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Example 2.38. $I=\int_{-1}^{1} \frac{d x}{\sqrt{1-x^{2}}}$.
| Solution. According to formulas (2.16) and (2.17), we have
$$
\begin{gathered}
I=\lim _{\substack{\varepsilon \rightarrow+0 \\
\alpha \rightarrow+0}}\left[\int_{0}^{1-\varepsilon} \frac{d x}{\sqrt{1-x^{2}}}+\int_{-1+\alpha}^{0} \frac{d x}{\sqrt{1-x^{2}}}\right]= \\
=\lim _{\substack{\varepsilon \rightarrow+0 \\
\alpha... | \pi | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,875 |
Example 2.39. $I=\int_{0}^{1} \frac{d x}{x^{2}-5 x+4}$.
Translating the text above into English, while preserving the original text's line breaks and format, yields the following result:
Example 2.39. $I=\int_{0}^{1} \frac{d x}{x^{2}-5 x+4}$. | Solution. Decompose the integrand into a sum of the simplest rational fractions of the first type, and then use formula (2.17)
$$
\begin{gathered}
I=\int_{0}^{1} \frac{d x}{(x-1)(x-4)}=\frac{1}{3} \int_{0}^{1}\left(\frac{1}{x-4}-\frac{1}{x-1}\right) d x= \\
=\lim _{\varepsilon \rightarrow+0} \frac{1}{3} \int_{0}^{1-\v... | \infty | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,876 |
Example 2.40. $I=\int_{0}^{0.5} \frac{d x}{x \ln ^{2} x}$. | Solution.
$$
\begin{aligned}
& I=\lim _{\varepsilon \rightarrow+0} \int_{\varepsilon}^{0.5} \frac{d(\ln x)}{\ln ^{2} x}=\left.\lim _{\varepsilon \rightarrow+0} \frac{-1}{\ln x}\right|_{\varepsilon} ^{0.5}= \\
& =-\frac{1}{\ln 0.5}+\lim _{\varepsilon \rightarrow+0} \frac{1}{\ln \varepsilon}=\frac{1}{\ln 2} \approx 1.44... | \frac{1}{\ln2}\approx1.443 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,877 |
Example 2.41. $I=\int_{0}^{1} \ln x d x$. | Solution.
$$
\begin{gathered}
I=\left|\begin{array}{l}
u=\ln x, d u=\frac{d x}{x} \\
d v=d x, v=x
\end{array}\right|=\lim _{a \rightarrow+0}\left(\left.x \ln x\right|_{a} ^{1}-\int_{a}^{1} x \frac{d x}{x}\right)= \\
=-\lim _{a \rightarrow+0} \frac{\ln a}{1 / a}-1=[\text{ apply L'Hôpital's rule] }= \\
=\lim _{a \righta... | -1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,878 |
Example 2.42. Compute the integral $I=\int_{0}^{1} \frac{x^{m}-x^{n}}{\ln x} d x$, $(m>0, n>0)$. | Solution. Let $F(m)=\int_{0}^{1} \frac{x^{m}-x^{n}}{\ln x} d x$, then the derivative
$$
\frac{d F(m)}{d m}=\int_{0}^{1} \frac{x^{m} \ln x d x}{\ln x}=\int_{0}^{1} x^{m} d x=\left.\frac{x^{m+1}}{m+1}\right|_{0} ^{1}=\frac{1}{m+1}
$$
Hence, $F(m)=\int_{0}^{m} \frac{d m}{m+1}=\ln |m+1|+C$, but for $m=$ $=n F(n)=0$, so $... | \ln|\frac{+1}{n+1}| | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,879 |
Example 2.43. Investigate the absolute and conditional convergence of the improper integral $\int_{0}^{1} \frac{(-1)^{[1 / x]} d x}{x}$, where $[1 / x]$ denotes the integer part of the number $1 / x$, i.e., the greatest integer not exceeding $1 / x$. | Solution. The integral of the modulus of the integrand
$$
\int_{0}^{1}\left|(-1)^{[1 / x]} \frac{d x}{x}\right|=\int_{0}^{1} \frac{d x}{x}=\left.\lim _{a \rightarrow+0} \ln x\right|_{a} ^{1}=\infty
$$
diverges.
We will prove the conditional convergence of this integral. First, we make the substitution $\frac{1}{t}=x... | proof | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,880 |
Example 2.44. Investigate the conditional and absolute convergence of the improper integral
$$
I=\int_{0}^{2}\left[2 x \sin \left(\frac{\pi}{x^{2}}\right)-\frac{2 \pi}{x} \cos \left(\frac{\pi}{x^{2}}\right)\right] d x
$$ | Solution.
$$
\begin{gathered}
I=\lim _{a \rightarrow+0} \int_{a}^{2}\left(2 x \sin \left(\frac{\pi}{x^{2}}\right)-\frac{2 \pi}{x} \cos \left(\frac{\pi}{x^{2}}\right)\right) d x= \\
=\lim _{a \rightarrow+0} \int_{a}^{2}\left(x^{2} \sin \left(\frac{\pi}{x^{2}}\right)\right)^{\prime} d x= \\
=\lim _{a \rightarrow+0}\left... | 2\sqrt{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,881 |
Example 2.45. $I=\mathrm{V} . \mathrm{p} . \int_{1 / e}^{e} \frac{d x}{x \ln x}$. Calculate the improper integral of the second kind in the sense of the principal value. | Solution. According to formula (2.18)
$$
\begin{gathered}
I=\lim _{a \rightarrow+0}\left(\int_{1 / e}^{1-a} \frac{d(\ln x)}{\ln x}+\int_{1+a}^{e} \frac{d(\ln x)}{\ln x}\right)= \\
=\lim _{a \rightarrow+0}\left(\ln \left|\ln x\left\|_{1 / e}^{1-a}+\ln \mid \ln x\right\|_{1+a}^{e}\right)=\right. \\
=\lim _{a \rightarrow... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,882 |
Example 2.46. Find the average value of the function $u(x)=$ $=1 / \sqrt{x}$ on the half-interval $x \in(0,1]$. | Solution.
$$
M(x)=\lim _{a \rightarrow+0}\left(\int_{a}^{1} \frac{d x / \sqrt{x}}{1-0}\right)=\lim _{a \rightarrow+0} \frac{2 \sqrt{1}-2 \sqrt{a}}{1}=2
$$
## 2.4. GEOMETRIC APPLICATIONS OF DEFINITE INTEGRALS
## Area of a Plane Curve
The area of a plane figure bounded by curves given by their equations in Cartesian ... | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,883 |
Example 2.47. Calculate the area of the figure bounded by the parabolas $y=x^{2}$ and $y=8-x^{2}$ (Fig. 2.2). | Solution. To find the points of intersection of the parabolas, we will solve the system of equations $y=x^{2}$ and $y=8-x^{2}$
$$
\Rightarrow x_{1}=-2, y_{1}=4, x_{2}=2, y_{2}=4
$$
We will use formula (2.20) due to the evenness of the integrand:
$$
\begin{gathered}
S=\int_{-2}^{2}\left[\left(8-x^{2}\right)-x^{2}\rig... | \frac{64}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,884 |
Example 2.48. Calculate the area of the figure bounded by the parabola $y=-x^{2}+3 x-2$ and the coordinate axes. | Solution. Since the figure $y=-x^{2}+3 x-2$ is located in the regions with $y \geqslant 0$ and $y<0$ (Fig. 2.3), the area $S$ should be calculated separately for the part $y \geqslant 0$ and the part $y<0$, and then the absolute values of the obtained integrals should be added:
$$
\begin{gathered}
S=\left|\int_{0}^{1}... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,885 |
Example 2.49. Compute the area of the figure bounded by the curve $y^{2}=x(x-2)^{2}$. | Solution. Due to the symmetry of the figure (Fig. 2.4), we can calculate half of the area and then double the result.
Fig. 2.2.
