\frac{625}{2 \sqrt{3}} \sqrt{3}(3-1)+\frac{625}{2 \sqrt{3}} \sqrt{3}(3-1)=625+625=1250
\end{aligned}
$$
Answer: $V=1250$
Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\� \%D0\%9A\%D1\%83\%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+20-21$ »
Categories: Kuznetsov's Problem Book Integrals Problem 20 | Integrals | Problems for Checking
Ukrainian Banner Network
- Last modified: 09:08, 24 June 2010.
- Content is available under CC-BY-SA 3.0.
|
1250
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the volumes of the bodies bounded by the surfaces.
$$
\frac{x^{2}}{27}+y^{2}=1, z=\frac{y}{\sqrt{3}}, z=0(y \geq 0)
$$
|
## Solution
The base of the considered area is a semi-ellipse, in which
$$
\begin{aligned}
& x=0 \text { when } y=1 \\
& y=0 \text { when } x=3 \sqrt{3}
\end{aligned}
$$
That is,
$x \in[-3 \sqrt{3}, 3 \sqrt{3}], y \in[0,1]$
Therefore, the volume will be
$$
V=\int_{-3 \sqrt{3}}^{3 \sqrt{3}} d x \int_{0}^{\sqrt{1-x^{2} / 27}} d y \int_{0}^{y / \sqrt{3}} d z
$$
Consider the surface $z=\frac{y}{\sqrt{3}}:$
$$
V_{z}=\int_{0}^{\sqrt{1-x^{2} / 27}} z d y=\int_{0}^{\sqrt{1-x^{2} / 27}} \frac{y}{\sqrt{3}} d y=\left.\frac{y^{2}}{2 \sqrt{3}}\right|_{0} ^{\sqrt{1-x^{2} / 27}}=\frac{1}{2 \sqrt{3}} \cdot\left(1-\frac{x^{2}}{27}\right)
$$
Now the volume of the given body:
$$
\begin{aligned}
& V=\frac{1}{2 \sqrt{3}} \int_{-3 \sqrt{3}}^{3 \sqrt{3}}\left(1-\frac{x^{2}}{27}\right) d x=\left.\frac{1}{2 \sqrt{3}}\left(x-\frac{x^{3}}{27 \cdot 3}\right)\right|_{-3 \sqrt{3}} ^{3 \sqrt{3}} d x= \\
& =\frac{1}{2 \sqrt{3}}\left(3 \sqrt{3}-\frac{(3 \sqrt{3})^{3}}{81}\right)-\frac{1}{2 \sqrt{3}}\left(-3 \sqrt{3}-\frac{(-3 \sqrt{3})^{3}}{81}\right)= \\
& =\frac{\sqrt{3}}{2 \sqrt{3}}\left(3-\frac{27 \cdot 3}{81}\right)+\frac{\sqrt{3}}{2 \sqrt{3}}\left(3-\frac{27 \cdot 3}{81}\right)= \\
& =\frac{1}{2}(3-1)+\frac{1}{2}(3-1)=1+1=2
\end{aligned}
$$
Answer: $V=2$
Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\� \%D0\%9A\%D1\%83\%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+20-26$ »
Categories: Kuznetsov's Problem Book Integrals Problem $20 \mid$ Integrals
Ukrainian Banner Network
- Last modified: 10:57, June 24, 2010.
- Content is available under CC-BY-SA 3.0.
