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543. Find the integrals:
1) $\int \frac{d x}{2 \sin x - \cos x}$
2) $\int \frac{d x}{5 + 4 \cos a x}$
3) $\int \frac{\tan x \, d x}{1 - \cot^2 x}$;
4) $\int \frac{e^{3 x} \, d x}{e^{2 x} + 1}$. | Solution. 1) Assuming $\operatorname{tg} \frac{x}{2}=z$ and replacing $\sin x, \cos x$ and $d x$ with their expressions in terms of $z$ derived from this substitution, we get
$$
\begin{gathered}
\int \frac{d x}{2 \sin x-\cos x}=\int \frac{2 d z}{z^{2}+4 z-1}=2 \int \frac{d(z+2)}{(z+2)^{2}-5}= \\
=\frac{1}{\sqrt{5}} \l... | \begin{aligned}1)&\frac{1}{\sqrt{5}}\ln|\frac{2-\sqrt{5}+\operatorname{tg}\frac{x}{2}}{2+\sqrt{5}+\operatorname{tg}\frac{x}{2}}|+C\\2)&\frac{2}{3}\operatorname{arctg}(\frac{1}{3}\operator | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,262 |
582. Compute the integrals:
1) $\int_{2}^{3} 3 x^{2} d x$
2) $\int_{0}^{1}\left(1+e^{\frac{x}{4}}\right) d x$
3) $\int_{-1}^{7} \frac{d t}{\sqrt{3 t+4}}$
4) $\int_{0}^{\frac{\pi}{2 a}}(x+3) \sin a x d x$ | Solution. Applying the Newton-Leibniz formula (*) and properties of the definite integral, we get:
1) $\int_{2}^{3} 3 x^{2} d x=3 \int_{2}^{3} x^{2} d x=\left.x^{3}\right|_{2} ^{3}=3^{3}-2^{3}=19$.
2) $\int_{0}^{4}\left(1+e^{\frac{x}{4}}\right) d x=\int_{0}^{4} d x+4 \int_{0}^{4} e^{-\frac{x}{4}} d \frac{x}{4}=x+\left... | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,263 | |
591. Calculate the integrals:
1) $\int_{0}^{5} \frac{x d x}{\sqrt{1+3 x}}$
2) $\int_{\ln 2}^{\ln 9} \frac{d x}{e^{x}-e^{-x}}$;
3) $\int_{1}^{\sqrt{3}} \frac{\left(x^{3}+1\right) d x}{x^{2} \sqrt{4-x^{2}}}$;
4) $\int_{0}^{\frac{\pi}{2}} \frac{d x}{2+\cos x}$. | Solution. 1) Introduce a new variable of integration, setting $\sqrt{1+3 x}=t$. From this, we find $x=\frac{t^{2}-1}{3}, d x=\frac{2}{3} t d t$ and the new limits of integration: $t_{1}=1$ when $x_{1}=0, t_{2}=4$ when $x_{2}=5$. Substituting, we get
$$
\begin{aligned}
\int_{0}^{5} \frac{x d x}{\sqrt{1+3 x}} & =\frac{2... | 4,\frac{\ln1.5}{2},\frac{7}{2\sqrt{3}}-1,\frac{\pi}{3\sqrt{3}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,264 |
592. Prove that for an even function $f(x)$:
$$
\int_{-a}^{a} f(x) d x=2 \int_{0}^{a} f(x) d x
$$
and for an odd function $f(x)$:
$$
\int_{-a}^{a} f(x) d x=0
$$ | Solution. Dividing the interval of integration $[-a, a]$ at the point $x=0$ into two parts, according to property 3, we obtain the identity
$$
\int_{-a}^{a} f(x) d x=\int_{0}^{a} f(x) d x+\int_{-a}^{0} f(x) d x
$$
By changing the variable in the last integral according to the formula $x=-z$, we have $d x=-d z ; z_{1}... | proof | Calculus | proof | Yes | Yes | olympiads | false | 34,265 |
619. Find the volume of the part of the cylinder cut off by a plane passing through the diameter $2 R$ of its base at an angle $x$ to the base plane. | Solution. Depicting the given body's section, Fig. 97, we notice that any section of it by a plane parallel to the plane $A B C$ represents a right-angled triangle.
Let's find the area of the section at a distance $O P=x$ from point $O$. From the right-angled triangle $A M P$ we have $M P^{2}=R^{2}-(R-x)^{2}$. From th... | \frac{2}{3}R^{3}\operatorname{tg}\alpha | Geometry | math-word-problem | Yes | Yes | olympiads | false | 34,267 |
620. Find the volume of a triaxial ellipsoid $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+\frac{z^{2}}{c^{2}}=1$. | Solution. A plane section of the ellipsoid, parallel to the $x O z$ plane and at a distance $y=h$ from it, as shown in Fig. 98, represents an ellipse
$$
\frac{x^{2}}{a^{2}}+\frac{z^{2}}{c^{2}}=1-\frac{h^{2}}{b^{2}}, \quad \frac{x^{2}}{a_{1}^{2}}+\frac{z^{2}}{c_{1}^{2}}=1
$$
with semi-axes
$$
a_{1}=\frac{a}{b} \sqrt{... | \frac{4}{3}\pi | Geometry | math-word-problem | Yes | Yes | olympiads | false | 34,268 |
621. Find the volume common to two cylinders: $x^{2}+y^{2}=a^{2}$ and $y^{2}+z^{2}=a^{2}$ (bounded by these cylindrical surfaces). | Solution. We will construct one-eighth of the body located in the first octant, see Fig. 99.
Any section of the body by a plane parallel to the $x O z$ plane is a square. The area of the section $P Q N M$, at a distance $O M=h$ from
 for applying... | 3.18\mathrm{MN} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,274 |
658. Find the pressure of water on the surface of a sphere with a diameter of 4 m, if its center is located at a depth of 3 m from the water surface. | Solution. Let us draw a vertical plane through the center of the sphere and choose a rectangular coordinate system $x O y$ on it, as shown in Fig. 119.
We will cut the sphere at a depth $h$ with a horizontal plane. Then the pressure of the water on the cut part of the sphere's surface will be some function $p(h)$.
Wh... | 48\pi | Geometry | math-word-problem | Yes | Yes | olympiads | false | 34,277 |
661. A ball lies at the bottom of a pool with a depth of $H=14 \partial$ m. Determine the work required to extract the ball from the water, if its radius $R=3 \partial$, and specific weight $\delta=2$. | Solution. When lifting the ball to the water surface, the force $P_{1}$, which performs work, is constant and equal to the difference between the weight of the ball and the weight of the displaced water:
$$
P_{1}=\frac{4}{3} \pi R^{3} \delta-\frac{4}{3} \pi R^{3}=\frac{4}{3} \pi R^{3}(\delta-1)
$$
Therefore, the work... | 600.4\pi | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,279 |
662. Determine the work required to launch a rocket weighing $P=1.5 T$ from the Earth's surface to a height of $H=2000$ km. | Solution. The force $F$ of attraction of a body by the Earth or the weight of the body depends on its distance $x$ from the center of the Earth: $F(x)=\frac{\lambda}{x^{2}}$, where $\lambda$ is a constant.
If $P$ is the weight of the body when it is on the surface of the Earth, i.e., at a distance of the Earth's radiu... | 2285714000\kappa\Gamma\approx22422854340\partialN | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,280 |
663. A cylinder with a height of $H=1.5$ m and a radius of $R=0.4$ m, filled with gas at atmospheric pressure ( $10330 \kappa \Gamma / \boldsymbol{M}^{2}$ ), is closed with a piston. Determine the work spent on isothermal compression of the gas when the piston is moved a distance of $h=1.2$ m inside the cylinder. | Solution. During an isothermal change of state of a gas, when its temperature remains constant, the dependence between the volume $v$ and the pressure $p$ of the gas is expressed by the formula $p v = c = \text{const}$. (Law of
, where $k$ is a constant for the given gas, greater than one. (For air, $k \approx 1.4$.) | Solution. Repeating the same reasoning and using the same notation as in the solution of the previous problem, we find the following expression for the differential of work:
$$
d q(x)=\frac{c d x}{S^{k-1}(H-x)^{k}}
$$
Integrating from $x=0$ to $x=h$, we obtain the entire sought work
$$
\begin{aligned}
& Q=\frac{c}{S... | 172591.3\partial\varkappa | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,282 |
665. A rectangular reservoir with a horizontal cross-sectional area of $S=6 \mathrm{~m}^{2}$ is filled with water to a height of $H=5$ m. Determine the time it takes for all the water to flow out of the reservoir through a small hole in its bottom with an area of $s=0.01 \mu^{2}$, assuming that the velocity of the wate... | Solution. According to the general scheme (1), divide the desired time $T$ into a large number $n$ of small intervals $\Delta t_{1}, \Delta t_{2}, \ldots, \Delta t_{n}$, and let the water level in the reservoir decrease by an amount $\Delta x=\frac{H}{n}$ during each such interval (Fig. 124).[^20]
If we assume that du... | 1010 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,283 |
668. Determine the mass of a sphere of radius $r$, if the density at each of its points is proportional to the distance from the center of the sphere. | Solution. Let the mass of a sphere of arbitrary radius $x$ be some function $m(x)$.