.
$$
\begin{gathered}
S=\int_{-1}^{1}\left[\left(1-4 y^{2}\right)-\left(-3 y^{2}\right)\right] d y=2 \int_{0}^{1}\left(1-y^{2}\right) d y... | \frac{4}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,887 |
Example 2.51. Calculate the area of the figure bounded by the ellipse $x=4 \cos t, y=3 \sin t$. | Solution. We will use the last equality from (2.22).
$$
\begin{gathered}
S=\frac{1}{2} \int_{0}^{2 \pi}\left[x(t) y^{\prime}(t)-y(t) x^{\prime}(t)\right] d t= \\
=\frac{1}{2} \int_{0}^{2 \pi}[4 \cos t \cdot 3 \cos t-3 \sin t(-4 \sin t)] d t=6 \int_{0}^{2 \pi} d t=12 \pi
\end{gathered}
$$
, y=$ $=2(1-\cos t)$. | Solution. One arch of the cycloid is obtained for \(0 \leqslant t \leqslant 2 \pi\) (Fig. 2.6). The required area is conveniently calculated using the first formula from (2.22).
\[
\begin{aligned}
S= & \left|-\int_{0}^{2 \pi} y(t) x^{\prime}(t) d t\right|=\left|-\int_{0}^{2 \pi} 2(1-\cos t) 2(1-\cos t) d t\right|= \\
... | 12\pi | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,889 |
Example 2.53. Compute the area bounded by the cardioid $\rho=1+\cos \varphi,(0 \leqslant \varphi \leqslant 2 \pi)$. | Solution. The cardioid is depicted in Fig. 2.7. We will use formula (2.23):
$$
\begin{aligned}
S & =0.5 \int_{0}^{2 \pi}(\rho(\varphi))^{2} d \varphi=0.5 \int_{0}^{2 \pi}(1+\cos \varphi)^{2} d \varphi= \\
& =0.5 \int_{0}^{2 \pi}\left(1+2 \cos \varphi+\cos ^{2} \varphi\right) d \varphi= \\
& =0.5 \int_{0}^{2 \pi}(1+2 \... | \frac{3\pi}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,890 |
Example 2.54. Calculate the area of one petal of the curve $\rho=\sin ^{2} \varphi$. | Solution. When $\varphi$ changes from 0 to $\pi$, one petal of the given curve is obtained (the shaded area in Fig. 2.8)
$$
\begin{aligned}
S= & 0.5 \int_{0}^{\pi}(\sin \varphi)^{2} d \varphi=0.125 \int_{0}^{\pi}(1-\cos 2 \varphi)^{2} d \varphi= \\
& =0.125 \int_{0}^{\pi}\left(1-2 \cos 2 \varphi+\cos ^{2} 2 \varphi\ri... | \frac{3\pi}{16} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,891 |
Example 2.55. Calculate the length of the arc of the semicubical parabola $y=x^{3 / 2}$ from the point $x=0$ to the point $x=9$. | Solution. We will use formula (2.24).
$$
\begin{gathered}
L=\int_{0}^{9} \sqrt{1+\frac{9}{4}} x d x=\frac{4}{9} \int_{0}^{9} \sqrt{1+\frac{9}{4} x} d\left(1+\frac{4}{9} x\right)= \\
=\left.\frac{8}{9}\left(1+\frac{9}{4} x\right)^{3 / 2}\right|_{0} ^{9}=\frac{8}{9}\left(\frac{85}{4}-1\right)
\end{gathered}
$$ | \frac{8}{9}(\frac{85}{4}-1) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,892 |
Example 2.56. Calculate the length of the arc of the curve $x=\ln (\cos y)$ between the points $y=0$ and $y=\frac{\pi}{3}$. | Solution. We will use formula (2.25).
\[
\begin{gathered}
L=\int_{0}^{\pi / 3} \sqrt{1+\left(-\frac{\sin y}{\cos y}\right)^{2}} d y=\int_{0}^{\pi / 3} \sqrt{1+\operatorname{tg}^{2} y} d y=\int_{0}^{\pi / 3} \frac{d y}{\cos y}= \\
=\left.\ln \left|\operatorname{tg}\left(\frac{\pi}{4}+\frac{y}{2}\right)\right|\right|_{0... | \ln\operatorname{tg}(\frac{5\pi}{12})-\ln(\frac{\pi}{4}) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,893 |
Example 2.57. Calculate the length of the arc of the curve $x=0.25 y^{2}-$ $-0.5 \ln y$, enclosed between the points $y=1$ and $y=1.5$. | Solution. According to formula (2.25)
$$
\begin{aligned}
& L=\int_{1}^{1.5} \sqrt{1+\left(0.5 y-\frac{0.5}{y}\right)^{2}} d y=\int_{1}^{1.5} \sqrt{\left(0.5 y+\frac{0.5}{y}\right)^{2}} d y= \\
& =\int_{1}^{1.5}\left(0.5 y+\frac{0.5}{y}\right)^{2} d y=\left.(y-0.5 \ln |y|)\right|_{1} ^{1.5}=0.5(1-\ln 1.5) .
\end{aligne... | 0.5(1-\ln1.5) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,894 |
Example 2.58. Compute the length of the arc of the astroid $x=\cos ^{3} t$, $y=\sin ^{3} t, 0 \leqslant t \leqslant 2 \pi$. | Solution. From Fig. 2.9, it is clear that due to symmetry, it is sufficient to compute only a quarter of the astroid for
Fig. 2.9.
$0 \leqslant t \leqslant \pi / 2$. We will use formula (2... | 12 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,895 |
Example 2.59. Calculate the length of the arc of the cardioid $\rho=$ $=2(1+\cos \varphi), 0 \leqslant \varphi \leqslant 2 \pi$ (see Fig. 2.7). | Solution. We will use formula (2.28):
$$
\begin{aligned}
L & =\int_{0}^{2 \pi} \sqrt{(2(1+\cos \varphi))^{2}+(-2 \sin \varphi)^{2}} d \varphi=\int_{0}^{2 \pi} \sqrt{16 \cos ^{2}(\varphi / 2)} d \varphi= \\
& =4 \int_{0}^{2 \pi}|\cos (\varphi / 2)| d \varphi=4 \cdot 2 \int_{0}^{\pi} \cos (\varphi / 2) d \varphi=\left.1... | 16 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,896 |
Example 2.60. Calculate the surface area formed by the rotation of the arc of the circle $x^{2}+(y-a)^{2}=R^{2}$ around the $O Y$ axis over the segment $0<y_{1} \leqslant y \leqslant y_{2}<R$. | Solution. Let's find the derivative of the implicit function
$$
2 x x_{y}^{\prime}+2(y-a)=0
$$
We will use the formula (2.31) $S=2 \pi \int_{y_{1}}^{y_{2}} x \sqrt{1+\left(x_{y}^{\prime}\right)^{2}} d y$ (bring $x$ under the square root)
$$
\begin{gathered}
2 \pi \int_{y_{1}}^{y_{2}} \sqrt{x^{2}+\left(x x_{y}^{\prim... | 2\piR(y_{2}-y_{1}) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,897 |
Example 2.61. Calculate the surface area formed by the rotation of the arc of the circle $x^{2}+y^{2}=16(y>0)$ over the segment $-1 \geqslant x \geqslant 1$ around the $O X$ axis. | Solution. We will use formula (2.30).
$$
\begin{gathered}
S=2 \pi \int_{-1}^{1} \sqrt{16-x^{2}} \sqrt{1+\left(-x / \sqrt{16-x^{2}}\right)^{2}} d x= \\
=2 \pi \int_{-1}^{1} 4 d x=16 \pi
\end{gathered}
$$ | 16\pi | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,898 |
Example 2.63. Calculate the surface area formed by the rotation around the polar axis of the lemniscate $\rho=a \sqrt{\cos 2 \varphi}$ over the interval $0 \leqslant \varphi \leqslant \pi / 4$. | Solution. The general view of the rotating curve is shown in Fig. 2.10. We will use formula (2.34):
$$
\begin{aligned}
& \sqrt{\rho^{2}+\left(\frac{d \rho}{d \varphi}\right)^{2}}=\sqrt{a^{2} \cos 2 \varphi+\left(\frac{a \sin 2 \varphi}{\sqrt{\cos 2 \varphi}}\right)^{2}}=\frac{a}{\sqrt{\cos 2 \varphi}} \\
S= & \left|2 ... | \pi^{2}(2-\sqrt{2}) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,900 |
Example 2.64. Calculate the volume of the body obtained by rotating the part of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ over the segment $0 \leqslant x \leqslant a$ around the $O X$ axis. | Solution. We will use formula (2.35).