Created by GeeTeatoo
## Problem Kuznetsov Integrals 20-27
## Material from Plusi
|
2
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Based on the definition of the derivative, find $f^{\prime}(0)$:
$f(x)=\left\{\begin{array}{c}\tan\left(2^{x^{2} \cos (1 /(8 x))}-1+x\right), x \neq 0 ; \\ 0, x=0\end{array}\right.$
|
## Solution
By definition, the derivative at the point $x=0$:
$f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}$
Based on the definition, we find:
$$
\begin{aligned}
& f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim _{\Delta x \rightarrow 0} \frac{\tan\left(2^{\Delta x^{2} \cos (1 /(8 \Delta x))}-1+\Delta x\right)-0}{\Delta x}= \\
& =\lim _{\Delta x \rightarrow 0} \frac{\tan\left(2^{\Delta x^{2} \cos (1 /(8 \Delta x))}-1+\Delta x\right)}{\Delta x}=
\end{aligned}
$$
We use the substitution of equivalent infinitesimals:
$\tan\left(2^{\Delta x^{2} \cos (1 /(8 \Delta x))}-1+\Delta x\right) \sim 2^{\Delta x^{2} \cos (1 /(8 \Delta x))}-1+\Delta x$, as
$\Delta x \rightarrow 0\left(2^{\Delta x^{2} \cos (1 /(8 \Delta x))}-1+\Delta x \rightarrow 0\right)$
We get:
$$
\begin{aligned}
& =\lim _{\Delta x \rightarrow 0} \frac{2^{\Delta x^{2} \cos (1 /(8 \Delta x))}-1+\Delta x}{\Delta x}=\lim _{\Delta x \rightarrow 0} \frac{2^{\Delta x^{2} \cos (1 /(8 \Delta x))}-1}{\Delta x}+\lim _{\Delta x \rightarrow 0} \frac{\Delta x}{\Delta x}= \\
& =\lim _{\Delta x \rightarrow 0} \frac{\left(e^{\ln 2}\right)^{\Delta x^{2} \cos (1 /(8 \Delta x))}-1}{\Delta x}+\lim _{\Delta x \rightarrow 0} 1=1+\lim _{\Delta x \rightarrow 0} \frac{e^{\ln 2 \cdot \Delta x^{2} \cos (1 /(8 \Delta x))}-1}{\Delta x}=
\end{aligned}
$$
We use the substitution of equivalent infinitesimals:
$e^{\ln 2 \cdot \Delta x^{2} \cos (1 /(8 \Delta x))}-1 \sim \ln 2 \cdot \Delta x^{2} \cos (1 /(8 \Delta x))$, as
$\Delta x \rightarrow 0\left(\ln 2 \cdot \Delta x^{2} \cos (1 /(8 \Delta x)) \rightarrow 0\right)$
We get:
$$
=1+\lim _{\Delta x \rightarrow 0} \frac{\ln 2 \cdot \Delta x^{2} \cos (1 /(8 \Delta x))}{\Delta x}=1+\lim _{\Delta x \rightarrow 0} \ln 2 \cdot \Delta x \cos (1 /(8 \Delta x))=
$$
Since $\cos (1 /(8 \Delta x))$ is bounded, then
$\Delta x \cos (1 /(8 \Delta x)) \rightarrow 0$, as $\Delta x \rightarrow 0$
Then:
$=1+\ln 2 \cdot 0=1$
Therefore, $f^{\prime}(0)=1$
## Problem Kuznetsov Differentiation 2-12
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the lengths of the arcs of the curves given by the equations in polar coordinates.
$$
\rho=1-\sin \varphi, -\frac{\pi}{2} \leq \varphi \leq -\frac{\pi}{6}
$$
|
## Solution
The length of the arc of a curve given by an equation in polar coordinates is determined by the formula
$$
L=\int_{\phi_{1}}^{\phi_{2}} \sqrt{\rho^{2}+\left(\frac{d \rho}{d \phi}\right)^{2}} d \phi
$$
Let's find $\frac{d \rho}{d \phi}$:
$$
\frac{d \rho}{d \phi}=(-\cos \phi)
$$
We get:
$$
\begin{aligned}
L & =\int_{-\pi / 2}^{-\pi / 6} \sqrt{(1-\sin \phi)^{2}+(-\cos \phi)^{2}} d \phi= \\
& =\int_{-\pi / 2}^{-\pi / 6} \sqrt{1-2 \sin \phi+\sin ^{2} \phi+\cos ^{2} \phi} d \phi= \\
& =\int_{-\pi / 2}^{-\pi / 6} \sqrt{2-2 \sin \phi} d \phi= \\
& =\int_{-\pi / 2}^{-\pi / 6} \sqrt{2-2 \cos \left(\frac{\pi}{2}-\phi\right)} d \phi= \\
& =\int_{-\pi / 2}^{-\pi / 6} \sqrt{4 \sin ^{2}\left(\frac{\pi}{4}-\frac{\phi}{2}\right)} d \phi= \\
& =\int_{-\pi / 2}^{-\pi / 6} 2 \sin \left(\frac{\pi}{4}-\frac{\phi}{2}\right) d \phi=\left.4 \cdot \cos \left(\frac{\pi}{4}-\frac{\phi}{2}\right)\right|_{-\pi / 2} ^{-\pi / 6}= \\
& =4\left(\cos \left(\frac{\pi}{4}-\frac{1}{2} \cdot \frac{-\pi}{6}\right)-\cos \left(\frac{\pi}{4}-\frac{1}{2} \cdot \frac{-\pi}{2}\right)\right)=4\left(\cos \left(\frac{\pi}{3}\right)-\cos \left(\frac{\pi}{2}\right)\right)=4\left(\frac{1}{2}-0\right)=2
\end{aligned}
$$
Categories: Kuznetsov Problem Book Integrals Problem 19 | Integrals
Ukrainian Banner Network
- Last modified: 13:05, 27 May 2010.
