Curvature and Torsion ~ Bed Prasad Dhakal

0,0

The radius at P is 1.10|| The curvature at P is 0.91

One of the basic problem in geometry is to determine geometric quantities exactly to distinguish from one another. For example, line segments are uniquely determined by their lengths, circles by their radii, triangles by side-angle-side axiom etc. We will now see that a space curve is also uniquely determined by two scalar quantities: called curvature and torsion

जसरी

दुईवटा line segment बराबर छ वा छैन थाहा पाउन line segment को lengths चाहिन्छ ( line segments are uniquely determined by their lengths),
दुईवटा circles बराबर छ वा छैन थाहा पाउन circles को radius चाहिन्छ (circles are uniquely determined by their radii),
दुईवटा triangle बराबर छ वा छैन थाहा पाउन side-angle-side को magnitude चाहिन्छ ( triangles are uniquely determined by their side-angle-side axiom)

त्यसैगरी,
दुईवटा space curve बराबर छ वा छैन थाहा पाउन curvature र torsion चाहिन्छ । Space curve are uniquely determined by two scalar quantities, called curvature and torsion

Curvature

Show/Hide 👉 Click Here

Let

C : \vec{r} = \vec{r} (s)

be a space curve and P be a point on C, then curvature at P is defined as rate of rotation of tangent (change in the direction of tangent) at P. It measures deviation of the curve from lying along a straight line
The magnitude of curvature is denoted by

κ

(kappa) and defined by

κ = lim_{δ s \to 0} \frac{δ θ}{δ s} = \frac{d θ}{d s}

where

δ θ

is angle between tangents at two neighboring points P and Q on the curve along arc length

δ s

0,0

–o+←↓↑→

az = 1.00

el = 0.30

rotate = 4.18

The curvature at t=4.18 is 0.23

More precisely, the curvature is

Rate of change in the direction of tangent line
Degree of curving along tangent line
Scalar measure of bending nature of the curve along tangent direction
Change in principal normal along tangent direction

0,0

–o+←↓↑→

az = 1.00

el = 0.30

rotate = -5.63

No change in tangent vector, so κ=0

Note

Straight line has zero curvature
A circle has constant curvature
The curvature of small circle is large and vice versa

Expression of Curvature

Show/Hide 👉 Click Here

Let

C : \vec{r} = \vec{r} (s)

be a space curve in which P and Q are two neighboring points. Assume that arc length s of C be measured from suitable base point A such that
arcAP=s, arcAQ=s+𝛿s
Then
arcPQ=𝛿s
Also, we denote

\vec{t}

and

\vec{t} + δ \vec{t}

as unit vectors along tangent lines at P and Q respectively and 𝛿θ be the angle between tangents at P and Q.

0,0

–o+←↓↑→

az = 1.00

el = 0.30

Then

\vec{Q M} = \vec{t}

with

| \vec{Q M} | = 1

where

\vec{Q N} = \vec{t} + δ \vec{t}

with

| \vec{Q N} | = 1

and

\vec{M N} = δ \vec{t}

with

∠ M Q N = δ θ

Now, from isosceles triangle ∆QMN, we have

s i n (\frac{δ θ}{2}) = \frac{\frac{M N}{2}}{Q N} = \frac{| δ \vec{t} |}{2}

| δ \vec{t} | = 2 s i n (\frac{δ θ}{2})

| \frac{δ \vec{t}}{δ θ} | = \frac{s i n (δ θ / 2)}{δ θ / 2}

Taking limit as

δ s \to 0

, we get

lim_{δ s \to 0} | \frac{δ \vec{t}}{δ θ} | = lim_{δ θ \to 0} \frac{s i n (δ θ / 2)}{δ θ / 2}

| \frac{d \vec{t}}{d θ} | = 1

Hence, curvature at P is

κ = \frac{d θ}{d s} = \frac{d θ}{| d \vec{t} |} \frac{| d \vec{t} |}{d s} = | \frac{d θ}{d \vec{t}} | | \frac{d \vec{t}}{d s} | = | {\vec{t}}^{'} |

Since

{\vec{t}}^{'}

lies in osculating plane and also

{\vec{t}}^{'}

is perpendicular to

\vec{t}

, we claim that

{\vec{t}}^{'} = κ \vec{n}

This is the required expression of curvature

NOTE

The vector ${\vec{t}}^{'}$ is called curvature vector. The magnitude of ${\vec{t}}^{'}$ s called curvature. It is denoted by $| {\vec{t}}^{'} |$ and defined by $| {\vec{t}}^{'} | = κ$ . The reciprocal of curvature is denoted by $ρ = \frac{1}{κ}$ , and it is called radius of curvature. A point in which $κ = 0$ is called point of inflection. Thus at a point of inflection curvature is zero and thus radius of curvature is infinite.

The curvature is tendency of a curve to change their direction from point to point along tangent line. Thus along a curve which has rapidly changing tangent direction with respect to arc length, such as a circle with small radius, the curvature is relatively large, or equivalently, the radius of curvature is small.

Torsion

Show/Hide 👉 Click Here

As the curvature measures deviation of the curve from lying along a straight line, the torsion measures deviation of the curve from lying in a plane. The precise definition is given below.

Let $C : \vec{r} = \vec{r} (s)$ be a space curve and P be a point on C, then torsion at P is defined as rate of rotation of binormal (change in the direction of binormal) at P. It measures deviation of the curve from lying along a straight line

The magnitude of torsion is denoted by $τ$ (tau) and defined by
$τ = lim_{δ s \to 0} \frac{δ θ}{δ s} = \frac{d θ}{d s}$
where $δ θ$ is angle between binormals at two neighboring points P and Q on the curve along arc length $δ s$ .