$$
\begin{gathered}
V_{O X}=\pi \int_{x_{1}}^{x_{2}} \frac{b^{2}}{a^{2}}\left(a^{2}-x^{2}\right) d x=\pi \frac{b^{2}}{a^{2}} \int_{0}^{a}\left(a^{2}-x^{2}\right) d x= \\
\quad=\left.\pi \frac{b^{2}}{a^{2}}\left(a^{2} x-\frac{x^{3}}{3}\right)\right|_{0} ^{a}=\frac{2 \pi}{3} a b^{2}... | \frac{2\pi}{3}^{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,901 |
Example 2.65. Calculate the volume of the body formed by the rotation around the $O Y$ axis of the curvilinear trapezoid bounded by the hyperbola $x y=2$ and the lines $y_{1}=1, y_{2}=4$ and $y_{3}=0$. | Solution.
$$
V_{O Y}=\pi \int_{y_{1}}^{y_{2}}(x(y))^{2} d y=\pi \int_{0}^{4} \frac{4}{y^{2}} d y=4 \pi[-0.25+1]=3 \pi
$$
## 2.5 . MECHANICAL AND PHYSICAL APPLICATIONS OF DEFINITE INTEGRALS | 3\pi | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,902 |
Example 2.66. The velocity $v$ of a point moving in a straight line changes with time $t$ according to the law $v(t)=$ $=t \sin 2 t$. Determine the path $s$ traveled by the point from the start of the motion to the moment of time $t=\frac{\pi}{4}$ units of time. | Solution. Since the path $d s=v(t) d t$, then
$$
\begin{aligned}
s=\int_{t_{\text {nav }}}^{t_{\text {toon }}} v(t) d t & =\int_{0}^{\pi / 4} t \sin 2 t d t=\left|\begin{array}{l}
u=t, d u=d t, d v=\sin 2 t d t \\
v=\int \sin 2 t \frac{d(2 t)}{2}=-0.5 \cos 2 t
\end{array}\right|= \\
& =-0.5 t \cos 2 t\left|_{0}^{\pi /... | 0.25 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,903 |
Example 2.67. Calculate the mass of the plane curve $y=\frac{x^{3}}{3}$ over the segment from $x=0$ to $x=0.1$ units of length, if the linear density $\rho$ depends on the coordinate $x$ as follows: $\rho(x)=1+x^{2}$. | Solution. The elementary mass of a curve segment from $x=$ $=0$ to $x=0.1$ is determined by the formula $d m=\rho(x) d l=$ $=\rho(x) \sqrt{1+\left(\frac{d y}{d x}\right)^{2}} d x$, hence the total mass
$$
\begin{gathered}
m=\int_{0}^{0.1} \rho(x) \sqrt{1+\left(\frac{d y}{d x}\right)^{2}} d x=\int_{0}^{0.1}\left(1+x^{2... | 0.099985655 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,904 |
Example 2.69. Find the center of gravity of the arc of the circle $x^{2}+$ $+y^{2}=R^{2}$, located in the first quadrant, if the density $\rho=\rho_{0} x y$. | Solution. Differentiating the equation of the circle as an implicit function, we find
$$
\begin{gathered}
2 x+2 y y^{\prime}=0 \Rightarrow y^{\prime}=-\frac{x}{y}, \Rightarrow \\
\Rightarrow d l=\sqrt{1+\left(y^{\prime}\right)^{2}} d x=\sqrt{1+\left(-\frac{x}{y}\right)^{2}} d x= \\
=\frac{R}{y} d x=\frac{R}{\sqrt{R^{2... | x_{}=y_{}=\frac{2R}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,905 |
Example 2.70. Calculate the coordinates of the center of gravity of a plane figure located in the first quadrant and bounded by the lines $y=a x^{3}$ and $y=a$, if the density is constant, $\rho=1$. | Solution. It is easy to obtain the coordinates of the intersection point of the lines $y=a x^{3}$ and $y=a: x=1, y=a$. According to formulas (2.46), (2.47), we get:
$$
\begin{gathered}
M_{x}=0.5 \int_{0}^{1}\left(a^{2}-a^{2} x^{6}\right) d x=0.5\left(a^{2} x-\frac{a^{2} x^{7}}{7}\right)\bigg|_{0} ^{1}=\frac{3 a^{2}}{7... | x_{}=0.4,y_{}=\frac{4}{7} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,906 |
Example 2.71. Find the moment of inertia of the arc of the parabola $y=x^{2}(0 \leqslant x \leqslant 1)$ with respect to the origin, if the linear density $\rho=\rho_{0} \sqrt{1+4 x}$. | Solution. We will use formula (2.48):
$$
\begin{aligned}
I_{0}= & \int_{0}^{1} \rho_{0} \sqrt{1+4 x}\left(x^{2}-\left(x^{2}\right)^{2}\right) \sqrt{1+(2 x)^{2}} d x= \\
& =\rho_{0} \int_{0}^{1}\left(x^{2}+4 x^{3}+x^{4}+4 x^{5}\right) d x= \\
= & \left.\rho_{0}\left(\frac{x^{3}}{3}+4 x^{3}+\frac{x^{5}}{5}+\frac{2 x^{6}... | 2.2\rho_{0} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,907 |
Example 2.72. Calculate the moment of inertia of a plane figure bounded by the lines $y^{2}=4 a^{2} x$ and the line $x=a^{2}$ with respect to the $O Y$ axis. The surface density of the material of the figure is $\rho=\rho_{0}+k x$. | Solution. Divide the interval $\left[0, a^{2}\right]$ into elementary intervals of length $d x:[x, x+d x]$. Highlight the elementary strips cut from the figure by lines passing through the ends of each interval $[x, x+d x]$ and parallel to the $O Y$ axis (Fig. 2.12). The elementary strip can be approximately considered... | 8^{9/2}(\frac{\rho_{0}}{7}+\frac{k}{9}) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,908 |
Example 2.74. Calculate the work done by a body of mass $m$ when falling to the surface of the Earth from a height $h$. | Solution. The force $F$ of attraction of a mass $m$, located at a distance $x$ from the Earth's surface is $F=\frac{\gamma m M}{(R+x)^{2}}$, where $\gamma$ is the gravitational constant, $R$ is the radius of the Earth. The elementary work $d A=F(x) d x$ (the $O X$ axis is directed to the center of the Earth along the r... | \frac{}{R+} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,910 |
Example 2.75. Determine the work that needs to be expended to compress an elastic spring by $\Delta h$ units of length. Perform the calculations assuming Hooke's law $F=-k x$ is valid. | Solution.
$$
A=\int_{0}^{\Delta h}(-k x) d x=-\left.k \frac{x^{2}}{2}\right|_{0} ^{\Delta h}=-\frac{k \Delta h^{2}}{2}
$$ | -\frac{k\Delta^{2}}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,911 |
Example 2.76. Calculate the work required to pump out a liquid of density $\rho$ from a fully filled tank of height $H$, which has the shape of a paraboloid of revolution defined by the equation $z=$ $=a^{2}\left(x^{2}+y^{2}\right)$. | Solution. Let us mentally slice the paraboloid with planes parallel to the XY plane (Fig. 2.14). Suppose at a depth of $H-z$, the thickness of the layer between two adjacent parallel planes is $d z$, then the volume element of the liquid between the specified planes is calculated by the formula $d V=\pi x^{2} d z$.
=2\left(b+\frac{(a-b)(h-x)}{h}\right) d x$. According to Pascal's law, we obtain that the elementary force of pressure
$$
d F=\rho g x d S(x)=2 \rho g x\left(b+\frac{(a-b)... | \rho\frac{^2}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,918 |
Example 2.83. The conductor has the shape of a truncated cone. The diameter of its arbitrary cross-section, located at a distance $x$ from the smaller cross-section, is $y=a+\frac{x(b-a)}{h}$, where $a$ is the diameter of the smaller, and $b$ is the diameter of the larger cross-section, $h$ is the height of the cone. K... | \rho\frac{}{\pi}(\frac{1}{}-\frac{1}{b}) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,919 | |
Example 3.2. Find the domain of the function $z=$ $=\sqrt{8-|x-8|-|y|}$. | Solution. The domain of the given function is the set of points $P(x, y) \in E^{2}$, satisfying the condition $8-|x-8|-|y| \geqslant 0$. From this, it follows that $|y| \leqslant 8-|x-8|$.
