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## Problem Statement
Calculate the volumes of the bodies bounded by the surfaces.
$$
\frac{x^{2}}{16}+\frac{y^{2}}{9}+\frac{z^{2}}{100}=1, z=5, z=0
$$ | ## Solution
In the section of the given figure by the plane $z=$ const, there is an ellipse:
$$
\frac{x^{2}}{16}+\frac{y^{2}}{9}=\frac{100-z^{2}}{100}
$$
The area of an ellipse with radii $a$ and $b$ is $\pi \cdot a \cdot b$
By the definition of the radius of an ellipse:
$$
\begin{aligned}
& \frac{x^{2}}{16 \cdot ... | 55\pi | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,252 |
## Problem Statement
Calculate the volumes of the bodies bounded by the surfaces.
$$
\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}
$$
That is,
$x \in[-3 \sqrt{3}, 3 \sqrt{3}], y \in[0,1]$
Therefore, the volume will be
![](https://cdn.mathpix.com/cropped/2024_05_22_348f4289c7b5f... | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,253 |
## Task Condition
Calculate the volumes of bodies bounded by surfaces.
$$
z=2 x^{2}+18 y^{2}, z=6
$$ | ## Solution
$$
\begin{aligned}
& a=\frac{\sqrt{z}}{\sqrt{2}} ; b=\frac{\sqrt{z}}{\sqrt{18}} \\
& V=\frac{1}{6} \pi \int_{0}^{6} z d z=\left.\frac{1}{12} \pi z^{2}\right|_{0} ^{6}=3 \pi
\end{aligned}
$$
Source — "http://pluspi.org/wiki/index.php/?\?\?\?\?\?\?\?\?\?\?\? \%D0\%9A\%D1\%83\%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\... | 3\pi | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,254 |
## Problem Statement
Calculate the volumes of the bodies bounded by the surfaces.
$$
\frac{x^{2}}{25}+\frac{y^{2}}{9}-z^{2}=1, z=0, z=2
$$ | ## Solution
In the section of the given figure by the plane $z=$ const, there is an ellipse:
$$
\frac{x^{2}}{25}+\frac{y^{2}}{9}=z^{2}+1
$$
The area of the ellipse described by the formula: $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ is $\pi \cdot a \cdot b$
Let's find the radii of the ellipse:
$:
$f(x)=\left\{\begin{array}{c}\tan\left(2^{x^{2} \cos (1 /(8 x))}-1+x\right), x \neq 0 ; \\ 0, x=0\end{array}\right.$ | ## Solution
By definition, the derivative at the point $x=0$:
$f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}$
Based on the definition, we find:
$$
\begin{aligned}
& f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim _{\Delta x \rightarrow 0} \fra... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,259 |
## Task Condition
Compose the equation of the normal to the given curve at the point with abscissa $x_{0}$.
$$
y=\frac{x^{3}+2}{x^{3}-2}, x_{0}=2
$$ | ## Solution
Let's find $y^{\prime}:$
$$
\begin{aligned}
& y^{\prime}=\left(\frac{x^{3}+2}{x^{5}-2}\right)^{\prime}=\frac{\left(x^{3}+2\right)^{\prime}\left(x^{5}-2\right)-\left(x^{3}+2\right)\left(x^{5}-2\right)^{\prime}}{\left(x^{5}-2\right)^{2}}=1+\frac{3}{2} \cdot x^{\frac{1}{2}}= \\
& =\frac{3 x^{2}\left(x^{5}-2\... | \frac{3}{4}\cdotx+\frac{1}{6} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,260 |
## Problem Statement
Find the differential $d y$.
$$
y=\ln \left(e^{x}+\sqrt{e^{2 x}-1}\right)+\arcsin e^{x}
$$ | ## Solution
$$
\begin{aligned}
& d y=y^{\prime} \cdot d x=\left(\ln \left(e^{x^{x}}+\sqrt{\epsilon^{2 x}-1}\right)+\arcsin \epsilon^{e^{x}}\right)^{\prime} d x= \\
& =\left(\frac{1}{e^{x}+\sqrt{\epsilon^{2 x}-1}} \cdot\left(e^{x}+\frac{1}{2 \sqrt{\epsilon^{2 x}-1}} \cdot \epsilon^{2 x} \cdot 2\right)+\frac{1}{\sqrt{1-... | (\frac{e^{x}}{\sqrt{e^{2x}-1}}+\frac{e^{x}}{\sqrt{1-e^{2x}}}) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,261 |
## Task Condition
Approximately calculate using the differential.
$y=\sqrt[3]{x^{2}}, x=1.03$ | ## Solution
If the increment $\Delta x = x - x_{\text{argument}}$ is small in absolute value, then
$$
f(x) = f\left(x_{0} + \Delta x\right) \approx f\left(x_{\overline{\mathrm{U}}}\right) + f^{\prime}\left(x_{\overline{\mathrm{U}}}\right) \cdot \Delta x
$$
Choose:
$x_{0} = 1$
Then
$\Delta x = 0.03$
Calculate:
$... | 1.02 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,262 |
## Task Condition
Find the derivative.
$$
y=\frac{\left(x^{2}-2\right) \sqrt{1+x^{2}}}{24 x^{3}}
$$ | ## Solution
$y^{\prime}=\left(\frac{\left(x^{2}-2\right) \sqrt{4+x^{2}}}{24 x^{\prime 3}}\right)^{\prime}=\frac{\left(\left(x^{2}-2\right) \sqrt{4+x^{2}}\right)^{\prime} \cdot x^{3}-\left(x^{2}-2\right) \sqrt{4+x^{2}}\left(x^{3}\right)^{\prime}}{24 x^{6}}=$
$=\frac{\left(2 x \sqrt{4+x^{2}}+\left(x^{2}-2\right) \frac{... | \frac{-x^{4}+x^{3}+2x^{2}-2x+24}{24x^{4}\sqrt{4+x^{2}}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,263 |
## Task Condition
Find the derivative.
$$
y=\frac{\epsilon^{a x}(3 \cdot \sin 3 x-\alpha \cos 3 x)}{a^{2}+3^{2}}
$$ | ## Solution
$y^{\prime}=\left(\frac{e^{a x}(3 \cdot \sin 3 x-\alpha \cos 3 x)}{a^{2}+3^{2}}\right)^{\prime}=$
$=\frac{1}{a^{2}+\beta^{2}}\left(\alpha \cdot e^{a x}(3 \cdot \sin 3 x-\alpha \cos 3 x)+\epsilon^{\alpha x}\left(3^{2} \cdot \cos 3 x+a \cdot 3 \cdot \sin 3 x\right)\right)=$
$=\frac{\epsilon^{a x}}{a^{2}+3^... | \frac{\epsilon^{-\alphax}(3^{2}\cdot\cos3x+2\beta\sin3x-\alpha^{2}\cdot\cos3x)}{^{2}+3^{2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,264 |
## Problem Statement
Find the derivative.
$$
y=x+\frac{1}{\sqrt{2}} \ln \frac{x-\sqrt{2}}{x+\sqrt{2}}+a^{\pi^{\sqrt{2}}}
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(x+\frac{1}{\sqrt{2}} \ln \frac{x-\sqrt{2}}{x+\sqrt{2}}+a^{\pi^{\sqrt{2}}}\right)^{\prime}=\left(x+\frac{1}{\sqrt{2}} \ln \frac{x-\sqrt{2}}{x+\sqrt{2}}\right)^{\prime}= \\
& =1+\frac{1}{\sqrt{2}} \cdot \frac{x+\sqrt{2}}{x-\sqrt{2}} \cdot\left(\frac{x-\sqrt{2}}{x+\sqrt{2... | \frac{x^{2}}{x^{2}-2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,265 |
## Task Condition
Find the derivative.