- Content is available under CC-BY-SA 3.0.
Created by GeeTeatoo
## Problem Kuznetsov Integrals 19-12
## Material from PlusPi
|
2
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Problem Statement
Calculate the lengths of the arcs of the curves given by the equations in polar coordinates.
$$
\rho=8(1-\cos \varphi),-\frac{2 \pi}{3} \leq \varphi \leq 0
$$
|
## Solution
The length of the arc of a curve given by an equation in polar coordinates is determined by the formula
$L=\int_{\varphi_{0}}^{\varphi_{1}} \sqrt{(\rho(\varphi))^{2}+\left(\rho^{\prime}(\varphi)\right)^{2}} d \varphi$
For the curve given by the equation $\rho=8(1-\cos \varphi)$, we find: $\rho^{\prime}=8 \sin \varphi$
We get:
$$
\begin{aligned}
L & =\int_{-2 \pi / 3}^{0} \sqrt{(8(1-\cos \varphi))^{2}+(8 \sin \varphi)^{2}} d \varphi= \\
& =\int_{-2 \pi / 3}^{0} \sqrt{64(1-\cos \varphi)^{2}+64 \sin ^{2} \varphi} d \varphi= \\
& =8 \int_{-2 \pi / 3}^{0} \sqrt{1-2 \cos \varphi+\cos ^{2} \varphi+\sin ^{2} \varphi} d \varphi= \\
& =8 \int_{-2 \pi / 3}^{0} \sqrt{2-2 \cos \varphi} d \varphi= \\
& =8 \sqrt{2} \int_{-2 \pi / 3}^{0} \sqrt{1-\cos \varphi} d \varphi= \\
& =8 \sqrt{2} \int_{-2 \pi / 3}^{0} \sqrt{2 \sin ^{2} \frac{\varphi}{2}} d \varphi= \\
& =16 \int_{-2 \pi / 3}^{0} \sin ^{0} \frac{\varphi}{2} d \varphi= \\
& =16 \cdot 2 \int_{-2 \pi / 3}^{0} \sin \frac{\varphi}{2} d \frac{\varphi}{2}= \\
& =-\left.32 \cos \frac{\varphi}{2}\right|_{-2 \pi / 3} ^{0}=-32\left(\cos 0-\cos \frac{-\pi}{3}\right)=-32\left(1-\frac{1}{2}\right)=16
\end{aligned}
$$
Source — «http://pluspi.org/wiki/index.php/%D0%9A%D1%83%D0%B7%D0%BD%D0%B5%D1%86%D0%BE%D0%B2_%D0%98%D0%BD%D1%82%D0%B5%D0%B3%D1%80%D0%B0%D0%BB%D1%8B_19-18 »$ Categories: Kuznetsov's Problem Book Integrals Problem 19 | Integrals | Problems for Checking
- Last edited on this page: 06:01, 28 May 2010.
- Content is available under CC-BY-SA 3.0.
Created by GeeTeatoo
## Problem Kuznetsov Integrals 19-19
## Material from PlusPi
|
16
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## problem statement
Find the cosine of the angle between vectors $\overrightarrow{A B}$ and $\overrightarrow{A C}$.