This inequality is equivalent to the double inequality $-8+|x-8| \leqslant y \leqslant 8-|x-8|$. Solving it:
\[
\begin{aligned}
&... | D(f)={(x,y):y\leqslantx\leqslant16-y,0\leqslanty\leqslant8}\cup{(x,y):-y\leqslantx\leqslanty+16,-8\leqslanty\leqslant0} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,920 |
Example 3.3. Find the level surfaces of the function $y=$ $=\sqrt{36-x_{1}^{2}-x_{2}^{2}-x_{3}^{2}}$ and the value of the function at the point $P(1,1,3)$. | Solution. According to the definition of level surfaces, we have: $\sqrt{36-x_{1}^{2}-x_{2}^{2}-x_{3}^{2}}=C$, where $C \geqslant 0$. From this, it follows that $36-x_{1}^{2}-x_{2}^{2}-x_{3}^{2}=C^{2}$, that is, $x_{1}^{2}+x_{2}^{2}+x_{3}^{2}=$ $=36-C^{2}$ (obviously, $0 \leqslant C \leqslant 6$). The obtained equation... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,921 |
Example 3.7. Is the function $z=$ $=f(x, y)=\sqrt{4-x^{2}-y^{2}}$ bounded above (below)? | Solution. The function $f$ is bounded from above and below, as $0 \leqslant f(x, y) \leqslant 2$. Moreover, the function has the maximum and minimum values on $D(f)$: $\max _{(x, y) \in D} f=2, \min _{(x, y) \in D} f=0$
## 3.2 . LIMIT AND CONTINUITY
The definition of a limit is generalized for the case of functions ... | 0\leqslantf(x,y)\leqslant2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,925 |
Example 3.9. Find the limit of the function $f(x, y)=\left(x^{2}+\right.$ $\left.+y^{2}\right)^{2} x^{2} y^{2}$ as $x \rightarrow 0$ and $y \rightarrow 0$. | Solution. For the calculation of the specified limit, it is more convenient to switch to polar coordinates $x=r \cos \varphi, y=r \sin \varphi$. We obtain
$$
\begin{gathered}
\lim _{\substack{x \rightarrow 0 \\
y \rightarrow 0}}\left(x^{2}+y^{2}\right)^{2 x^{2} y^{2}}=\lim _{r \rightarrow 0}\left(r^{2}\right)^{2 r^{4}... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,927 |
Example 3.10. Investigate the continuity of the functions:
$$
\begin{aligned}
\text { 1) } f(x, y)= & \frac{2 x^{2} y^{2}}{x^{4}+y^{4}} ; \quad \text { 2) } f(x, y) = \frac{\sin \left(x^{2}+y^{2}\right)}{x^{2}+y^{2}} \\
& \text { 3) } f(x, y)=\sin \left(\frac{3}{x y}\right)
\end{aligned}
$$ | Solution. 1) The function $f(x, y)=\frac{2 x^{2} y^{2}}{x^{4}+y^{4}}$ is not defined at the point $O(0,0)$ and does not have a limit at this point (see Example 3.8).
In all other points of the plane $X O Y$, the function is continuous, as it represents the ratio of continuous functions.
2) The function $f(x, y)$ is c... | proof | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,928 |
Example 3.11. Investigate the continuity of the function
$$
f(x, y, z)=\left\{\begin{array}{lr}
x^{4}+\frac{2 x y z}{y^{2}+z^{2}}, & y^{2}+z^{2} \neq 0 \\
x^{4}, & y^{2}+z^{2}=0
\end{array}\right.
$$ | Solution. If $y^{2}+z^{2} \neq 0$, then the function is continuous due to the fact that $\lim _{P(x, y, z) \rightarrow P\left(x_{0}, y_{0}, z_{0}\right)} f(x, y, z)=f\left(x_{0}, y_{0}, z_{0}\right)$.
For points where $y^{2}+z^{2}=0$, consider two cases:
1) Point $P_{1}\left(x_{0}, 0,0\right)$, where $x_{0} \neq 0$. ... | proof | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,929 |
Example 3.12. Investigate the continuity of the function
$$
f(x, y)= \begin{cases}-\frac{x y^{2}}{y^{2}+y^{4}}, & \text { when } x^{2}+y^{4} \neq 0 \\ 0, & \text { when } x^{2}+y^{4}=0\end{cases}
$$
at the point $O(0,0)$. | Solution. The given function is continuous with respect to each of the variables $x$ and $y$ at the point $O(0,0)$, since
$$
\begin{gathered}
\lim _{\Delta x \rightarrow 0} \Delta_{x} f(0,0)=\lim _{\Delta x \rightarrow 0}(f(\Delta x, 0)-f(0,0))= \\
=\left.\lim _{\Delta x \rightarrow 0}\left(\frac{\Delta x y^{2}}{\Delt... | proof | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,930 |
Example 3.13. Find the partial derivatives of the functions:
$$
\text { 1) } z=x^{2} y^{3} \text {; 2) } z=x^{5}+y^{6} \text {; 3) } z=x^{y}+y^{x} \text {; 4) } z=\log _{y} x
$$ | Solution. 1) The given function is a product of power functions of $x$ and $y$. Therefore, when computing the partial derivative with respect to one of the variables, the other variable is fixed and can be factored out of the derivative according to the rules of differentiation. The derivative is found using the rules ... | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,931 |
Example 3.14. Find the partial derivatives of the function
$$
u=\left(x^{2}+y^{2}+z^{2}\right)^{\sin (x y z)}
$$ | Solution. In this case, both the base and the exponent depend on independent variables. One way to find partial derivatives in this case is to logarithmize both sides of the equation defining the function (i.e., the same technique used when finding the derivative of a single-variable function in a similar situation). W... | \begin{aligned}u_{x}^{\}&=(x^{2}+y^{2}+z^{2})^{\sin(xyz)}\times(yz\cos(xyz)\ln(x^{2}+y^{2}+z^{2})+\frac{2x\sin(xyz)}{x^{2}+y^{2}+z^{} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,932 |
Example 3.15. Find the partial derivatives of the function
$$
f(x, y, z)=\left\{\begin{array}{lr}
x^{4}+\frac{2 x y z}{y^{2}+z^{2}}, & y^{2}+z^{2} \neq 0 \\
x^{4}, & y^{2}+z^{2} = 0
\end{array}\right.