0,0

–o+←↓↑→

az = 1.00

el = 0.30

slide = -3.26

The curvature at t=-3.26 is 0.29

The torsion at t=-3.26 is 0.11

More precisely, the torsion is

Rate of change in the direction of binormal line
Degree of curving along binormal line
Signed scalar measure of bending nature of the curve along binormal direction
Change in principal normal along binormal direction

0,0

–o+←↓↑→

az = 1.00

el = 0.30

rotate = -5.77

show plane

No change in binormal vector, so τ=0

Note

Plane curve has zero torsion
Circle has zero torsion.
A circular helix has constant torsion
Torsion at any point on the curve is signed scalar

Expression of Torsion

Show/Hide 👉 Click Here

Let

C : \vec{r} = \vec{r} (s)

be a space curve in which P and Q are two neighboring points. Assume that arc length s of C be measured from suitable base point A such that

a r c A P = s, a r c A Q = s + δ s

Then

a r c P Q = δ s

Also, we denote

\vec{b}

and

\vec{b} + δ \vec{b}

as unit vectors along binormal lines at P and Q respectively and

δ θ

be the angle between binormals at P and Q.

0,0

–o+←↓↑→

az = 1.00

el = 0.30

Then

\vec{Q M} = \vec{b}

with

| \vec{Q M} | = 1

\vec{Q N} = \vec{b} + δ \vec{b}

with

| \vec{Q N} | = 1

\vec{M N} = δ \vec{b}

∠ M Q N = δ θ

Now, from isosceles triangle QMN, we have

s i n (\frac{δ θ}{2}) = \frac{\frac{M N}{2}}{Q N} = \frac{| δ \vec{b} |}{2}

| δ \vec{b} | = 2 s i n (\frac{δ θ}{2})

| \frac{δ \vec{b}}{δ θ} | = \frac{s i n (δ θ / 2)}{δ θ / 2}

Taking limit as

δ s \to 0

, we get

lim_{δ s \to 0} | \frac{δ \vec{b}}{δ θ} | = lim_{δ θ \to 0} \frac{s i n (δ θ / 2)}{δ θ / 2}

| \frac{d \vec{b}}{d θ} | = 1

Hence, torsion at P is

τ = \frac{d θ}{d s} = \frac{d θ}{| d \vec{b} |} \frac{| d \vec{b} |}{d s} = | \frac{d θ}{d \vec{b}} | | \frac{d \vec{b}}{d s} | = | {\vec{b}}^{'} |

Since

{\vec{b}}^{'}

lies in osculating plane and also

{\vec{b}}^{'}

is perpendicular to

\vec{b}

, we claim that

{\vec{b}}^{'} = - τ \vec{n}

This is the required expression of torsion. NOTE

The vector ${\vec{b}}^{^{'}}$ is called torsion vector. The magnitude of ${\vec{b}}^{^{'}}$ is called torsion. It is denoted by $| {\vec{b}}^{'} |$ and defined by $| {\vec{b}}^{'} | = - τ$ .

The reciprocal of torsion is denoted by $σ = \frac{1}{τ}$ , and it is called radius of torsion.

The torsionis tendency of a curve to change their direction from point to point along principal normal line. Thus it measures the bending along a curve which has changing principal normal direction with respect to arc length

Expression of Screw-Curvature:

Show/Hide 👉 Click Here

By the definition of fundamental unit vectors, we have

\vec{n} = \vec{b} \times \vec{t}

Differentiation w. r. to. s, we get

{\vec{n}}^{'} = {\vec{b}}^{'} \times \vec{t} + \vec{b} \times {\vec{t}}^{'}

{\vec{n}}^{'} = (- τ \vec{n}) \times \vec{t} + \vec{b} \times (κ \vec{n})

{\vec{n}}^{'} = τ \vec{b} - κ \vec{t}

which is the required expression of screw curvature

Serret-frenet formula.

Given a space curve

C : \vec{r} = \vec{r} (s)

, we can define a local trihedron

\vec{t}, \vec{n}, \vec{b}

at each point on the curve unless the curve is a straight line, on which the trihedrons will change as we move them along the curve. This movement naturally leads us to investigate how exactly the trihedron change along the curve. This is most concisely expressed using the Frenet formulas.
The Serret Frenet formula is a matrix equation representing the fundamental unit vectors along with their derivatives, which give the derivatives of the trihedron vectors:

[\begin{matrix} {\vec{t}}^{'} \\ {\vec{n}}^{'} \\ {\vec{b}}^{'} \end{matrix}] = [\begin{matrix} κ \vec{n} \\ - κ \vec{t} & τ \vec{b} \\ - τ \vec{n} \end{matrix}]

In summary, it is written as

[\begin{matrix} {\vec{t}}^{'} \\ {\vec{n}}^{'} \\ {\vec{b}}^{'} \end{matrix}] = [\begin{matrix} 0 & κ & 0 \\ - κ & 0 & τ \\ 0 & - τ & 0 \end{matrix}] [\begin{matrix} \vec{t} \\ \vec{n} \\ \vec{b} \end{matrix}]

It is to note that the coefficient matrix appearing on the right of (i) is skew-symmetric, also it is notifying that the curvature

κ

is non negative scalar where as the torsion is signed scalar.

Summary Table

Fundamental Vector	Expression	Magnitude
Curvature Change in tangent line	${\vec{t}}^{'} = κ \vec{n}$	$\| {\vec{t}}^{'} \| = κ$
Torsion Change in binormal line	${\vec{b}}^{'} = - τ \vec{n}$	$\| {\vec{b}}^{'} \| = τ \vec{n}$
Screw-curvature Change in principal normal line	${\vec{n}}^{'} = τ \vec{b} - κ \vec{t}$	$\| {\vec{n}}^{'} \| = \sqrt{κ^{2} + τ^{2}}$

Theorems

Show/Hide 👉 Click Here

A necessary and sufficient condition for a curve to be a straight line is that curvature $κ = 0$ at all points on the curve.
Proof

Let $C : \vec{r} = \vec{r} (s)$ be a space curve.
Then, necessary and sufficient condition for a curve to be a straight line is that, the position of arbitrary point on C is expressed as
$\vec{r} = s \vec{a} + \vec{b}$ ...(i)
where $\vec{a}$ and $\vec{b}$ are constant vectors
⟺ ${\vec{r}}^{'} = \vec{a}$
⟺ ${\vec{r}}^{″} = 0$
⟺ $| {\vec{r}}^{″} | = 0$
⟺ $κ = 0$ at all points of the curve
This completes the proof
A necessary and sufficient condition for a curve to be a plane curve is that $τ = 0$ at all points on the plane curve.

