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## Task Condition
Calculate the area of the parallelogram constructed on vectors $a$ and $b$.
$a=2 p-3 q$
$b=3 p+q$
$|p|=4$
$|q|=1$
$(\widehat{p, q})=\frac{\pi}{6}$ | ## 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=(2 p-3 q) \times(3 p+q)=2 \cdot 3 \cdot p \times p+2 \cdot p \times q-3 \cdot 3 \... | 22 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,527 |
## Task Condition
Are the vectors $a, b$ and $c$ coplanar?
$a=\{3 ; 10 ; 5\}$
$b=\{-2 ;-2 ;-3\}$
$c=\{2 ; 4 ; 3\}$ | ## 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.
$$
\begin{aligned}
& (a, b, c)=\left|\begin{array}{ccc}
3 & 10 & 5 \\
-2 & -2 & -3 \\
2 & 4 & 3
\end{array}\right|= \\
& =3 \cdot\left|... | -2\neq0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,528 |
## Problem Statement
Calculate the volume of the tetrahedron with vertices at points \( A_{1}, A_{2}, A_{3}, A_{4} \) and its height dropped from vertex \( A_{4} \) to the face \( A_{1} A_{2} A_{3} \).
\( A_{1}(1 ; 5 ;-7) \)
\[
\begin{aligned}
& A_{2}(-3 ; 6 ; 3) \\
& A_{3}(-2 ; 7 ; 3) \\
& A_{4}(-4 ; 8 ;-12)
\end{a... | ## Solution
From vertex $A_{1 \text {, we draw vectors: }}$
$$
\begin{aligned}
& \overrightarrow{A_{1} A_{2}}=\{-3-1 ; 6-5 ; 3-(-7)\}=\{-4 ; 1 ; 10\} \\
& A_{1} A_{3}=\{-2-1 ; 7-5 ; 3-(-7)\}=\{-3 ; 2 ; 10\} \\
& \overrightarrow{A_{1} A_{4}}=\{-4-1 ; 8-5 ;-12-(-7)\}=\{-5 ; 3 ;-5\}
\end{aligned}
$$
According to the ge... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,529 |
## Problem Statement
Find the distance from point $M_{0}$ to the plane passing through three points $M_{1}, M_{2}, M_{3}$.
$M_{1}(1 ; 3 ; 6)$
$M_{2}(2 ; 2 ; 1)$
$M_{3}(-1 ; 0 ; 1)$
$M_{0}(5 ;-4 ; 5)$ | ## Solution
Find the equation of the plane passing through three points $M_{1}, M_{2}, M_{3}$:
$\left|\begin{array}{ccc}x-1 & y-3 & z-6 \\ 2-1 & 2-3 & 1-6 \\ -1-1 & 0-3 & 1-6\end{array}\right|=0$
Perform transformations:
$$
\begin{aligned}
& \left|\begin{array}{ccc}
x-1 & y-3 & z-6 \\
1 & -1 & -5 \\
-2 & -3 & -5
\e... | 2\sqrt{14} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,530 |
## Problem Statement
Write the equation of the plane passing through point $A$ and perpendicular to vector $\overrightarrow{B C}$.
$A(-10 ; 0 ; 9)$
$B(12 ; 4 ; 11)$
$C(8 ; 5 ; 15)$ | ## Solution
Let's find the vector $\overrightarrow{BC}$:
$\overrightarrow{BC}=\{8-12 ; 5-4 ; 15-11\}=\{-4 ; 1 ; 4\}$
Since the vector $\overrightarrow{BC}$ is perpendicular to the desired plane, it can be taken as the normal vector. Therefore, the equation of the plane will be:
$-4 \cdot(x+10) + (y-0) + 4 \cdot(z-9... | -4x+y+4z-76=0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,531 |
## Task Condition
Find the angle between the planes:
$$
3 x+2 y-3 z-1=0
$$
$x+y+z-7=0$ | ## Solution
The dihedral angle between planes is equal to the angle between their normal vectors. The normal vectors of the given planes are:
$\overrightarrow{n_{1}}=\{3 ; 2 ;-3\}$
$\overrightarrow{n_{2}}=\{1 ; 1 ; 1\}$
The angle $\phi$ between the planes is determined by the formula:
$$
\begin{aligned}
& \cos \ph... | 7544^{\}54^{\\} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,532 |
## Problem Statement
Find the coordinates of point $A$, which is equidistant from points $B$ and $C$.
$A(0 ; y ; 0)$
$B(2 ; 2 ; 4)$
$C(0 ; 4 ; 2)$ | ## Solution
Let's find the distances $A B$ and $A C$:
$$
\begin{aligned}
& A B=\sqrt{(2-0)^{2}+(2-y)^{2}+(4-0)^{2}}=\sqrt{4+4-4 y+y^{2}+16}=\sqrt{y^{2}-4 y+24} \\
& A C=\sqrt{(0-0)^{2}+(4-y)^{2}+(2-0)^{2}}=\sqrt{0+16-8 y+y^{2}+4}=\sqrt{y^{2}-8 y+20}
\end{aligned}
$$
Since according to the problem's condition $A B=A ... | A(0;-1;0) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,533 |
## 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 $a$?
$A(5 ; 0 ;-1)$
$a: 2x - y + 3z - 1 = 0$
$k = 3$ | ## 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 $a$:
$a^{\prime}: 2 x-y+3 z-3=0$
Substitute the coordinates of point $A$ into the equat... | 4\neq0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,534 |
## Task Condition
Write the canonical equations of the line.
$x+5 y-z-5=0$
$2 x-5 y+2 z+5=0$ | ## Solution
Canonical equations of a line:
$\frac{x-x_{0}}{m}=\frac{y-y_{0}}{n}=\frac{z-z_{0}}{p}$
where $\left(x_{0} ; y_{0} ; z_{0}\right)_{\text {- are the coordinates of some point on the line, and }} \vec{s}=\{m ; n ; p\}_{\text {- is its direction }}$ vector.
Since the line belongs to both planes simultaneous... | \frac{x}{5}=\frac{y-1}{-4}=\frac{z}{-15} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,535 |
Condition of the problem
Find the point of intersection of the line and the plane.
$\frac{x-3}{-1}=\frac{y-4}{5}=\frac{z-4}{2}$
$7 x+y+4 z-47=0$ | ## Solution
Let's write the parametric equations of the line.
$$
\begin{aligned}
& \frac{x-3}{-1}=\frac{y-4}{5}=\frac{z-4}{2}=t \Rightarrow \\
& \left\{\begin{array}{l}
x=3-t \\
y=4+5 t \\
z=4+2 t
\end{array}\right.