$$
y=\frac{3+x}{2} \cdot \sqrt{x(2-x)}+3 \arccos \sqrt{\frac{x}{2}}
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\frac{3+x}{2} \cdot \sqrt{x(2-x)}+3 \arccos \sqrt{\frac{x}{2}}\right)^{\prime}= \\
& =\frac{1}{2} \cdot \sqrt{x(2-x)}+\frac{3+x}{2} \cdot \frac{1}{2 \sqrt{x(2-x)}} \cdot(2-2 x)+3 \cdot \frac{-1}{\sqrt{1-\left(\sqrt{\frac{x}{2}}\right)^{2}}} \cdot \frac{1}{2 \sqrt{\frac... | -\frac{x^{2}}{\sqrt{x(2-x)}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,267 |
## Problem Statement
Find the derivative.
$$
y=\frac{\operatorname{sh} x}{1+\operatorname{ch} x}
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\frac{\sinh x}{1+\cosh x}\right)^{\prime}=\frac{\cosh x \cdot(1+\cosh x)-\sinh x \cdot \sinh x}{(1+\cosh x)^{2}}= \\
& =\frac{\cosh x+\cosh^{2} x-\sinh^{2} x}{(1+\cosh x)^{2}}=\frac{\cosh x+1}{(1+\cosh x)^{2}}=\frac{1}{1+\cosh x}
\end{aligned}
$$
## Problem Kuznetsov ... | \frac{1}{1+\coshx} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,268 |
## Problem Statement
Find the derivative.
$$
y=\frac{x+2}{x^{2}+4 x+6}+\frac{1}{\sqrt{2}} \cdot \operatorname{arctg} \frac{x+2}{\sqrt{2}}
$$ | ## Solution
$y^{\prime}=\left(\frac{x+2}{x^{2}+4 x+6}+\frac{1}{\sqrt{2}} \cdot \operatorname{arctg} \frac{x+2}{\sqrt{2}}\right)^{\prime}=\frac{x^{2}+4 x+6-(x+2) \cdot(2 x+4)}{\left(x^{2}+4 x+6\right)^{2}}+\frac{1}{\sqrt{2}} \cdot \frac{1}{1+\left(\frac{x+2}{\sqrt{2}}\right)^{2}} \cdot \frac{1}{\sqrt{2}}=$
$$
\begin{a... | \frac{4}{(x^{2}+4x+6)^{2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,270 |
## Problem Statement
Find the derivative.
$$
y=\ln \frac{\sqrt{x^{2}-x+1}}{x}+\sqrt{3} \cdot \operatorname{arctg} \frac{2 x-1}{\sqrt{3}}
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\ln \frac{\sqrt{x^{2}-x+1}}{x}+\sqrt{3} \cdot \operatorname{arctg} \frac{2 x-1}{\sqrt{3}}\right)^{\prime}= \\
& =\frac{x}{\sqrt{x^{2}-x+1}} \cdot\left(\frac{\sqrt{x^{2}-x+1}}{x^{2}}\right)^{\prime}+\sqrt{3} \cdot \frac{1}{1+\left(\frac{2 x-1}{\sqrt{3}}\right)^{2}} \cdo... | \frac{2x-1}{x\cdot(x^{2}-x+1)} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,271 |
## Task Condition
Find the derivative.
$$
y=\frac{\operatorname{ctg} x+x}{1-x \cdot \operatorname{ctg} x}
$$ | ## Solution
$y^{\prime}=\left(\frac{\operatorname{ctg} x+x}{1-x \cdot \operatorname{ctg} x}\right)^{\prime}=\frac{(\operatorname{ctg} x+x)^{\prime} \cdot(1-x \cdot \operatorname{ctg} x)-(\operatorname{ctg} x+x) \cdot(1-x \cdot \operatorname{ctg} x)^{\prime}}{(1-x \cdot \operatorname{ctg} x)^{2}}=$
$$
\begin{aligned}
... | -\frac{x^{2}}{(\sinx-x\cdot\cosx)^{2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,272 |
## Problem Statement
Find the derivative $y_{x}^{\prime}$.
$$
\left\{\begin{array}{l}
x=\sqrt{1-t^{2}} \\
y=\frac{t}{\sqrt{1-t^{2}}}
\end{array}\right.
$$ | ## Solution
$x_{t}^{\prime}=\left(\sqrt{1-t^{2}}\right)^{\prime}=\frac{1}{2 \sqrt{1-t^{2}}} \cdot(-2 t)=-\frac{t}{\sqrt{1-t^{2}}}$
$y_{t}^{\prime}=\left(\frac{t}{\sqrt{1-t^{2}}}\right)^{\prime}=\frac{\sqrt{1-t^{2}}-t \cdot \frac{1}{2 \sqrt{1-t^{2}}} \cdot(-2 t)}{1-t^{2}}=$
$=\frac{\sqrt{1-t^{2}}+\frac{t^{2}}{\sqrt{1... | \frac{1}{(^{2}-1)} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,273 |
## Problem Statement
Derive the equations of the tangent and normal lines to the curve at the point corresponding to the parameter value $t=t_{0}$.
\[
\left\{\begin{array}{l}
x=\sin t \\
y=\cos t
\end{array}\right.
\]
$t_{0}=\frac{\pi}{6}$ | ## Solution
Since $t_{0}=\frac{\pi}{6}$, then
$x_{0}=\sin \frac{\pi}{6}=\frac{1}{2}$
$y_{0}=\cos \frac{\pi}{6}=\frac{\sqrt{3}}{2}$
Let's find the derivatives:
$x_{t}^{\prime}=(\sin t)^{\prime}=\cos t$
$y_{t}^{\prime}=(\cos t)^{\prime}=-\sin t$
$y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{t}^{\prime}}=\frac{-\sin t}{... | -\frac{1}{\sqrt{3}}\cdotx+\frac{2}{\sqrt{3}}\sqrt{3}\cdotx | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,274 |
## Task Condition
Find the $n$-th order derivative.
$y=\sin (x+1)+\cos 2 x$ | ## Solution
$y^{\prime}=(\sin (x+1)+\cos 2 x)^{\prime}=\cos (x+1)-2 \sin 2 x$
$y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}=(\cos (x+1)-2 \sin 2 x)^{\prime}=-\sin (x+1)-4 \cos 2 x$
$y^{\prime \prime \prime}=\left(y^{\prime \prime}\right)^{\prime}=(\sin (x+1)-4 \cos 2 x)^{\prime}=-\cos (x+1)+8 \sin 2 x$
$y^{(I... | y^{(n)}=\sin(\frac{3\pi}{2}\cdotn+x+1)+2^{n}\cdot\cos(\frac{3\pi}{2}\cdotn+2x) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,275 |
## Task Condition
Find the derivative of the specified order.
$$
y=(4 x+3) \cdot 2^{-x}, y^{V}=?
