Question

U = sqrt(z) * sin(y/x) ;du| M 0 (2:0:4) =

Answer 1

Ответ:

Найти дифференциал функции в точке М₀  .

Формула :   $\bf du=u'_{x}\, dx+u'_{y}\, dy+u'_{z}\, dz\ \ ,\ \ u=u(x,y,z)$  .

$\bf u=\sqrt{z}\cdot sin\dfrac{y}{x}\ \ ,\ \ \ M_0(2;0;4)\\\\u'_{x}=\sqrt{z}\cdot cos\dfrac{y}{x}\cdot \dfrac{-1}{x^2}\ \ ,\ \ u'_{x}(M_0)=-2\cdot cos0\cdot \dfrac{1}{4}=-\dfrac{1}{2}\\\\u'_{y}=\sqrt{z}\cdot cos\dfrac{y}{x}\cdot \dfrac{1}{x}\ \ ,\ \ \ \ u'_{y}(M_0)=2\cdot cos0\cdot \dfrac{1}{2}=1\\\\u'_{z}=\dfrac{1}{2\sqrt{z}}\cdot sin\dfrac{y}{x}\ \ ,\ \ \ \ \ \ \ u'_{z}(M_0)=\dfrac{1}{4}\cdot sin0=0$  

$\bf du=-\sqrt{z}\cdot cos\dfrac{y}{x}\cdot \dfrac{1}{x^2}\cdot dx+\sqrt{z}\cdot cos\dfrac{y}{x}\cdot \dfrac{1}{x}\cdot dy+\dfrac{1}{2\sqrt{z}}\cdot sin\dfrac{y}{x}\cdot dz\\\\\\du\, \Big|_{M_0}=-\dfrac{1}{2}\, dx+dy$              

Answer 2

To find the partial derivative of the function $U = \sqrt{z} \cdot \sin\left(\frac{y}{x}\right)$ with respect to $u$ at the point $(2, 0, 4)$, we need to compute the partial derivative with respect to each of the variables $x$, $y$, and $z$, and then evaluate these derivatives at the given point $(2, 0, 4)$. Let's do that step by step:

1. Partial Derivative with respect to $x$, denoted as $\frac{\partial U}{\partial x}$:

$\frac{\partial U}{\partial x} = \frac{\partial}{\partial x} \left(\sqrt{z} \cdot \sin\left(\frac{y}{x}\right)\right)$

Using the product rule, we get:

$\frac{\partial U}{\partial x} = \sqrt{z} \cdot \cos\left(\frac{y}{x}\right) \cdot \frac{-y}{x^2}$

2. Partial Derivative with respect to $y$, denoted as $\frac{\partial U}{\partial y}$:

$\frac{\partial U}{\partial y} = \frac{\partial}{\partial y} \left(\sqrt{z} \cdot \sin\left(\frac{y}{x}\right)\right)$

Using the product rule, we get:

$\frac{\partial U}{\partial y} = \sqrt{z} \cdot \cos\left(\frac{y}{x}\right) \cdot \frac{1}{x}$

3. Partial Derivative with respect to $z$, denoted as $\frac{\partial U}{\partial z}$:

$\frac{\partial U}{\partial z} = \frac{\partial}{\partial z} \left(\sqrt{z} \cdot \sin\left(\frac{y}{x}\right)\right)$

This derivative is simpler:

$\frac{\partial U}{\partial z} = \sqrt{z} \cdot \sin\left(\frac{y}{x}\right)$

Now, we evaluate these partial derivatives at the point $(2, 0, 4)$:

$x = 2,$ $y = 0,$ $z = 4.$

1. $\frac{\partial U}{\partial x} \Big|_{(2, 0, 4)}$:

$\frac{\partial U}{\partial x} \Big|_{(2, 0, 4)} = 4 \cdot \cos\left(\frac{0}{2}\right) \cdot \frac{-0}{2^2} = 0$