\end{aligned}
$$
Substitute into the equation of the plane:
$7(3-t)+(4+5 t)+4(4+2 t)-47=0$
$21-7 t+... | (2;9;6) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,536 |
## Problem Statement
Find the point $M^{\prime}$ symmetric to the point $M$ with respect to the plane.
$$
\begin{aligned}
& M(1 ; 0 ; 1) \\
& 4 x+6 y+4 z-25=0
\end{aligned}
$$ | ## Solution
Let's find the equation of the line that is perpendicular to the given plane and passes through point $M$. Since the line is perpendicular to the given plane, we can take the normal vector of the plane as its direction vector:
$\vec{s}=\vec{n}=\{4 ; 6 ; 4\}$
Then the equation of the desired line is:
$\f... | M^{\}(3;3;3) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,537 |
## Problem Statement
Write the decomposition of vector $x$ in terms of vectors $p, q, r$:
$x=\{1 ;-4 ; 4\}$
$p=\{2 ; 1 ;-1\}$
$q=\{0 ; 3 ; 2\}$
$r=\{1 ;-1 ; 1\}$ | ## Solution
The desired decomposition of vector $x$ is:
$x=\alpha \cdot p+\beta \cdot q+\gamma \cdot r$
Or in the form of a system:
$$
\left\{\begin{array}{l}
\alpha \cdot p_{1}+\beta \cdot q_{1}+\gamma \cdot r_{1}=x_{1} \\
\alpha \cdot p_{2}+\beta \cdot q_{2}+\gamma \cdot r_{2}=x_{2} \\
\alpha \cdot p_{3}+\beta \c... | -p+3r | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,538 |
## Problem Statement
Are the vectors $c_{1 \text { and }} c_{2}$, constructed from vectors $a \text{ and } b$, collinear?
$a=\{-2 ; 4 ; 1\}$
$b=\{1 ;-2 ; 7\}$
$c_{1}=5 a+3 b$
$c_{2}=2 a-b$ | ## Solution
Vectors are collinear if there exists a number $\gamma$ such that $c_{1}=\gamma \cdot c_{2}$. That is, vectors are collinear if their coordinates are proportional.
We find:
$$
\begin{aligned}
& c_{1}=5 a+3 b=\{5 \cdot(-2)+3 \cdot 1 ; 5 \cdot 4+3 \cdot(-2) ; 5 \cdot 1+3 \cdot 7\}=\{-7 ; 14 ; 26\} \\
& c_{... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,539 |
## Problem Statement
Find the cosine of the angle between vectors $\overrightarrow{A B}$ and $\overrightarrow{A C}$.
$A(3 ; 3 ;-1), B(5 ; 5 ;-2), C(4 ; 1 ; 1)$ | ## Solution
Let's find $\overrightarrow{A B}$ and $\overrightarrow{A C}$:
$\overrightarrow{A B}=(5-3 ; 5-3 ;-2-(-1))=(2 ; 2 ;-1)$
$\overrightarrow{A C}=(4-3 ; 1-3 ; 1-(-1))=(1 ;-2 ; 2)$
We find the cosine of the angle $\phi$ between the vectors $\overrightarrow{A B}$ and $\overrightarrow{A C}$:
$\cos (\overrightar... | -\frac{4}{9} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,540 |
## Problem Statement
Calculate the area of the parallelogram constructed on vectors $a$ and $b$.
$a=p-3q$
$b=p+2q$
$|p|=\frac{1}{5}$
$|q|=1$
$(\widehat{p, q})=\frac{\pi}{2}$ | ## 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:
$$
\begin{aligned}
& a \times b=(p-3 q) \times(p+2 q)=p \times p+2 \cdot p \times q-3 \cd... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,541 |
## Task Condition
Are the vectors $a, b$ and $c$ coplanar?
$a=\{1 ; 5 ; 2\}$
$b=\{-1 ; 1 ;-1\}$
$c=\{1 ; 1 ; 1\}$ | ## Solution
For three vectors to be coplanar (lie in the same plane or parallel planes), it is necessary and sufficient that their scalar triple product $(a, b, c)$ be equal to zero.
$$
(a, b, c)=\left|\begin{array}{ccc}
1 & 5 & 2 \\
-1 & 1 & -1 \\
1 & 1 & 1
\end{array}\right|=
$$
$=1 \cdot\left|\begin{array}{cc}1 &... | -2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,542 |
## Problem Statement
Calculate the volume of the tetrahedron with vertices at points \( A_{1}, A_{2}, A_{3}, A_{4} \) and its height dropped from vertex \( A_{4} \) to the face \( A_{1} A_{2} A_{3} \).
\( A_{1}(7 ; 2 ; 4) \)
\( A_{2}(7 ;-1 ;-2) \)
\( A_{3}(3 ; 3 ; 1) \)
\( A_{4}(-4 ; 2 ; 1) \) | ## Solution
From vertex $A_{1 \text { we draw vectors: }}$
$\overrightarrow{A_{1} A_{2}}=\{7-7 ;-1-2 ;-2-4\}=\{0 ;-3 ;-6\}$
$\overrightarrow{A_{1} A_{3}}=\{3-7 ; 3-2 ; 1-4\}=\{-4 ; 1 ;-3\}$
$\overrightarrow{A_{1} A_{4}}=\{-4-7 ; 2-2 ; 1-4\}=\{-11 ; 0 ;-3\}$
According to the geometric meaning of the scalar triple p... | \frac{43}{\sqrt{105}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,543 |
## Problem Statement
Find the distance from point $M_{0}$ to the plane passing through three points $M_{1}, M_{2}, M_{3}$.
$$
\begin{aligned}
& M_{1}(-3 ;-1 ; 1) \\
& M_{2}(-9 ; 1 ;-2) \\
& M_{3}(3 ;-5 ; 4) \\
& M_{0}(-7 ; 0 ;-1)
\end{aligned}
$$ | ## Solution
Find the equation of the plane passing through three points $M_{1}, M_{2}, M_{3}$:
$$
\left|\begin{array}{ccc}
x-(-3) & y-(-1) & z-1 \\
-9-(-3) & 1-(-1) & -2-1 \\
3-(-3) & -5-(-1) & 4-1
\end{array}\right|=0
$$
Perform transformations:
$$
\begin{aligned}
& \left|\begin{array}{ccc}
x+3 & y+1 & z-1 \\
-6 &... | 0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,544 |
## Problem Statement
Write the equation of the plane passing through point $A$ and perpendicular to vector $\overrightarrow{B C}$.
$A(4, -2, 0)$
$B(1, -1, -5)$
$C(-2, 1, -3)$ | ## Solution
Let's find the vector $\overrightarrow{BC}$:
$\overrightarrow{BC}=\{-2-1 ; 1-(-1) ;-3-(-5)\}=\{-3 ; 2 ; 2\}$
Since the vector $\overrightarrow{BC}$ is perpendicular to the desired plane, it can be taken as the normal vector. Therefore, the equation of the plane will be:
$$
\begin{aligned}
& -3 \cdot(x-4... | -3x+2y+2z+16=0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,545 |
## problem statement
Find the angle between the planes:
$$
\begin{aligned}
& 4 x-5 y+3 z-1=0 \\
& x-4 y-z+9=0
\end{aligned}
$$ | ## Solution
The dihedral angle between planes is equal to the angle between their normal vectors. The normal vectors of the given planes are:
$$
\begin{aligned}
& \overrightarrow{n_{1}}=\{4 ;-5 ; 3\} \\
& \overrightarrow{n_{2}}=\{1 ;-4 ;-1\}
\end{aligned}
$$
The angle $\phi_{\text{between the planes is determined by... | 4534^{\}23^{\\} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,546 |
## problem statement
Find the coordinates of point $A$, which is equidistant from points $B$ and $C$.