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left((4 x+3) \cdot 2^{-x}\right)^{\prime}=4 \cdot 2^{-x}+(4 x+3) \cdot 2^{-x} \cdot \ln 2 \cdot(-1)= \\
& =(-\ln 2 \cdot(4 x+3)+4) 2^{-x} \\
& y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}=\left((-\ln 2 \cdot(4 x+3)+4) 2^{-x}\right)^{\prime}= \\
& =\left((-\ln 2 \cdot(... | (-\ln^{5}2\cdot(4x+3)+20\ln^{4}2)2^{-x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,276 |
## Problem Statement
Find the second-order derivative $y_{x x}^{\prime \prime}$ of the function given parametrically.
$$
\left\{\begin{array}{l}
x=\frac{\cos t}{1+2 \cos t} \\
y=\frac{\sin t}{1+2 \cos t}
\end{array}\right.
$$ | ## Solution
$$
\begin{aligned}
& x_{t}^{\prime}=\left(\frac{\cos t}{1+2 \cos t}\right)^{\prime}=\frac{-\sin t \cdot(1+2 \cos t)-\cos t \cdot(-2 \sin t)}{(1+2 \cos t)^{2}}=-\frac{\sin t}{(1+2 \cos t)^{2}} \\
& y_{t}^{\prime}=\left(\frac{\sin t}{1+2 \cos t}\right)^{\prime}=\frac{\cos t \cdot(1+2 \cos t)-\sin t \cdot(-2 ... | -\frac{(1+2\cos)^{3}}{\sin^{3}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,277 |
## Problem Statement
Calculate the area of the figure bounded by the lines given by the equations.
$$
\begin{aligned}
& \left\{\begin{array}{l}
x=4 \sqrt{2} \cdot \cos ^{3} t \\
y=2 \sqrt{2} \cdot \sin ^{3} t
\end{array}\right. \\
& x=2(x \geq 2)
\end{aligned}
$$ | ## Solution
Let's find the points of intersection:
$x=4 \sqrt{2} \cos ^{3} t=2 ; \Rightarrow$
$\cos ^{3} t=\frac{2}{4 \sqrt{2}}=\frac{1}{2 \sqrt{2}}$
$\cos t=\frac{1}{\sqrt{2}}$
$t= \p... | \frac{3}{2}\pi-2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,279 |
## Problem Statement
Calculate the area of the figure bounded by the lines given by the equations.
$$
\begin{aligned}
& \left\{\begin{array}{l}
x=3(t-\sin t) \\
y=3(1-\cos t)
\end{array}\right. \\
& y=3(0<x<6 \pi, y \geq 3)
\end{aligned}
$$ | ## Solution
Let's find the points of intersection:
$$
\begin{aligned}
& y=3(1-\cos t)=3 \\
& 1-\cos t=1 \\
& \cos t=0 \\
& t=\frac{\pi}{2}+\pi n, n \in \mathbb{Z}
\end{aligned}
$$
We are interested in the interval \(0 < x < 6\pi\). Then the abscissas of the points of intersection will be:
\end{aligned}
$$ | ## Solution
Let's find the points of intersection:
$$
\begin{aligned}
& x=8 \sqrt{2} \cdot \cos ^{3} t=4 \\
& \cos ^{3} t=\frac{1}{2 \sqrt{2}} \\
& \cos t=\frac{1}{\sqrt{2}} \\
& t= \pm \frac{\pi}{4}+2 \pi k, k \in \mathbb{Z}
\end{aligned}
$$
Since the functions
$x=8 \sqrt{2} \cdot \cos ^{3} t, y=\sqrt{2} \cdot \si... | \frac{3\pi}{2}+2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,288 |
## Problem Statement
Calculate the area of the figure bounded by the lines given by the equations.
$$
\begin{aligned}
& \left\{\begin{array}{l}
x=2 \sqrt{2} \cdot \cos ^{3} t \\
y=\sqrt{2} \cdot \sin ^{3} t
\end{array}\right. \\
& x=1(x \geq 1)
\end{aligned}
$$ | ## Solution
Let's find the points of intersection:
$$
\begin{aligned}
& x=2 \sqrt{2} \cos ^{3} t=1 ; \Rightarrow \\
& \cos ^{3} t=\frac{1}{2 \sqrt{2}} \\
& \cos t=\frac{1}{\sqrt{2}} \\
& ... | \frac{3}{8}\pi-\frac{1}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,297 |
## Problem Statement
Calculate the area of the figure bounded by the lines given by the equations.
$$
\begin{aligned}
& \left\{\begin{array}{l}
x=t-\sin t \\
y=1-\cos t
\end{array}\right. \\
& y=1(0<x<2 \pi, y \geq 1)
\end{aligned}
$$ | ## Solution
Let's find the points of intersection:
$$
\begin{aligned}
& y=1-\cos t=1 \\
& \cos t=1-1=0 \\
& t=\frac{\pi}{2}+\pi n, n \in \mathbb{Z}
\end{aligned}
$$
From the problem stat... | \frac{\pi}{2}+2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,299 |
## Problem Statement
Calculate the area of the figure bounded by the lines given by the equations.
$$
\begin{aligned}
& \left\{\begin{array}{l}
x=8 \cos ^{3} t \\
y=8 \sin ^{3} t
\end{array}\right. \\
& x=1(x \geq 1)
\end{aligned}
$$ | ## Solution
Let's find the points of intersection:
$$
\begin{aligned}
& x=8 \cos ^{3} t=1 ; \Rightarrow \\
& \cos ^{3} t=\frac{1}{8} \\
& \cos t=\frac{1}{2} \\
& t= \pm \frac{\pi}{3}+2 \pi k, k \in \mathbb{Z}
\end{aligned}
$$
Since the functions
$x=8 \cdot \cos ^{3} t, y=8 \cdot \sin ^{3} t$
are periodic (with per... | 8\pi | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,300 |
## Problem Statement
Calculate the area of the figure bounded by the lines given by the equations.
$$
\begin{aligned}
& \left\{\begin{array}{l}
x=4 \sqrt{2} \cdot \cos ^{3} t \\
y=\sqrt{2} \cdot \sin ^{3} t
\end{array}\right. \\
& x=2(x \geq 2)
\end{aligned}
$$ | ## Solution
Let's find the points of intersection:
$$
\begin{aligned}
& \text { Integrals 15-28 } \\
& \left\{\begin{array}{l}
x=4 \sqrt{2} \cdot \cos ^{3} t \\
y=\sqrt{2} \cdot \sin ^{3} t
\end{array}\right.
\end{aligned}
$$
^{2}} d \phi
$$
Let's find $\frac{d \rho}{d \phi}$:
$$
\frac{d \rho}{d \phi}=3 \cdot \frac{3}{4} e^{3 \phi / 4}=\frac{9... | 10\cdot\operatorname{sh}\frac{3\pi}{8} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,310 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by the equations in polar coordinates.
$$
\rho=2 e^{4 \varphi / 3},-\frac{\pi}{2} \leq \varphi \leq \frac{\pi}{2}
$$ | ## 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 $\frac{d \rho}{d \phi}$:
$$
\frac{d \rho}{d \phi}=2 \cdot \frac{4}{3} e^{4 \phi / 3}=\frac{8... | 5\cdot\operatorname{sh}\frac{2\pi}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,311 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by the equations in polar coordinates.
$$
\rho=\sqrt{2} e^{\varphi},-\frac{\pi}{2} \leq \varphi \leq \frac{\pi}{2}
$$ | ## Solution
The length of the arc of a curve given by an equation in polar coordinates is determined by the formula
$$
L=\int_{\phi_{1}}^{\phi_{2}} \sqrt{\rho^{2}+\left(\frac{d \rho}{d \phi}\right)^{2}} d \phi
$$
Find $\frac{d \rho}{d \phi}$:
$$
\frac{d \rho}{d \phi}=\sqrt{2} e^{\phi}
$$
We get:
$$
\begin{aligned... | 4\cdot\operatorname{sh}\frac{\pi}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,312 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by the equations in polar coordinates.
$$
\rho=5 e^{5 \varphi / 12},-\frac{\pi}{2} \leq \varphi \leq \frac{\pi}{2}
$$ | ## 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 $\frac{d \rho}{d \phi}$:
$$
\frac{d \rho}{d \phi}=5 \cdot \frac{5}{12} e^{5 \phi / 12}=\frac... | 26\cdot\operatorname{sh}\frac{5\pi}{24} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,313 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by the equations in polar coordinates.