$A(-2 ; 1 ; 1), B(2 ; 3 ;-2), C(0 ; 0 ; 3)$
|
## Solution
Let's find $\overrightarrow{A B}$ and $\overrightarrow{A C}$:
$$
\begin{aligned}
& \overrightarrow{A B}=(2-(-2) ; 3-1 ;-2-1)=(4 ; 2 ;-3) \\
& \overrightarrow{A C}=(0-(-2) ; 0-1 ; 3-1)=(2 ;-1 ; 2)
\end{aligned}
$$
We find the cosine of the angle $\phi_{\text {between vectors }} \overrightarrow{A B}$ and $\overrightarrow{A C}$:
$\cos (\widehat{A B,} \overrightarrow{A C})=\frac{(\overrightarrow{A B}, \overrightarrow{A C})}{|\overrightarrow{A B}| \cdot|\overrightarrow{A C}|}=$
$=\frac{4 \cdot 2+2 \cdot(-1)+(-3) \cdot 2}{\sqrt{4^{2}+2^{2}+(-3)^{2}} \cdot \sqrt{2^{2}+(-1)^{2}+2^{2}}}=$
$=\frac{8-2-6}{\sqrt{16+4+9} \cdot \sqrt{4+1+4}}=\frac{0}{\sqrt{29} \cdot \sqrt{9}}=0$
Thus, the cosine of the angle:
$\cos (\overrightarrow{A B,} \overrightarrow{A C})=0$
and consequently the angle
$\widehat{A B,} \overrightarrow{A C}=\frac{\pi}{2}$
Problem Kuznetsov Analytic Geometry 4-27
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## Task Condition
Are the vectors $a, b$ and $c$ coplanar?
$a=\{6 ; 3 ; 4\}$
$b=\{-1 ;-2 ;-1\}$
$c=\{2 ; 1 ; 2\}$
|
## Solution
For three vectors to be coplanar (lie in the same plane or parallel planes), it is necessary and sufficient that their scalar triple product $(a, b, c)$ be equal to zero.
$(a, b, c)=\left|\begin{array}{ccc}6 & 3 & 4 \\ -1 & -2 & -1 \\ 2 & 1 & 2\end{array}\right|=$
$=6 \cdot\left|\begin{array}{cc}-2 & -1 \\ 1 & 2\end{array}\right|-3 \cdot\left|\begin{array}{cc}-1 & -1 \\ 2 & 2\end{array}\right|+4 \cdot\left|\begin{array}{cc}-1 & -2 \\ 2 & 1\end{array}\right|=$
$=6 \cdot(-3)-3 \cdot 0+4 \cdot 3=-18-0+12=-6$
Since $(a, b, c)=-6 \neq 0$, the vectors $a, b$ and $c$ are not coplanar.
## Problem Kuznetsov Analytic Geometry 6-27
|
-6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## problem statement
Find the angle between the planes:
$2 x-6 y+14 z-1=0$
$5 x-15 y+35 z-3=0$
|
## Solution
The dihedral angle between planes is equal to the angle between their normal vectors. The normal vectors of the given planes:
$\overrightarrow{n_{1}}=\{2 ;-6 ; 14\}$
$\overrightarrow{n_{2}}=\{5 ;-15 ; 35\}$
The angle $\phi_{\text{between the planes is determined by the formula: }}$
$$
\begin{aligned}
& \cos \phi=\frac{\left(\overrightarrow{n_{1}}, \overrightarrow{n_{2}}\right)}{\left|\overrightarrow{n_{1}}\right| \cdot\left|\overrightarrow{n_{2}}\right|}=\frac{2 \cdot 5+(-6) \cdot(-15)+14 \cdot 35}{\sqrt{2^{2}+(-6)^{2}+14^{2}} \cdot \sqrt{5^{2}+(-15)^{2}+35^{2}}}= \\
& =\frac{10+90+490}{\sqrt{4+36+196} \cdot \sqrt{25+225+1225}}=\frac{590}{\sqrt{236} \cdot \sqrt{1475}}=\frac{590}{590}=1 \\
& \phi=\arccos 1=0
\end{aligned}
$$
## Problem Kuznetsov Analytic Geometry 10-27
|
0
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
## problem statement
Calculate the limit of the function:
$\lim _{x \rightarrow-1} \frac{\left(x^{3}-2 x-1\right)(x+1)}{x^{4}+4 x^{2}-5}$
|
## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow-1} \frac{\left(x^{3}-2 x-1\right)(x+1)}{x^{4}+4 x^{2}-5}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow-1} \frac{\left(x^{3}-2 x-1\right)(x+1)}{\left(x^{3}-x^{2}+5 x-5\right)(x+1)}= \\
& =\lim _{x \rightarrow-1} \frac{x^{3}-2 x-1}{x^{3}-x^{2}+5 x-5}=\frac{(-1)^{3}-2(-1)-1}{(-1)^{3}-(-1)^{2}+5(-1)-5}= \\
& =\frac{-1+2-1}{-1-1-5-5}=\frac{0}{-12}=0
\end{aligned}
$$
## Problem Kuznetsov Limits 10-1
|
0
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
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