$$
at the point $A(1,0,0)$. | Solution. Consider the function $f(x, 0,0)=x^{4}$. Let's find the derivative of this function: $f_{x}^{\prime}(x, 0,0)=4 x^{3}$ and $f_{x}^{\prime}(1,0,0)=4$. Next, consider the functions $f(1, y, 0)=1$ and $f(1,0, z)=1$. We have: $f_{x}^{\prime}(1, y, 0)=0, f_{z}^{\prime}(1,0, z)=0$ and, in particular, $f_{y}^{\prime}... | f_{x}^{\}(1,0,0)=4,f_{y}^{\}(1,0,0)=0,f_{z}^{\}(1,0,0)=0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,933 |
Example 3.16. Find $u_{x x}^{\prime \prime}, u_{y y}^{\prime \prime}, u_{x y}^{\prime \prime}, u_{y x}^{\prime \prime}, u_{x y z}^{\prime \prime \prime}, u_{x y^{2} z}^{I V}$, if $u=e^{x y z}$. | Solution. According to the definition of higher-order partial derivatives (3.2), it is necessary to first find the corresponding partial derivatives of a lower order. In this case, it is necessary to find the first-order partial derivatives
$$
u_{x}^{\prime}, u_{y}^{\prime}, u_{z}^{\prime} \Rightarrow u_{x}^{\prime}=e... | u_{xy^{2}z}^{IV}=e^{xyz}(x^{3}y^{2}z^{3}+5x^{2}yz^{2}+4xz) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,934 |
Example 3.17. Does $f_{x y}^{\prime \prime}(0,0)$ exist if
$$
f(x, y, z)= \begin{cases}\frac{x y^{2}}{x^{2}+y^{4}}, & \text { when } x^{2}+y^{4} \neq 0 \\ 0, & \text { when } x^{2}+y^{4}=0\end{cases}
$$
(see example 3.12). | Solution. The given second-order partial derivative represents the partial derivative with respect to the variable $y$ of the first-order partial derivative with respect to the variable $x$. Therefore, we first need to compute $f_{x}^{\prime}$. We have
$$
f_{x}^{\prime}=\frac{y^{2}\left(x^{2}+y^{4}\right)-x y^{2} 2 x}... | proof | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,935 |
Example 3.19. Find $d^{2} f$ :
$$
\text { 1) } f(x, y)=x^{6} y^{8}, \text { 2) } f(x, y)=x^{6}+y^{8}
$$ | Solution. 1) To compute using formula (3.5), we need to find the necessary second-order partial derivatives. We have:
$$
\begin{gathered}
\frac{\partial f}{\partial x}=6 x^{5} y^{8} ; \frac{\partial f}{\partial y}=8 x^{6} y^{7} ; \frac{\partial^{2} f}{\partial x^{2}}=\left(6 x^{5} y^{8}\right)_{x}^{\prime}=30 x^{4} y^... | ^{2}f=^{2}(x^{6}y^{8})=30x^{4}y^{8}^{2}+96x^{5}y^{7}y+56x^{6}y^{6}^{2},\quad^{2}f=^{2}(x^{6}+y^{8})=30x^{4} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,937 |
Example 3.21. Show that the function $f(x, y)=\frac{x}{x^{2}-a y^{2}}$ satisfies the equation $\frac{\partial^{2} f}{\partial x^{2}}=a \frac{\partial^{2} f}{\partial y^{2}}$. | Solution. Let's find the necessary partial derivatives:
$$
\begin{aligned}
& \frac{\partial f}{\partial x}=\frac{x^{2}-a y^{2}-2 x^{2}}{\left(x^{2}-a y^{2}\right)^{2}}=-\frac{x^{2}+a y^{2}}{\left(x^{2}-a y^{2}\right)^{2}} ; \frac{\partial f}{\partial y}=\frac{2 a x}{\left(x^{2}-a y^{2}\right)^{2}} \\
& \frac{\partial^... | proof | Calculus | proof | Yes | Yes | olympiads | false | 30,938 |
Example 3.23. Calculate $\Delta f\left(P_{0}\right), d f\left(P_{0}\right)$, if $f(x, y)=$ $=x^{2} y, x_{0}=5, y_{0}=4, \Delta x=0.1, \Delta y=-0.2$. | Solution. By the definition of the total increment, we have
$$
\begin{gathered}
\Delta f\left(x_{0}, y_{0}\right)=f\left(x_{0}+\Delta x, y_{0}+\Delta y\right)-f\left(x_{0}, y_{0}\right)= \\
=\left(x_{0}+\Delta x\right)^{2}\left(y_{0}+\Delta y\right)-x_{0}^{2} y_{0}= \\
=\left(x_{0}^{2}+2 x_{0} \Delta x+\Delta x^{2}\ri... | -0.802 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,940 |
Example 3.24. Replacing the increment of the function with its differential, approximately calculate $a=\frac{2.04^{2}}{\sqrt[3]{0.97 \sqrt[5]{1.02^{2}}}}$. | Solution. For the approximate calculation of this number, consider the function of three independent variables $f(x, y, z)=\frac{x^{2}}{\sqrt[3]{y \sqrt[5]{z^{2}}}}$. Its value at $x=2.04, y=0.97$, $z=1.02$ is the desired number $a$. To find $a$, note that at the point $P_{0}(2,1,1)$, that is, when $x_{0}=2, y_{0}=1$, ... | 4.16 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,941 |
Example 3.25. Verify that the expression
$$
\left(3 x^{2} y+4 y^{4}-5\right) d x+\left(x^{3}+8 x y\right) d y
$$
is a total differential and find the function given the differential. | Solution. 1) We have $P(x, y)=3 x^{2} y+4 y^{4}-5, Q(x, y)=$ $=x^{3}+8 x y$ - continuous and continuously differentiable in $E^{2}$.
2) Check the equality $\frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x}: \frac{\partial P}{\partial y}=$ $=3 x^{2}+8 y ; \frac{\partial Q}{\partial x}=3 x^{2}+8 y \Rightarrow ... | f(x,y)=x^{3}y+4xy^{2}-5x+C,C=\text{const.} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,942 |
Example 3.26. Verify that the expression
$$
y\left(e^{x y}+6\right) d x+x\left(e^{x y}+6\right) d y
$$
is a total differential of some function $f(x, y)$, and find this function. | Solution.
$$
\begin{gathered}
\text { 1) } P(x, y)=y\left(e^{x y}+6\right) ; \frac{\partial P}{\partial y}=\left(1+x y e^{x y}\right)+6 \\
Q(x, y)=x\left(e^{x y}+6\right)
\end{gathered} \begin{aligned}
\frac{\partial Q}{\partial x}=6+\left(1+x y e^{x y}\right) ; \frac{\partial P}{\partial y}=\frac{\partial Q}{\partial... | f(x,y)=e^{xy}+6xy+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,943 |
Example 3.27. Find $\frac{d z}{d t}$, if $z=\arcsin \left(x^{2}+y^{2}+t^{2}\right)$, where $x=2 t, y=4 t^{2}$. | Solution. The required derivative is found using formula (3.13). We find the necessary partial derivatives
$$
\begin{gathered}
\frac{\partial z}{\partial x}=\frac{2 x}{\sqrt{1-\left(x^{2}+y^{2}+t^{2}\right)^{2}}} ; \frac{\partial z}{\partial y}=\frac{2 y}{\sqrt{1-\left(x^{2}+y^{2}+t^{2}\right)^{2}}} \\
\frac{\partial ... | \frac{2(1+4+32^{2})}{\sqrt{1-(x^{2}+y^{2}+^{2})^{2}}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,944 |
Example 3.29. Find $\frac{\partial^{2} z}{\partial y \partial x}$ if $Z=f\left(x^{2}+y^{2}, x^{2} y^{2}\right)$. | Solution. The desired partial derivative will be found as the partial derivative with respect to the variable of the partial derivative $\frac{\partial z}{\partial x}$, which has already been found in problem $3.28(2)$. We have
$$
\begin{gathered}
\frac{\partial^{2} z}{\partial y \partial x}=\frac{\partial}{\partial y... | \frac{\partial^{2}f}{\partialu^{2}}4x^{2}+8x^{2}y^{2}\frac{\partial^{2}f}{\partialu\partialv}+\frac{\partial^{2}f}{\partialv^{2}}4x^{2}y^{4}+\frac{\partialf}{\partialv}4xy | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,946 |
Example 3.31. Find the derivatives $y^{\prime}$ and $y^{\prime \prime}$ of the function given implicitly by the equation: $x^{y}-y^{x}=0$. | Solution. The derivative of an implicit function of one variable will be found using formula (3.19). In the considered case, $F(x, y)=x^{y}-y^{x}$. We find the partial derivatives $F_{x}^{\prime}$ and $F_{y}^{\prime}$ according to formula (3.19). We have:
$$
F_{x}^{\prime}=y x^{y-1}-y^{x} \ln y ; F_{y}^{\prime}=x^{y} ... | y^{\}=\frac{y^{2}}{x^{2}}\frac{\lnx-1}{\lny-1},\quady^{\\}=\frac{x(3-2\lnx)(\lny-1)^{2}+y(\lnx-1)^{2}(2\lny-3)}{x^{4}(\lny-1)^{3}}y^{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,948 |
Example 3.32. Find the partial derivatives $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$ of the function $z=f(x, y)$, given implicitly by the equation $e^{z^{2}}-$ $-x^{2} y^{2} z^{2}=0$ | Solution. The partial derivatives $z_{x}^{\prime}$ and $z_{y}^{\prime}$ are found in this case using formulas (3.20). We need to find the necessary partial derivatives of the function $F(x, y, z)$, which is defined, obviously, by the expression
$$
F(x, y, z)=e^{z^{2}}-x^{2} y^{2} z^{2}
$$
## We have
$$
F_{x}^{\prime... | \begin{aligned}\frac{\partialz}{\partialx}&=\frac{z}{x(z^{2}-1)}\\\frac{\partialz}{\partialy}&=\frac{z}{y(z^{2}-1)}\end{aligned} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,949 |
Example 3.33. Find $\frac{\partial^{2} z}{\partial x^{2}}$ and $\frac{\partial^{2} z}{\partial y \partial x}$ of the implicit function from problem 3.32. | Solution. By the definition of the second-order partial derivative, we obtain
$$
\begin{aligned}
& \frac{\partial^{2} z}{\partial x^{2}}=\frac{\partial}{\partial x}\left(\frac{z}{x\left(z^{2}-1\right)}\right)=\frac{z_{x}^{\prime} x\left(z^{2}-1\right)-z\left(z^{2}-1+2 x z z_{x}^{\prime}\right)}{x^{2}\left(z^{2}-1\righ... | \frac{\partial^{2}z}{\partialx^{2}}=z\frac{z^{4}-z^{2}+2}{x^{2}(z^{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,950 |
Example 3.34. Functions $u$ and $v$ of independent variables $x$ and $y$ are given by the system of equations
$$
\left\{\begin{array}{l}
x u - y v - 1 = 0 \\
x - y + u - v = 0
\end{array}\right.