$A(0 ; 0 ; z)$
$B(3 ; 1 ; 3)$
$C(1 ; 4 ; 2)$ | ## Solution
Let's find the distances $A B$ and $A C$:
$$
\begin{aligned}
& A B=\sqrt{(3-0)^{2}+(1-0)^{2}+(3-z)^{2}}=\sqrt{9+1+9-6 z+z^{2}}=\sqrt{z^{2}-6 z+19} \\
& A C=\sqrt{(1-0)^{2}+(4-0)^{2}+(2-z)^{2}}=\sqrt{1+16+4-4 z+z^{2}}=\sqrt{z^{2}-4 z+21}
\end{aligned}
$$
$$
\begin{aligned}
& \sqrt{z^{2}-6 z+19}=\sqrt{z^{2... | A(0;0;-1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,547 |
## problem statement
Let $k$ be the coefficient of similarity transformation with the center at the origin. Is it true that point $A$ belongs to the image of plane $a$?
$A(-1 ; 1 ; 1)$
$a: 3 x-y+2 z+4=0$
$k=\frac{1}{2}$ | ## Solution
When transforming similarity with the center at the origin of the plane
$a: A x+B y+C z+D=0_{\text{and coefficient }} k$ transitions to the plane
$a^{\prime}: A x+B y+C z+k \cdot D=0$. We find the image of the plane $a$:
$a^{\prime}: 3 x-y+2 z+2=0$
Substitute the coordinates of point $A$ into the equat... | 0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,548 |
## problem statement
Write the canonical equations of the line.
$$
\begin{aligned}
& x-2 y+z-4=0 \\
& 2 x+2 y-z-8=0
\end{aligned}
$$ | ## Solution
Canonical equations of a line:
$\frac{x-x_{0}}{m}=\frac{y-y_{0}}{n}=\frac{z-z_{0}}{p}$
where $\left(x_{0} ; y_{0} ; z_{0}\right)_{\text {- coordinates of some point on the line, and }} \vec{s}=\{m ; n ; p\}$ - its direction vector.
Since the line belongs to both planes simultaneously, its direction vecto... | \frac{x-4}{0}=\frac{y}{3}=\frac{z}{6} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,549 |
## Problem Statement
Find the point of intersection of the line and the plane.
$$
\begin{aligned}
& \frac{x-1}{-1}=\frac{y+5}{4}=\frac{z-1}{2} \\
& x-3 y+7 z-24=0
\end{aligned}
$$ | ## Solution
Let's write the parametric equations of the line.
$$
\begin{aligned}
& \frac{x-1}{-1}=\frac{y+5}{4}=\frac{z-1}{2}=t \Rightarrow \\
& \left\{\begin{array}{l}
x=1-t \\
y=-5+4 t \\
z=1+2 t
\end{array}\right.
\end{aligned}
$$
Substitute into the equation of the plane:
$$
\begin{aligned}
& (1-t)-3(-5+4 t)+7(... | (0,-1,3) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,550 |
## Task Condition
Find the point $M^{\prime}$ symmetric to the point $M$ with respect to the line.
$M(1 ; 1 ; 1)$
$$
\frac{x-2}{1}=\frac{y+1.5}{-2}=\frac{z-1}{1}
$$ | ## Solution
We find the equation of the plane that is perpendicular to the given line and passes through the point $M$. Since the plane is perpendicular to the given line, we can use the direction vector of the line as its normal vector:
$\vec{n}=\vec{s}=\{1 ;-2 ; 1\}$
Then the equation of the desired plane is:
$1 ... | M^{\}(1;0;-1) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,551 |
## 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:
$$
\begin{aligned}
& f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim _{\Delta x \rightarrow 0} \fra... | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,552 |
## Task Condition
Derive the equation of the tangent line to the given curve at the point with abscissa $x_{0}$.
$$
y=\frac{x^{2}}{10}+3, x_{0}=2
$$ | ## Solution
Let's find $y^{\prime}:$
$y^{\prime}=\left(\frac{x^{2}}{10}+3\right)^{\prime}=\frac{2 x}{10}=\frac{x}{5}$
Then:
$y_{0}^{\prime}=y^{\prime}\left(x_{0}\right)=\frac{2}{5}$
Since the function $y^{\prime}$ at the point $x_{0}$ has a finite derivative, the equation of the tangent line is:
$y-y_{0}=y_{0}^{\... | \frac{2}{5}\cdotx+\frac{13}{5} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,553 |
## Condition of the problem
Find the differential $d y$.
$y=\sqrt{x}-(1+x) \operatorname{arctg} \sqrt{x}$ | ## Solution
$d y=y^{\prime} \cdot d x=(\sqrt{x}-(1+x) \operatorname{arctg} \sqrt{x})^{\prime} d x=$
$$
\begin{aligned}
& =\left(\frac{1}{2 \sqrt{x}}-\left(\operatorname{arctg} \sqrt{x}+(1+x) \cdot \frac{1}{1+(\sqrt{x})^{2}} \cdot \frac{1}{2 \sqrt{x}}\right)\right) d x= \\
& =\left(\frac{1}{2 \sqrt{x}}-\operatorname{a... | -\operatorname{arctg}\sqrt{x}\cdot | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,554 |
## Task Condition
Approximately calculate using the differential.
$y=\sqrt[4]{2 x-\sin \frac{\pi x}{2}}, x=1.02$ | ## Solution
If the increment $\Delta x=x-x_{0}$ of the argument $x$ is small in absolute value, then
$f(x)=f\left(x_{0}+\Delta x\right) \approx f\left(x_{0}\right)+f^{\prime}\left(x_{0}\right) \cdot \Delta x$
Choose:
$x_{0}=1$
Then:
$\Delta x=0.02$
Calculate:
$$
\begin{aligned}
& y(1)=\sqrt[4]{2 \cdot 1-\sin \f... | 1.01 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,555 |
## Task Condition
Find the derivative.
$$
y=\frac{(x+3) \sqrt{2 x-1}}{2 x+7}
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\frac{(x+3) \sqrt{2 x-1}}{2 x+7}\right)^{\prime}=\frac{\left(\sqrt{2 x-1}+(x+3) \frac{1}{2 \sqrt{2 x-1}} \cdot 2\right) \cdot(2 x+7)-(x+3) \sqrt{2 x-1} \cdot 2}{(2 x+7)^{2}}= \\
& =\frac{(2 x-1+(x+3)) \cdot(2 x+7)-(x+3) \cdot(2 x-1) \cdot 2}{(2 x+7)^{2} \sqrt{2 x-1}}= ... | \frac{2x^{2}+15x+20}{(2x+7)^{2}\sqrt{2x-1}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,556 |
## Task Condition
Find the derivative.
