Question

Задание по математическому анализу прикреплено ниже

Answer 1

Ответ:

Формулы:

$\frac{dz}{du} = \frac{dz}{dx} \times \frac{dx}{du} + \frac{dz}{dy} \times \frac{dy}{du} \\ \\ \frac{dz}{dv} = \frac{dz}{dz} \times \frac{dx}{dv} + \frac{dz}{dy} \times \frac{dy}{dv}$

(везде закругленная d)

$\frac{dx}{dx} = \frac{1}{1 + \frac{ {x}^{2} }{ {y}^{2} } } \times \frac{1}{y} = \frac{ {y}^{2} }{ {x}^{2} + {y}^{2} } \times \frac{1}{y} = \\ = \frac{y}{ {x}^{2} + {y}^{2} } \\ \frac{dz}{dy} = \frac{ {y}^{2} }{ {x}^{2} + {y}^{2} } \times ( - \frac{x}{ {y}^{2} } ) = - \frac{x}{ {x}^{2} + {y}^{2} }$

$\frac{dx}{du} = \sin(v) \\ \frac{dy}{du} = \cos(v) \\ \\ \frac{dx}{dv} = u \cos(v) \\ \frac{dy}{dv} = - u \sin(v)$

Собираем:

первая:

$\frac{dz}{du} = \frac{y}{ {x}^{2} + {y}^{2} } \times \sin(v) - \frac{x}{ {y}^{2} + {y}^{2} } \times \cos(v) = \\ = \frac{y \sin(v) - x \cos(v) }{ {x}^{2} + {y}^{2} }$

вторая:

$\frac{dz}{dv} = \frac{y}{ {x}^{2} + {y}^{2} } \times u \cos(v) + ( - \frac{x}{ {x}^{2} + {y}^{2} } ) \times ( - u \sin(v) ) = \\ = \frac{yu \cos(v) + xu \sin(v) }{ {x}^{2} + {y}^{2} }$