Example 7. Find the angle $\Theta$ between the gradients of the functions

$$
u=\sqrt{x^{2}+y^{2}} \text { and } v=x+y+2 \sqrt{x y}
$$

at the point $M_{0}(1, 1)$.
Solution. We find the gradients of the given functions at the point $M_{1}(1,1)$. We have

$$
\begin{aligned}
& \left.\operatorname{grad} u\right|_{M_{1}}=\left.\frac{x i+y j}{\sqrt{x^{2}+y^{2}}}\right|_{M_{0}} \pm \frac{1}{\sqrt{2}} I+\frac{1}{\sqrt{2}} j \\
& \left.\operatorname{grad} v\right|_{M_{1}}=\left.\left[\left(I+\sqrt{\frac{y}{x}}\right) 1+\left(I+\sqrt{\frac{x}{y}}\right)\right]\right|_{M_{0}}=2 i+2 j
\end{aligned}
$$

The angle $\theta$ between $\operatorname{grad} u$ and $\operatorname{grad} v$ at the point $M_{0}$ is determined by the equality

![](https://cdn.mathpix.com/cropped/2024_05_22_267d965536514099194eg-16.jpg?height=79&width=473&top_left_y=1185&top_left_x=272)

From this,

$$
\Theta=0
$$