$$
\rho=6 e^{12 \varphi / 5},-\frac{\pi}{2} \leq \varphi \leq \frac{\pi}{2}
$$ | ## Solution
The length of the arc of a curve given by an equation in polar coordinates is determined by the formula
$$
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}=6 \cdot \frac{12}{5} e^{12 \phi / 5}=\frac... | 13\cdot\operatorname{sh}\frac{6\pi}{5} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,314 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by the equations in polar coordinates.
$$
\rho=3 e^{3 \varphi / 4}, 0 \leq \varphi \leq \frac{\pi}{3}
$$ | ## 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 $\frac{d \rho}{d \phi}$:
$$
\frac{d \rho}{d \phi}=3 \cdot \frac{3}{4} e^{3 \phi / 4}=\frac{9... | 5\cdot(e^{\pi/4}-1) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,315 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by the equations in polar coordinates.
$$
\rho=4 e^{4 \varphi / 3}, 0 \leq \varphi \leq \frac{\pi}{3}
$$ | ## Solution
The length of the arc of a curve given by an equation in polar coordinates is determined by the formula
$$
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}=4 \cdot \frac{4}{3} e^{4 \phi / 3}=\frac{1... | \frac{5}{3}\cdot(e^{4\pi/9}-1) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,316 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by the equations in polar coordinates.
$$
\rho=\sqrt{2} e^{\varphi}, 0 \leq \varphi \leq \frac{\pi}{3}
$$ | ## 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 $\frac{d \rho}{d \phi}$:
$$
\frac{d \rho}{d \phi}=\sqrt{2} e^{\phi}
$$
We get:
$$
\begin{a... | 2\cdot(e^{\pi/3}-1) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,317 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by the equations in polar coordinates.
$$
\rho=5 e^{5 \varphi / 12}, 0 \leq \varphi \leq \frac{\pi}{3}
$$ | ## Solution
The length of the arc of a curve given by an equation in polar coordinates is determined by the formula
$$
L=\int_{\phi_{1}}^{\phi_{2}} \sqrt{\rho^{2}+\left(\frac{d \rho}{d \phi}\right)^{2}} d \phi
$$
Let's find \(\frac{d \rho}{d \phi}\):
$$
\frac{d \rho}{d \phi}=5 \cdot \frac{5}{12} e^{5 \phi / 12}=\fr... | 13\cdot(e^{5\pi/36}-1) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,318 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by the equations in polar coordinates.
$$
\rho=12 e^{12 \varphi / 5}, 0 \leq \varphi \leq \frac{\pi}{3}
$$ | ## Solution
The length of the arc of a curve given by an equation in polar coordinates is determined by the formula
$$
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}=12 \cdot \frac{12}{5} e^{12 \phi / 5}=\fra... | 13\cdot(e^{4\pi/5}-1) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,319 |
## 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:
$$
\begin{aligne... | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,320 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by the equations in polar coordinates.
$$
\rho=2(1-\cos \varphi),-\pi \leq \varphi \leq-\frac{\pi}{2}
$$ | ## Solution
The length of the arc of a curve given by an equation in polar coordinates is determined by the formula
$L=\int_{\varphi_{0}}^{\varphi_{1}} \sqrt{(\rho(\varphi))^{2}+\left(\rho^{\prime}(\varphi)\right)^{2}} d \varphi$
For the curve given by the equation $\rho=2(1-\cos \varphi)$, we find: $\rho^{\prime}=2... | -4\sqrt{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,321 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by the equations in polar coordinates.
$$
\rho=3(1+\sin \varphi),-\frac{\pi}{6} \leq \varphi \leq 0
$$ | ## Solution
The length of the arc of a curve given by an equation in polar coordinates is determined by the formula
$L=\int_{\varphi_{0}}^{\varphi_{1}} \sqrt{(\rho(\varphi))^{2}+\left(\rho^{\prime}(\varphi)\right)^{2}} d \varphi$
For the curve given by the equation $\rho=3(1+\sin \varphi)$, we find: $\rho^{\prime}=3... | 6(\sqrt{3}-\sqrt{2}) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,322 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by the equations in polar coordinates.
$$
\rho=4(1-\sin \varphi), 0 \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}=4(-\cos \phi)
$$
We get:
$$
\begin{align... | 8(\sqrt{3}-\sqrt{2}) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,323 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by the equations in polar coordinates.
$$
\rho=5(1-\cos \varphi),-\frac{\pi}{3} \leq \varphi \leq 0
$$ | ## Solution
As is known, the length of the arc of a curve given by the equation $\rho=\rho(\varphi)$ in polar coordinates, where $\varphi \in\left[\varphi_{0}, \varphi_{1}\right]$, is calculated using the formula
$l=\int_{\varphi_{0}}^{\varphi_{1}} \sqrt{(\rho(\varphi))^{2}+\left(\rho^{\prime}(\varphi)\right)^{2}} d ... | 20(1-\sqrt{\frac{3}{4}}) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,324 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by the equations in polar coordinates.
$$
\rho=6(1+\sin \varphi),-\frac{\pi}{2} \leq \varphi \leq 0
$$ | ## 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 $\rho=6(1+\sin \varphi)$, we find: $\rho^{\prime}=6... | 12(2-\sqrt{2}) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,325 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by the equations in polar coordinates.
$$
\rho=7(1-\sin \varphi),-\frac{\pi}{6} \leq \varphi \leq \frac{\pi}{6}
$$ | ## 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 $\frac{d \rho}{d \phi}:$
$$
\frac{d \rho}{d \phi}=-7 \cos \phi
$$
We get:
$$
\begin{aligne... | 14(\sqrt{3}-1) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,326 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by the equations in polar coordinates.
$$
\rho=8(1-\cos \varphi),-\frac{2 \pi}{3} \leq \varphi \leq 0
$$ | ## Solution
The length of the arc of a curve given by an equation in polar coordinates is determined by the formula
$L=\int_{\varphi_{0}}^{\varphi_{1}} \sqrt{(\rho(\varphi))^{2}+\left(\rho^{\prime}(\varphi)\right)^{2}} d \varphi$
For the curve given by the equation $\rho=8(1-\cos \varphi)$, we find: $\rho^{\prime}=8... | 16 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,327 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by the equations in polar coordinates.
$$
\rho=2 \varphi, 0 \leq \varphi \leq \frac{3}{4}
$$ | ## Solution
The length of the arc of a curve given by an equation in polar coordinates is determined by the formula
$L=\int_{\varphi_{0}}^{\varphi_{1}} \sqrt{(\rho(\varphi))^{2}+\left(\rho^{\prime}(\varphi)\right)^{2}} d \varphi$
For the curve given by the equation $\rho=2 \varphi$, we find: $\rho^{\prime}=2$
We ob... | \frac{15}{16}+\ln2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,328 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by the equations in polar coordinates.
$$
\rho=2 \varphi, 0 \leq \varphi \leq \frac{4}{3}
$$ | ## Solution
The length of the arc of a curve given by an equation in polar coordinates is determined by the formula
$L=\int_{\varphi_{0}}^{\varphi_{1}} \sqrt{(\rho(\varphi))^{2}+\left(\rho^{\prime}(\varphi)\right)^{2}} d \varphi$
For the curve given by the equation $\rho=2 \varphi$, we find: $\rho^{\prime}=2$
We ge... | \frac{20}{9}+\ln3 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,329 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by the equations in polar coordinates.