When $x$ is increased by a small amount $d x$, the volume $v$ of this sphere increases by an amount $\Delta v$, equal to the difference in volumes of spheres with radii $x$ and $x+d x$:
$$
\begin{gathered}
\Delta v=\frac{4}{3} \pi\lef... | k\pir^{4} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,286 |
680. Find the center of gravity of a quarter circle $x^{2}+y^{2}=a^{2}$, located in the first quadrant, if the linear density at each of its points is proportional to the product of the coordinates of the point. | Solution. From the equation of the circle, we find $y^{\prime}$, then $d l$:
$$
\begin{gathered}
2 x+2 y y^{\prime}=0 ; \quad y^{\prime}=-\frac{x}{y} \\
d l=\sqrt{1+\left(y^{\prime}\right)^{2}} d x=\sqrt{1+\frac{x^{2}}{y^{2}}} d x=\sqrt{\frac{y^{2}+x^{2}}{y^{2}}} d x=\frac{a}{y} d x \\
-223-
\end{gathered}
$$
Next, w... | x_{}=y_{}=\frac{2}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,287 |
681. Find the center of gravity of a homogeneous cycloid arch
$$
x=a(t-\sin t), \quad y=a(1-\cos t), \quad \text{fig. } 126
$$ | Solution. The given homogeneous arc is symmetric with respect to the line $x=\pi a$. Therefore, the center of gravity of the arc lies on this line, i.e., $x_{c}=\pi a$. To determine $y_{c}$, we find the differential of the arc of the cycloid
$$
d l=\sqrt{x^{2}+y^{2}} d t=\sqrt{a^{2}(1-\cos t)^{2}+a^{2} \sin ^{2} t} d ... | y_{C}=-\frac{1}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,288 |
682. Find the center of gravity of a homogeneous figure (plate) bounded by the parabola $\sqrt{x}+\sqrt{y}=\sqrt{a}$ and the coordinate axes. | Solution. The given homogeneous figure is symmetric with respect to the bisector of the first coordinate angle (Fig. 127), therefore $x_{c}=y_{c}$.
Fig. 126
 $\int_{0}^{+\infty} e^{-x} d x$
2) $\int_{-\infty}^{+\infty} \frac{d x}{x^{2}+1}$
3) $\int_{0}^{1} \frac{d x}{x}$
4) $\int_{-1}^{2} \frac{d x}{\sqrt[3]{(x-1)^{2}}}$
Explain the solutions geometrically. | Solution. 1) Using equality (1), we have
$$
\int_{0}^{+\infty} e^{-x} d x=\lim _{\beta \rightarrow+\infty} \int_{0}^{\beta} e^{-x} d x=\left.\lim \left(-e^{-x}\right)\right|_{0} ^{\beta}=\lim \left(e^{0}-e^{-\beta}\right)=1
$$
Therefore, the given improper integral converges.
Geometrically, in a rectangular coordina... | 1,\pi,+\infty,3(\sqrt[3]{2}+1) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,290 |
701. Calculate the integral $\int_{1}^{9} \sqrt{6 x-5} d x$ using the Newton-Leibniz formula and approximate formulas for rectangles, trapezoids, and Simpson's method, dividing the integration interval into 8 equal parts. Then estimate the percentage error of the results obtained using the approximate formulas. | Solution. By the Newton-Leibniz formula
$$
I=\int_{1}^{9} \sqrt{6 x-5} d x=\frac{1}{6} \int_{1}^{9}(6 x-5)^{\frac{1}{2}} d(6 x-5)=\left.\frac{1}{9}(6 x-5)^{\frac{3}{2}}\right|_{1} ^{9}=38
$$
Next, we divide the integration interval $[1 ; 9]$ into 8 equal parts, find the length of one part $h=1$, the division points $... | 38 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,291 |
702. Using Simpson's formula, calculate the approximate value of the integral $\int_{0} \cos x d x$ with an accuracy of 0.00001. | Solution. First, we determine the number $n$ of parts into which the integration interval $\left[0, \frac{\pi}{2}\right]$ should be divided to achieve the required accuracy of the computation.
Assuming the error $\delta(n)$ of Simpson's formula is less than $10^{-5}$, we have
$$
\frac{(b-a)^{5}}{180 n^{4}} y_{H L}^{(... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,292 |
708. Construct the region $D$ of variation of the variables $x$ and $y$, defined by the following inequalities:
1) $2 \leqslant x \leqslant 6, 1 \leqslant y \leqslant 3$
2) $\frac{x^{2}}{9}+\frac{y^{2}}{4}<1$
3) $4 \leqslant x^{2}+y^{2} \leqslant 9$
4) $0<y<x$ | Solution. 1) These inequalities are satisfied by the coordinates of any point located inside and on the boundary of the rectangle whose sides lie on the lines $x=2, x=6$, $y=1$ and $y=3$. This rectangle is the domain $D$ of variation
 $\lim _{\substack{x \rightarrow 3 \\ y \rightarrow 0}} \frac{\tan(x y)}{y}$
2) $\lim _{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{x}{x+y}$. | Solution. Having made sure that the function is not defined at the limit point, we perform transformations, guided by the instructions in § 7, Chapter I:
1) $\lim _{\substack{x \rightarrow 3 \\ y \rightarrow 0}} \frac{\tan(x y)}{y}=\lim x \cdot \lim \frac{\tan(x y)}{x y}=3 \cdot 1=3$, since $\lim _{a \rightarrow 0} \f... | 3 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,295 |
718. Find the partial derivatives of the functions:
1) $z=x^{3}+5 x y^{2}-y^{3}$
2) $u=\frac{x}{y}+\frac{y}{z}-\frac{z}{x}$
3) $v=\sqrt[x]{e^{y}}$ | Solution. 1) Considering $z$ as a function of only one argument $x$, using the formulas from Chapter II, we find $\frac{\partial z}{\partial x}=3 x^{2}+5 y^{2}$.
Similarly, considering $z$ as a function of only $y$, we get $\frac{\partial z}{\partial y}=$ $=10 x y-3 y^{2}$.
2) Considering $u$ as a function of only $x... | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,297 |
719. Calculate the values of the partial derivatives of the given functions at the specified values of the arguments:
1) $f(\alpha, \beta)=\cos (m \alpha-n \beta) ; \alpha=\frac{\pi}{2 m}, \beta=0$;
2) $z=\ln \left(x^{2}-y^{2}\right) ; x=2, y=-1$. | Solution. 1) Using the differentiation formulas (Chapter II), we find the partial derivatives:
$$
f_{\alpha}^{\prime}=-m \sin (m \alpha-n \beta) ; \quad f_{\beta}^{\prime}=n \sin (m \alpha-n \beta) .
$$
Setting $\alpha=\frac{\pi}{2 m}, \beta=0$, we get $f_{\alpha}^{\prime}\left(\frac{\pi}{2 m}, 0\right)=-m ; f_{\beta... | f_{\alpha}^{\}(\frac{\pi}{2},0)=-;f_{\beta}^{\}(\frac{\pi}{2},0)=n;z_{x}^{\}(2;-1)=\frac{4}{3};z_{y}^{\}(2;-1)=\frac{2}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,298 |
720. Check that the function $z=x \ln \frac{y}{x}$ satisfies the equation $x \frac{\partial z}{\partial x}+y \frac{\partial z}{\partial y}=z$. | Solution. We will identically transform the given function and find its partial derivatives with respect to $x$ and $y$:
$$
z=x(\ln y-\ln x) ; \frac{\partial z}{\partial x}=\ln y-\ln x-1=\ln \frac{y}{x}-1 ; \quad \frac{\partial z}{\partial y}=\frac{x}{y} .
$$
Substituting $z, \frac{\partial z}{\partial x}$, and $\fra... | proof | Calculus | proof | Yes | Yes | olympiads | false | 34,299 |
729. Find the total differentials of the functions:
1) $z=3 x^{2} y^{5}$;
2) $u=2 x^{y z}$
3)^{*} $p=\arccos \frac{1}{u v}$. | Solution.
