Найти d^2z, если а) z=sin(x^2+y^2) б) z=√(x^2+y^2). Created by kjuliaaa333. matematika-ru

Найти d^2z, если а) z=sin(x^2+y^2) б) z=√(x^2+y^2)...

kjuliaaa333 @kjuliaaa333

July 2022 1 10 Report

Найти d^2z, если а) z=sin(x^2+y^2)
б) z=√(x^2+y^2)

Answers & Comments

Miroslava227

Ответ:

$z = \sin( {x}^{2} + {y}^{2} )$

${d}^{2} z = z''_{xx}dx {}^{2} + z''_{xy}dxdy + z''_{yy}dy {}^{2} \\$

$z'_x = \cos( {x}^{2} + {y}^{2} ) \times 2x$

$z'_y = \cos( {x}^{2} + {y}^{2} ) \times 2y$

$z''_{xx} = ( \cos( {x}^{2} + {y}^{2} ) )'_x\times 2x + (2x)' \times \cos( {x}^{2} + {y}^{2} ) = \\ = - \sin( {x}^{2} + {y}^{2} ) \times 2x \times 2x + 2 \cos( {x}^{2} + {y}^{2} ) = \\ = 2 \cos( {x}^{2} + {y}^{2} ) - 4 {x}^{2} \sin( {x}^{2} + {y}^{2} )$

$z''_{yy }= ( \cos( {x}^{2} + {y}^{2} ))'_y \times 2y + (2y)' \times \cos( {x}^{2} + {y}^{2} ) = \\ = - \sin( {x}^{2} + {y}^{2} ) \times 2y \times 2y + 2 \cos( {x}^{2} + {y}^{2} ) = \\ = 2 \cos( {x}^{2} + {y}^{2} ) - 4 {y}^{2} \sin( {x}^{2} + {y}^{2} )$

$z''_{xy} = 2x \times ( \cos( {x}^{2} + {y}^{2} ) )'_y = \\ = 2x \times ( - \sin( {x}^{2} + {y}^{2} ) ) \times 2{y}^{} = \\ = - 4{xy}^{} \sin( {x}^{2} + {y}^{2} )$

$d {}^{2} z = (2 \cos( {x}^{2} + {y}^{2} ) - 4 {x}^{2} \sin( {x}^{2} + {y}^{2} ) ) dx - \\ - 4xy \sin( {x}^{2} + {y}^{2} ) dxdy + (2 \cos( {x}^{2} + {y}^{2} ) - \\ - 4 {y}^{2} \sin( {x}^{2} + {y}^{2} ) )dy {}^{2} \\$

$z = \sqrt{ {x}^{2} + {y}^{2} } \\$

$z'_x = \frac{1}{2 \sqrt{ {x}^{2} + {y}^{2} } } \times 2x = \frac{x}{ \sqrt{ {x}^{2} + {y}^{2} } } \\$

$z'_y = \frac{y}{ \sqrt{ {x}^{2} + {y}^{2} } } \\$

$z''_{xx }= \frac{ \sqrt{ {x}^{2} + {y}^{2} } - x \times \frac{1}{2 \sqrt{ {x}^{2} + {y}^{2} } } \times 2x}{ {x}^{2} + {y}^{2} } = \\ = \frac{ \sqrt{ {x}^{2} + {y}^{2} } - \frac{ {x}^{2} }{ \sqrt{ {x}^{2} + {y}^{2} } } }{ {x}^{2} + {y}^{2} } = \frac{ {x}^{2} + {y}^{2} - {x}^{2} }{ \sqrt{( {x}^{2} + {y}^{2} ) {}^{3} } } = \\ = \frac{ {y}^{2} }{ \sqrt{ {( {x}^{2} + {y}^{2} )}^{3} } }$

$z''_{yy }= \frac{ \sqrt{ {x}^{2} + {y}^{2} } - y \times \frac{y}{ \sqrt{ {x}^{2} + {y}^{2} } } }{ {x}^{2} + {y}^{2} } = \\ = \frac{ {x}^{2} + {y}^{2} - {y}^{2} }{ \sqrt{ ({x}^{2} + {y}^{2} ) {}^{3} } } = \frac{ {x}^{2} }{ \sqrt{( {x}^{2} + {y}^{2}) {}^{3} } }$

$z''_{xy} = x \times ( - \frac{1}{2} ) \times \frac{1}{ \sqrt{ {( {x}^{2} + {y}^{2} )}^{3} } } \times 2y = \\ = - \frac{xy}{ \sqrt{ {( {x}^{2} + {y}^{2}) }^{3} } }$

${d}^{2}z = \frac{ {y}^{2} }{ \sqrt{ {( {x}^{2} + {y}^{2}) }^{3} } } dx {}^{2} - \frac{xy}{ \sqrt{ {( {x}^{2} + {y}^{2} )}^{3} } } dxdy + \frac{ {x}^{2} }{ \sqrt{ {( {x}^{2} + {y}^{2}) }^{3} } } dy {}^{2} = \\ = \frac{ {y}^{2} dx {}^{2} - xydxdy + {x}^{2} dy {}^{2} }{ \sqrt{ {( {x}^{2} + {y}^{2} ) }^{3} } }$

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kjuliaaa333 Спасибо большое)

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