$$
\rho=2 \varphi, 0 \leq \varphi \leq \frac{5}{12}
$$ | ## Solution
The length of the arc of a curve given by an equation in polar coordinates is determined by the formula
$$
L=\int_{\varphi_{0}}^{\varphi_{1}} \sqrt{(\rho(\varphi))^{2}+\left(\rho^{\prime}(\varphi)\right)^{2}} d \varphi
$$
For the curve given by the equation $\rho=2 \varphi$, we find: $\rho^{\prime}=2$
W... | \frac{65}{144}+\ln\frac{3}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,330 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by the equations in polar coordinates.
$$
\rho=2 \varphi, 0 \leq \varphi \leq \frac{12}{5}
$$ | ## Solution
The length of an arc of a curve given by an equation in polar coordinates is determined by the formula
$L=\int_{\varphi_{0}}^{\varphi_{1}} \sqrt{(\rho(\varphi))^{2}+\left(\rho^{\prime}(\varphi)\right)^{2}} d \varphi$
For the curve given by the equation $\rho=2 \varphi$, we find: $\rho^{\prime}=2$
We obt... | \frac{156}{25}+\ln5 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,331 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by the equations in polar coordinates.
$$
\rho=4 \varphi, 0 \leq \varphi \leq \frac{3}{4}
$$ | ## Solution
The length of the arc of a curve given by an equation in polar coordinates is determined by the formula
$L=\int_{\varphi_{0}}^{\varphi_{1}} \sqrt{(\rho(\varphi))^{2}+\left(\rho^{\prime}(\varphi)\right)^{2}} d \varphi$
For the curve given by the equation $\rho=4 \varphi$, we find: $\rho^{\prime}=4$
We ge... | \frac{15}{8}+\ln4 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,332 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by the equations in polar coordinates.
$$
\rho=3 \varphi, 0 \leq \varphi \leq \frac{4}{3}
$$ | ## Solution
The length of the arc of a curve given by an equation in polar coordinates is determined by the formula
$L=\int_{\varphi_{0}}^{\varphi_{1}} \sqrt{(\rho(\varphi))^{2}+\left(\rho^{\prime}(\varphi)\right)^{2}} d \varphi$
For the curve given by the equation $\rho=3 \varphi$, we find: $\rho^{\prime}=3$
We ob... | \frac{10}{3}+\frac{3}{2}\ln3 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,333 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by the equations in polar coordinates.
$$
\rho=5 \varphi, 0 \leq \varphi \leq \frac{12}{5}
$$ | ## Solution
The length of the arc of a curve given by an equation in polar coordinates is determined by the formula
$$
L=\int_{\varphi_{0}}^{\varphi_{1}} \sqrt{(\rho(\varphi))^{2}+\left(\rho^{\prime}(\varphi)\right)^{2}} d \varphi
$$
For the curve given by the equation $\rho=5 \varphi$, we find: $\rho^{\prime}=5$
W... | \frac{78}{5}+\frac{5}{2}\ln5 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,334 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by the equations in polar coordinates.
$$
\rho=2 \cos \varphi, 0 \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_{\varphi_{0}}^{\varphi_{1}} \sqrt{(\rho(\varphi))^{2}+\left(\rho^{\prime}(\varphi)\right)^{2}} d \varphi
$$
For the curve given by the equation $\rho=2 \cos \varphi$, we find: $\rho^{\prime}=... | \frac{\pi}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,335 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by the equations in polar coordinates.
$$
\rho=8 \cos \varphi, 0 \leq \varphi \leq \frac{\pi}{4}
$$ | ## Solution
The length of the arc of a curve given by an equation in polar coordinates is determined by the formula
$$
L=\int_{\varphi_{0}}^{\varphi_{1}} \sqrt{(\rho(\varphi))^{2}+\left(\rho^{\prime}(\varphi)\right)^{2}} d \varphi
$$
For the curve given by the equation $\rho=8 \cos \varphi$, we find: $\rho^{\prime}=... | 2\pi | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,336 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by the equations in polar coordinates.
$$
\rho=6 \cos \varphi, 0 \leq \varphi \leq \frac{\pi}{3}
$$ | ## Solution
The length of the arc of a curve given by an equation in polar coordinates is determined by the formula
$$
L=\int_{\varphi_{0}}^{\varphi_{1}} \sqrt{(\rho(\varphi))^{2}+\left(\rho^{\prime}(\varphi)\right)^{2}} d \varphi
$$
For the curve given by the equation $\rho=6 \cos \varphi$, we find: $\rho^{\prime}=... | 2\pi | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,337 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by the equations in polar coordinates.
$$
\rho=2 \sin \varphi, 0 \leq \varphi \leq \frac{\pi}{6}
$$ | ## Solution
$$
\begin{aligned}
\rho & =2 \sin \varphi, 0 \leq \varphi \leq \frac{\pi}{6} \\
L & =\int_{0}^{\pi / 6} \sqrt{\rho^{2}+\left(\rho^{\prime}\right)^{2}} d x \\
\rho^{\prime} & =2(\sin \varphi)^{\prime}=2 \cos \varphi \\
& =\int_{0}^{\pi / 6} \sqrt{(2 \sin \varphi)^{2}+(2 \cos \varphi)^{2}} d \varphi=2 \int_{... | \frac{\pi}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,338 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by the equations in polar coordinates.
$$
\rho=8 \sin \varphi, 0 \leq \varphi \leq \frac{\pi}{4}
$$ | ## Solution
$$
\begin{aligned}
L & =\int_{0}^{\pi / 4} \sqrt{\rho^{2}+\left(\rho^{\prime}\right)^{2}} d x \\
\rho^{\prime} & =8(\sin \varphi)^{\prime}=8 \cos \varphi \\
L & =\int_{0}^{\pi / 4} \sqrt{(8 \sin \varphi)^{2}+(8 \cos \varphi)^{2}} d \varphi=8 \int_{0}^{\pi / 4} \sqrt{(\sin \varphi)^{2}+(\cos \varphi)^{2}} d... | 2\pi | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,339 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by the equations in polar coordinates.
$$
\rho=6 \sin \varphi, 0 \leq \varphi \leq \frac{\pi}{3}
$$ | ## Solution
$$
\begin{aligned}
L & =\int_{0}^{\pi / 3} \sqrt{\rho^{2}+\left(\rho^{\prime}\right)^{2}} d x \\
\rho^{\prime} & =6(\sin \varphi)^{\prime}=6 \cos \varphi \\
L & =\int_{0}^{\pi / 3} \sqrt{(6 \sin \varphi)^{2}+(6 \cos \varphi)^{2}} d \varphi=6 \int_{0}^{\pi / 3} \sqrt{(\sin \varphi)^{2}+(\cos \varphi)^{2}} d... | 2\pi | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,340 |
## Problem Statement
Write the decomposition of vector $x$ in terms of vectors $p, q, r$:
$x=\{-13 ; 2 ; 18\}$
$p=\{1 ; 1 ; 4\}$
$q=\{-3 ; 0 ; 2\}$
$r=\{1 ; 2 ;-1\}$ | ## 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:
$$
\left\{\begin{array}{l}
\alpha \cdot p_{1}+\beta \cdot q_{1}+\gamma \cdot r_{1}=x_{1} \\
\alpha \cdot p_{2}+\beta \cdot q_{2}+\gamma \cdot r_{2}=x_{2} \\
\alpha \cdot p_{3}+\beta \c... | 2p+5q | Algebra | math-word-problem | Yes | Yes | olympiads | false | 47,341 |
## Problem Statement
Are the vectors $c_{1 \text { and }} c_{2}$, constructed from vectors $a \text{ and } b$, collinear?