1) a. Find the partial derivatives of the given function:
$$
\frac{\partial z}{\partial x}=6 x y^{5} ; \quad \frac{\partial z}{\partial y}=15 x^{2} y^{4}
$$ | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,300 | |
730. Calculate the value of the total differential of the function $z=\operatorname{arcctg} \frac{x}{y}$ at $x=1, y=3, d x=0.01, d y=-0.05$. | Solution. We find the partial derivatives, then the partial differentials and the total differential of the given function:
$$
\frac{\partial z}{\partial x}=-\frac{y}{x^{2}+y^{2}} ; \quad \frac{\partial z}{\partial y}=\frac{x}{x^{2}+y^{2}} ; \quad d z=\frac{x d y-y d x}{x^{2}+y^{2}}
$$
Substituting the given values o... | -0.008 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,301 |
731. Calculate the approximate value:
1) $1.08^{3.96}$;
2) $\frac{\sin 1.49 \cdot \operatorname{arctg} 0.07}{2^{2.95}}$. | Solution. If it is required to calculate the value of the function $f(x, y, \ldots, t)$ at the point $M_{1}\left(x_{1}, y_{1}, \ldots, t_{1}\right)$ and if it is easier to calculate the values of this function and its partial derivatives at the point $M_{0}\left(x_{0}, y_{0}, \ldots, t_{0}\right)$, then for sufficientl... | 0.01 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,302 |
745. Find the derivative of the implicit function $y$, given by the equation: 1) $x^{2}+y^{2}+2 x-6 y+2=0$; 2) $x^{y}=y^{x}$, and calculate its value at $x=1$. | Solution. 1) Denoting the left-hand side of the given equation by $f(x, y)$, we find the partial derivatives $f_{x}^{\prime}=2 x+2, f_{y}^{\prime}=2 y-6$ and, substituting them into formula (B), we get $y^{\prime}=\frac{x+1}{3-y}$.
Next, substituting $x=1$ into the original equation, we find two corresponding values o... | y_{1}^{\}(1)=1,y_{2}^{\}(1)=-1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,304 |
746. Find the partial derivatives of the implicit function $z(x, y)$ given by the equations: 1) $x^{2}+y^{2}+z^{2}-z=0$; 2) $a x+b y-c z=$ $=k \cos (a x+b y-c z)$. | Solution. 1) Denoting the left-hand side of the equation by $\varphi(x, y, z)$ and using formulas (A), we get
$$
\frac{\partial z}{\partial x}=-\frac{\varphi_{x}^{\prime}}{\varphi_{z}^{\prime}}=-\frac{2 x}{2 z-1} ; \quad \frac{\partial z}{\partial y}=-\frac{\varphi_{y}^{\prime}}{\varphi_{z}^{\prime}}=-\frac{2 y}{2 z-1... | \frac{\partialz}{\partialx}=-\frac{2x}{2z-1},\quad\frac{\partialz}{\partialy}=-\frac{2y}{2z-1},\quad\frac{\partialz}{\partialx}=\frac{}{},\quad\frac{\partialz}{\partialy}=\frac{b}{} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,305 |
754. Find the second-order partial derivatives of the following functions: 1) $z=x^{3}-2 x^{2} y+3 y^{2}$; 2) $u(x, y, t)=e^{x y t}$. | Solution. 1) First, we find the first-order partial derivatives, then the required second-order partial derivatives:
$$
\begin{aligned}
& z_{x}^{\prime}=3 x^{\prime \prime}-4 x y ; \quad z_{y,}^{\prime}=-2 x^{2}+6 y ; \\
& z_{x x}^{\prime \prime}=6 x-4 y ; \quad z_{x y}^{\prime \prime}=z_{y x}^{\prime \prime}=-4 x ; \... | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,306 | |
756. Check that the function $z=2 \cos ^{2}\left(y-\frac{x}{2}\right)$ satisfies the differential equation $2 \frac{\partial^{2} z}{\partial x^{2}}+\frac{\partial^{2} z}{\partial x \partial y}=0$. | Solution. Let's find the second-order partial derivatives contained in the given equation:
$$
\begin{gathered}
\frac{\partial z}{\partial x}=2 \cdot 2 \cos \left(y-\frac{x}{2}\right) \cdot\left[-\sin \left(y-\frac{x}{2}\right)\right] \cdot\left(-\frac{1}{2}\right)=\sin (2 y-x) \\
\frac{\partial^{2} z}{\partial x^{2}}=... | proof | Calculus | proof | Yes | Yes | olympiads | false | 34,307 |
765. Find the equations of the tangent plane and normal to the elliptic paraboloid $z=2 x^{2}+y^{2}$ at point A $(1, -1, 3)$. | Solution. Transform the surface equation to the form $2 x^{2}+$ $+y^{2}-z=0$ and, denoting its left side by $F(x, y, z)$, find the partial derivatives $F_{x}^{\prime}=4 x, F_{y}^{\prime}=2 y, F_{z}^{\prime}=-1$, calculate their numerical values at the given point $F_{x}^{\prime}(A)=4, F_{y}^{\prime}(A)=-2$, $F_{2}^{\pr... | 4x-2y-z-3=0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,308 |
766. On the sphere $x^{2}+y^{2}+z^{2}=676$ find the points where the tangent plane is parallel to the plane $3 x-12 y+4 z=0$. | Solution. Using the general equation (I), we will form the equation of the tangent plane to the given sphere at its point $\left(x_{0}, y_{0}, z_{0}\right)$:
$$
x_{0}\left(x-x_{0}\right)+y_{0}\left(y-y_{0}\right)+z_{0}\left(z-z_{0}\right)=0
$$
$$
x_{6} x+y_{0} y+z_{0} z=x_{0}^{2}+y_{0}^{2}+z_{j}^{4}=676
$$
According... | (6,-24,8)(-6,24,-8) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 34,309 |
767. Show that the tangent planes to the surface $x y z = m^{3}$ form a tetrahedron of constant volume with the coordinate planes.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Solution. The equation of the tangent plane to the given surface at the point $P\left(x_{0}, y_{0}, z_{0}\right)$ will be $y_{0} z_{11} x + x_{0} z_{0} y + x_{i 1} y_{0} z = 3 x_{0} y_{0} z_{0}$. It intercepts segments on the coordinate axes $a = 3 x_{0}, b = 3 y_{1}, c = 3 z_{0}$. These segments are mutually perpendic... | proof | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,310 |
784. Find such a point in an isosceles right triangle for which the sum of the squares of the distances to its vertices is the smallest. | Solution. Let's choose a rectangular coordinate system $x O y$ as shown in Fig. 144, then the coordinates of the vertices of the triangle will be $A(0,0), B(a, 0), C(0, a)$. We take an arbitrary point $M(x, y)$ inside the triangle
 $\left.I_{1}=\int_{-2}^{2} d x \int_{x^{2}}^{4} f(x, y) d y ; 2\right) I_{2}=\int_{1}^{3} d y \int_{0}^{2 y} u d x$.
 | Solution. 1) First, we determine the region of integration from the limits of integration. By setting $x$ equal to the limits of integration with the variable $x$, and $y$ equal to the limits of integration with the variable $y$, we obtain the equations of the lines bounding this region: $x=-2, x=2, y=x^{2}, y=4$.
![]... | \int_{0}^{4}\int_{-\sqrt{y}}^{\sqrt{y}}f(x,y) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,315 |
831. Find the mass of a circular ring if at each of its points the surface density is inversely proportional to the square of the distance from it to the center of the ring. | Solution. Let the radii of the circles bounding the ring be denoted by $r_{1}$ and $r_{2}\left(r_{1}<r_{2}\right)$, and place the pole of the polar coordinate system at the center of the ring; then the equations of the circles will be $\rho=r_{1}$ and $\rho=r_{2}$, and the surface density at point $M(\varphi, \rho)$ of... | 2k\pi\ln\frac{r_{2}}{r_{1}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,319 |
832. Find the mass of a plate having the shape of an ellipse, if the surface density at each point of the plate is proportional to its distance $r$ from the minor axis of the ellipse and at $r=1$ it is equal to $\lambda$. | Solution. Let the semi-axes of the ellipse be denoted by $a$ and $b (a > b)$, and choose the axes $O x$ and $O y$ of the rectangular coordinate system to coincide with the axes of the ellipse. Then the equation of the ellipse is $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$.
