### Space Curve

What is parameterParameter भनेको variable हो जसले function लाई represent गर्दछ। In another word, parameter is a value that is already "built in" to define a function.

For example, if we define \(t \in [ a,b]\) by

\( x=f(t), y=g(t), z=h(t) \)

For example, say

\( x=1+t,y=t^3, z=t^2 \)

then,

- putting \( t=1\), we get \(x=1+t,=t^3,z=t^2 \), which is a a point \( ( 2,1,1) \) in 3D space.
- putting \(t=2\), we get \(x=1+t,=t^3,z=t^2 \), which is a a point \( ( 3,8,4) \) in 3D space.

This curve is called space curve.

wrte x,y,z in terms of t in the applet below

### Space curve in Parametric form

Define curves in space: Explicit formLet \(t \in [ a,b]\) then a curve in space is defined as locus of a point \( P (x,y,z) \) whose Cartesian coordinates \( x,y,z \) are function of single parameter t, it is given by

\( x=f(t), y=g(t), z=h(t) \) (1)

The equation given in (1) is called explicit form or parametric form of space curve.

For example,

\( x=2t, y=t^2, z=t^3-1 \) is a space curve.

###### Visualization of space curve: an example

In the example below, a space curve

\( x=t \cos t, y=t \sin t, z=t \) for the interval \(-7 \le t \le 8 \).

is visualized.

### Space curve in Implicit form

Define curve in space: Vector formVector भनेको directed magnitude हो। For example, if we define

\( (1,2,3) \) then \( \vec{p} \) is a vector.

Now,

In the vector form, a curve in space is denoted by \( C:\vec{r}=\vec{r}(s) \) , and defined as locus of a point \( P (x,y,z) \) whose position vector with respect to origin O, is function of single parameter t.

It is given by

\( \vec{r} = (x,y,z) \) यहाँ, \( x=f(t), y=g(t), z=h(t) \) भएकोले,

\( \vec{r} = ( f(t), g(t), h(t)) \) \( x=f(t), y=g(t), z=h(t) \) सबै function को मान \( t\) मा आउने भएकोले

\( \vec{r} =\vec{r}(t) \) (2)

The equation given in (2) is called Gaussian form (or vector form) of space curve.

For example,

\( \vec{r} =(2t, t^2, t^3-1) \) is a space curve. Note:

- Parametric form मा space curve लाई \( x=t, y=t+2 , z=t^3 \) को रुपमा लेखिन्छ
- Vector form मा space curve लाई \( \vec{r} = (t, t+2 , t^3) \) को रुपमा लेखिन्छ ।

Surface: Implicit form

A surface is given by the equation

\( f(x,y,z) =0 \)

उदाहरणको लागी,

- \( x+2=0 \) is a surface. [degree 1 भएकोले यो plane surface हो ]
- \( x^2+2=0\) is a surface. [degree 2 भएकोले यो curved surface हो ]
- \( x+y+2=0 \) is a surface. [degree 1 भएकोले यो plane surface हो ]
- \( x^2+y^2+2=0 \) is a surface. [degree 2 भएकोले यो curved surface हो ]
- \( x+y+2 z=0 \) is a surface. [degree 1 भएकोले यो plane surface हो ]
- \( x^2+y^2+2z^3=0 \) is a surface. [degree 3 भएकोले यो curved surface हो ]

Plot your surface

Define curve in space: Implicit formA space curve is also defined by intersection of two surfaces of the form

\(f_1(x,y,z) =0; f_2(x,y,z) =0 \) (3)

The equation given in (3) is called implicit form of space curve.

For example,

\( z^2=y^3 , z=x^3 \) is a space curve.

which is obtained in the parametric form by

\( x=t, y=t^2,z=t^3 \)

Note

The implicit representation of a spacer curve is obtained by eliminating parameter t from parametric representations.

Another implict form of space curve is given as below

This curve is obtained by intersection of sphere and a cylinder.