Proof
Let $C : \vec{r} = \vec{r} (s)$ be a space curve.
Then, necessary and sufficient condition for a curve to be a straight line is that, binormal tvector is constant ${\vec{b}}^{'} = 0$
⟺ $- τ \vec{n} = 0$
⟺ $τ = 0$
Also
$\frac{d}{d s} (\vec{r} . \vec{b})$
or $\vec{r} . {\vec{b}}^{'} + \vec{t} . \vec{b}$
or $\vec{r} .0 + 0$
or $0$
It shows that projection of arbitrary point of the curve on binormal is constant and this implies that the curve is a plane curve.
This completes the proof
Prove that $κ = | {\vec{r}}^{'} \times {\vec{r}}^{″} |, τ = \frac{[{\vec{r}}^{'}, {\vec{r}}^{″}, {\vec{r}}^{‴}]}{{| {\vec{r}}^{'} \times {\vec{r}}^{″} |}^{2}}$ and $[{\vec{r}}^{'}, {\vec{r}}^{″}, {\vec{r}}^{‴}] = κ^{2} τ$ .
Solution
Given that
${\vec{r}}^{'} = \vec{t}$ ...(i)
and
${\vec{r}}^{″} = κ \vec{n}$ ...(ii)
and
${\vec{r}}^{‴} = κ^{'} \vec{n} + κ τ \vec{b} - κ^{2} \vec{t}$
or ${\vec{r}}^{‴} = - κ^{2} \vec{t} + κ^{'} \vec{n} + κ τ \vec{b}$ ...(iii)
To find the curvature, we operate cross product between (i) and (ii), then
${\vec{r}}^{'} \times {\vec{r}}^{″} = \vec{t} \times κ \vec{n}$
or ${\vec{r}}^{'} \times {\vec{r}}^{″} = κ \vec{b}$
Taking magnitude, we get
$κ = | {\vec{r}}^{'} \times {\vec{r}}^{″} |$ ...(A)
which is the required curvature
To find the torsion, we operate scalar triple product among (i),(ii) and (iii), then
$[{\vec{r}}^{'}, {\vec{r}}^{″}, {\vec{r}}^{‴}] = | \begin{matrix} 1 & 0 & 0 \\ 0 & κ & 0 \\ - κ^{2} & κ^{'} & κ τ \end{matrix} |$
or $[{\vec{r}}^{'}, {\vec{r}}^{″}, {\vec{r}}^{‴}] = κ^{2} τ$
or $τ = \frac{[{\vec{r}}^{'}, {\vec{r}}^{″}, {\vec{r}}^{‴}]}{κ^{2}}$
or $τ = \frac{[{\vec{r}}^{'}, {\vec{r}}^{″}, {\vec{r}}^{‴}]}{{| {\vec{r}}^{'} \times {\vec{r}}^{″} |}^{2}}$ from (A)
which is the required torsion
Show that $κ = \frac{| \dot{\vec{r}} \times \ddot{\vec{r}} |}{{| \dot{\vec{r}} |}^{3}}, τ = \frac{[\dot{\vec{r}}, \ddot{\vec{r}}, \overset{⃛}{\vec{r}}]}{{| \dot{\vec{r}} \times \ddot{\vec{r}} |}^{2}}$ and $[\dot{\vec{r}}, \ddot{\vec{r}}, \overset{⃛}{\vec{r}}] = κ^{2} τ {\dot{s}}^{6}$
Solution
Given that
$\dot{\vec{r}} = \frac{d \vec{r}}{d t}$
or $\dot{\vec{r}} = \frac{d \vec{r}}{d s} \frac{d s}{d t}$
or $\dot{\vec{r}} = \vec{t} \dot{s}$ with $\dot{s} = | \dot{\vec{r}} |$ ...(i)
and
$\ddot{\vec{r}} = \vec{t} \ddot{s} + κ \vec{n} {\dot{s}}^{2}$ ...(ii)
and
$\overset{⃛}{\vec{r}} = \vec{t} + κ \vec{n} \dot{s} \ddot{s} + κ^{'} \vec{n} {\dot{s}}^{3} + κ (τ \vec{b} - κ \vec{t}) {\dot{s}}^{3} + 2 κ \vec{n} \dot{s} \ddot{s}$
or $\overset{⃛}{\vec{r}} = (- κ^{2} {\dot{s}}^{3}) \vec{t} + (3 κ \dot{s} \ddot{s} + κ^{'} {\dot{s}}^{3}) \vec{n} + κ τ \vec{b} {\dot{s}}^{3}$ ...(iii)
To find the curvature, we operate cross product between (i) and (ii) then
$\dot{\vec{r}} \times \ddot{\vec{r}} = \vec{t} \dot{s} \times (\vec{t} \ddot{s} + κ \vec{n} {\dot{s}}^{2})$
or $\dot{\vec{r}} \times \ddot{\vec{r}} = κ {\dot{s}}^{3} \vec{b}$
or $| \dot{\vec{r}} \times \ddot{\vec{r}} | = κ {\dot{s}}^{3}$ ...(A)
or $κ = \frac{| \dot{\vec{r}} \times \ddot{\vec{r}} |}{{| \dot{\vec{r}} |}^{3}}$ since $\dot{s} = | \dot{\vec{r}} |$ from (ii)
which is the required curvature
To find the torsion, we operate scalar triple product among (i), (ii), and (iii), then
$[\dot{\vec{r}}, \ddot{\vec{r}}, \overset{⃛}{\vec{r}}] = | \begin{matrix} \dot{s} & 0 & 0 \\ \ddot{s} & κ {\dot{s}}^{2} & 0 \\ - κ^{2} {\dot{s}}^{3} & 3 κ \dot{s} \ddot{s} + κ^{'} {\dot{s}}^{3} & κ τ {\dot{s}}^{3} \end{matrix} |$
or $[\dot{\vec{r}}, \ddot{\vec{r}}, \overset{⃛}{\vec{r}}] = κ^{2} τ {\dot{s}}^{6}$
or $τ = \frac{[\dot{\vec{r}}, \ddot{\vec{r}}, \overset{⃛}{\vec{r}}]}{κ^{2} {\dot{s}}^{6}}$
or $τ = \frac{[\dot{\vec{r}}, \ddot{\vec{r}}, \overset{⃛}{\vec{r}}]}{{| \dot{\vec{r}} \times \ddot{\vec{r}} |}^{2}}$ since $κ {\dot{s}}^{3} = | \dot{\vec{r}} \times \ddot{\vec{r}} |$ from (A)
A necessary and sufficient condition for a curve to be a plane curve is that $[{\vec{r}}^{'}, {\vec{r}}^{″}, {\vec{r}}^{‴}] = 0$ at all points.

