Please someone help me!!!

You are flying a kite on a line that is 350 feet long. Let's suppose the line is perfectly straight (it never really is) and it

aliina [53]

Answer:

We can think that the line of the kite is the hypotenuse of a triangle rectangle, and the altitude is one of the cathetus of the triangle.

And we know that it makes an angle of 65° with the horizontal (i guess this is measured between the hypotenuse and the horizontal adjacent to the kite.

This angle is complementary to the top angle of our triangle rectangle, such that A + 65° = 90°

A = 90° - 65° = 25°

Then the altitude of the kite is the adjacent cathetus to this angle.

We can use the relation:

sin(A) = Adjacent cathetus/hypotenuse.

Sin(25°) = X/350ft

Sin(25°)*350ft = X = 147.9m

4 0

3 years ago

Let φ(x, y) = arctan (y/x) .

Alexandra [31]

Answer:

a) $\large F(x,y)=(-\frac{y}{x^2+y^2},\frac{x}{x^2+y^2})$

b) $\large \mathbb{R}^2-\{(0,0)\}$

c) the points of the form (x, -x) for x≠0

Step-by-step explanation:

a)

If φ(x, y) = arctan (y/x), the vector field F = ∇φ would be

$\large F(x,y)=(\frac{\partial \phi}{\partial x},\frac{\partial \phi}{\partial y})$

On one hand we have,

$\large \frac{\partial \phi}{\partial x}=\frac{\partial arctan(y/x)}{\partial x}=\frac{-y/x^2}{1+(y/x)^2}=-\frac{y/x^2}{1+y^2/x^2}=\\\\=-\frac{y/x^2}{(x^2+y^2)/x^2}=-\frac{y}{x^2+y^2}$

On the other hand,

$\large \frac{\partial \phi}{\partial y}=\frac{\partial arctan(y/x)}{\partial y}=\frac{1/x}{1+(y/x)^2}=\frac{1/x}{1+y^2/x^2}=\\\\=\frac{1/x}{(x^2+y^2)/x^2}=\frac{x}{x^2+y^2}$