Mathematics: How do you do this question?

How do you do this question? - id8274467, dpazmembreno

Ответ:

nadiarose6366

05.02.2022

Part 1

$x = g(y), \ c \le y \le d\\\\\\x = 3+4y^2, \ 1 \le y \le 2\\\\\\\frac{dx}{dy} = 8y, \ 1 \le y \le 2\\\\\\\displaystyle A = 2\pi \int_{c}^{d}y\sqrt{1+[g'(y)]^2}dy\\\\\\\displaystyle A = 2\pi \int_{c}^{d}y\sqrt{1+\left[\frac{dx}{dy}\right]^2}dy\\\\\\\displaystyle A = 2\pi \int_{1}^{2}y\sqrt{1+(8y)^2}dy\\\\\\\displaystyle A = 2\pi \int_{1}^{2}y\sqrt{1+64y^2}dy\\\\\\\displaystyle A = 2\pi \int_{1}^{2}\sqrt{1+64y^2}*ydy\\\\\\\displaystyle A = 2\pi \int_{65}^{257}\sqrt{u}*\frac{du}{128}\\\\\\$

In the last two steps I let u = 1+64y^2, so du/dy = 128y which leads to ydy = du/128.

Note the change in the limits of integration. If y = 1, then u = 1+64y^2 = 65. If y = 2, then 1+64y^2 = 257

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Part 2

$\displaystyle A = 2\pi \int_{65}^{257}\sqrt{u}*\frac{du}{128}\\\\\\\displaystyle A = 2\pi*\frac{1}{128}\int_{65}^{257}u^{1/2}*du\\\\\\\displaystyle A = \frac{\pi}{64}\int_{65}^{257}u^{1/2}*du\\\\\\\displaystyle A = \frac{\pi}{64}\left(\frac{2}{3}u^{3/2}+C\right)\Bigg]_{65}^{257}\\\\\\\displaystyle A = \frac{\pi}{64}\left(\left(\frac{2}{3}(257)^{3/2}+C\right)-\left(\frac{2}{3}(65)^{3/2}+C\right)\right)\\\\\\\displaystyle A = \frac{\pi}{64}*\frac{2}{3}\left((257)^{3/2}-(65)^{3/2}\right)\\\\\\$

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Part 3

$\displaystyle A = \frac{\pi}{96}\left(\sqrt{257^3}-\sqrt{65^3}\right)\\\\\\\displaystyle A = \frac{\pi}{96}\left(\sqrt{257^2*257}-\sqrt{65^2*65}\right)\\\\\\\displaystyle A = \frac{\pi}{96}\left(\sqrt{257^2}*\sqrt{257}-\sqrt{65^2}*\sqrt{65}\right)\\\\\\\displaystyle A = \frac{\pi}{96}\left(257\sqrt{257}-65\sqrt{65}\right)\\\\\\$

$\displaystyle A = \frac{\pi}{96}*257\sqrt{257}-\frac{\pi}{96}*65\sqrt{65}\\\\\\\displaystyle A = \frac{257\pi\sqrt{257}}{96}-\frac{65\pi\sqrt{65}}{96}\\\\\\\displaystyle A = \frac{-65\sqrt{65}\pi}{96}+\frac{257\sqrt{257}\pi}{96}\\\\\\$

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Ответ:

Jasminehenry123

15.02.2023

Mathematics

x = 5, y = 4

Step-by-step explanation:

$\left\{\begin{array}{ccc}4x+5y=40\\x-y=1\end{array}\right\\\\A=\left[\begin{array}{ccc}4&5\\1&-1\end{array}\right]\\\\detA=\left|\begin{array}{ccc}4&5\\1&-1\end{array}\right|=(4)(-1)-(1)(5)=-4-5=-9\\\\A^D=\left[\begin{array}{ccc}-1&-1\\-5&4\end{array}\right]\\\\\left(A^D\right)^T=\left[\begin{array}{ccc}-1&-5\\-1&4\end{array}\right]$

$A^{-1}=\dfrac{1}{detA}\left(A^D\right)^T\\\\A^{-1}=\dfrac{1}{-9}\left[\begin{array}{ccc}-1&-5\\-1&4\end{array}\right]=\left[\begin{array}{ccc}\dfrac{1}{9}&\dfrac{5}{9}\\\dfrac{1}{9}&-\dfrac{4}{9}\end{array}\right]$

$A\cdot\left[\begin{array}{ccc}x\\y\end{array}\right] =\left[\begin{array}{ccc}40\\1\end{array}\right] \\\\A^{-1}A\cdot\left[\begin{array}{ccc}x\\y\end{array}\right] =A^{-1}\cdot\left[\begin{array}{ccc}40\\1\end{array}\right] \\\\\left[\begin{array}{ccc}x\\y\end{array}\right] =A^{-1}\cdot\left[\begin{array}{ccc}40\\1\end{array}\right]$

$\left[\begin{array}{ccc}x\\y\end{array}\right] =\left[\begin{array}{ccc}\dfrac{1}{9}&\dfrac{5}{9}\\\dfrac{1}{9}&-\dfrac{4}{9}\end{array}\right]\cdot\left[\begin{array}{ccc}40\\1\end{array}\right] \\\\\left[\begin{array}{ccc}x\\y\end{array}\right] =\left[\begin{array}{ccc}\left(\frac{1}{9}\right)(40)+\left(\frac{5}{9}\right)(1)\\\\\left(\frac{1}{9}\right)(40)+\left(-\frac{4}{9}\right)(1)\end{array}\right]\\\\\left[\begin{array}{ccc}x\\y\end{array}\right] =\left[\begin{array}{ccc}5\\4\end{array}\right]\Rightarrow x=5,\ y=4$

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