Proof
Let $C : \vec{r} = \vec{r} (s)$ be a space curve.
Then, necessary and sufficient condition for a curve to be a plane curve is that
$τ = 0$ at all points of the curve
or $\frac{[{\vec{r}}^{'}, {\vec{r}}^{″}, {\vec{r}}^{‴}]}{{| {\vec{r}}^{'} \times {\vec{r}}^{″} |}^{2}} = 0$ at all points of the curve
or $[{\vec{r}}^{'}, {\vec{r}}^{″}, {\vec{r}}^{‴}] = 0$ at all points of the curve
This completes the proof.
A necessary and sufficient condition for a curve to be a plane curve is that $[\dot{\vec{r}}, \ddot{\vec{r}}, \overset{⃛}{\vec{r}}] = 0$ at all points.

Proof
Let $C : \vec{r} = \vec{r} (s)$ be a space curve.
Then, necessary and sufficient condition for a curve to be a plane curve is that
$τ = 0$ at all points of the curve
or $\frac{[\dot{\vec{r}}, \ddot{\vec{r}}, \overset{⃛}{\vec{r}}]}{{| \dot{\vec{r}} \times \ddot{\vec{r}} |}^{2}} = 0$ at all points on the curve
or $[\dot{\vec{r}}, \ddot{\vec{r}}, \overset{⃛}{\vec{r}}] = 0$ at all points on the curve
This completes the proof.
Show that principal normal at consecutive points on a curve do not intersect unless $τ = 0$
Solution
Let $C : \vec{r} = \vec{r} (s)$ be a space curve, also we consider two consecutive points P and Q on C with positions vectors
$\vec{O P} = \vec{r}, \vec{O Q} = \vec{r} + d \vec{r}$
Let $\vec{n}, \vec{n} + d \vec{n}$ be unit vectors along principal normal at P and Q respectively, then these principal normals will intersect if and only if
$\vec{n}, \vec{n} + d \vec{n}, d \vec{r}$ are coplanar
or $[\vec{n}, \vec{n} + d \vec{n}, d \vec{r}] = 0$
or $[\vec{n}, d \vec{n}, d \vec{r}] = 0$
or $[\vec{n}, {\vec{n}}^{'}, {\vec{r}}^{'}] = 0$
or $[\vec{n}, τ \vec{b} - κ \vec{t}, \vec{t}] = 0$
or $τ [\vec{n}, \vec{b}, \vec{t}] = 0$
or $τ = 0$
Hence, principal normal at consecutive points do not intersect unless $τ = 0$
If tangent and binormal on a space curve makes angles $θ$ and $ϕ$ respectively with a fixed direction, then show that $\frac{s i n θ}{s i n ϕ} \frac{d θ}{d ϕ} = \frac{- κ}{τ}$
Solution
Let $\vec{r} = \vec{r} (s)$ a space curve on which $\vec{t}$ and $\vec{b}$ denote the unit vectors along tangent and binormal respectively.
Assume that, fixed direction (say $\vec{a}$ ) make angles $θ$ and $ϕ$ with $\vec{t}$ and $\vec{b}$ respectively, then
$\vec{t} . \vec{a} = a \cos θ$
${\vec{t}}^{'} . \vec{a} = - a \sin θ \frac{d θ}{d s}$
$κ \vec{n} . \vec{a} = - a \sin θ \frac{d θ}{d s}$
Again
$\vec{b} . \vec{a} = a \cos ϕ$
${\vec{b}}^{'} . \vec{a} = - a \sin ϕ \frac{d ϕ}{d s}$
$τ \vec{n} . \vec{a} = - a \sin ϕ \frac{d ϕ}{d s}$
Dividing the equations , we get
$\frac{\sin θ}{\sin ϕ} \frac{d θ}{d ϕ} = \frac{- κ}{τ}$
If there is one-one correspondence between the points of two curves and tangents at corresponding points on these curves are parallel, show that the principal normal at these points are parallel and, therefore the binormal, also prove that $\frac{κ_{1}}{κ} = \frac{d s}{d_{1}} = \frac{τ_{1}}{τ}$
Solution
Let C and C1 be two curves such there is one-one correspondence between their points and the tangents at these corresponding points are parallel.
$\vec{t} = \pm {\vec{t}}_{1}$
[we use suffix-unity to distinguish quantities belonging to curve c1]
Differentiating (i) w.r.to. s, we get
${\vec{t}}^{'} = \pm \frac{d {\vec{t}}_{1}}{d s_{1}} \frac{d s_{1}}{d s}$
${\vec{t}}^{'} = \pm κ_{1} {\vec{n}}_{1} \frac{d s_{1}}{d s}$
$κ \vec{n} = \pm κ_{1} {\vec{n}}_{1} \frac{d s_{1}}{d s}$
Taking magnitude we get
$\frac{d s}{d s_{1}} = \frac{κ_{1}}{κ}$ ...(A)
Substituting the value of $\frac{d s}{d s_{1}}$ from (i) we get that
$\vec{n} = \pm {\vec{n}}_{1}$
Thus principal normals at corresponding points of C and C1 are parallel.
1st relation established