According to the problem, at the point $M(x, ... | \frac{4}{3}^{2}b\lambda | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,320 |
833. Find the center of gravity of an isosceles right triangle if the surface density at each of its points is proportional to the distance from it to the hypotenuse. | Solution. Let in a right isosceles triangle $ABC$ the hypotenuse $AB=2a$. Then, relative to the coordinate system depicted in Fig. 169, the equations of the legs $AC$ and $BC$ will be $y=x+a$ and $y=a-x$.
According to the problem's condition, the density $\delta=ky$ at the point $(x, y)$ of the triangle.
Next, using ... | y_{}=\frac{}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 34,321 |
835. A homogeneous plate is bounded by two concentric ellipses with coinciding axes (elliptical ring). Find the moments of inertia of this plate relative to its axes. | Solution. Let the outer ellipse have semi-axes $a_{1}$ and $b_{1}$, and the inner ellipse have semi-axes $a_{2}$ and $b_{2}$. Suppose the axes of the rectangular coordinate system are directed along the symmetry axes of the plate. Then the equations of the ellipses will be
$$
\frac{x^{2}}{a_{1}^{2}}+\frac{y^{2}}{b_{1}... | I_{x}=\frac{1}{4}\pi\delta(a_{1}b_{1}^{3}-a_{2}b_{2}^{3}),\quadI_{y}=\frac{1}{4}\pi\delta(b_{1}a_{1}^{3}-b_{2}a_{2}^{3}) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,323 |
836. Find the center of gravity of a homogeneous truncated prism bounded by the coordinate planes and the planes $x+y+z=4, x=1, y=1$. | Solution. By constructing the given planes (Fig. 170), we observe that the truncated prism bounded by them is symmetric with respect to the plane $x=y$. Consequently,
Fig. 170 this $x_{c}=y... | x_{}=y_{}=\frac{17}{36};\quadz_{}=\frac{55}{36} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 34,324 |
849. Compute the triple integral $I=\int_{-1}^{1} d x \int_{x^{2}}^{1} d y \int_{0}^{z}(4+z) d z$ and sketch its region of integration. | Solution. We sequentially compute three ordinary (single) definite integrals, starting with the inner one:
$$
\begin{gathered}
I_{1}=\int_{0}^{2}(4+z) d z=\left.\frac{(4+z)^{2}}{2}\right|_{0} ^{2}=\frac{36-16}{2}=10 \\
I_{2}=\int_{x^{2}}^{1} I_{1} d y=10 \int_{x^{2}}^{1} d y=\left.10 y\right|_{x^{2}} ^{1}=10\left(1-x^... | \frac{40}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,325 |
858. Find the mass of the body bounded by the cylindrical surface $x^{2}=2 y$ and the planes $y+z=1, 2 y+z=2$, if at each point of the body the volumetric density is numerically equal to the ordinate of that point. | Solution. According to the condition, at point $M(x, y, z)$ of the body, the volume density $\delta(M)=y$. By formula (2), the mass of this body is
$$
m=\iint_{G} \delta \delta(M) d v=\iiint_{G} y d x d y d z
$$
where $G$ is the region occupied by the given body (Fig. 182).[^31]
Evaluating the triple integral using ... | \frac{8\sqrt{2}}{35} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,329 |
859. Find the center of gravity of a spherical segment, if at each (y) point the volumetric density is proportional to its distance from the base of the segment. | Solution. Let the radius of the sphere be denoted by $R$, and the height of the segment by $h$; we place the origin of the rectangular coordinate system at the center of the sphere and direct the z-axis along the axis of the segment (Fig. 183). The equations of the sphere and the plane that limit the segment ($G$) will... | z_{C}=\frac{20R^{2}-15R+3^{2}}{5(4R-)} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,330 |
872. Compute the line integral \( I=\int_{L}(x y-1) d x + x^{2} y d y \) from point \( A(1 ; 0) \) to point \( B(0 ; 2) \):
1) along the line \( 2 x+y=2 \);
2) along the arc of the parabola \( 4 x+y^{2}=4 \);
3) along the arc of the ellipse \( x=\cos t ; y=2 \sin t \) (Fig. 185). | Solution. 1) Using the given equation of the line of integration, we transform the curvilinear integral into an ordinary definite integral with variable $x$, then compute it:
$$
y=2-2 x, \quad d y=-2 d x
$$
$$
\begin{aligned}
& i_{1}=\int_{x_{A}}^{x_{R}}[x(2-2 x)-1] d x+x^{2}(2-2 x)(-2 d x)= \\
= & \int_{1}^{0}\left(... | 1,-\frac{1}{5},\frac{4}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,332 |
874. Given points $A(3, -6, 0)$ and $B(-2, 4, 5)$. Compute the line integral $I=\int_{C} x y^{2} d x + y z^{2} d y - z x^{2} d z$:
1) along the straight line segment $O B$ and
2) along the arc $A B$ of the circle defined by the equations $x^{2} + y^{2} + z^{2} = 45, 2 x + y = 0$. | Solution. 1) First, we write the equation of the line passing through two points $\frac{x-x_{1}}{x_{2}-x_{1}}=\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{z-z_{1}}{z_{2}-z_{1}}$, obtaining $\frac{x}{-2}=\frac{y}{4}=\frac{z}{5}$. By equating these equal ratios to the parameter $t$, we transform the canonical equations of the line ... | 91 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,333 |
875. Compute the line integrals:
1) $\oint_{-i} 2 x d x-(x+2 y) d y \quad$ and
2) $\oint_{+l} y \cos x d x+\sin x d y$
along the perimeter of the triangle with vertices $A(-1 ; 0), B(0 ; 2)$ and $C(2 ; 0)$. | Solution. 1) Here (Fig. 186) the integration line (closed) consists of three segments lying on different lines (with different equations). Accordingly, the line integral along the broken line $A B C A$ is calculated as the sum of integrals taken over the segments $A B, B C$, and $C A$.
By formulating the equation of t... | 3 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,334 |
884. Find the length of the cardioid $x=2 a \cos t-a \cos 2 t, y=$ $-2 a \sin t-a \sin 2 t$. | Solution. We apply formula (1); based on the given parametric equations of the cardioid and the formula for the differential of the arc of a plane curve (Chapter 5, §), we transform the line integral in formula (1) into an ordinary integral with the variable $t$.
$$
\begin{gathered}
\dot{x}=-2 a \sin t+2 a \sin 2 t, \... | 16a | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,335 |
886. Find the mass of the arc $A B$ of the curve $y=\ln x$, if at each point of it the linear density is proportional to the square of the abscissa of the point; $x_{A}=1, x_{B}=3$. | Solution. We apply formula (3). Based on the given equation of the curve, we transform the curvilinear integral in formula (3) into an ordinary one with the variable $x$:
$$
\begin{gathered}
y^{\prime}=\frac{1}{x}, \quad d l=\sqrt{1+\left(y^{\prime}\right)^{2}} d x=\sqrt{1+\frac{1}{x^{2}}} d x, \quad \delta=k x^{3} \\... | \frac{k}{3}(10\sqrt{10}-2\sqrt{2}) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,337 |
887. Find the coordinates of the center of gravity of the arc $A B$ of the helical line $x=a \cos t, y=a \sin t, z=b t$, if at each point of it the linear density is proportional to the ordinate of that point; $t_{A}=0, t_{B}=\pi$. | Solution. We apply formulas (4). We will compute the curvilinear integrals contained in these formulas, transforming them into ordinary integrals with the variable $t$:
$$
\begin{aligned}
& \dot{x}=-a \sin t, \dot{y}=a \cos t, \quad \dot{z}=b ; d l=\sqrt{\dot{x}^{2}+\dot{y}^{2}+\dot{z}^{2}} d t= \\
& =\sqrt{a^{2}+b^{2... | x_{C}=-\frac{4}{\pi^{2}};\quady_{C}=\frac{2}{\pi};\quadz_{C}=\frac{2}{3}b\pi | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,338 |
889. Find the work of the force field, in each point $(x, y)$ of which the intensity (force acting on a unit mass) $\bar{p}=$ $=(x+y) \bar{i}-x \bar{j}$, when a point mass $m$ describes the circle $x=a \cos t, y=-a \sin t$, moving in the clockwise direction. | Solution. Substituting into formula (5) the projections of the force $\bar{F}=m \vec{p}$ acting on the point: $F_{x}=m(x+y), F_{y}=-m x$, and transforming the curvilinear integral into an ordinary one with variable $t$, we get
$$
\begin{gathered}
E=\oint_{-c} P d x+Q d y=\oint_{-c} m(x+y) d x-m x d y= \\
=\int_{0}^{-2... | 2\pi^{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,340 |
899. Check that the given expression is a total differential of the function $u(x, y)$, and find $u$:
1) $\left.\left(2 x-3 y^{2}+1\right) d x+(2-6 x y) d y ; \quad 2\right)\left(e^{x y}+5\right)(x d y+y d x)$;
2) $(1-\sin 2 x) d y-(3+2 y \cos 2 x) d x$. | Solution. 1) Let's denote the coefficients at differentials as $P=2 x-3 y^{2}+1, Q=2-6 x y$ and find $P_{y}^{\prime}=-6 y$ and $Q_{x}^{\prime}=-6 y$. Since here $P_{y}^{\prime}=Q_{x}^{\prime}$ and $P, Q, P_{y}^{\prime}, Q_{x}^{\prime}$ are continuous, the given expression is a total differential of some function $u$.