$a=\{5 ; 0 ; 8\}$
$b=\{-3 ; 1 ; 7\}$
$c_{1}=3 a-4 b$
$c_{2}=12 b-9 a$ | ## Solution
Vectors are collinear if there exists a number $\gamma$ such that $c_{1}=\gamma \cdot c_{2}$. That is, vectors are collinear if their coordinates are proportional.
It is not hard to notice that $c_{2}=-3(3 a-4 b)=-3 c_{1}$ for any $a$ and $b$.
That is, $c_{1}=-\frac{1}{3} \cdot c_{2}$, which means the ve... | c_{1}=-\frac{1}{3}\cdotc_{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 47,342 |
## problem statement
Find the cosine of the angle between vectors $\overrightarrow{A B}$ and $\overrightarrow{A C}$.
$A(-2 ; 1 ; 1), B(2 ; 3 ;-2), C(0 ; 0 ; 3)$ | ## Solution
Let's find $\overrightarrow{A B}$ and $\overrightarrow{A C}$:
$$
\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 $... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 47,343 |
## problem statement
Calculate the area of the parallelogram constructed on vectors $a_{\text {and }} b$.
$a=3 p-4 q$
$b=p+3 q$
$|p|=2$
$|q|=3$
$(\widehat{p, q})=\frac{\pi}{4}$ | ## 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=(3 p-4 q) \times(p+3 q)=3 \cdot p \times p+3 \cdot 3 \cdot p \times q-4 \cdot q \... | 39\sqrt{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 47,344 |
## 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 \... | -6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 47,345 |
## problem statement
Calculate the volume of the tetrahedron with vertices at points $A_{1}, A_{2}, A_{3}, A_{4_{\text {and }}}$ its height dropped from vertex $A_{4 \text { to the face }} A_{1} A_{2} A_{3}$.
$A_{1}(1 ; 3 ; 0)$
$A_{2}(4 ;-1 ; 2)$
$A_{3}(3 ; 0 ; 1)$
$A_{4}(-4 ; 3 ; 5)$ | ## Solution
From vertex $A_{1 \text {, we draw vectors: }}$
$\overrightarrow{A_{1} A_{2}}=\{4-1 ;-1-3 ; 2-0\}=\{3 ;-4 ; 2\}$
$\overrightarrow{A_{1} A_{3}}=\{3-1 ; 0-3 ; 1-0\}=\{2 ;-3 ; 1\}$
$\overrightarrow{A_{1} A_{4}}=\{-4-1 ; 3-3 ; 5-0\}=\{-5 ; 0 ; 5\}$
According to the geometric meaning of the scalar triple pr... | 2.5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 47,346 |
## Problem Statement
Find the distance from point $M_{0}$ to the plane passing through three points $M_{1}, M_{2}, M_{3}$.
$M_{1}(2 ; -1 ; 2)$
$M_{2}(1 ; 2 ; -1)$
$M_{3}(3 ; 2 ; 1)$
$M_{0}(-5 ; 3 ; 7)$ | ## Solution
Find the equation of the plane passing through three points $M_{1}, M_{2}, M_{3}$:
$$
\left|\begin{array}{llc}
x-2 & y-(-1) & z-2 \\
1-2 & 2-(-1) & -1-2 \\
3-2 & 2-(-1) & 1-2
\end{array}\right|=0
$$
Perform transformations:
$$
\begin{aligned}
& \left|\begin{array}{ccc}
x-2 & y+1 & z-2 \\
-1 & 3 & -3 \\
... | 2\sqrt{22} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 47,347 |
## Problem Statement
Write the equation of the plane passing through point $A$ and perpendicular to vector $\overrightarrow{B C}$.
$A(5 ; 3 ;-1)$
$B(0 ; 0 ;-3)$
$C(5 ;-1 ; 0)$ | ## Solution
Let's find the vector $\overrightarrow{BC}:$
$$
\overrightarrow{BC}=\{5-0 ;-1-0 ; 0-(-3)\}=\{5 ;-1 ; 3\}
$$
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:
$5 \cdot(x-5)-(y-3)+3 \cdot(z-(-1))... | 5x-y+3z-19=0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 47,348 |
## 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 $\phi_{\text{between the planes is determined by the formula: }}$
$$
\begin{aligned}
&... | 0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 47,349 |
## Problem Statement
Find the coordinates of point $A$, which is equidistant from points $B$ and $C$.
$A(x ; 0 ; 0)$
$B(-2 ; 0 ; 6)$
$C(0 ;-2 ;-4)$ | ## Solution
Let's find the distances $A B$ and $A C$:
$$
\begin{aligned}
& A B=\sqrt{(-2-x)^{2}+(0-0)^{2}+(6-0)^{2}}=\sqrt{4+4 x+x^{2}+0+36}=\sqrt{x^{2}+4 x+40} \\
& A C=\sqrt{(0-x)^{2}+(-2-0)^{2}+(-4-0)^{2}}=\sqrt{x^{2}+4+16}=\sqrt{x^{2}+20}
\end{aligned}
$$
Since according to the problem's condition $A B=A C$, the... | A(-5;0;0) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 47,350 |
## Problem Statement
Let $k$ be the coefficient of similarity transformation with the center at the origin. Is it true that point $A$ belongs to the image of plane $a$?
$A(5 ; 0 ;-6)$
$a: 6x - y - z + 7 = 0$
$k = \frac{2}{7}$ | ## Solution
When transforming similarity with the center at the origin of the plane
$a: A x+B y+C z+D=0_{\text{and coefficient }} k$ transitions to the plane
$a^{\prime}: A x+B y+C z+k \cdot D=0$. We find the image of the plane $a$:
$a^{\prime}: 6 x-y-z+2=0$
Substitute the coordinates of point $A$ into the equatio... | 38\neq0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 47,351 |
## Task Condition
Write the canonical equations of the line.
$$
\begin{aligned}
& 2 x+3 y-2 z+6=0 \\
& x-3 y+z+3=0
\end{aligned}
$$ | ## Solution
Canonical equations of a line:
$\frac{x-x_{0}}{m}=\frac{y-y_{0}}{n}=\frac{z-z_{0}}{p}$
where $\left(x_{0} ; y_{0} ; z_{0}\right)_{\text {- are coordinates of some point on the line, and }} \vec{s}=\{m ; n ; p\}$ - its direction vector.
Since the line belongs to both planes simultaneously, its direction ... | \frac{x+3}{-3}=\frac{y}{-4}=\frac{z}{-9} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 47,352 |
## Problem Statement
Find the point of intersection of the line and the plane.
$\frac{x-2}{4}=\frac{y-1}{-3}=\frac{z+3}{-2}$
$3 x-y+4 z=0$ | ## Solution
Let's write the parametric equations of the line.
$$
\begin{aligned}
& \frac{x-2}{4}=\frac{y-1}{-3}=\frac{z+3}{-2}=t \Rightarrow \\
& \left\{\begin{array}{l}
x=2+4 t \\
y=1-3 t \\
z=-3-2 t
\end{array}\right.
\end{aligned}
$$
Substitute into the equation of the plane:
$3(2+4 t)-(1-3 t)+4(-3-2 t)=0$
$6+1... | (6,-2,-5) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 47,353 |
## Condition of the problem
Prove that $\lim _{n \rightarrow \infty} a_{n}=a_{\text {(specify }} N(\varepsilon)$.