Now taking cross product, we get
$\vec{t} \times \vec{n} = \pm {\vec{t}}_{1} \times {\vec{n}}_{1}$
$\vec{b} = \pm {\vec{b}}_{1}$ ...(i)
Hence binormals at corresponding points of the curves C and $C_{1}$ are also parallel.
2nd relation established

Again differentiating (i) w. r. to. s, we get
${\vec{b}}^{'} = \pm \frac{d {\vec{b}}_{1}}{d s_{1}} \frac{d s_{1}}{d s}$
${\vec{b}}^{'} = \pm τ_{1} {\vec{n}}_{1} \frac{d s_{1}}{d s}$
$- τ \vec{n} = \pm τ_{1} {\vec{n}}_{1} \frac{d s_{1}}{d s}$
Taking magnitude we get
$\frac{d s_{1}}{d s} = \frac{τ_{1}}{τ}$ ...(B)
Hence equating (A) and (B) we get
$\frac{κ_{1}}{κ} = \frac{d s}{d_{1}} = \frac{τ_{1}}{τ}$
3rd relation established

Hence, if tangents at corresponding points on the curves are parallel, then principal normals are parallel and, therefore, the binormals along with the given relation.

Exercise

Show/Hide 👉 Click Here

Find the curvature and torsion of a following curve
1. $x = a (3 t - t^{3}), y = 3 a t^{2}, z = a (3 t + t^{3})$
  
  Show/Hide 👉 Click Here
  
  Solution
  Given space curve is
  $\vec{r} = (3 a - 3 a t^{2}, 3 a t^{2}, 3 a + 3 a t^{2})$
  By successive differentiation w. r. to. t, we get
  $\dot{\vec{r}} = (3 a - 3 a t^{2}, 6 a t, 3 a + 3 a t^{2})$
  $\ddot{\vec{r}} = (- 6 a t, 6 a, 6 a t)$
  $\overset{⃛}{\vec{r}} = (- 6 a, 0, 6 a)$
  Since, curvature and torsion of space curve is given by the formula
  $κ = \frac{| \dot{\vec{r}} \times \ddot{\vec{r}} |}{| \dot{\vec{r}} |^{3}}, τ = \frac{[\dot{\vec{r}}, \ddot{\vec{r}}, \overset{⃛}{\vec{r}}]}{| \dot{\vec{r}} \times \ddot{\vec{r}} |^{2}}$
  Thus, taking magnitude of (i), we get
  $| \dot{\vec{r}} | = 3 a \sqrt{(1 - t^{2})^{2} + (2 t)^{2} + (1 + t^{2})^{2}}$
  or $| \dot{\vec{r}} | = 3 a \sqrt{2} (1 + t^{2})$
  Taking cross product between (ii) and (iii), we get
  $\dot{\vec{r}} \times \ddot{\vec{r}} = 18 a^{2} [\begin{matrix} \vec{i} & \vec{j} & \vec{k} \\ 1 - t^{2} & 2 t & 1 + t^{2} \\ - t & 1 & t \end{matrix}]$
  or $\dot{\vec{r}} \times \ddot{\vec{r}} = 18 a^{2} (t^{2} - 1, - 2 t, t^{2} + 1)$
  Taking magnitude, we get
  $| \dot{\vec{r}} \times \ddot{\vec{r}} | = 18 a^{2} \sqrt{2} (1 + t^{2})$
  Operating scalar triple product among (i) ,(ii) (iii), we get
  $[\dot{\vec{r}}, \ddot{\vec{r}}, \overset{⃛}{\vec{r}}] = | \begin{matrix} 3 a - 3 a t^{2} & 6 a t & 3 a + 3 a t^{2} \\ - 6 a t & 6 a & 6 a t \\ - 6 a & 0 & 6 a \end{matrix} |$
  or $[\dot{\vec{r}}, \ddot{\vec{r}}, \overset{⃛}{\vec{r}}] = 108 a^{3} | \begin{matrix} 2 & 2 t & 1 + t^{2} \\ 0 & 1 & t \\ 0 & 0 & 1 \end{matrix} |$
  or $[\dot{\vec{r}}, \ddot{\vec{r}}, \overset{⃛}{\vec{r}}] = 216 a^{3}$
  Hence, curvature and torsion of the curve are
  $κ = \frac{| \dot{\vec{r}} \times \ddot{\vec{r}} |}{| \dot{\vec{r}} |^{3}}$
  or $κ = \frac{18 a^{2} \sqrt{2} (1 + t^{2})}{3^{3} a^{3} \sqrt{2^{3}} (1 + t^{2})^{3}}$
  or $κ = \frac{1}{3 a (1 + t^{2})^{2}}$
  and
  $τ = \frac{[\dot{\vec{r}}, \ddot{\vec{r}}, \overset{⃛}{\vec{r}}]}{| \dot{\vec{r}} \times \ddot{\vec{r}} |^{2}}$
  or $τ = \frac{216 a^{3}}{18^{2} a^{4} \sqrt{2^{2}} (1 + t^{2})^{2}}$
  or $τ = \frac{1}{3 a (1 + t^{2})^{2}}$
  This completes the solution.
2. $x = t, y = t^{2}, z = t^{3}$
  