... | y-3x-y\sin2x+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,341 |
906. Compute surface integrals of the first kind (over the area of the surface):
1) $I=\iint_{\sigma}(6 x+4 y+3 z) d s$, where $\sigma$ is the part of the plane $x+2 y+$ $+3 z=6$ located in the first octant.
2) $K=\iint_{W}\left(y+z+V \sqrt{a^{2}-x^{2}}\right) d s$, where $W-$ is the surface of the cylinder $x^{2}+y^{2... | Solution. 1) The surface of integration $\sigma$ is the triangle $ABC$ (Fig. 190). Using its equation and formula (1), we transform the given surface integral into a double integral with variables $x$ and $y$:
$$
\begin{aligned}
& z=\frac{1}{3}(6-x-2 y), d s=\sqrt{1+\left(z_{x}^{\prime}\right)^{2}+\left(z_{4}^{\prime}... | 54\sqrt{14} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,342 |
920. Find the area of the part of the surface:
1) of the cone $z^{2}=2 x y$, located in the first octant between the planes $x=2, y=4$;
2) of the sphere $x^{2}+y^{2}+z^{2}=R^{2}$, located inside the cylinder $x^{2}+y^{2}=R x$
3) of the cylinder $x^{2}+y^{2}=R x$, located inside the sphere $x^{2}+$ $+y^{2}+z^{2}=R^{2}$ | Solution. 1) We apply formula (1). Using the equation of the cone, we transform the surface integral into a double integral with variables $x$ and $y$:
$$
\begin{gathered}
S=\iint_{\boldsymbol{\sigma}} d s=\iint_{\sigma_{x y}} \sqrt{1+\left(z_{x}^{\prime}\right)^{2}+\left(z_{y}^{\prime}\right)^{2}} d x d y=\frac{1}{\s... | 16,2R^2(\pi-2),4R^2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,346 |
921. Find the mass of a hemisphere if the surface density at each of its points is numerically equal to the distance of this point from the radius perpendicular to the base of the hemisphere. | Solution. Place the origin of the rectangular coordinate system at the center of the base of the hemisphere and direct the z-axis perpendicular to this base. Then the equation of the hemisphere will be $z=\sqrt{R^{2}-x^{2}-y^{2}}$, where $R$ is the radius of the hemisphere; the surface density at point $M(x, y, z)$ of ... | \frac{\pi^{2}R^{3}}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,347 |
923. Find the center of gravity of a homogeneous surface of the paraboloid $y^{2}+z^{2}=10 x$, cut off by the plane $x=10$. | Solution. The given homogeneous surface ($\sigma$) is symmetric with respect to the x-axis (Fig. 196). Therefore, $y_{C}=z_{C}=0$.
Fig. 196
To find the abscissa of the center of gravity, ... | \frac{25\sqrt{5}+1}{5\sqrt{5}-1} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,348 |
934. Find the derivative of the function $u=\sqrt{x^{2}+y^{2}}$ at the point $A(3 ; 4):$
1) in the direction of the bisector of the first coordinate angle
2) in the direction of the radius vector of the point $A$;
3) in the direction of the vector $\bar{q}\{4 ;-3\}$. | Solution. We find the partial derivatives of the function $u$ and calculate their values at point $A$:
$$
\begin{gathered}
\frac{\partial u}{\partial x}=\frac{x}{\sqrt{x^{2}+y^{2}}} ;\left.\frac{\partial u}{\partial x}\right|_{A}=\frac{3}{5} ; \\
\frac{\partial u}{\partial y}=\frac{y}{\sqrt{x^{2}+y^{2}}} ;\left.\frac{... | \frac{7\sqrt{2}}{10} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,349 |
935. Find the derivative of the function $u=x y+y z+1$ in the direction of the vector $\bar{l}\{12 ;-3 ;-4\}$ at any point and at points $A(0 ;-2 ;-1)$ and $B(3 ; 3 ; 5)$. | Solution. Let's find the partial derivatives of the function and the direction cosines of the vector $\bar{l}$:
$$
\begin{gathered}
u_{x}^{\prime}=y ; u_{y}^{\prime}=x+z ; u_{z}^{\prime}=y \\
\cos \alpha=\frac{12}{13} ; \cos \beta=-\frac{3}{13} ; \quad \cos \gamma=-\frac{4}{13}
\end{gathered}
$$
Substituting into for... | u_{}^{\}(A)=-1;\,u_{}^{\}(B)=0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,350 |
936. What is the greatest rate at which the function $u(M)=\frac{10}{x^{2}+y^{2}+z^{2}+1}$ can increase as the point $M(x, y, z)$ passes through the point $M_{0}(-1 ; 2 ;-2)$? In what direction should the point $M$ move as it passes through the point $M_{1}(2 ; 0 ; 1)$, so that the function $u(M)$ decreases at the grea... | Solution. The greatest in absolute value the rate of change (increase or decrease) of the function and (M) when the point $M$ passes through the point $P$ numerically equals the modulus of the gradient of the function at point $P$. In this case, the function will increase or decrease at the greatest rate, depending on ... | \frac{3}{5} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,351 |
937. Find the points at which the function $z=e^{x}\left(x-y^{3}+3 y\right)$ is stationary (i.e., points where the derivative in any direction is zero). | Solution. For the derivative of a function in any direction to be zero at some point $P$, it is necessary and sufficient that all the first-order partial derivatives of the function simultaneously become zero at this point. [According to formula $(a) .1$]
Therefore, by finding the partial derivatives: $z_{x}^{\prime}=... | (-3,1)(1,-1) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,352 |
945. Find the flux of the vector field $\bar{p}=x \bar{i}-y^{2} \bar{j}+\left(x^{2}+z^{2}-1\right) \bar{k}$ through the surface $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+\frac{z^{3}}{c^{2}}=1$ (ellipsoid) from inside this surface. | Solution. According to formula (1)
$$
K=\oiint_{+\sigma} x d y d z-y^{2} d x d z+\left(x^{2}+z^{2}-1\right) d x d y
$$
We decompose this surface integral (of the second type) into three component integrals and, using the given equation of the ellipsoid (b), reduce their computation to the computation of double integr... | \frac{4}{3}\pi | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,353 |
946. Find the divergence of the vector field:
1) $\bar{r}=x i+y \bar{j}+z k$
2) $\bar{p}=\frac{\bar{i}+\bar{i}+\bar{k}}{\sqrt[3]{(x+y+z)^{2}}}$
3) $\vec{q}=e^{x y}(y \bar{j}-x \bar{i}+x y \bar{k})$. | Solution. Applying formula (2):
1) $\operatorname{div} \dddot{r}(M)=\frac{\partial r_{x}}{\partial x}+\frac{\partial^{r} y}{\partial y}+\frac{\partial r_{z}}{\partial z}=1+1+1=3$.
2) $p_{x}=p_{y}=p_{z}=(x+y+z)^{-\frac{2}{3}}$;
$$
\frac{\partial \rho_{x}}{\partial x}=\frac{\partial \rho_{y}}{\partial y}=\frac{\partial... | 3,-2(x+y+z)^{-} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,354 |
960. Investigate the convergence of the series by the integral test:
1) $\sum_{n=1}^{+\infty} \frac{2 n}{n^{2}+1}$;
2) $\sum_{n=2}^{+\infty} \frac{1}{n \ln ^{3} n} ;$
3) $\sum_{n=0}^{+\infty} \frac{1}{\sqrt{4 n+1}}$
4) $\sum_{n=3}^{+\infty} \frac{n}{n^{4}-9}$. | Solution. We replace the general term of the given series $a_{n}=f(n)$ with the number $n$ by a continuous variable $x$ and ensure that the resulting function $f(x)$ is continuous and decreasing over the entire infinite interval of $x$. Then we find the improper integral of $f(x)$ with an infinite upper limit.
