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3 years ago

11

If the side length of the base of a square pyramid is divided by 4 and the slant height is divided by 3, what is the surface are

a formula that represents these changes?

1 answer:

faust18 [17]3 years ago

7 0

The answer might be 4/3*4, 4 representing the sides<span />

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Find the Area of the figure below, composed of a rectangle and a semicircle. Round to the nearest tenths place.

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Find the form of the general solution of y^(4)(x) - n^2y''(x)=g(x)

Dennis_Churaev [7]

The differential equation

$y^{(4)}-n^2y'' = g(x)$

has characteristic equation

<em>r</em> ⁴ - <em>n </em>² <em>r</em> ² = <em>r</em> ² (<em>r</em> ² - <em>n </em>²) = <em>r</em> ² (<em>r</em> - <em>n</em>) (<em>r</em> + <em>n</em>) = 0

with roots <em>r</em> = 0 (multiplicity 2), <em>r</em> = -1, and <em>r</em> = 1, so the characteristic solution is

$y_c=C_1+C_2x+C_3e^{-nx}+C_4e^{nx}$

For the non-homogeneous equation, reduce the order by substituting <em>u(x)</em> = <em>y''(x)</em>, so that <em>u''(x)</em> is the 4th derivative of <em>y</em>, and

$u''-n^2u = g(x)$

Solve for <em>u</em> by using the method of variation of parameters. Note that the characteristic equation now only admits the two exponential solutions found earlier; I denote them by <em>u₁ </em>and <em>u₂</em>. Now we look for a particular solution of the form

$u_p = u_1z_1 + u_2z_2$

where

$\displaystyle z_1(x) = -\int\frac{u_2(x)g(x)}{W(u_1(x),u_2(x))}\,\mathrm dx$

$\displaystyle z_2(x) = \int\frac{u_1(x)g(x)}{W(u_1(x),u_2(x))}\,\mathrm dx$

where <em>W</em> (<em>u₁</em>, <em>u₂</em>) is the Wronskian of <em>u₁ </em>and <em>u₂</em>. We have

$W(u_1(x),u_2(x)) = \begin{vmatrix}e^{-nx}&e^{nx}\\-ne^{-nx}&ne^{nx}\end{vmatrix} = 2n$

and so

$\displaystyle z_1(x) = -\frac1{2n}\int e^{nx}g(x)\,\mathrm dx$

$\displaystyle z_2(x) = \frac1{2n}\int e^{-nx}g(x)\,\mathrm dx$

So we have

$\displaystyle u_p = -\frac1{2n}e^{-nx}\int_0^x e^{n\xi}g(\xi)\,\mathrm d\xi + \frac1{2n}e^{nx}\int_0^xe^{-n\xi}g(\xi)\,\mathrm d\xi$

and hence

$u(x)=C_1e^{-nx}+C_2e^{nx}+u_p(x)$

Finally, integrate both sides twice to solve for <em>y</em> :

$\displaystyle y(x)=C_1+C_2x+C_3e^{-nx}+C_4e^{nx}+\int_0^x\int_0^\omega u_p(\xi)\,\mathrm d\xi\,\mathrm d\omega$

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3 years ago