  Show/Hide 👉 Click Here
  
  Solution
  Given space curve is
  $\vec{r} = (t, t^{2}, t^{3})$
  By successive differentiation w. r. to. t, we get
  $\dot{\vec{r}} = (1, 2 t, 3 t^{2})$
  $\ddot{\vec{r}} = (0, 2, 6 t)$
  $\overset{⃛}{\vec{r}} = (0, 0, 6)$
  Since, curvature and torsion of space curve is given by the formula
  $κ = \frac{| \dot{\vec{r}} \times \ddot{\vec{r}} |}{| \dot{\vec{r}} |^{3}}, τ = \frac{[\dot{\vec{r}}, \ddot{\vec{r}}, \overset{⃛}{\vec{r}}]}{| \dot{\vec{r}} \times \ddot{\vec{r}} |^{2}}$
  Thus, taking magnitude of (i), we get
  $| \dot{\vec{r}} | = \sqrt{1 + (2 t)^{2} + (3 t^{2})^{2}}$
  or $| \dot{\vec{r}} | = \sqrt{1 + 4 t^{2} + 9 t^{4}}$
  Taking cross product between (ii) and (iii), we get
  $\dot{\vec{r}} \times \ddot{\vec{r}} = | \begin{matrix} \vec{i} & \vec{j} & \vec{k} \\ 1 & 2 t & 3 t^{2} \\ 0 & 2 & 6 t \end{matrix} |$
  or $\dot{\vec{r}} \times \ddot{\vec{r}} = \vec{i} (2 t \cdot 6 t - 3 t^{2} \cdot 2) - \vec{j} (1 \cdot 6 t - 0 \cdot 3 t^{2}) + \vec{k} (1 \cdot 2 - 0 \cdot 2 t)$
  or $\dot{\vec{r}} \times \ddot{\vec{r}} = \vec{i} (12 t^{2} - 6 t^{2}) - \vec{j} (6 t) + \vec{k} (2)$
  or $\dot{\vec{r}} \times \ddot{\vec{r}} = (6 t^{2}, - 6 t, 2)$
  Taking magnitude, we get
  $| \dot{\vec{r}} \times \ddot{\vec{r}} | = \sqrt{(6 t^{2})^{2} + (- 6 t)^{2} + 2^{2}}$
  or $| \dot{\vec{r}} \times \ddot{\vec{r}} | = \sqrt{36 t^{4} + 36 t^{2} + 4}$
  Operating scalar triple product among (i), (ii), (iii), we get
  $[\dot{\vec{r}}, \ddot{\vec{r}}, \overset{⃛}{\vec{r}}] = | \begin{matrix} 1 & 2 t & 3 t^{2} \\ 0 & 2 & 6 t \\ 0 & 0 & 6 \end{matrix} |$
  or $[\dot{\vec{r}}, \ddot{\vec{r}}, \overset{⃛}{\vec{r}}] = 6 | \begin{matrix} 1 & 2 t & 3 t^{2} \\ 0 & 2 & 6 t \\ 0 & 0 & 1 \end{matrix} |$
  or $[\dot{\vec{r}}, \ddot{\vec{r}}, \overset{⃛}{\vec{r}}] = 6 (1 \cdot 2 \cdot 6 - 0 \cdot 6 t)$
  or $[\dot{\vec{r}}, \ddot{\vec{r}}, \overset{⃛}{\vec{r}}] = 12$
  Hence, curvature and torsion of the curve are
  $κ = \frac{| \dot{\vec{r}} \times \ddot{\vec{r}} |}{| \dot{\vec{r}} |^{3}}$
  or $κ = \frac{\sqrt{36 t^{4} + 36 t^{2} + 4}}{{(\sqrt{1 + 4 t^{2} + 9 t^{4}})}^{3}}$
  or $κ = \frac{\sqrt{36 t^{4} + 36 t^{2} + 4}}{(1 + 4 t^{2} + 9 t^{4})^{3 / 2}}$
  and
  $τ = \frac{[\dot{\vec{r}}, \ddot{\vec{r}}, \overset{⃛}{\vec{r}}]}{| \dot{\vec{r}} \times \ddot{\vec{r}} |^{2}}$
  or $τ = \frac{12}{(36 t^{4} + 36 t^{2} + 4)}$
  or $τ = \frac{3}{9 t^{4} + 9 t^{2} + 1}$
  This completes the solution.
3. $x = a \cos t, y = a \sin t, z = a t \cot α$
  