1) $\in... | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,356 |
1000. Determine the domain of convergence of the functional series:
$$
\text { 1) } \sum_{n=1}^{+\infty} \frac{1}{n(x+2)^{n}} ; \quad \text { 2) } \sum_{n=1}^{+\infty} n \sqrt[3]{\sin ^{n} x}
$$ | Solution: 1) Using the D'Alembert's criterion:
$$
\begin{aligned}
& u_{n}=\frac{1}{n(x+2)^{n}} ; u_{n+1}=\frac{1}{(n+1)(x+2)^{n+1}} ; \\
& \rho=\lim _{n \rightarrow+\infty}\left|\frac{u_{n+1}}{u_{n}}\right|=\lim _{n \rightarrow+\infty} \frac{n}{(n+1)|x+2|}=\frac{1}{|x+2|} . \\
& \frac{1}{|x+2|}<1 ; \quad |x+2|>1 ; \qu... | \mathbb{R}\setminus{\frac{\pi}{2}+\pik\midk\in\mathbb{Z}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,359 |
1011. Expand the functions $e^{x}, \sin x, \cos x$ into Maclaurin series. | Solution. A) The values of these functions and their derivatives of any order at $x=0$ were calculated earlier in the solution of problem 297 (Chapter II, § 1). Substituting these values into the general expression of the Maclaurin series (M) for an arbitrary function, we obtain the Maclaurin series for the given funct... | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,360 |
1012. Expand the functions in a Maclaurin series: 1) $(1+x)^{m}$, 2) $\ln (1+x)$. | Solution. 1) A. Based on the solution of problem 298 and according to the definition of the Maclaurin series for an arbitrary function, we get
$$
\begin{aligned}
(1+x)^{m}= & 1+\frac{m}{1!} x+\frac{m(m-1)}{2!} x^{2}+\frac{m(m-1)(m-2)}{3!} x^{3}+\ldots \\
& \ldots+\frac{m(m-1) \ldots(m-n+1)}{n!} x^{n}+\ldots
\end{align... | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,361 |
1013. Expand the functions into Taylor series:
1) $\frac{1}{x}$ at $\left.a=-2 ; 2\right) \cos x$ at $a=\frac{\pi}{4}$. | Solution. 1) A. We calculate the values of the given function and its derivatives at $x=a:=-2$:
$$
\begin{array}{rl}
f(x)=x^{-1} & f(-2)=-\frac{1}{2} \\
f^{\prime}(x)=-1 \cdot x^{-2} & f^{\prime}(-2)=-\frac{1!}{2^{2}} \\
f^{\prime \prime}(x)=1 \cdot 2 x^{-3} & f^{\prime \prime}(-2)=-\frac{2!}{2^{3}} \\
f^{\prime \prim... | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,362 | |
1018. Using the Maclaurin series for the functions $e^{x}, \sin x, \cos x$, $(1+x)^{\prime \prime}, \ln (1+x)$ and the rules for multiplying and adding power series, find the expansions in series of powers of $x$ for the following functions: 1) $(1+x) e^{x}$; 2) $\sin ^{2} x$; 3) $\frac{x-3}{(x+1)^{2}}$; 4) $e^{-x} \si... | Solution. 1) Consider the binomial $1+x$ as a power series, whose coefficients of all terms, except the first two, are zero and which converges over the entire number line. By multiplying this series term by term with the Maclaurin series for the function $e^{x}$, which also converges over the entire number line, we ob... | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,363 |
1019. Using appropriate series, calculate with an accuracy of $0.0001: 1) \ln 1.1 ; 2) \sqrt[4]{17}$. | Solution. For calculating approximate values of a function with a given accuracy, it is convenient to use series when the corresponding series is alternating; for an alternating convergent series, it is easy to estimate the error of the approximate value of the sum - it is less than the absolute value of the first 13 d... | \ln1.1\approx0.0953,\sqrt[4]{17}\approx2.0305 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,364 |
1020. Find the series expansion of the function $\operatorname{arctg} x$, starting from its expression as an integral: $\operatorname{arctg} x=\int_{0}^{x} \frac{d t}{1+t^{2}}$, by expanding the integrand in a Maclaurin series and integrating it term by term. | Solution. Transform the integrand $\frac{1}{1+t^{2}}=$ $=\left(1+t^{2}\right)^{-1}$ and expand it into a binomial series (B), setting $x=t^{2}, m=-1$:
$$
\left(1+t^{2}\right)^{-1}=1-t^{2}+t^{4}-t^{6}+\ldots+(-1)^{n-1} t^{2 n-2}+\ldots
$$
Integrating within the limits from 0 to $x$, we obtain the desired series
$$
\o... | \operatorname{arctg}x-\frac{x^{3}}{3}+\frac{x^{5}}{5}-\frac{x^{7}}{7}+\ldots+(-1)^{n-1}\frac{x^{2n-1}}{2n-1}+\ldots | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,365 |
1021. Find the Maclaurin series for the function $\arcsin x$, starting from its representation as an integral: $\arcsin x=\int_{0}^{x} \frac{d t}{\sqrt{1-t^{2}}}$. | Solution. As in the solution of the previous problem, we transform the integrand function $\frac{1}{\sqrt{1-t^{2}}}=\left(1-t^{2}\right)^{-\frac{1}{2}}$, expand it into a binomial series (B), setting $x=-t^{2}, m=-\frac{1}{2}$ :
$$
\begin{aligned}
\left(1-t^{2}\right)^{-\frac{1}{2}} & =1+\frac{1}{2} t^{2}+\frac{1 \cdo... | \begin{aligned}\arcsinx&=x+\frac{1}{2}\frac{x^{3}}{3}+\frac{1\cdot3}{2\cdot4}\frac{x^{5}}{5}+\frac{1\cdot3\cdot5}{2\cdot4\cdot6}\frac{x^{7}}{7}+\ldots+\\&+\frac{1} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,366 |
1022. By expanding the integrand into a Maclaurin series and integrating it term by term, find the series expansions of the following integrals:
1) $\int \sin x^{2} d x$; 2) $\int \sqrt{x} e^{x} d x$; 3) $\int \sqrt{1-x^{3}} d x$. | Solution. 1) Using the Maclaurin series for $\sin x$, replacing $x$ with $x^{2}$, we have
$$
\sin x^{2}=x^{2}-\frac{x^{8}}{3!}+\frac{x^{10}}{5!}-\frac{x^{14}}{7!}+\ldots+(-1)^{n-1} \frac{x^{4 n-2}}{(2 n-1)!}+\ldots
$$
Integrating term by term, we obtain the desired expansion
$$
\int \sin x^{2} d x=\frac{x^{3}}{3}-\f... | \begin{aligned}1)&\quad\int\sinx^{2}=\frac{x^{3}}{3}-\frac{x^{7}}{3\cdot7}+\frac{x^{11}}{5!\cdot11}-\frac{x^{15}}{7!\cdot15}+\ldots+C\\2)&\quad\int\sqrt{x}e | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,367 |
1023. Using series, calculate the approximate values of the following integrals with an accuracy of 0.0001:
$$
I_{1}=\int_{0}^{\frac{1}{3}} \frac{d t}{\sqrt{1+t^{2}}} ; \quad I_{2}=\int_{0}^{1} \cos \sqrt{x} d x ; \quad I_{3}=\int_{1}^{1.5} \frac{1}{v} \operatorname{arctg} \frac{v}{4} d v
$$ | Solution. 1) We expand the integrand into a binomial series (B), setting $x=t^{4}, m=-\frac{1}{2}$:
$$
\frac{1}{\sqrt{1+t^{4}}}=\left(1+t^{4}\right)^{-\frac{1}{2}}=1-\frac{1}{2} t^{4}+\frac{1 \cdot 3}{2 \cdot 4} t^{8}-\frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6} t^{12}+\ldots
$$
This series converges to the binomial $... | I_{1}\approx0.4969,\quadI_{2}\approx0.7635,\quadI_{3}\approx0.1211 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,368 |
1059. Check that the given function is an integral (solution) of the given differential equation:
1) $y=\sqrt{x}, 2 y y^{\prime}=1$
2) $\ln x \ln y=c, y \ln y d x+x \ln x d y=0$.