$$
Find $d u, d v, u_{x}^{\prime}, u_{y}^{\prime}, v_{x}^{\prime}, v_{y}^{\prime}, d^{2} u$. | Solution. According to the condition,
$$
F(x, y, u, v)=x u-y v-1=0, G(x, y, u, v)=x-y+u-v=0
$$
Differentiating, we get $\left\{\begin{array}{l}d \hat{F}=u d x-v d y+x d u-y d v=0 \\ d G=d x-d y+d u-d v=0\end{array}\right.$
The Jacobian $J(u, v)=\left|\begin{array}{ll}x & -y \\ 1 & -1\end{array}\right|=-x+y$ is non-z... | 2\frac{u-y}{(y-x)^{2}}^{2}+2\frac{(x+y-v-u)}{(y-x)^{2}}y+2\frac{v-x}{(y-x)^{2}}^{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,951 |
Example 3.35. Find the derivative of the function $u=x^{3} y^{3} z^{3}$ at the point $M(1,1,1)$ in the direction that forms angles of $60^{\circ}, 45^{\circ}, 60^{\circ}$ with the coordinate axes. | Solution. According to (3.26) and (3.29), it is necessary to find the partial derivatives of the function $u$ at point $M$. We have
$$
\frac{\partial u}{\partial x}=3 x^{2} y^{3} z^{3}; \frac{\partial u}{\partial y}=3 x^{3} y^{2} z^{3} ; \frac{\partial u}{\partial z}=3 x^{3} y^{3} z^{2}
$$
At point $M$, the values of... | \frac{3}{2}(2+\sqrt{2}) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,952 |
Example 3.36. Find the directional derivative of the function $u=x^{2}+y^{3}+$ $+z^{3}$ at the point $A(1,2,3)$ in the direction leading from this point to the point $B(13,5,7)$. | Solution. Unlike the previous problem, the direction cosines of the unit vector of the given direction are not specified here. They can be easily found if the coordinates of the vector are known: $\overrightarrow{A B}: \overrightarrow{A B}=12 \vec{i}+3 \vec{j}+4 \vec{k}$, then
$$
\begin{gathered}
\cos \alpha=\frac{12}... | \frac{180}{13} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,953 |
Example 3.37. Find the derivative of the function $u=\frac{1}{r^{2}}$, where $r^{2}=x^{2}+y^{2}+z^{2}$, in the direction of its fastest growth. | Solution. In this case, it is necessary to find the directional derivative along the gradient. This derivative, according to (3.27), is equal to $\frac{\partial u}{\partial \ell}=|\operatorname{grad} u|$ since $\cos (\angle \operatorname{grad} f, \vec{\ell})=1$. We determine the coordinates of the vector $\operatorname... | \frac{4}{r^{2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,954 |
Example 3.38. Write the equation of the tangent plane and the normal line to the surface $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+\frac{z^{2}}{c^{2}}=1$ at the point $M\left(\frac{a}{\sqrt{2}} \cdot \frac{b}{2}, \frac{c}{2}\right)$. | Solution. The surface is given implicitly, so the corresponding equation of the tangent plane has the form (3.30). We find the partial derivatives involved in this equation, taking into account that in this case $F(x, y, z)=\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+\frac{z^{2}}{c^{2}}-1$. We have: $F_{x}^{\prime}=\frac{2... | \frac{}{\sqrt{2}}(x-\frac{}{\sqrt{2}})=b(y-\frac{b}{2})=(z-\frac{}{2}) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,955 |
Example 3.39. Write the equation of the tangent plane and the normal line to the surface $z=2 x^{2}+3 y^{2}$ at the point $M(1,2,14)$. | Solution. The required equations are found using formulas (3.32) and (3.33):
$$
f(x, y)=2 x^{2}+3 y^{2}, f_{x}^{\prime}(M)=4 x=4, f_{y}^{\prime}(M)=6 y=12
$$
The equation of the tangent plane is
$$
z-14=4(x-1)+12(y-2) \Rightarrow 4 x+12 y-z-14=0
$$
The equation of the normal line $\frac{x-1}{4}=\frac{y-2}{12}=\frac... | 4x+12y-z-14=0,\frac{x-1}{4}=\frac{y-2}{12}=\frac{z}{-1} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,956 |
Example 3.40. Find the equation of the tangent planes to the surface $x^{2}+4 y^{2}+9 z^{2}=1$, parallel to the plane $x+y+2 z=1$. | Solution. Since the tangent plane must be parallel to the plane $x+y+2z=1$, the normal to the tangent plane must be collinear with the normal of the given plane. This means that, according to the condition of collinearity, the following condition holds:
$$
\frac{F_{x}^{\prime}(M)}{1}=\frac{F_{y}^{\prime}(M)}{1}=\frac{... | x+y+2z\\frac{109}{6\sqrt{61}}=0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,957 |
Example 3.41. Find the stationary points of the following functions:
1) $f(x, y)=(x-3)^{2}+(y-2)^{2}$
2) $f(x, y, z)=x^{2}+4 y^{2}+9 z^{2}-4 x+16 y+18 z+1$. | Solution. 1) Find the partial derivatives $f_{x}^{\prime}=2(x-3)$; $f_{y}^{\prime}=2(y-2)$. Setting the partial derivatives to zero, we get the system of equations: $x-3=0$ and $y-2=0$. From this, it follows that $x=3, y=2$. Thus, the stationary point is $(3,2)$. The partial derivatives are
$$
f_{x}^{\prime}=2 x-4, f_... | (3,2)(2,2,-1) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,958 |
Example 3.42. Investigate the function for extremum
$$
z=x^{2}+2 y^{2}-2 x y-x-2 y
$$ | Solution. To find the stationary points of the given function, we solve the following system of equations:
$$
\left\{\begin{array}{l}
f_{x}^{\prime}=2 x-2 y-1=0 \\
f_{y}^{\prime}=4 y-2 x-2=0
\end{array}\right.
$$
From which it follows that $x=2, y=\frac{3}{2}$. We will investigate the sufficient conditions for an ext... | (2,\frac{3}{2}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,959 |
Example 3.43. Investigate the function for extremum
$$
u=x^{2}+2 y^{2}+4 z^{2}+4 x+2 y-8
$$ | Solution. We find the stationary points of the function from the system of equations
$$
\begin{aligned}
& u_{x}^{\prime}=2 x+4=0 \\
& u_{y}^{\prime}=4 y+2=0 \\
& u_{z}^{\prime}=8 z-8=0
\end{aligned}
$$
From which $x=-2, y=-\frac{1}{2}, z=1$. We find the second-order partial derivatives:
$$
\begin{gathered}
u_{x x}^{... | (-2,-\frac{1}{2},1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,960 |
Example 3.44. Investigate the function for local extremum
$$
f\left(x_{1}, x_{2}\right)=3 x_{1}^{2} x_{2}-x_{1}^{3}-\frac{4}{3} x_{2}^{3}
$$ | Solution. We find the derivatives $f_{x_{1}}^{\prime}$ and $f_{x_{2}}^{\prime}$; solving the obtained system
$$
\left\{\begin{array}{l}
6 x_{1} x_{2}-3 x_{1}^{2}=0 \\
3 x_{1}^{2}-4 x_{2}^{2}=0
\end{array}\right.
$$
we find two points $P_{0}(6,3)$ and $P_{1}(0,0)$.