  Show/Hide 👉 Click Here
  
  Solution
  Given space curve is
  $\vec{r} = (a \cos t, a \sin t, a t \cot α)$
  By successive differentiation w. r. to. t, we get
  $\dot{\vec{r}} = (- a \sin t, a \cos t, a \cot α)$
  $\ddot{\vec{r}} = (- a \cos t, - a \sin t, 0)$
  $\overset{⃛}{\vec{r}} = (a \sin t, - a \cos t, 0)$
  Since, curvature and torsion of space curve is given by the formula
  $κ = \frac{| \dot{\vec{r}} \times \ddot{\vec{r}} |}{| \dot{\vec{r}} |^{3}}, τ = \frac{[\dot{\vec{r}}, \ddot{\vec{r}}, \overset{⃛}{\vec{r}}]}{| \dot{\vec{r}} \times \ddot{\vec{r}} |^{2}}$
  Thus, taking magnitude of (i), we get
  $| \dot{\vec{r}} | = \sqrt{(- a \sin t)^{2} + (a \cos t)^{2} + (a \cot α)^{2}}$
  or $| \dot{\vec{r}} | = \sqrt{a^{2} \sin^{2} t + a^{2} \cos^{2} t + a^{2} \cot^{2} α}$
  or $| \dot{\vec{r}} | = a \sqrt{1 + \cot^{2} α}$
  or $| \dot{\vec{r}} | = \frac{a}{\sin α}$
  Taking cross product between (ii) and (iii), we get
  $\dot{\vec{r}} \times \ddot{\vec{r}} = | \begin{matrix} \vec{i} & \vec{j} & \vec{k} \\ - a \sin t & a \cos t & a \cot α \\ - a \cos t & - a \sin t & 0 \end{matrix} |$
  or $\dot{\vec{r}} \times \ddot{\vec{r}} = \vec{i} (a^{2} \sin t \cot α) - \vec{j} (a^{2} \cos t \cot α) + \vec{k} (a^{2} \sin^{2} t + a^{2} \cos^{2} t)$
  or $\dot{\vec{r}} \times \ddot{\vec{r}} = (a^{2} \sin t \cot α, - a^{2} \cos t \cot α, a^{2})$
  Taking magnitude, we get
  $| \dot{\vec{r}} \times \ddot{\vec{r}} | = \sqrt{(a^{2} \sin t \cot α)^{2} + (- a^{2} \cos t \cot α)^{2} + a^{2}}$
  or $| \dot{\vec{r}} \times \ddot{\vec{r}} | = \sqrt{a^{4} \sin^{2} t \cot^{2} α + a^{4} \cos^{2} t \cot^{2} α + a^{4}}$
  or $| \dot{\vec{r}} \times \ddot{\vec{r}} | = a^{2} \sqrt{\cot^{2} α (\sin^{2} t + \cos^{2} t) + 1}$
  or $| \dot{\vec{r}} \times \ddot{\vec{r}} | = a^{2} \sqrt{\cot^{2} α + 1}$
  or $| \dot{\vec{r}} \times \ddot{\vec{r}} | = a^{2} \csc α$
  Operating scalar triple product among (i), (ii), (iii), we get
  $[\dot{\vec{r}}, \ddot{\vec{r}}, \overset{⃛}{\vec{r}}] = | \begin{matrix} - a \sin t & a \cos t & a \cot α \\ - a \cos t & - a \sin t & 0 \\ a \sin t & - a \cos t & 0 \end{matrix} |$
  or $[\dot{\vec{r}}, \ddot{\vec{r}}, \overset{⃛}{\vec{r}}] = a^{3} \cot α | \begin{matrix} - \sin t & \cos t \\ - \cos t & - \sin t \end{matrix} |$
  or $[\dot{\vec{r}}, \ddot{\vec{r}}, \overset{⃛}{\vec{r}}] = a^{3} \cot α ((- \sin t) (- \sin t) - (\cos t) (- \cos t))$
  or $[\dot{\vec{r}}, \ddot{\vec{r}}, \overset{⃛}{\vec{r}}] = a^{3} \cot α (\sin^{2} t + \cos^{2} t)$
  or $[\dot{\vec{r}}, \ddot{\vec{r}}, \overset{⃛}{\vec{r}}] = a^{3} \cot α$
  Hence, curvature and torsion of the curve are
  $κ = \frac{| \dot{\vec{r}} \times \ddot{\vec{r}} |}{| \dot{\vec{r}} |^{3}}$
  or $κ = \frac{a^{2} \csc α}{{(\frac{a}{\sin α})}^{3}}$
  or $κ = \frac{a^{2} \csc α}{\frac{a^{3}}{\sin^{3} α}}$
  or $κ = \frac{\sin^{3} α}{a \sin α}$
  or $κ = \frac{\sin^{2} α}{a}$
  and
  $τ = \frac{[\dot{\vec{r}}, \ddot{\vec{r}}, \overset{⃛}{\vec{r}}]}{| \dot{\vec{r}} \times \ddot{\vec{r}} |^{2}}$
  or $τ = \frac{a^{3} \cot α}{(a^{2} \csc α)^{2}}$
  or $τ = \frac{a^{3} \cot α}{a^{4} \csc^{2} α}$
  or $τ = \frac{a^{3} \cot α \sin^{2} α}{a^{4}}$
  or $τ = \frac{\cos α \sin α}{a}$
  This completes the solution.
4. $x = a \cos t, y = a \sin t, z = c t$
  
  Show/Hide 👉 Click Here
  
  Solution
  Given space curve is
  $\vec{r} = (a \cos t, a \sin t, c t)$
  By successive differentiation w. r. to. t, we get
  $\dot{\vec{r}} = (- a \sin t, a \cos t, c)$
  $\ddot{\vec{r}} = (- a \cos t, - a \sin t, 0)$
  $\overset{⃛}{\vec{r}} = (a \sin t, - a \cos t, 0)$
  Since, curvature and torsion of space curve is given by the formula
  $κ = \frac{| \dot{\vec{r}} \times \ddot{\vec{r}} |}{| \dot{\vec{r}} |^{3}}, τ = \frac{[\dot{\vec{r}}, \ddot{\vec{r}}, \overset{⃛}{\vec{r}}]}{| \dot{\vec{r}} \times \ddot{\vec{r}} |^{2}}$
  Thus, taking magnitude of (i), we get
  $| \dot{\vec{r}} | = \sqrt{(- a \sin t)^{2} + (a \cos t)^{2} + c^{2}}$
  or $| \dot{\vec{r}} | = \sqrt{a^{2} \sin^{2} t + a^{2} \cos^{2} t + c^{2}}$
  or $| \dot{\vec{r}} | = \sqrt{a^{2} + c^{2}}$
  Taking cross product between (ii) and (iii), we get
  $\dot{\vec{r}} \times \ddot{\vec{r}} = | \begin{matrix} \vec{i} & \vec{j} & \vec{k} \\ - a \sin t & a \cos t & c \\ - a \cos t & - a \sin t & 0 \end{matrix} |$
  or $\dot{\vec{r}} \times \ddot{\vec{r}} = \vec{i} (a c \sin t) - \vec{j} (a c \cos t) + \vec{k} (a^{2} (\sin^{2} t + \cos^{2} t))$
  or $\dot{\vec{r}} \times \ddot{\vec{r}} = (a c \sin t, - a c \cos t, a^{2})$
  Taking magnitude, we get
  $| \dot{\vec{r}} \times \ddot{\vec{r}} | = \sqrt{(a c \sin t)^{2} + (- a c \cos t)^{2} + a^{2}}$
  or $| \dot{\vec{r}} \times \ddot{\vec{r}} | = \sqrt{a^{2} c^{2} (\sin^{2} t + \cos^{2} t) + a^{4}}$
  or $| \dot{\vec{r}} \times \ddot{\vec{r}} | = \sqrt{a^{2} c^{2} + a^{4}}$
  or $| \dot{\vec{r}} \times \ddot{\vec{r}} | = a \sqrt{a^{2} + c^{2}}$
  Operating scalar triple product among (i), (ii), (iii), we get
  $[\dot{\vec{r}}, \ddot{\vec{r}}, \overset{⃛}{\vec{r}}] = | \begin{matrix} - a \sin t & a \cos t & c \\ - a \cos t & - a \sin t & 0 \\ a \sin t & - a \cos t & 0 \end{matrix} |$
  or $[\dot{\vec{r}}, \ddot{\vec{r}}, \overset{⃛}{\vec{r}}] = c | \begin{matrix} - a \sin t & a \cos t \\ - a \cos t & - a \sin t \end{matrix} |$
  or $[\dot{\vec{r}}, \ddot{\vec{r}}, \overset{⃛}{\vec{r}}] = c (a^{2} \sin^{2} t + a^{2} \cos^{2} t)$
  or $[\dot{\vec{r}}, \ddot{\vec{r}}, \overset{⃛}{\vec{r}}] = c a^{2} (\sin^{2} t + \cos^{2} t)$
  or $[\dot{\vec{r}}, \ddot{\vec{r}}, \overset{⃛}{\vec{r}}] = c a^{2}$
  Hence, curvature and torsion of the curve are
  $κ = \frac{| \dot{\vec{r}} \times \ddot{\vec{r}} |}{| \dot{\vec{r}} |^{3}}$
  or $κ = \frac{a \sqrt{a^{2} + c^{2}}}{(a^{2} + c^{2})^{3 / 2}}$
  or $κ = \frac{a}{(a^{2} + c^{2})}$
  and
  $τ = \frac{[\dot{\vec{r}}, \ddot{\vec{r}}, \overset{⃛}{\vec{r}}]}{| \dot{\vec{r}} \times \ddot{\vec{r}} |^{2}}$
  or $τ = \frac{c a^{2}}{(a \sqrt{a^{2} + c^{2}})^{2}}$
  or $τ = \frac{c a^{2}}{a^{2} (a^{2} + c^{2})}$
  or $τ = \frac{c}{a^{2} + c^{2}}$
  This completes the solution.
5. $\vec{r} = (a - a \sin t, a - a \cos t, b t)$
  