3) $s=-t-\frac{1}{2} \sin 2 t, \frac{d^{2} s}{d t^{2}}+\operatorname{tg} t \frac{d s}{d t}=\sin 2 t$. | Solution. 1) Let's find the derivative of the given function $y'=\frac{1}{2 \sqrt{x}}$. Substituting the given equation $y=\sqrt{x}$ and $y'=\frac{1}{2 \sqrt{x}}$, we can verify that it becomes an identity: $2 \sqrt{x} \times \frac{1}{2 \sqrt{x}}=1 ; 1=1$.
2) Differentiate the given implicit function: $\ln y \frac{dx}... | proof | Calculus | proof | Yes | Yes | olympiads | false | 34,375 |
1067. Find the particular integral of the equation satisfying the given initial condition:
1) $y d x+\operatorname{ctg} x d y=0 ; \quad y\left(\frac{\pi}{3}\right)=-1$.
2) $s=s^{\prime} \cos ^{2} t \ln s ; s(\pi)=1$. | Solution. 1) By separating variables and integrating, we first find the general integral of the given equation:
$$
\begin{gathered}
\operatorname{tg} x d x+\frac{d y}{y}=0 ; \quad-\ln |\cos x|+\ln |y|=\ln C \\
|y|=C|\cos x| ; \quad y= \pm C \cos x=C_{1} \cos x
\end{gathered}
$$
Then, using the specified initial condi... | \ln^{2}-2\operatorname{tg}=0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,377 |
1077. Integrate the following equations:
1) $\left(x^{2}+y^{2}\right) d x-2 x y d y=0$
2) $y-x y^{\prime}=y \ln \frac{x}{y}$
3) $x d y-y d x=y d y$ with the condition $y(-1)=1$. | Solution. 1) Solving the given equation for the derivative
$$
\frac{d y}{d x}=y^{\prime}=\frac{x^{2}+y^{2}}{2 x y}=\frac{1+\left(\frac{y}{x}\right)^{2}}{2 \frac{y}{x}}=\varphi\left(\frac{y}{x}\right)
$$
we establish that it is a function only of the ratio of the variables $\frac{y}{x}$, i.e., we establish that the gi... | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,378 |
1084. Solve the equations:
1) $y^{\prime}-y \cot x=\sin x$
2) $x^{2} y^{2} y^{\prime}+x y^{3}=1$
3) $y d x-(3 x+1+\ln y) d y=0$ with the condition $y\left(-\frac{1}{3}\right)=1$. | Solution: 1) Having made sure that the given equation is linear, we assume $y=u v$; then $y^{\prime}=u^{\prime} v+v^{\prime} u$ and the given equation transforms into
$$
u^{\prime} v+v^{\prime} u-u v \operatorname{ctg} x=\sin x \text { or } u^{\prime} v+u\left(v^{\prime}-v \operatorname{ctg} x\right)=\sin x
$$
Since ... | \begin{aligned}1)&\quady=(x+C)\sinx\\2)&\quady=\sqrt[3]{\frac{3}{2x}+\frac{C}{x^{3}}}\\3)&\quadx=\frac{y^{3}-4}{9}-\frac{1}{3}\lny\end{aligned} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,379 |
1091. Solve the equations:
1) $(2 y-3) d x+\left(2 x+3 y^{2}\right) d y=0$
2) $(x+\ln |y|) d x+\left(1+\frac{x}{y}+\sin y\right) d y=0$. | Solution. 1) First, we ensure that the given equation is an equation in total differentials:
$$
P_{y}^{\prime}=(2 y-3)_{y}^{\prime}=2 ; \quad Q_{x}^{\prime}=\left(2 x+3 y^{2}\right)_{x}^{\prime}=2 ; \quad P_{y}^{\prime}=Q_{x}^{\prime}
$$
Then we find the indefinite integrals:
$\int P d x=\int(2 y-3) d x=2 x y-3 x+\v... | 2xy-3x+y^{3}=C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,380 |
1107. Solve the equations:
1) $y^{\prime \prime}-5 y^{\prime}-6 y=0$
2) $y^{\prime \prime \prime}-6 y^{\prime \prime}+13 y^{\prime}=0$
3) $\frac{d^{2} S}{d t^{2}}+4 \frac{d S}{d t}+4 S=0$;
4) $\frac{d^{4} y}{d x^{4}}-y=0$;
5) $y^{(4)}+13 y^{(2)}+36 y=0$
6) $y^{(2)}+2 y^{(5)}+y^{(3)}=0$. | Solution. 1) Replacing the function $y$ with one and its derivatives with the corresponding powers of $r$ in the given differential equation, we write its characteristic equation: $r^{2}-5 r-6=0$.
The roots of this equation $r_{1}=6, r_{2}=-1$ are real and distinct. Therefore, according to Rule 1, the sought general i... | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,382 |
1108. Find the particular integral of the equation satisfying the given initial conditions:
1) $y^{\prime \prime}+4 y^{\prime}+5 y=0 ; \quad y(0)=-3, \quad y^{\prime}(0)=0$.
2) $y^{\prime \prime \prime}+3 y^{\prime \prime}+3 y^{\prime}+y=0 ; y(0)=-1, y^{\prime}(0)=2, y^{\prime \prime}(0)=3$. | Solution. 1) First, we find the general integral of the given equation. Its characteristic equation $r^{2}+4 r+5=0$ has roots $r_{1,2}=-2 \pm i$. Therefore, according to rule 2, the general integral is $y=e^{-2 x}\left(C_{1} \cos x+C_{2} \sin x\right)$.
Next, using the initial conditions, we determine the values of th... | e^{-x}(3x^{2}+x-1) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,383 |
1162. For which curve is the segment of any tangent, enclosed between the point of tangency and the x-axis, bisected by the y-axis? | Solution. The equation of the tangent at any point $(x, y)$ of the desired curve will be $Y-y=y^{\prime}(X-x)$, where $X, Y-$ are the coordinates of any point on the tangent (Chapter II, § 11).
By setting $Y=0$ in this equation, we find the abscissa $X_{0}$ of the point of intersection of the tangent with the $O x$ ax... | y^{2}=Cx | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,387 |
1163. What shape should a homogeneous vertical column with a circular cross-section have so that the pressure from the load $P$ it supports and its own weight, per unit area of horizontal cross-section, is the same everywhere? (Column of equal pressure.) The specific weight of the column material is $\delta$, and the r... | Solution. Let the section of the column by a vertical plane passing through its axis of symmetry have the shape shown in Fig. 210.
Choosing a rectangular coordinate system $x O y$, we intersect the column with a horizontal plane passing through an arbitrary point $M(x, y)$ of the desired curve $A A_{1}$ and determine ... | r\approx3.09\partial\mu,\quadr_{1}\approx3.24\mu | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,388 |
1167. A locomotive is moving along a horizontal section of the track c) at a speed of 72 km/hour. In what time and at what distance will it be stopped by the brake, if the resistance to motion after the start of braking is equal to 0.2 of its weight. | Solution. According to Newton's second law in mechanics, the differential equation of motion for the locomotive will be
$$
m \frac{d^{2} s}{d t^{2}}=-0.2 m g
$$
where $s$ is the distance traveled in time $t$, $m$ is the mass of the locomotive, and $g$ is the acceleration due to gravity.
Multiplying both sides of thi... | 102 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,390 |
1168. A bullet enters a board 10 cm thick at a speed of $200 \mathrm{m} /$ sec, and exits the board, having penetrated it, at a speed of 50 m/sec. Find how long the bullet's movement through the board lasted, if the resistance of the board to the bullet's motion is proportional to the square of its speed. | Solution. Let $m$ be the mass of the bullet, $s$ - the distance traveled by it in time $t$, measured from the moment it enters the board. Then the differential equation of the bullet's motion through the board is
$$
m \frac{d^{2} s}{d t^{2}}=-k\left(\frac{d s}{d t}\right)^{2} \text { or } \frac{d^{2} s}{d t^{2}}=-a\le... | 0.001 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,391 |
1169. A chain hanging on a smooth hook slides down. At the beginning of the movement, 10 m of the chain hangs on one side of the hook, and 8 m on the other. Neglecting resistances, find: 1) in what time the entire chain will slide off the hook and 2) what will be the speed of the chain at the initial moment of its free... | Solution. If at time $t$ the length of the part of the chain moving downward is $s$ (m), then at this moment the force $F$ driving the chain is the difference between the weights of the parts of the chain hanging on either side of the hook, $F=\delta g s-\delta g(18-s)=2 \delta g(s-9)$; where $\delta$ is the mass of 1 ... | =\frac{3}{\sqrt{}}\ln(9+\sqrt{80})\approx2.9\text{},\quadv=\frac{\sqrt{80}}{3}\approx9.3\mathrm{}/\mathrm{} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,392 |
1191. Find the general integral of the equation $\frac{d y}{d x}=y^{2}$ in the form of a power series. | Solution. Let the desired integral be a power series
$$
y=a_{11}+a_{1} x+a_{2} x^{2}+\ldots+a_{n} x^{n}+\ldots
$$
where $a_{0}, a_{1}, \ldots, a_{n}, \ldots$ are unknown constants to be determined.
