Curves in Space

---

---

---

Space Curve

Show/Hide 👉 Click Here

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

x(t):
y(t):
z(t):

---

---

---

Space curve in Parametric form

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

Note:

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

---

---

---

Surface: Implicit form

Plot your surface

---

---

---

Define curve in space: Implicit form

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

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

Show/Hide 👉 Click Here

Define class of space curve

Let \( C:\vec{r}=\vec{r}( t )\) 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

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

up to mth order at each point on I.

Example of class of space curve

1. 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.
2. 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

1. Describe the curve
   1. \( \vec{𝐫}=(\sin 𝑡,\cos \cos 8𝑡)\)
   2. \( \vec{𝐫}=( 𝑡 \cos 𝑡,𝑡 \sin 𝑡,𝑡)\)
   3. \( \vec{𝐫}=( 𝑡,𝑡^2,\cos 𝑡 ) \)
   4. \( \vec{𝐫}=( \cos(20𝑡) \sqrt{1−𝑡^2},\sin(20𝑡)\sqrt{1−𝑡^2},𝑡 ) \)
2. Find a vector function for the curve of intersection of \( 𝑥^2+𝑦^2=9\) and \(𝑦+𝑧=2\)
3. What is the difference between the parametric curves \(𝑓(𝑡)=⟨𝑡,𝑡,𝑡^2⟩, 𝑔(𝑡)=⟨𝑡^2,𝑡^2,𝑡^4⟩\) , and \(ℎ(𝑡)=⟨\sin (𝑡),\sin (𝑡),\sin 2(𝑡)⟩\) ?
4. 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

---

---

---