$a_{n}=\frac{3 n-2}{2 n-1}, a=\frac{3}{2}$ | ## Solution
By the definition of the limit:
$$
\begin{aligned}
& \forall \varepsilon>0: \exists N(\varepsilon) \in \mathbb{N}: \forall n: n \geq N(\varepsilon):\left|a_{n}-a\right| \\
& \left|\frac{6 n-4-6 n+3}{2(2 n-1)}\right| \\
& \left|\frac{-1}{2(2 n-1)}\right| \\
& \left|\frac{1}{2(2 n-1)}\right|
\end{aligned}
$... | N(\varepsilon)=[\frac{1+6\varepsilon}{4\varepsilon}] | Calculus | proof | Yes | Yes | olympiads | false | 47,355 |
## Task Condition
Calculate the limit of the numerical sequence:
$$
\lim _{n \rightarrow \infty} \frac{(3-n)^{2}+(3+n)^{2}}{(3-n)^{2}-(3+n)^{2}}
$$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{(3-n)^{2}+(3+n)^{2}}{(3-n)^{2}-(3+n)^{2}}=\lim _{n \rightarrow \infty} \frac{9-6 n+n^{2}+9+6 n+n^{2}}{9-6 n+n^{2}-\left(9+6 n+n^{2}\right)}= \\
& =\lim _{n \rightarrow \infty} \frac{2\left(9+n^{2}\right)}{9-6 n+n^{2}-9-6 n-n^{2}}=\lim _{n \rightarrow ... | -\infty | Algebra | math-word-problem | Yes | Yes | olympiads | false | 47,356 |
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{n \sqrt[3]{5 n^{2}}+\sqrt[4]{9 n^{8}+1}}{(n+\sqrt{n}) \sqrt{7-n+n^{2}}}$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{n \sqrt[3]{5 n^{2}}+\sqrt[4]{9 n^{8}+1}}{(n+\sqrt{n}) \sqrt{7-n+n^{2}}}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n^{2}}\left(n \sqrt[3]{5 n^{2}}+\sqrt[4]{9 n^{8}+1}\right)}{\frac{1}{n^{2}}(n+\sqrt{n}) \sqrt{7-n+n^{2}}}= \\
& =\lim _{n \rightarrow \... | \sqrt{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,357 |
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty}\left(\frac{1}{n^{2}}+\frac{2}{n^{2}}+\frac{3}{n^{2}}+\ldots+\frac{n-1}{n^{2}}\right)$ | Solution
$\lim _{n \rightarrow \infty}\left(\frac{1}{n^{2}}+\frac{2}{n^{2}}+\frac{3}{n^{2}}+\ldots+\frac{n-1}{n^{2}}\right)=\lim _{n \rightarrow \infty}\left(\frac{1+2+3+\ldots+(n-1)}{n^{2}}\right)=$
$=\lim _{n \rightarrow \infty} \frac{1}{n^{2}}\left(\frac{n(n-1)}{2}\right)=\lim _{n \rightarrow \infty} \frac{n-1}{2 ... | \frac{1}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,359 |
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty}\left(\frac{n+1}{n-1}\right)^{n}$ | ## Solution
$\lim _{n \rightarrow \infty}\left(\frac{n+1}{n-1}\right)^{n}=\lim _{n \rightarrow \infty}\left(\frac{n-1+2}{n-1}\right)^{n}=\lim _{n \rightarrow \infty}\left(1+\frac{2}{n-1}\right)^{n}=$ $=\lim _{n \rightarrow \infty}\left(1+\frac{1}{\left(\frac{n-1}{2}\right)}\right)^{\left(\frac{n-1}{2}\right)\left(\fra... | e^2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,360 |
## Condition of the problem
Prove that (find $\delta(\varepsilon)$ ):
$\lim _{x \rightarrow-3} \frac{2 x^{2}+5 x-3}{x+3}=-7$ | ## 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 | 47,361 |
## problem statement
Prove that the function $f(x)_{\text {is continuous at the point }} x_{0}$ (find $\delta(\varepsilon)$ ):
$$
f(x)=5 x^{2}-1, x_{0}=6
$$ | ## 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 | 47,362 |
## 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
$$
\begin{aligned}
& \lim _{x \rightarrow-1} \frac{\left(x^{3}-2 x-1\right)(x+1)}{x^{4}+4 x^{2}-5}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow-1} \frac{\left(x^{3}-2 x-1\right)(x+1)}{\left(x^{3}-x^{2}+5 x-5\right)(x+1)}= \\
& =\lim _{x \rightarrow-1} \frac{x^{3}-2 x-1}{x^{3}-x^{2}+5 x-5}=\frac{(-1)^{3}... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,363 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 4} \frac{\sqrt{1+2 x}-3}{\sqrt{x}-2}$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 4} \frac{\sqrt{1+2 x}-3}{\sqrt{x}-2}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow 4} \frac{(\sqrt{1+2 x}-3)(\sqrt{1+2 x}+3)}{(\sqrt{x}-2)(\sqrt{1+2 x}+3)}= \\
& =\lim _{x \rightarrow 4} \frac{1+2 x-9}{(\sqrt{x}-2)(\sqrt{1+2 x}+3)}=\lim _{x \rightarrow 4} \frac{2... | \frac{4}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,364 |
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow 0} \frac{\ln (1+\sin x)}{\sin 4 x}
$$ | ## Solution
Let's use the substitution of equivalent infinitesimals:
\(\sin x \sim x\), as \(x \rightarrow 0\)
\(\sin 4x \sim 4x\), as \(x \rightarrow 0\) (or \(4x \rightarrow 0\))
\(\ln (1+x) \sim x\), as \(x \rightarrow 0\)
We get:
\[
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{\ln (1+\sin x)}{\sin 4 x}=\le... | \frac{1}{4} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,365 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 1} \frac{x^{2}-1}{\ln x}$ | ## Solution
Substitution:
$x=y+1 \Rightarrow y=x-1$
$x \rightarrow 1 \Rightarrow y \rightarrow 0$
We get:
$\lim _{x \rightarrow 1} \frac{x^{2}-1}{\ln x}=\lim _{y \rightarrow 0} \frac{(y+1)^{2}-1}{\ln (y+1)}=$
Using the substitution of equivalent infinitesimals:
$\ln (1+y) \sim y$, as $y \rightarrow 0$
We get:
... | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,366 |
Problem condition
Calculate the limit of the function:
$\lim _{x \rightarrow \frac{\pi}{2}} \frac{2^{\cos ^{2} x}-1}{\ln (\sin x)}$ | 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:
$$
\begin{aligned}
& \lim _{x \rightarrow \frac{\pi}{2}} \frac{2^{\cos ^{2} x}-1}{\ln (\sin x)}=\lim _{y \rightarrow 0} \frac{2^{\cos ^... | -2\ln2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,367 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{7^{2 x}-5^{3 x}}{2 x-\operatorname{arctg} 3 x}$ | ## Solution
$\lim _{x \rightarrow 0} \frac{7^{2 x}-5^{3 x}}{2 x-\operatorname{arctg} 3 x}=\lim _{x \rightarrow 0} \frac{\left(49^{x}-1\right)-\left(125^{x}-1\right)}{2 x-\operatorname{arctg} 3 x}=$
$$
\begin{aligned}
& =\lim _{x \rightarrow 0} \frac{\left(\left(e^{\ln 49}\right)^{x}-1\right)-\left(\left(e^{\ln 125}\r... | \ln\frac{125}{49} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,368 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{e^{x}+e^{-x}-2}{\sin ^{2} x}$ | ## Solution