### Class of Space Curve

Define class of space curveLet \( C:\vec{r}=\vec{r}( t )\) be a space curve defined in an interval I. Also let \( m \in Z^+ \).

Now,\( C:\vec{r}=\vec{r}( t )\) is called \(C^m\)-curve if

- \( \vec{r}\) possess non-vanishing derivatives
- \( \vec{r}\) possess continuous derivatives

Example of class of space curve

- Example 1

Show that a space curve \( x=1+t, y=t^3, z=t^2 ;-2 \leq t \leq 2 \) has class 3.

Solution

The space curve is

\( x=1+t, y=t^3, z=t^2 \)

In the vector form, it is

\( \vec{r}( t )=( 1+t,t^3,t^2 )\) (1)

Diff. (i) w. r. to. t, we get

\( \dot{\vec{r}}( t )=( 1,3t^2,2t )\)

or \( \ddot{\vec{r}}( t )=( 0,6t,2 )\) exist and continuous for all t

or \( \overset{...}{\vec{r}}( t )=( 0,6,0 )\) exist and continuous for all t

or \( \overset{....}{\vec{r}}( t )=( 0,0,0 )\) vanished

Here, \( \vec{r}( t )\) possesses non-vanishing and continuous derivatives up to third order at each point in I, so the curve \( C:\vec{r}=\vec{r}( t )\) has class 3. -
Example 2

Show that a space curve \( x=t^3,y=\sin t,z=t^{\frac{8}{3}};0 \le t \le 2\) has class 2.

Solution

The space curve is

\( x=t^3,y=\sin t,z=t^{\frac{8}{3}}\)

The vector form of the curve is

\( \vec{r}( t )=( t^3, \sin t,t^{\frac{8}{3}} )\) (1)

Diff. (i) w. r. to. t, we get

\( \dot{\vec{r}}( t )=( 3t^2,\cos t,\frac{8}{3} t^{\frac{5}{3}} )\) exists and continuous for all t

or\( \ddot{\vec{r}}( t )=( 6t,- \sin t,\frac{40}{9}t^{\frac{2}{3}} )\) exists and continuous for all t

or\( \overset{...}{\vec{r}}( t )=( 6,- \cos t,\frac{80}{27}t^{-\frac{1}{3}} )\) does not exist at t=0.

Here, \( \vec{r}( t )\) possess non-vanishing continuous derivatives up to second order only at each point in I, so the curve \( C:\vec{r}=\vec{r}( t )\) has class-2.

### Exercise

- Describe the curve
- \( \vec{𝐫}=(\sin 𝑡,\cos \cos 8𝑡)\)
- \( \vec{𝐫}=( 𝑡 \cos 𝑡,𝑡 \sin 𝑡,𝑡)\)
- \( \vec{𝐫}=( 𝑡,𝑡^2,\cos 𝑡 ) \)
- \( \vec{𝐫}=( \cos(20𝑡) \sqrt{1−𝑡^2},\sin(20𝑡)\sqrt{1−𝑡^2},𝑡 ) \)

- Find a vector function for the curve of intersection of \( 𝑥^2+𝑦^2=9\) and \(𝑦+𝑧=2\)
- What is the difference between the parametric curves \(𝑓(𝑡)=⟨𝑡,𝑡,𝑡^2⟩, 𝑔(𝑡)=⟨𝑡^2,𝑡^2,𝑡^4⟩\) , and \(ℎ(𝑡)=⟨\sin (𝑡),\sin (𝑡),\sin 2(𝑡)⟩\) ?
- Plot each of the curves below in 2 dimensions, projected onto three rectangular plane (a) \( 𝑓(𝑡)=⟨𝑡,𝑡^3,𝑡^2⟩\) (b) \(𝑓(𝑡)=⟨𝑡^2,𝑡−1,𝑡^2+5⟩\) for 0≤𝑡≤3

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