$y=\arcsin e^{-x}-\sqrt{1-e^{2 x}}$ | ## Solution
$y^{\prime}=\left(\arcsin e^{-x}-\sqrt{1-e^{2 x}}\right)^{\prime}=\frac{1}{\sqrt{1-e^{-2 x}}} \cdot\left(-e^{-x}\right)-\frac{1}{2 \sqrt{1-e^{2 x}}} \cdot\left(-2 e^{2 x}\right)=$
$=\frac{-e^{-x}}{\sqrt{1-e^{-2 x}}}+\frac{e^{2 x}}{\sqrt{1-e^{2 x}}}=\frac{-e^{-x} \sqrt{1-e^{2 x}}+e^{2 x} \sqrt{1-e^{-2 x}}}... | \frac{e^{x}\sqrt{e^{2x}-1}-\sqrt{e^{-2x}-1}}{\sqrt{1-e^{-2x}}\cdot\sqrt{1-e^{2x}}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,557 |
## Task Condition
Find the derivative.
$y=\ln \ln \sin \left(1+\frac{1}{x}\right)$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\ln \ln \sin \left(1+\frac{1}{x}\right)\right)^{\prime}=\frac{1}{\ln \sin \left(1+\frac{1}{x}\right)} \cdot\left(\ln \sin \left(1+\frac{1}{x}\right)\right)^{\prime}= \\
& =\frac{1}{\ln \sin \left(1+\frac{1}{x}\right)} \cdot \frac{1}{\sin \left(1+\frac{1}{x}\right)} \cd... | -\frac{\operatorname{ctg}(1+\frac{1}{x})}{x^{2}\cdot\ln\sin(1+\frac{1}{x})} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,558 |
## Task Condition
Find the derivative.
$$
y=\cos ^{2}(\sin 3)+\frac{\sin ^{2} 29 x}{29 \cos 58 x}
$$ | ## Solution
$y^{\prime}=\left(\cos ^{2}(\sin 3)+\frac{\sin ^{2} 29 x}{29 \cos 58 x}\right)^{\prime}=0+\left(\frac{\sin ^{2} 29 x}{29 \cos 58 x}\right)^{\prime}=$
$=\frac{2 \sin 29 x \cdot \cos 29 x \cdot 29 \cdot \cos 58 x-\sin ^{2} 29 x \cdot(-\sin 58 x) \cdot 58}{29 \cos ^{2} 58 x}=$
$=\frac{\sin 58 x \cdot \cos 5... | \frac{\operatorname{tg}58x}{\cos58x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,559 |
## Task Condition
Find the derivative.
$$
y=(x+2 \sqrt{x}+2) \operatorname{arctg} \frac{\sqrt{x}}{\sqrt{x}+2}-\sqrt{x}
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left((x+2 \sqrt{x}+2) \operatorname{arctg} \frac{\sqrt{x}}{\sqrt{x}+2}-\sqrt{x}\right)^{\prime}= \\
& =\left(1+2 \cdot \frac{1}{2 \sqrt{x}}\right) \operatorname{arctg} \frac{\sqrt{x}}{\sqrt{x}+2}+(x+2 \sqrt{x}+2) \cdot \frac{1}{1+\left(\frac{\sqrt{x}}{\sqrt{x}+2}\right)^{2}... | (1+\frac{1}{\sqrt{x}})\operatorname{arctg}\frac{\sqrt{x}}{\sqrt{x}+2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,560 |
## problem statement
Find the derivative.
$$
y=\frac{1}{2} \cdot\left(\frac{\operatorname{sh} x}{\operatorname{ch}^{2} x}+\operatorname{arctg}(\operatorname{sh} x)\right)
$$ | ## Solution
$y=\left(\frac{1}{2} \cdot\left(\frac{\operatorname{sh} x}{\operatorname{ch}^{2} x}+\operatorname{arctg}(\operatorname{sh} x)\right)\right)^{\prime}=\left(\frac{\operatorname{sh} x}{2 \operatorname{ch}^{2} x}+\frac{1}{2} \cdot \operatorname{arctg}(\operatorname{sh} x)\right)^{\prime}=$
$=\frac{\operatorna... | \frac{1}{\operatorname{ch}^{3}x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,561 |
## Task Condition
Find the derivative.
$y=x^{29^{x}} \cdot 29^{x}$ | ## Solution
$y=x^{29^{x}} \cdot 29^{x}$
$\ln y=\ln \left(x^{29^{x}} \cdot 29^{x}\right)=\ln \left(x^{29^{x}}\right)+x \ln 29=29^{x} \cdot \ln (x)+x \ln 29$
$\frac{y^{\prime}}{y}=\left(29^{x} \cdot \ln (x)+x \ln 29\right)^{\prime}=29^{x} \cdot \ln 29 \cdot \ln (x)+29^{x} \cdot \frac{1}{x}+\ln 29$
$y^{\prime}=y \cdot\... | y^{\}=x^{29^{x}}\cdot29^{x}\cdot(29^{x}\cdot\ln29\cdot\ln(x)+\frac{29^{x}}{x}+\ln29) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,562 |
## Task Condition
Find the derivative.
$$
y=\sqrt{49 x^{2}+1} \cdot \operatorname{arctg} 7 x-\ln \left(7 x+\sqrt{49 x^{2}+1}\right)
$$ | ## Solution
$y^{\prime}=\left(\sqrt{49 x^{2}+1} \cdot \operatorname{arctg} 7 x-\ln \left(7 x+\sqrt{49 x^{2}+1}\right)\right)^{\prime}=$
$=\frac{1}{2 \sqrt{49 x^{2}+1}} \cdot 7 \cdot \operatorname{arctg} 7 x+\sqrt{49 x^{2}+1} \cdot \frac{1}{1+49 x^{2}} \cdot 7-$
$-\frac{1}{7 x+\sqrt{49 x^{2}+1}} \cdot\left(7+\frac{1}... | \frac{7\operatorname{arctg}7x}{2\sqrt{49x^{2}+1}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,563 |
## Task Condition
Find the derivative.