$\lim _{x \rightarrow 0} \frac{e^{x}+e^{-x}-2}{\sin ^{2} x}=\lim _{x \rightarrow 0} \frac{e^{-x}\left(e^{2 x}-2 e^{x}+1\right)}{\sin ^{2} x}=$
$=\lim _{x \rightarrow 0} \frac{e^{-x}\left(e^{x}-1\right)^{2}}{\sin ^{2} x}=$
Using the substitution of equivalent infinitesimals:
$e^{x}-1 \sim x$, as $x \righ... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,369 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0}\left(1-\ln \left(1+x^{3}\right)\right)^{\frac{3}{x^{2} \arcsin x}}$ | ## Solution
$\lim _{x \rightarrow 0}\left(1-\ln \left(1+x^{3}\right)\right)^{\frac{3}{x^{2} \arcsin x}}=$
$=\lim _{x \rightarrow 0}\left(e^{\ln \left(1-\ln \left(1+x^{3}\right)\right)}\right)^{\frac{3}{x^{2} \arcsin x}}=$
$=\lim _{x \rightarrow 0} e^{3 \cdot \ln \left(1-\ln \left(1+x^{3}\right)\right) /\left(x^{2} \... | e^{-3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,370 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0}\left(\frac{\sin 2 x}{x}\right)^{1+x}$ | ## Solution
$\lim _{x \rightarrow 0}\left(\frac{\sin 2 x}{x}\right)^{1+x}=\left(\lim _{x \rightarrow 0}\left(\frac{\sin 2 x}{x}\right)\right)^{\lim _{x \rightarrow 0} 1+x}=$
$=\left(\lim _{x \rightarrow 0}\left(\frac{\sin 2 x}{x}\right)\right)^{1}=\lim _{x \rightarrow 0} \frac{\sin 2 x}{x}=$
Using the substitution o... | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,371 |
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow 1}\left(\frac{3 x-1}{x+1}\right)^{\frac{1}{\sqrt[3]{x}-1}}
$$ | ## Solution
$\lim _{x \rightarrow 1}\left(\frac{3 x-1}{x+1}\right)^{\frac{1}{\sqrt[3]{x}-1}}=\lim _{x \rightarrow 1}\left(e^{\ln \left(\frac{3 x-1}{x+1}\right)}\right)^{\frac{1}{\sqrt[3]{x}-1}}=$
$=\lim _{x \rightarrow 1} e^{\frac{1}{\sqrt[3]{x}-1} \cdot \ln \left(\frac{3 x-1}{x+1}\right)}=\exp \left\{\lim _{x \right... | e^3 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,372 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow e}\left(\frac{\ln x-1}{x-e}\right)^{\sin \left(\frac{\pi}{2 e} x\right)}$ | ## Solution
$\lim _{x \rightarrow e}\left(\frac{\ln x-1}{x-e}\right)^{\sin \left(\frac{\pi}{2 e} x\right)}=\lim _{x \rightarrow e}\left(\frac{\ln x-\ln e}{x-e}\right)^{\sin \left(\frac{\pi}{2 e} x\right)}=$
$=\left(\lim _{x \rightarrow e} \frac{\ln \frac{x}{e}}{x-e}\right)^{\lim _{x \rightarrow e} \sin \left(\frac{\p... | \frac{1}{e} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,373 |
## Condition of the problem
Calculate the limit of the function:
$$
\lim _{x \rightarrow 0} \sqrt{4 \cos 3 x+x \cdot \operatorname{arctg}\left(\frac{1}{x}\right)}
$$ | ## Solution
Since $\operatorname{arctg}\left(\frac{1}{x}\right)_{\text { is bounded, then }}$
$$
x \cdot \operatorname{arctg}\left(\frac{1}{x}\right) \rightarrow 0 \underset{\text { as } x \rightarrow 0}{ }
$$
Then:
$\lim _{x \rightarrow 0} \sqrt{4 \cos 3 x+x \cdot \operatorname{arctg}\left(\frac{1}{x}\right)}=\sqr... | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,374 |
## Problem Statement
Based on the definition of the derivative, find $f^{\prime}(0)$:
$$
f(x)=\left\{\begin{array}{c}
6 x+x \sin \frac{1}{x}, x \neq 0 \\
0, x=0
\end{array}\right.
$$ | ## 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:
$$
\begin{aligned}
& f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim _{\Delta x \rightarrow 0} ... | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,375 |
## Problem Statement
Derive the equation of the tangent line to the given curve at the point with abscissa $x_{0}$.
$y=\frac{x^{16}+9}{1-5 x^{2}}, x_{0}=1$ | ## Solution
Let's find $y^{\prime}:$
$$
\begin{aligned}
& y^{\prime}=\left(\frac{x^{16}+9}{1-5 x^{2}}\right)^{\prime}=\frac{\left(x^{16}+9\right)^{\prime}\left(1-5 x^{2}\right)-\left(x^{16}+9\right)\left(1-5 x^{2}\right)^{\prime}}{\left(1-5 x^{2}\right)^{2}}= \\
& =\frac{16 x^{15}\left(1-5 x^{2}\right)-\left(x^{16}+9... | \frac{9}{4}\cdotx-\frac{19}{4} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,376 |
## Task Condition
Find the differential $d y$.
$y=\sqrt[3]{\frac{x+2}{x-2}}$ | ## Solution
$d y=y^{\prime} \cdot d x=\left(\sqrt[3]{\frac{x+2}{x-2}}\right)^{\prime} d x=\left(\left(\frac{x+2}{x-2}\right)^{\frac{1}{3}}\right)^{\prime} d x=$
$$
\begin{aligned}
& =\frac{1}{3} \cdot\left(\frac{x+2}{x-2}\right)^{-\frac{2}{3}} \cdot\left(\frac{x+2}{x-2}\right)^{\prime} d x=\frac{1}{3} \cdot \sqrt[3]{... | -\frac{4}{3(x-2)\sqrt[3]{(x+2)^{2}\cdot(x-2)}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,377 |
## Problem Statement
Calculate approximately using the differential.
$$
y=\frac{1}{\sqrt{2 x^{2}+x+1}}, x=1.016
$$ | ## 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:
$\Delta x = 0.016$
Calculate:
\[
\begin{aligned}
& y(1) = \frac{1}{... | 0.495 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,378 |
## Task Condition
Find the derivative.
$$
y=\left(1-x^{2}\right) \sqrt[5]{x^{3}+\frac{1}{x}}
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\left(1-x^{2}\right) \sqrt[5]{x^{3}+\frac{1}{x}}\right)^{\prime}= \\
& =-2 x \cdot \sqrt[5]{x^{3}+\frac{1}{x}}+\left(1-x^{2}\right) \cdot \frac{1}{5} \cdot \frac{1}{\sqrt[5]{\left(x^{3}+\frac{1}{x}\right)^{4}}} \cdot\left(3 x^{2}-\frac{1}{x^{2}}\right)= \\
& =\frac{1}{... | \frac{1}{5\sqrt[5]{(x^{3}+\frac{1}{x})^{4}}}\cdot(-13x^{4}+3x^{2}-9-\frac{1}{x^{2}}) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,379 |
## Task Condition
Find the derivative.
$$
y=x-e^{-x} \arcsin e^{x}-\ln \left(1+\sqrt{1-e^{2 x}}\right)
$$ | ## Solution
$y^{\prime}=\left(x-e^{-x} \arcsin e^{x}-\ln \left(1+\sqrt{1-e^{2 x}}\right)\right)^{\prime}=$
$=1-\left(-e^{-x} \arcsin e^{x}+e^{-x} \frac{1}{\sqrt{1-e^{2 x}}} \cdot e^{x}\right)-\frac{1}{\left(1+\sqrt{1-e^{2 x}}\right)} \cdot \frac{1}{2 \sqrt{1-e^{2 x}}} \cdot\left(-e^{2 x} \cdot 2\right)=$
$=1+e^{-x} ... | e^{-x}\arcsine^{x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,380 |
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