To establish the nature of the found stationary poin... | f_{\text{max}}=72 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,961 |
Example 3.45. Investigate the function for local extremum
$$
f\left(x_{1}, x_{2}\right)=1+x_{1}^{4} \sqrt[3]{x_{2}}
$$ | Solution. Let's find the derivatives $f_{x_{1}}^{\prime}=4 x_{1}^{3} \sqrt[3]{x_{2}}$ and $f_{x_{2}}^{\prime}=$ $=\frac{1+x_{1}^{4}}{3 \sqrt[3]{x_{2}^{2}}}$. These derivatives do not simultaneously become zero at any point. For all $x_{2}=0$, $f_{x_{2}}^{\prime}$ does not exist, while the function $f\left(x_{1}, x_{2}\... | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,962 |
Example 3.46. Investigate the function $f\left(x_{1}, x_{2}\right)=\frac{x_{1} x_{2}}{1+x_{1}^{2} x_{2}^{2}}$ for extrema. | Solution. We have
$$
\begin{aligned}
& f_{x_{1}}^{\prime}=\frac{x_{2}\left(1-x_{1}^{2} x_{2}^{2}\right)}{\left(1+x_{1}^{2} x_{2}^{2}\right)^{2}}=0, f_{x_{2}}^{\prime}=\frac{x_{1}\left(1-x_{1}^{2} x_{2}^{2}\right)}{\left(1+x_{1}^{2} x_{2}^{2}\right)^{2}}=0 \Rightarrow \\
\Rightarrow & P_{0}(0,0), P_{1}\left(x_{1}, x_{2... | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,963 |
Example 3.48. Find the extremum of the function
\[
f(x, y)=x^{2}+y^{2}-2 x-y
\]
subject to the condition that the variables \(x\) and \(y\) are related by the equation \(\varphi(x, y)=x+y-1=0\). | Solution. For the Lagrange function
$$
L(x, y, \lambda)=x^{2}+y^{2}-2 x-y+\lambda(x+y-1)
$$
considering (3.44), we have a system of three equations:
$$
\left\{\begin{array}{l}
L_{x}^{\prime}=2 x-2+\lambda=0 \\
L_{y}^{\prime}=2 y-1+\lambda=0 \\
x+y-1=0
\end{array}\right.
$$
The solution is \( x_{0}=0.75, y_{0}=0.25,... | -1.125 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,965 |
Example 3.49. Investigate the function $f\left(x_{1}, x_{2}\right)=x_{2}^{2}-x_{1}^{2}$ for the presence of a local extremum subject to the condition $x_{1}-2 x_{2}+3=0$. | Solution. Let's form the Lagrange function
$$
L(x, y, \lambda)=x_{2}^{2}+x_{1}^{2}+\lambda\left(x_{1}-2 x_{2}+3\right)
$$
We will find the points where the first-order partial derivatives of this function are zero. We have
$$
\left\{\begin{array}{l}
\frac{\partial L}{\partial x_{1}}=-2 x_{1}+\lambda=0 \\
\frac{\part... | x_{1}=1,x_{2}=2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,966 |
Example 3.50. Find the conditional extremum of the function $f\left(x_{1}, x_{2}\right)=x_{1}^{2}+x_{2}^{2}-x_{1} x-2+x_{1}+x_{2}-4$ subject to the constraint $x_{1}+x_{2}+3=0$. | Solution. Let's form the Lagrange function
$$
L\left(x, x_{2}, \lambda\right)=f\left(x_{1}, x_{2}\right)+\lambda\left(x_{1}+x_{2}+3\right).
$$
We need to solve the system of equations
$$
\left\{\begin{array}{l}
\frac{\partial L}{\partial x_{1}}=2 x_{1}-x_{2}+1+\lambda=0 \\
\frac{\partial L}{\partial x_{2}}=2 x_{2}-x... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,967 |
Example 4.1. Investigate the convergence of the series:
$$
\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\ldots+\frac{1}{4 n^{2}-1}+\ldots
$$ | Solution. Let's transform the general term of this series:
$$
a_{n}=\frac{1}{4 n^{2}-1}=\frac{1}{(2 n-1)(2 n+1)}=\frac{1}{2}\left(\frac{1}{2 n-1}-\frac{1}{2 n+1}\right)
$$
After this, it is not difficult to calculate the partial sums of the series:
$$
S_{n}=\sum_{p=1}^{n} a_{p}=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{... | \frac{1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,968 |
Example 4.2. Prove the divergence of the harmonic series
$$
\sum_{p=1}^{\infty} \frac{1}{n}=1+\frac{1}{2}+\frac{1}{3}+\ldots
$$ | Solution. Let us set in the Cauchy criterion $p=n$, then
$$
S_{n+p}-S_{n}=S_{2 n}-S_{n}=\frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{n+n}
$$
Since $\frac{1}{n+p}>\frac{1}{2 n}$ for $p=1,2, \ldots, n-1$, we obtain the inequality
$$
S_{2 n}-S_{n}>\frac{n}{2 n}=\frac{1}{2}
$$
Thus, for the harmonic series, the Cauchy c... | proof | Calculus | proof | Yes | Yes | olympiads | false | 30,969 |
Example 4.3. Investigate the convergence of the series $\sum_{n=0}^{\infty} a^{n}$ (infinite geometric progression) $a \in R$. | Solution. Using the formula for the sum of the first terms of a geometric progression, we will calculate the partial sums of this series:
$$
\begin{gathered}
S_{n}=\frac{1-a^{n}}{1-a}=\frac{1}{1-a}-\frac{a^{n}}{1-a}(a \neq 1) \\
S_{n}=n(a=1)
\end{gathered}
$$
Thus, the question of the convergence of an infinite geome... | S_{n}=\frac{1}{1-}if||<1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,970 |
Example 4.4. Prove the divergence of the series $\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}$. | Solution. In Example 4.2, the divergence of the harmonic series $\sum_{n=1}^{\infty} \frac{1}{n}$ is proven, but since for all $n \geqslant 2$ the inequality $\frac{1}{\sqrt{n}}>\frac{1}{n}$ holds, by Theorem 4.5 the series $\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}$ diverges. | proof | Calculus | proof | Yes | Yes | olympiads | false | 30,971 |
Example 4.5. Establish the convergence for the reciprocals of squares $\sum_{n=1}^{\infty} \frac{1}{n^{2}}$. | Solution. Similarly to Example 4.1, it is not difficult to establish the convergence of the series $\sum_{n=1}^{\infty} \frac{1}{n(n+1)}$. The series under investigation can be represented differently as $\sum_{n=1}^{\infty} \frac{1}{n^{2}}=\sum_{n=0}^{\infty} \frac{1}{(n+1)^{2}}$. However, the terms of the latter seri... | proof | Calculus | proof | Yes | Yes | olympiads | false | 30,972 |
Example 4.6. Investigate the convergence of the series
$$
\sum_{n=1}^{\infty} \frac{n^{2}+n+1}{4 n^{4}+5 n^{3}+6 n^{2}+n+2}
$$ | Solution. In Example 4.5, the convergence of the series $\sum_{n=1}^{\infty} \frac{1}{n^{2}}$ is proven. Let's find the limit of the ratio of the terms of these series:
$$
\lim _{n \rightarrow \infty} \frac{n^{2}+n+1}{4 n^{4}+5 n^{3}+6 n^{2}+n+2}: \frac{1}{n^{2}}=\frac{1}{4}
$$
Therefore, based on Theorem 4.6, the se... | proof | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,973 |
Example 4.7. Prove the divergence of the series
$$
\sum_{n=1}^{\infty} \frac{2 n+3}{5 n^{2}-2 n}