$y=\arcsin \frac{1}{2 x+3}+2 \sqrt{x^{2}+3 x+2}, 2 x+3>0$ | ## Solution
$y^{\prime}=\left(\arcsin \frac{1}{2 x+3}+2 \sqrt{x^{2}+3 x+2}\right)^{\prime}=$
$=\frac{1}{\sqrt{1-\left(\frac{1}{2 x+3}\right)^{2}}} \cdot\left(-\frac{1}{(2 x+3)^{2}} \cdot 2\right)+2 \cdot \frac{1}{2 \sqrt{x^{2}+3 x+2}} \cdot(2 x+3)=$
$=-\frac{2 x+3}{\sqrt{(2 x+3)^{2}-1}} \cdot \frac{2}{(2 x+3)^{2}}+\... | \frac{4\sqrt{x^{2}+3x+2}}{2x+3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,564 |
## Task Condition
Find the derivative.
$$
y=\frac{3^{x}(\ln 3 \cdot \sin 2 x-2 \cos 2 x)}{\ln ^{2} 3+4}
$$ | ## Solution
$y^{\prime}=\left(\frac{3^{x}(\ln 3 \cdot \sin 2 x-2 \cos 2 x)}{\ln ^{2} 3+4}\right)^{\prime}=$
$=\frac{1}{\ln ^{2} 3+4} \cdot\left(3^{x} \cdot \ln 3 \cdot(\ln 3 \cdot \sin 2 x-2 \cos 2 x)+3^{x}(2 \ln 3 \cdot \cos 2 x+4 \sin 2 x)\right)=$
$=\frac{3^{x}}{\ln ^{2} 3+4} \cdot\left(\ln ^{2} 3 \cdot \sin 2 x-... | 3^{x}\cdot\sin2x | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,565 |
## Condition of the problem
Find the derivative $y_{x}^{\prime}$.
$$
\left\{\begin{array}{l}
x=e^{\sec ^{2} t} \\
y=\operatorname{tg} t \cdot \ln \cos t+\operatorname{tg} t-t
\end{array}\right.
$$ | ## Solution
$x_{t}^{\prime}=\left(e^{\sec ^{2} t}\right)^{\prime}=\left(e^{\frac{1}{\cos ^{2} t}}\right)^{\prime}=e^{\frac{1}{\cos ^{2} t}} \cdot\left(\frac{1}{\cos ^{2} t}\right)^{\prime}=e^{\frac{1}{\cos ^{2} t}} \cdot \frac{-2}{\cos ^{3} t} \cdot(-\sin t)=$ $=e^{\frac{1}{\cos ^{2} t}} \cdot \frac{2 \sin t}{\cos ^{3... | \frac{1}{2}\cdot\cot\cdot\ln\cos\cdote^{-\^{2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,566 |
## Problem Statement
Derive the equations of the tangent and normal lines to the curve at the point corresponding to the parameter value $t=t_{0}$.
$$
\begin{aligned}
& \left\{\begin{array}{l}
x=\sin t \\
y=a^{t}
\end{array}\right. \\
& t_{0}=0
\end{aligned}
$$ | ## Solution
Since $t_{0}=0$, then
$x_{0}=\sin 0=0$
$y_{0}=a^{0}=1$
Let's find the derivatives:
$x_{t}^{\prime}=(\sin t)^{\prime}=\cos t$
$y_{t}^{\prime}=\left(a^{t}\right)^{\prime}=a^{t} \cdot \ln a$
$y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{t}^{\prime}}=\frac{a^{t} \cdot \ln a}{\cos t}$
Then:
$y_{0}^{\prime}=\... | x\cdot\ln+1-\frac{x}{\ln}+1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,567 |
## Task Condition
Find the $n$-th order derivative.
$$
y=\frac{1+x}{1-x}
$$ | ## Solution
Let's calculate the pattern:
$$
\begin{aligned}
& y^{\prime}=\frac{(1+x)^{\prime}(1-x)-(1+x)(1-x)^{\prime}}{(1-x)^{2}}=\frac{1-x+1+x}{(1-x)^{2}}=\frac{2}{(1-x)^{2}} \\
& y^{\prime \prime}=\frac{2^{\prime}(1-x)^{2}-2\left((1-x)^{2}\right)^{\prime}}{(1-x)^{4}}=\frac{2 \cdot 2(1-x)}{(1-x)^{4}}=\frac{2 \cdot ... | y^{(n)}=\frac{2\cdotn!}{(1-x)^{n+1}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,568 |
## Task Condition
Find the derivative of the specified order.
$$
y=(5 x-1) \ln ^{2} x, y^{\prime \prime \prime}=?
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left((5 x-1) \ln ^{2} x\right)^{\prime}=5 \ln ^{2} x+(5 x-1) \cdot 2 \cdot \ln x \cdot \frac{1}{x}= \\
& =5 \ln ^{2} x+2 \cdot \frac{(5 x-1) \cdot \ln x}{x} \\
& y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}=\left(5 \ln ^{2} x+2 \cdot \frac{(5 x-1) \cdot \ln x}{x}\righ... | \frac{6-2(5x+2)\lnx}{x^{3}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,569 |
## Problem Statement
Find the second-order derivative $y_{x x}^{\prime \prime}$ of the function given parametrically.
$$
\left\{\begin{array}{l}
x=\frac{1}{t^{2}} \\
y=\frac{1}{t^{2}+1}
\end{array}\right.
$$ | ## Solution
$x_{t}^{\prime}=\left(\frac{1}{t^{2}}\right)^{\prime}=-\frac{2}{t^{3}}$
$y_{t}^{\prime}=\left(\frac{1}{1+t^{2}}\right)^{\prime}=-\frac{1}{\left(1+t^{2}\right)^{2}} \cdot 2 t=-\frac{2 t}{\left(1+t^{2}\right)^{2}}$
We obtain:
$y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{t}^{\prime}}=\frac{-\frac{2 t}{\left(1+t... | -\frac{2^{6}}{(1+^{2})^{3}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,570 |
## Problem Statement
Show that the function $y_{\text {satisfies equation (1). }}$.
\[
\begin{aligned}
& y=(x+1)^{n} \cdot\left(e^{x}-1\right) \\
& y^{\prime}-\frac{n \cdot y}{x+1}=e^{x}(1+x)^{n}
\end{aligned}
\] | ## Solution
$y^{\prime}=\left((x+1)^{n} \cdot\left(e^{x}-1\right)\right)^{\prime}=n(x+1)^{n-1} \cdot\left(e^{x}-1\right)+(x+1)^{n} \cdot e^{x}$
Substitute into equation (1):
$n(x+1)^{n-1} \cdot\left(e^{x}-1\right)+(x+1)^{n} \cdot e^{x}-\frac{n \cdot(x+1)^{n} \cdot\left(e^{x}-1\right)}{x+1}=e^{x}(1+x)^{n}$
Simplify:... | proof | Calculus | proof | Yes | Yes | olympiads | false | 46,571 |
## Condition of the problem
Prove that $\lim _{n \rightarrow \infty} a_{n}=a_{\text {(specify }} N(\varepsilon)$ ).
