on the circle x2+y2=1

 Find the point P(x0,y0)

 so the directed derivative dzdOP−→−

 of the function

z=y2+2⋅y−x2+3⋅x+2

at point O(0,0)

 and in the direction of OP−→−

 be maximal and calculate the cosine of the acute angle α

 between OP−→−

 to the line y=x

.

[x0,y0]

=

cosα

=




Please note - the coordinate of a point must be entered in the following way: to enter the point (1,2) you must type [1,2].

Let's solve the problem step by step to find the point \( P(x_0, y_0) \) on the circle \( x^2 + y^2 = 1 \) such that the directional derivative of the function \( z = y^2 + 2y - x^2 + 3x + 2 \) at the origin \( O(0,0) \) in the direction of \( \overrightarrow{OP} \) is maximal. Afterward, we'll calculate the cosine of the acute angle \( \alpha \) between \( \overrightarrow{OP} \) and the line \( y = x \).

### Step 1: Compute the Gradient of \( z \)
The gradient \( \nabla z \) is given by:
\[
\nabla z = \left( \frac{\partial z}{\partial x}, \frac{\partial z}{\partial y} \right)
\]
Given:
\[
z = y^2 + 2y - x^2 + 3x + 2
\]
Calculate the partial derivatives:
\[
\frac{\partial z}{\partial x} = -2x + 3, \quad \frac{\partial z}{\partial y} = 2y + 2
\]
At point \( O(0,0) \):
\[
\nabla z(O) = (3, 2)
\]

### Step 2: Directional Derivative in the Direction of \( \overrightarrow{OP} \)
The directional derivative of \( z \) at \( O \) in the direction \( \overrightarrow{OP} \) is:
\[
\frac{dz}{d\overrightarrow{OP}} = \nabla z(O) \cdot \frac{\overrightarrow{OP}}{|\overrightarrow{OP}|} = 3x_0 + 2y_0
\]

### Step 3: Maximize the Directional Derivative
We need to maximize \( 3x_0 + 2y_0 \) subject to the constraint \( x_0^2 + y_0^2 = 1 \).

This can be done by considering the direction \( \overrightarrow{OP} \) to be along the gradient \( \nabla z(O) = (3, 2) \).

Since the maximum value occurs when \( \overrightarrow{OP} \) is aligned with \( \nabla z \), we align \( (x_0, y_0) \) with \( (3, 2) \) while keeping \( x_0^2 + y_0^2 = 1 \).

Let \( k \) be a constant such that \( (x_0, y_0) = k(3, 2) \). Then:
\[
x_0 = 3k, \quad y_0 = 2k
\]
Substituting into the constraint:
\[
(3k)^2 + (2k)^2 = 1
\]
\[
9k^2 + 4k^2 = 1 \quad \Rightarrow \quad 13k^2 = 1 \quad \Rightarrow \quad k = \frac{1}{\sqrt{13}}
\]
Thus:
\[
x_0 = \frac{3}{\sqrt{13}}, \quad y_0 = \frac{2}{\sqrt{13}}
\]
So the point \( P(x_0, y_0) \) is:
\[
[x_0, y_0] = \left[\frac{3}{\sqrt{13}}, \frac{2}{\sqrt{13}}\right]
\]

### Step 4: Cosine of the Angle \( \alpha \)
The direction vector of the line \( y = x \) is \( \overrightarrow{d} = (1, 1) \).

The cosine of the angle \( \alpha \) between \( \overrightarrow{OP} \) and \( \overrightarrow{d} \) is:
\[
\cos \alpha = \frac{\overrightarrow{OP} \cdot \overrightarrow{d}}{|\overrightarrow{OP}| |\overrightarrow{d}|} = \frac{\frac{3}{\sqrt{13}} \cdot 1 + \frac{2}{\sqrt{13}} \cdot 1}{\sqrt{1} \cdot \sqrt{2}} = \frac{5/\sqrt{13}}{\sqrt{2}} = \frac{5}{\sqrt{26}}
\]

### Final Answers:
\[
[x_0, y_0] = \left[\frac{3}{\sqrt{13}}, \frac{2}{\sqrt{13}}\right]
\]
\[
\cos \alpha = \frac{5}{\sqrt{26}}
\]

These are your final results. Would you like any further details or explanations?

Here are 8 related questions you might find interesting:
1. What does the gradient vector represent geometrically?
2. How do you calculate directional derivatives in different coordinate systems?
3. Can the method used here be applied to other shapes like ellipses?
4. How do directional derivatives relate to the concept of slope in calculus?
5. How would the solution change if the function \( z \) were different?
6. What are some applications of maximizing directional derivatives in physics or engineering?
7. How can you extend this approach to functions of three variables?
8. What is the importance of the Lagrange multiplier in constrained optimization problems?

**Tip:** In problems involving gradients and directional derivatives, always verify the normalization of vectors to ensure accurate calculations.

Explore how to maximize the directional derivative of a function at a point on a circle and calculate the cosine of the angle between the direction vector and the line y=x. Detailed step-by-step solution provided.

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