Assuming that such a series exists and converges in some interval of $x$ values, we find the series for $\frac{d y}{d x... | \frac{a_{0}}{1-a_{0}x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,394 |
1193. Find the first four terms of the power series expansion of the particular integral of the equation $y^{\prime}+x y^{2}=2 \cos x$, satisfying the initial condition: $y(0)=1$. | Solution. As in the previous problems, we seek the integral in the form of a power series (1).
According to the initial condition $y(0)=a_{0}=1$.
Further, finding the series for $y^{2}$ and $y^{\prime}$ and substituting them and the series for $\cos x$
$$
\cos x=1-\frac{x^{2}}{2!}+\frac{x^{4}}{4!}-\frac{x^{6}}{6!}+\... | 1+2x-\frac{1}{2}x^{2}-\frac{5}{3}x^{3}+\ldots | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,395 |
1194. Find the power series expansion of the particular integral of the equation $y^{\prime \prime}+x y=0$, satisfying the initial conditions: $y(0)=1, y^{\prime}(0)=0$. | Solution. Let the desired function $y(x)$ be expanded into a Maclaurin series
$$
y(x)=y(0)+\frac{y^{\prime}(0)}{1!} x+\frac{y^{\prime \prime}(0)}{2!} x^{2}+\ldots+\frac{y^{(n)}(0)}{n!} x^{n}+\ldots
$$
where the quantities $y(0), y^{\prime}(0), y^{\prime \prime}(0), \ldots$ are the values of the function $y(x)$ and it... | 1-\frac{1}{3!}x^{3}+\frac{1\cdot4}{6!}x^{6}-\frac{1\cdot4\cdot7}{9!}x^{3}+\ldots+(-1)^{}\frac{1\cdot4\cdot7\ldots(3-2)}{(3)!}x^{3}+\ldots | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,396 |
1195. Find the first five terms of the power series expansion of the particular integral of the equation $y^{\prime \prime}-y e^{x}=0$, satisfying the initial conditions: $y(0)=2, y^{\prime}(0)=1$. | Solution. Applying the same method as in the solution of the previous problem, we get:
$$
\begin{aligned}
& y^{\prime \prime}=y e^{x}, \quad y^{\prime \prime}(0)=2 \\
& y^{(3)}=\left(y+y^{\prime}\right) e^{x}, \quad y^{(3)}(0)=3 \\
& y^{(4)}=\left(y+2 y^{\prime}+y^{\prime \prime}\right) e^{x}, \quad y^{(4)}(0)=6
\end{... | 2+x+x^{2}+\frac{x^{3}}{2}+\frac{x^{4}}{4}+\ldots | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,397 |
1200. Find the general solution of the system of linear differential equations with constant coefficients:
1) $\left\{\begin{array}{l}\frac{d y}{d x}+2 y-4 z=0 \\ \frac{d z}{d x}+y-3 z=3 x^{2}\end{array}\right.$
2) $\left\{\begin{array}{l}6 u^{\prime}-u-7 v+5 w=10 e^{x} \\ 2 v^{\prime}+u+v-w=0 \\ 3 w^{\prime}-u+2 v-w=e... | Solution. 1) Differentiate the first equation with respect to $x$:
$$
y^{\prime \prime}+2 y^{\prime}-4 z^{\prime}=0
$$
then eliminate $z$ and $z^{\prime}$ from the resulting equation and the two given equations. As a result, we obtain one second-order differential equation with one unknown function $y$:
$$
y^{\prime... | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,398 | |
1201. Find the particular solution of the system of differential equations
$$
\frac{d x}{d t}+2 x+y=\sin t ; \quad \frac{d y}{d t}-4 x-2 y=\cos t
$$
satisfying the initial conditions: $x(\pi)=1, y(\pi)=2$. | Solution. First, we find the general solution of the given system. Differentiating the first equation with respect to $t$: $x^{\prime \prime}+2 x^{\prime}+y^{\prime}=\cos t$ and, substituting in the result the derivative $y^{\prime}$ with its expression through $t$ and $x^{\prime}$, determined from the given system, we... | 1-2(-\pi)+2\sin;\quad4(-\pi)-2\cos-3\sin | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,399 |
1202. Solve the system $y^{\prime \prime} - z=0 ; z^{\prime} + 8 y=0$. | Solution. Differentiating the first equation, we determine $z'$, and substituting it into the second equation, we get $y''' + 8y = 0$. From this, we find (§ 7)
$$
y = C_1 e^{-2x} + e^x \left(C_2 \cos \sqrt{3} x + C_3 \sin \sqrt{3} x\right)
$$
Differentiating $y$ twice and substituting into the first equation, we find... | C_1e^{-2x}+e^x(C_2\cos\sqrt{3}x+C_3\sin\sqrt{3}x) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,400 |
1208. Find the particular solution $u(x, t)$ of the differential equation $\frac{\partial^{2} u}{\partial t^{2}}-a^{2} \frac{\partial^{2} u}{\partial x^{2}}=0$, satisfying the boundary conditions:
$\begin{array}{ll}\text { 1) } u(0, t)=0, & \text { 2) } u(l, t)=0,\end{array}$
3) $u(x, 0)=\varphi_{1}(x)$
4) $u_{t}(x, 0... | Solution. By the Fourier method, we first look for particular solutions of the given equation in the form of the product of two functions, one of which depends only on $x$, and the other only on $t$:
$$
u(x, t)=X(x) T(t)
$$
Finding the derivatives $u_{x x}^{\prime \prime}=T X_{x x}^{\prime \prime}, u_{t t}^{\prime \p... | u(x,)=\sum_{n=1}^{+\infty}(\alpha_{n}\cos\frac{\pi}{}+\beta_{n}\sin\frac{\pi}{})\sin\frac{n\pix}{} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,401 |
Example 1. Verify that the function $y=x+C$ is the general solution of the differential equation $y^{\prime}=1$ and find the particular solution that satisfies the initial condition $\left.y\right|_{x=0}=0$. Provide a geometric interpretation of the result. | Solution. The function $y=x+C$ satisfies the given equation for any value of the arbitrary constant $C$. Indeed, $y'=(x+C)'=1$.
Let us set an arbitrary initial condition $\left.y\right|_{x=x_{0}}=y_{0}$. Substituting $x=x_{0}$ and $y=y_{0}$ into the equation $y=x+C$, we find that $C=y_{0}-x_{0}$. Substituting this val... | x | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,403 |
Example 2. Verify that the function $y=C e^{x}$ is the general solution of the equation $y^{\prime}-y=0$ and find the particular solution that satisfies the initial condition $\left.y\right|_{x=1}=-1$.
$.
翻译结果如下：
Example 2. Using the isocline method, construct approximate integral curves of the differential equation $y^{\prime}=\sin (x+y)$. | Solution. Assuming $y^{\prime}=k$, where $k=$ const, we obtain the isocline equation $\sin (x+y)=k$, with $-1 \leqslant k \leqslant 1$. When $k=0$, we get $\sin (x+y)=0$, from which
$$
y=-x+\pi n \quad(n=0, \pm 1, \pm 2, \ldots)
$$
The integral curves have horizontal tangents at the points of intersection with these ... | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,406 |
Example 3. Using the method of isoclines, construct the integral curves of the equation $y^{\prime}=y-x^{2}+2 x-2$. | Solution. Let $y^{\prime}=k, k=$ const. Then the equation of isoclines will be
$$
y-x^{2}+2 x-2=k, \quad \text { or } \quad y=x^{2}-2 x+2+k
$$
The isoclines are parabolas with a vertical axis of symmetry $x=1$. Among the isoclines, there are no integral curves. Indeed, substituting into the given equation $y=x^{2}-2 ... | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,407 |
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