$a_{n}=\frac{2-2 n}{3+4 n}, a=-\frac{1}{2}$ | ## Solution
By the definition of the limit:
$$
\begin{aligned}
& \forall \varepsilon>0: \exists N(\varepsilon) \in \mathbb{N}: \forall n: n \geq N(\varepsilon):\left|a_{n}-a\right| \\
& \left|\frac{4-4 n+3+4 n}{2(3+4 n)}\right| \\
& \left.\frac{7}{2(3+4 n)} \right\rvert\, \\
& \frac{7}{2(3+4 n)} \\
& 3+4 n>\frac{7}{2... | N(\varepsilon)=[\frac{7+2\varepsilon}{8\varepsilon}] | Calculus | proof | Yes | Yes | olympiads | false | 46,572 |
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{(n+1)^{3}-(n-1)^{3}}{(n+1)^{2}-(n-1)^{2}}$ | ## Solution
$\lim _{n \rightarrow \infty} \frac{(n+1)^{3}-(n-1)^{3}}{(n+1)^{2}-(n-1)^{2}}=\lim _{n \rightarrow \infty} \frac{n^{3}+3 n^{2}+3 n+1-n^{3}+3 n^{2}-3 n+1}{n^{2}+2 n+1-n^{2}+2 n-1}=$
$=\lim _{n \rightarrow \infty} \frac{6 n^{2}+2}{4 n}=\lim _{n \rightarrow \infty}\left(\frac{3}{2} n+\frac{1}{2 n}\right)=+\i... | +\infty | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,573 |
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{\sqrt{n+2}-\sqrt[3]{n^{3}+2}}{\sqrt[7]{n+2}-\sqrt[5]{n^{5}+2}}$ | Solution
$$
\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{\sq... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,574 |
## 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
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} n^{3}\left(\sqrt[3]{n^{2}\left(n^{6}+4\right)}-\sqrt[3]{n^{8}-1}\right)= \\
& =\lim _{n \rightarrow \infty} \frac{n^{3}\left(\sqrt[3]{n^{2}\left(n^{6}+4\right)}-\sqrt[3]{n^{8}-1}\right)\left(\left(\sqrt[3]{n^{2}\left(n^{6}+4\right)}\right)^{2}+\sqrt[3]{n^{2... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,575 |
## Problem Statement
Calculate the limit of the numerical sequence:
$$
\lim _{n \rightarrow \infty}\left(\frac{1+5+9+13+\ldots+(4 n-3)}{n+1}-\frac{4 n+1}{2}\right)
$$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty}\left(\frac{1+5+9+13+\ldots+(4 n-3)}{n+1}-\frac{4 n+1}{2}\right)= \\
& =\lim _{n \rightarrow \infty}\left(\frac{\frac{(1+(4 n-3)) n}{2}}{n+1}-\frac{4 n+1}{2}\right)=\lim _{n \rightarrow \infty}\left(\frac{(4 n-2) n}{2(n+1)}-\frac{4 n+1}{2}\right)= \\
& =\lim... | -\frac{7}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,576 |
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty}\left(\frac{7 n^{2}+18 n-15}{7 n^{2}+11 n+15}\right)^{n+2}$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty}\left(\frac{7 n^{2}+18 n-15}{7 n^{2}+11 n+15}\right)^{n+2}= \\
& =\lim _{n \rightarrow \infty}\left(\frac{7 n^{2}+11 n+15+7 n-30}{7 n^{2}+11 n+15}\right)^{n+2}=
\end{aligned}
$$
$$
\begin{aligned}
& =\lim _{n \rightarrow \infty}\left(1+\frac{7 n-30}{7 n^{2}... | e | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,577 |
## Condition of the problem
Prove that (find $\delta(\varepsilon)$ :
$\lim _{x \rightarrow 8} \frac{3 x^{2}-40 x+128}{x-8}=8$ | ## Solution
According to the definition of the limit of a function by Cauchy:
If a function \( f: M \subset \mathbb{R} \rightarrow \mathbb{R} \) and \( a \in M' \) is a limit point of the set \( M \). A number \( A \in \mathbb{R} \) is called the limit of the function \( f \) as \( x \) approaches \( a (x \rightarrow... | proof | Calculus | proof | Yes | Yes | olympiads | false | 46,578 |
## Condition of the problem
Prove that the function $f(x)_{\text {is continuous at the point }} x_{0}$ (find $\delta(\varepsilon)$ ):
$f(x)=-4 x^{2}-8, x_{0}=2$ | ## Solution
By definition, the function $f(x)_{\text {is continuous at the point }} x=x_{0}$, if $\forall \varepsilon>0: \exists \delta(\varepsilon)>0:$. $\left|x-x_{0}\right|0_{\text {there exists such }} \delta(\varepsilon)>0$, that $\left|f(x)-f\left(x_{0}\right)\right|<\varepsilon_{\text {when }}$ $\left|x-x_{0}\r... | proof | Calculus | proof | Yes | Yes | olympiads | false | 46,579 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 1} \frac{2 x^{2}-x-1}{x^{3}+2 x^{2}-x-2}$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 1} \frac{2 x^{2}-x-1}{x^{3}+2 x^{2}-x-2}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow 1} \frac{(x-1)(2 x+1)}{(x-1)\left(x^{2}+3 x+2\right)}= \\
& =\lim _{x \rightarrow 1} \frac{2 x+1}{x^{2}+3 x+2}=\frac{2 \cdot 1+1}{1^{2}+3 \cdot 1+2}=\frac{3}{6}=\frac{1}{2}
\en... | \frac{1}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,580 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{\sqrt{1-2 x+3 x^{2}}-(1+x)}{\sqrt[3]{x}}$ | ## Solution
$$
\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... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,581 |
## 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:
$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{\sin ^{2} x-\operatorname{tg}^{2} x}{x^{4}... | -1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,582 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow \frac{\pi}{3}} \frac{1-2 \cos x}{\pi-3 x}$ | ## Solution
Substitution:
$$
\begin{aligned}
& x=y+\frac{\pi}{3} \Rightarrow y=x-\frac{\pi}{3} \\
& x \rightarrow \frac{\pi}{3} \Rightarrow y \rightarrow 0
\end{aligned}
$$
We get:
$$
\begin{aligned}
& \lim _{x \rightarrow \frac{\pi}{3}} \frac{1-2 \cos x}{\pi-3 x}=\lim _{y \rightarrow 0} \frac{1-2 \cos \left(y+\fra... | -\frac{\sqrt{3}}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,583 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow \pi} \frac{\ln (\cos 2 x)}{\ln (\cos 4 x)}$ | ## Solution
Substitution:
$$
\begin{aligned}
& x=y+\pi \Rightarrow y=x-\pi \\
& x \rightarrow \pi \Rightarrow y \rightarrow 0
