Biology, asked by phonixgaming49, 1 month ago

plz answer these question according to the image
I will mark brainliest to whoever answers it correctly​

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Answers

Answered by xXItzSujithaXx34
1

Explanation:

Notice that if u is held constant, then the resulting curve is a circle of radius u in plane z=u. Therefore, as u increases, the radius of the resulting circle increases. If v is held constant, then the resulting curve is a vertical parabola. Therefore, we expect the surface to be an elliptic paraboloid. To confirm this, notice that

\begin{array}{cc}\hfill {x}^{2}+{y}^{2}& ={\left(u\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}v\right)}^{2}+{\left(u\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}v\right)}^{2}\hfill \\ & ={u}^{2}{\text{cos}}^{2}v+{u}^{2}{\text{sin}}^{2}v\hfill \\ & ={u}^{2}\hfill \\ & =z.\hfill \end{array}

Therefore, the surface is elliptic paraboloid {x}^{2}+{y}^{2}=z ((Figure)).

(a) Circles arise from holding u constant; the vertical parabolas arise from holding v constant. (b) An elliptic paraboloid results from all choices of u and v in the parameter domain.

Two images in three dimensions. The first shows parallel circles on the z axis with radii increasing as z increases. Vertical parabolas opening up frame the circles, forming the skeleton of a paraboloid. The second shows the elliptic paraboloid, which is made of all the possible circles and vertical parabolas in the parameter domain.

Describe the surface parameterized by \text{r}\left(u,v\right)=〈u\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}v,u\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}v,u〉,\text{−}\infty <u<\infty ,0\le v<2\pi .

Cone {x}^{2}+{y}^{2}={z}^{2}

Hint

Hold u constant and see what kind of curves result. Imagine what happens as u increases or decreases.

Finding a Parameterization

Give a parameterization of the cone {x}^{2}+{y}^{2}={z}^{2} lying on or above the plane z=-2.

The horizontal cross-section of the cone at height z=u is circle {x}^{2}+{y}^{2}={u}^{2}. Therefore, a point on the cone at height u has coordinates \left(u\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}v,u\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}v,u\right) for angle v. Hence, a parameterization of the cone is \text{r}\left(u,v\right)=〈u\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}v,u\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}v,u〉. Since we are not interested in the entire cone, only the portion on or above plane z=-2, the parameter domain is given by -2\le u<\infty ,0\le v<2\pi ((Figure)).

Cone {x}^{2}+{y}^{2}={z}^{2} has parameterization r\left(u,v\right)=〈u\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}v,u\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}v,u〉,\text{−}\infty <u<\infty ,0\le v\le 2\pi .

A three-dimensional diagram of the cone x^2 + y^2 = z^2, which opens up along the z axis for positive z values and opens down along the z axis for negative z values. The center is at the origin.

Answered by llFollowll27
3

Explanation:

Notice that if u is held constant, then the resulting curve is a circle of radius u in plane z=u. Therefore, as u increases, the radius of the resulting circle increases. If v is held constant, then the resulting curve is a vertical parabola. Therefore, we expect the surface to be an elliptic paraboloid. To confirm this, notice that

\begin{array}{cc}\hfill {x}^{2}+{y}^{2}& ={\left(u\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}v\right)}^{2}+{\left(u\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}v\right)}^{2}\hfill \\ & ={u}^{2}{\text{cos}}^{2}v+{u}^{2}{\text{sin}}^{2}v\hfill \\ & ={u}^{2}\hfill \\ & =z.\hfill \end{array}

Therefore, the surface is elliptic paraboloid {x}^{2}+{y}^{2}=z ((Figure)).

(a) Circles arise from holding u constant; the vertical parabolas arise from holding v constant. (b) An elliptic paraboloid results from all choices of u and v in the parameter domain.

Two images in three dimensions. The first shows parallel circles on the z axis with radii increasing as z increases. Vertical parabolas opening up frame the circles, forming the skeleton of a paraboloid. The second shows the elliptic paraboloid, which is made of all the possible circles and vertical parabolas in the parameter domain.

Describe the surface parameterized by \text{r}\left(u,v\right)=〈u\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}v,u\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}v,u〉,\text{−}\infty <u<\infty ,0\le v<2\pi .

Cone {x}^{2}+{y}^{2}={z}^{2}

Hint

Hold u constant and see what kind of curves result. Imagine what happens as u increases or decreases.

Finding a Parameterization

Give a parameterization of the cone {x}^{2}+{y}^{2}={z}^{2} lying on or above the plane z=-2.

The horizontal cross-section of the cone at height z=u is circle {x}^{2}+{y}^{2}={u}^{2}. Therefore, a point on the cone at height u has coordinates \left(u\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}v,u\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}v,u\right) for angle v. Hence, a parameterization of the cone is \text{r}\left(u,v\right)=〈u\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}v,u\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}v,u〉. Since we are not interested in the entire cone, only the portion on or above plane z=-2, the parameter domain is given by -2\le u<\infty ,0\le v<2\pi ((Figure)).

Cone {x}^{2}+{y}^{2}={z}^{2} has parameterization r\left(u,v\right)=〈u\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}v,u\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}v,u〉,\text{−}\infty <u<\infty ,0\le v\le 2\pi .

A three-dimensional diagram of the cone x^2 + y^2 = z^2, which opens up along the z axis for positive z values and opens down along the z axis for negative z values. The center is at the origin.

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