$$ | Solution. In Example 4.2, the divergence of the harmonic series $\sum_{n=1}^{\infty} \frac{1}{n}$ is proven. Let's find the limit of the ratio of the terms of these series
$$
\lim _{n \rightarrow \infty} \frac{2 n+3}{5 n^{2}-2 n}: \frac{1}{n}=\frac{2}{5}
$$
By Theorem 4.6, the divergence of the harmonic series implie... | proof | Calculus | proof | Yes | Yes | olympiads | false | 30,974 |
Example 4.8. Investigate the convergence of the series $\sum_{n=1}^{\infty} \frac{100^{n}}{n!}$. | Solution. In Example 4.5, the convergence of the series of inverse squares $b_{n}=\frac{1}{n^{2}}$ was proven. In our case, $a_{n}=\frac{100^{n}}{n!}$. Let's form the ratios of subsequent terms to previous ones:
$$
\begin{aligned}
\frac{a_{n+1}}{a_{n}} & =\frac{100^{n+1}}{(n+1)!}: \frac{100^{n}}{n!}=\frac{100}{n+1} \\... | proof | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,975 |
Example 4.9. Investigate the convergence of the series $\sum_{n=1}^{\infty} \frac{n}{2^{n}}$. | Solution. For comparison, let's take the series $\sum_{n=1}^{\infty} a^{n}$, where $\frac{1}{2} < a < 1$. According to Example 4.3, this series converges. Let's investigate from which value of $n_{0}$ the inequality
$$
\frac{a_{n+1}}{a_{n}}=\frac{n+1}{2^{n+1}}: \frac{n}{2^{n}}=\frac{n+1}{2 n} \leqslant a
$$
will hold... | proof | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,976 |
Example 4.10. Investigate the convergence of the series $\sum_{n=1}^{\infty} \frac{2 n+1}{a^{n}}$, $(a>0)$. | Solution. In this case $a_{n}=\frac{2 n+1}{a^{n}}>0$. Let's find the limit (4.6)
$$
\lim _{n \rightarrow \infty} \frac{2 n+3}{a^{n+1}}: \frac{2 n+1}{a^{n}}=\lim _{n \rightarrow \infty} \frac{2 n+3}{2 n+1} \cdot \frac{1}{a}=\frac{1}{a}
$$
Based on Theorem 4.8, we conclude: if $a>1$ the series converges, if $0<a<1$ the... | if\>1\the\series\converges,\if\0<<1\or\=1\the\series\diverges | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,977 |
Example 4.11. Investigate the convergence of the series
$$
\sum_{n=1}^{\infty}(-1)^{n+1} \frac{1 \cdot 4 \cdot 7 \cdot \ldots \cdot(3 n-2)}{3 \cdot 5 \cdot 7 \cdot \ldots \cdot(2 n+1)}
$$ | Solution. Let's compute the limit (4.7)
$$
\begin{array}{r}
\lim _{n \rightarrow \infty} \frac{1 \cdot 4 \cdot 7 \cdot \ldots \cdot(3 n-2) \cdot(3 n+1)}{3 \cdot 5 \cdot 7 \cdot \ldots \cdot(2 n+1) \cdot(2 n+3)}: \frac{3 \cdot 5 \cdot 7 \cdot \ldots \cdot(2 n+1)}{1 \cdot 4 \cdot 7 \cdot \ldots \cdot(3 n-2)}= \\
=\lim _... | proof | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,978 |
Example 4.12. Investigate the convergence of the series $\sum_{n=1}^{\infty} 2^{(-1)^{n}-n}$. | Solution. Theorem 4.8 (the ratio test for positive series) cannot be applied to this series since the limit (4.6) does not exist. Even the strengthened ratio test (Theorem 4.10) does not yield a result, since
$$
\begin{aligned}
& \varlimsup_{n \rightarrow+\infty} \frac{a_{n+1}}{a_{n}}=2>1 \\
& \varliminf_{n \rightarro... | proof | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,979 |
Example 4.14. Investigate the series $\sum_{n=1}^{\infty}(-1)^{n} \frac{n!}{n}$. | Solution. According to Theorem 4.9
$$
\lim _{n \rightarrow \infty} \sqrt[n]{\frac{n!}{n^{n}}}=\lim _{n \rightarrow \infty} \frac{\sqrt[n]{n!}}{n}=\lim _{n \rightarrow \infty} \frac{(2 \pi n)^{1 / 2 n} \frac{n}{e} e^{\frac{\Theta}{12 n}}}{n}=\frac{1}{e}<1
$$
Therefore, the series converges absolutely. | \frac{1}{e}<1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,981 |
Example 4.15. Consider the series $\sum_{n=1}^{\infty}\left(\frac{7+(-1)^{n}}{2}\right)^{-n}$ and $\sum_{n=1}^{\infty}\left(\frac{7+(-1)^{n}}{2}\right)^{n}$ | Solution. Theorem 4.11 is not applicable to both these series since the limit $\lim _{n \rightarrow \infty} \sqrt[n]{a_{n}}$ does not exist. However, applying Theorem 4.13 yields certain results:
$$
\varlimsup_{n \rightarrow \infty} \sqrt[n]{\left(\frac{7+(-1)^{n}}{2}\right)^{-n}}=\frac{1}{3}1
$$
therefore the series... | Theseries\sum_{n=1}^{\infty}(\frac{7+(-1)^{n}}{2})^{n}diverges. | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,982 |
Example 4.16. Investigate the convergence of the series $\sum_{n=1}^{\infty} 3^{(-1)^{n}-n}$ | Solution. We apply the strengthened form of the D'Alembert's criterion (Theorem 4.10).
$$
\begin{aligned}
& \varlimsup_{n \rightarrow \infty} \frac{3^{(-1)^{n+1}-n-1}}{3^{(-1)^{n}-n}}=\varlimsup_{n \rightarrow \infty} 3^{(-1)^{n+1}-(-1)^{n}-1}=3 \\
& \varliminf_{n \rightarrow \infty} \frac{3^{(-1)^{n+1}-n-1}}{3^{(-1)^... | \frac{1}{3}<1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,983 |
Example 4.17. Investigate the convergence of the Dirichlet series $\sum_{n=1}^{\infty} \frac{1}{n^{p}}$. | Solution. Consider the function $f(x)=\frac{1}{x^{p}}$. This function is positive and monotonic for $x \geqslant 1$, and moreover, $\forall n \in N$ the equality $f(n)=\frac{1}{n^{p}}=a_{n}$ holds. Therefore, the question of the convergence of this series reduces, by Theorem 4.14, to the investigation of the improper i... | p>1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,984 |
Example 4.18. Investigate the convergence of the series $\sum_{n=2}^{\infty}-\frac{1}{n \sqrt{\ln n}}$. | Solution. Let us consider the function $f(x)=\frac{1}{x \sqrt{x}}$ for $x \geqslant 2$. This function is positive and monotonic. The improper integral $\int_{2}^{+\infty} \frac{d x}{x \sqrt{x}}$ diverges, since
$$
\int_{2}^{+\infty} \frac{d x}{x \sqrt{x}}=\left.\lim _{A \rightarrow+\infty} 2 \sqrt{\ln x}\right|_{2} ^{... | proof | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,985 |
Example 4.19. Investigate the convergence of the series
$$
\sum_{n=1}^{\infty} \frac{p(p+1) \ldots(p+n-1)}{n!} \frac{1}{n^{q}}
$$ | Solution. Apply Raabe's test:
$$
\begin{gathered}
\lim _{n \rightarrow \infty} n\left(\frac{a_{n}}{a_{n+1}-1}\right)= \\
=\lim _{n \rightarrow \infty} n\left(\frac{p(p+1) \ldots(p+n-1)}{n!} \frac{1}{n^{q}} \frac{(n+1)!(n+1)^{q}}{p(p+1) \ldots(p+n)}\right)= \\
=\lim _{n \rightarrow \infty} n\left(\frac{(n+1)^{q}}{n^{q}... | p | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,986 |
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