\end{aligned}
$$
We get:
$$
\begin{aligned}
& \lim _{x \rightarrow \pi} \frac{\ln (\cos 2 x)}{\ln (\cos 4 x)}=\lim _{y \rightarrow 0} \frac{\ln (\cos 2(y+\pi))}{\ln (\cos 4(y+\pi))}= \\
& =\... | \frac{1}{4} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,584 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 10} \frac{\lg x-1}{\sqrt{x-9}-1}$ | ## Solution
Substitution:
$$
\begin{aligned}
& x=10(y+1) \Rightarrow y=\frac{x}{10}-1 \\
& x \rightarrow 10 \Rightarrow y \rightarrow 0
\end{aligned}
$$
We obtain:
$$
\begin{aligned}
& \lim _{x \rightarrow 10} \frac{\lg x-1}{\sqrt{x-9}-1}=\lim _{y \rightarrow 0} \frac{\lg 10(y+1)-1}{\sqrt{10(y+1)-9}-1}= \\
& =\lim ... | \frac{1}{5\ln10} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,585 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0}\left(1+\ln \frac{1}{3} \cdot \operatorname{arctg}^{6} \sqrt{x}\right)^{\frac{1}{x^{3}}}$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 0}\left(1+\ln \frac{1}{3} \cdot \operatorname{arctg}^{6} \sqrt{x}\right)^{\frac{1}{x^{3}}}= \\
& =\lim _{x \rightarrow 0}\left(e^{\left.\ln \left(1+\ln \frac{1}{3} \cdot \operatorname{arctg}^{6} \sqrt{x}\right)\right)^{\frac{1}{x^{3}}}}\right. \\
& =\lim _{x \right... | \frac{1}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,586 |
## Problem Statement
Calculate the limit of the function:
$$
\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}
$$
## Problem Kuznetsov Limits 18-25 | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,587 |
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow 1}(2-x)^{\sin \left(\frac{\pi x}{2}\right) / \ln (2-x)}
$$ | ## Solution
Calculate the limit of the function:
$$
\begin{aligned}
& \lim _{x \rightarrow 1}(2-x)^{\sin \left(\frac{\pi x}{2}\right) / \ln (2-x)}= \\
& =\lim _{x \rightarrow 1}\left(e^{\ln (2-x)}\right)^{\sin \left(\frac{\pi x}{2}\right) / \ln (2-x)}= \\
& =\lim _{x \rightarrow 1} e^{\ln (2-x) \cdot \sin \left(\frac... | e | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,588 |
## 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 | 46,589 |
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow 0} \frac{\cos (1+x)}{\left(2+\sin \left(\frac{1}{x}\right)\right) \ln (1+x)+2}
$$ | ## Solution
Since $\sin \left(\frac{1}{x}\right)_{\text { is bounded, and }} \ln (1+x) \rightarrow 0$, as $x \rightarrow 0$, then
$$
\left(2+\sin \left(\frac{1}{x}\right)\right) \ln (1+x) \rightarrow 0 \quad \text {, as } x \rightarrow 0
$$
Then:
$\lim _{x \rightarrow 0} \frac{\cos (1+x)}{\left(2+\sin \left(\frac{1... | \frac{\cos1}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,590 |
## Problem Statement
Calculate the force with which water presses on a dam, the cross-section of which has the shape of an isosceles trapezoid (see figure). The density of water \(\rho = 1000 \text{ kg} / \text{m}^3\), and the acceleration due to gravity \(g\) is taken to be \(10 \text{ m} / \text{s}^2\).
Hint: The p... | ## Solution
$$
\begin{aligned}
& c=b-2 F B ; \triangle F B G \text { is similar to } \\
& \Delta E B C \Rightarrow c=b-x \frac{b-a}{h}
\end{aligned}
$$
. The density of water \(\rho = 1000 \mathrm{kg} / \mathrm{m}^{3}\), and the acceleration due to gravity \(g\) is taken to be \(10 \mathrm{m} / \mathrm{s}^{2}\).
H... | ## Solution
$$
\begin{aligned}
& c=b-2 F B ; \triangle F B G \text { is similar to } \\
& \Delta E B C \Rightarrow c=b-x \frac{b-a}{h}
\end{aligned}
$$
. The density of water \(\rho = 1000 \mathrm{kg} / \mathrm{m}^{3}\), and the acceleration due to gravity \(g\) is taken to be \(10 \mathrm{m} / \mathrm{s}^{2}\).
H... | ## Solution
$$
\begin{aligned}
& c=b-2 F B ; \triangle F B G \text { is similar to } \\
& \Delta E B C \Rightarrow c=b-x \frac{b-a}{h}
\end{aligned}
$$
. The density of water \(\rho = 1000 \mathrm{kg} / \mathrm{m}^{3}\), and the acceleration due to gravity \(g\) is taken to be \(10 \mathrm{m} / \mathrm{s}^{2}\).
H... | ## Solution
$$
\begin{aligned}
& c=b-2 F B ; \triangle F B G \text { is similar to } \\
& \Delta E B C \Rightarrow c=b-x \frac{b-a}{h}
\end{aligned}
$$
. The density of water \(\rho = 1000 \mathrm{kg} / \mathrm{m}^{3}\), and the acceleration due to gravity \(g\) is taken to be \(10 \mathrm{m} / \mathrm{s}^{2}\).
H... | ## Solution
$$
\begin{aligned}
& c=b-2 F B ; \triangle F B G \text { is similar to } \\
& \Delta E B C \Rightarrow c=b-x \frac{b-a}{h}
\end{aligned}
$$
. The density of water \(\rho = 1000 \text{ kg} / \text{m}^3\), and the acceleration due to gravity \(g\) is taken to be \(10 \text{ m} / \text{s}^2\).
Hint: The p... | ## Solution
$$
\begin{aligned}
& c=b-2 F B ; \triangle F B G \text { is similar to } \\
& \Delta E B C \Rightarrow c=b-x \frac{b-a}{h}
\end{aligned}
$$
. The density of water \(\rho = 1000 \mathrm{kg} / \mathrm{m}^{3}\), and the acceleration due to gravity \(g\) is taken to be \(10 \mathrm{m} / \mathrm{s}^{2}\).
H... | ## Solution
$$
\begin{aligned}
& c=b-2 F B ; \triangle F B G \text { is similar to } \\
& \Delta E B C \Rightarrow c=b-x \frac{b-a}{h}
\end{aligned}
$$
. The density of water \(\rho = 1000 \mathrm{kg} / \mathrm{m}^{3}\), and the acceleration due to gravity \(g\) is taken to be \(10 \mathrm{m} / \mathrm{s}^{2}\).
H... | ## Solution
$$
c=b-2 F B ; \triangle F B G \text { is similar to }
$$
$$
\Delta E B C \Rightarrow c=b-x \frac{b-a}{h}
$$
. The density of water \(\rho = 1000 \mathrm{kg} / \mathrm{m}^{3}\), and the acceleration due to gravity \(g\) is taken to be \(10 \mathrm{m} / \mathrm{s}^{2}\).
H... | ## Solution
$$
c=b-2 F B ; \Delta F B G \text { is similar to }
$$
$$
\Delta E B C \Rightarrow c=b-x \frac{b-a}{h}
$$
. The density of water \(\rho = 1000 \mathrm{kg} / \mathrm{m}^{3}\), and the acceleration due to gravity \(g\) is taken to be \(10 \mathrm{m} / \mathrm{s}^{2}\).
H... | ## Solution
$$
c=b-2 F B ; \triangle F B G \text { is similar to }
$$
$$
\Delta E B C \Rightarrow c=b-x \frac{b-a}{h}
$$
 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 $10 \mathrm{~m} / \mathrm{c}^{2}$... | ## 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 ... | 13574468085 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,601 |
## 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 $10 \mathrm{~m} / \mathrm{c}^{2}$... | ## Solution
By definition, the elementary work $d A=F(x) d x$, where $F(r)=G \frac{m \cdot M}{r^{2}} ; G=6.67 \cdot 10^{-11} \mathrm{H}^{*} \mathrm{m}^{*} \mathrm{m} /(\mathrm{kg}^{*} \mathrm{kg})$
\[
\begin{aligned}
& F_{0}=G \frac{m \cdot M}{R^{2}}=m g \\
& F_{x}=G \frac{m \cdot M}{(R+x)^{2}} \text{ - force of attr... | 1.68\cdot10^{10} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,602 |
## 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 $10 \mathrm{~m} / \mathbf{c}^{2}$... | ## 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 ... | 17191616766 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,603 |
## 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 $10 \mathrm{~m} / \mathbf{c}^{2}$... | ## 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 ... | 19907875186 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,604 |
## 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 $10 \mathrm{~m} / \mathbf{c}^{2}$... | ## 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 ... | 18820058997 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,605 |
## 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 $10 \mathrm{~m} / \mathrm{c}^{2}$... | ## 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 ... | 21017569546 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,606 |
## 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 $10 \mathrm{~m} / \mathbf{c}^{2}$... | ## 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 ... | 18546511628 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,607 |
## 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 $10 \mathrm{~m} / \mathbf{c}^{2}$... | ## 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 ... | 20253968254 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,608 |
## 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 $10 \mathrm{~m} / \mathbf{c}^{2}$... | ## 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 ... | 16452722063 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,609 |
## 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 $10 \mathrm{~m} / \mathbf{c}^{2}$... | ## 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 ... | 17697012802 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,610 |
## 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{con... | ## 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 $\Delta A=F(x) \Delta x \Rightarrow$
... | 2700 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,611 |
## 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 ... | ## 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 ... | 1800 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,612 |
## 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
$\rho V = \text{const}$,... | ## 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 $\Delta A=F(x) \Delta x \Rightarrow$
... | 900 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,613 |
## 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
$\rho V = \text{const}$,... | ## 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 $\Delta A=F(x) \Delta x \Rightarrow$
... | 21595[\mathrm{kJ}] | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,614 |
## 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, wh... | ## 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 $\Delta A=F(x) \Delta x \Rightarrow$
... | 14400 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,615 |
## 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 $... | ## 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... | 7200[kJ] | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,616 |
## 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 ... | ## 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 $\Delta A=F(x) \Delta x \Rightarrow$
... | 97200 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,617 |
## 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 ... | ## 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 $\Delta A=F(x) \Delta x \Rightarrow$
... | 64800 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,618 |
## 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 ... | ## 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 $\Delta A=F(x) \Delta x \Rightarrow$
... | 32400 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,619 |
## 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 $... | ## 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 $\Delta A=F(x) \Delta x \Rightarrow$
... | 144000 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,620 |
## 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, wh... | ## 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... | 72000 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,621 |
## 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:
$$
\begin{aligned}
& f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim _{\Delta x \rightarrow 0} \fra... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,622 |
## Task Condition
Compose the equation of the normal to the given curve at the point with abscissa $x_{0}$.
$y=x^{2}+8 \sqrt{x}-32, x_{\bar{u}}=4$ | ## Solution
Let's find $y^{\prime}:$
$$
y^{\prime}=\left(x^{2}+8 \sqrt{x}-32\right)^{\prime}=2 x+\frac{8}{2 \sqrt{x}}=2 x+\frac{4}{\sqrt{x}}
$$
Then
$$
y_{0}^{\prime}=y^{\prime}\left(x_{0}\right)=2 x_{\overline{0}}+\frac{4}{\sqrt{x_{0}}}=2 \cdot 4+\frac{4}{\sqrt{4}}=8+2=10
$$
Since $y^{\prime}\left(x_{\bar{u}}\rig... | -\frac{x}{10}+\frac{2}{5} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,623 |
Condition of the problem
Find the differential $d y$.
$$
y=x^{2} \cdot \operatorname{arctg}\left(\sqrt{x^{2}-1}\right)-\sqrt{x^{2}-1}
$$ | ## Solution
$$
\begin{aligned}
& d y=y^{\prime} \cdot d x=\left(x^{2} \cdot \operatorname{arctg}\left(\sqrt{x^{2}-1}\right)-\sqrt{x^{2}-1}\right)^{\prime} d x= \\
& =\left(2 x \cdot \operatorname{arctg}\left(\sqrt{x^{2}-1}\right)+x^{2} \cdot \frac{1}{1+\left(\sqrt{x^{2}-1}\right)^{2}} \cdot \frac{1}{2 \sqrt{x^{2}-1}} ... | 2x\cdot\operatorname{arctg}(\sqrt{x^{2}-1})\cdot | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,624 |
## Task Condition
Find the derivative.
$y=\frac{2 x^{2}-x-1}{3 \sqrt{2+4 x}}$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\frac{2 x^{2}-x-1}{3 \sqrt{2+4 x}}\right)^{\prime}=\frac{(4 x-1) \cdot \sqrt{2+4 x}-\left(2 x^{2}-x-1\right) \cdot \frac{1}{2 \sqrt{2+4 x}} \cdot 4}{3(2+4 x)}= \\
& =\frac{(4 x-1) \cdot(2+4 x)-\left(2 x^{2}-x-1\right) \cdot \frac{1}{3}}{3(2+4 x) \sqrt{2+4 x}}=\frac{8 x... | \frac{x}{\sqrt{2+4x}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,626 |
## Task Condition
Find the derivative.
$$
y=\frac{1}{\ln 4} \ln \frac{1+2^{x}}{1-2^{x}}
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\frac{1}{\ln 4} \ln \frac{1+2^{x}}{1-2 x}\right)^{\prime}=\frac{1}{\ln 4} \cdot \frac{1-2^{x}}{1+2^{x}} \cdot\left(\frac{1+2^{x}}{1-2^{x}}\right)^{\prime}= \\
& =\frac{1}{\ln 2^{2}} \cdot \frac{1-2^{x}}{1+2^{x}} \cdot \frac{2^{x} \cdot \ln 2 \cdot\left(1-2^{x}\right)-\... | \frac{2^{x}}{1-2^{2x}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,627 |
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