  Show/Hide 👉 Click Here
  
  Solution
  Given space curve is
  $\vec{r} = (a - a \sin t, a - a \cos t, b t)$
  By successive differentiation w. r. to. t, we get
  $\dot{\vec{r}} = (- a \cos t, a \sin t, b)$
  $\ddot{\vec{r}} = (a \sin t, a \cos t, 0)$
  $\overset{⃛}{\vec{r}} = (a \cos t, - a \sin t, 0)$
  Since, curvature and torsion of space curve is given by the formula
  $κ = \frac{| \dot{\vec{r}} \times \ddot{\vec{r}} |}{| \dot{\vec{r}} |^{3}}, τ = \frac{[\dot{\vec{r}}, \ddot{\vec{r}}, \overset{⃛}{\vec{r}}]}{| \dot{\vec{r}} \times \ddot{\vec{r}} |^{2}}$
  Thus, taking magnitude of (i), we get
  $| \dot{\vec{r}} | = \sqrt{(- a \cos t)^{2} + (a \sin t)^{2} + (b)^{2}}$
  or $| \dot{\vec{r}} | = \sqrt{a^{2} (\cos^{2} t + \sin^{2} t) + b^{2}} = \sqrt{a^{2} + b^{2}}$
  Taking cross product between (ii) and (iii), we get
  $\dot{\vec{r}} \times \ddot{\vec{r}} = [\begin{matrix} \vec{i} & \vec{j} & \vec{k} \\ - a \cos t & a \sin t & b \\ a \sin t & a \cos t & 0 \end{matrix}]$
  or $\dot{\vec{r}} \times \ddot{\vec{r}} = [\begin{matrix} (- a b \cos t) \vec{i} - (a b \sin t) \vec{j} + (a^{2} \cos^{2} t + a^{2} \sin^{2} t) \vec{k} \end{matrix}]$
  or $\dot{\vec{r}} \times \ddot{\vec{r}} = (- a b \cos t, - a b \sin t, a^{2})$
  Taking magnitude, we get
  $| \dot{\vec{r}} \times \ddot{\vec{r}} | = \sqrt{(- a b \cos t)^{2} + (- a b \sin t)^{2} + a^{2}} = a \sqrt{a^{2} + b^{2}}$
  Operating scalar triple product among (i) ,(ii) (iii), we get
  $[\dot{\vec{r}}, \ddot{\vec{r}}, \overset{⃛}{\vec{r}}] = | \begin{matrix} - a \cos t & a \sin t & b \\ a \sin t & a \cos t & 0 \\ a \cos t & - a \sin t & 0 \end{matrix} |$
  or $[\dot{\vec{r}}, \ddot{\vec{r}}, \overset{⃛}{\vec{r}}] = a^{2} b | \begin{matrix} - \cos t & \sin t & 1 \\ \sin t & \cos t & 0 \\ \cos t & - \sin t & 0 \end{matrix} |$
  or $[\dot{\vec{r}}, \ddot{\vec{r}}, \overset{⃛}{\vec{r}}] = a^{2} b (- \sin t (\sin t) - \cos t (\cos t))$
  or $[\dot{\vec{r}}, \ddot{\vec{r}}, \overset{⃛}{\vec{r}}] = - a^{2} b$
  Hence, curvature and torsion of the curve are
  $κ = \frac{| \dot{\vec{r}} \times \ddot{\vec{r}} |}{| \dot{\vec{r}} |^{3}}$
  or $κ = \frac{a \sqrt{a^{2} + b^{2}}}{(a^{2} + b^{2})^{3 / 2}}$
  or $κ = \frac{a}{(a^{2} + b^{2})}$
  and
  $τ = \frac{[\dot{\vec{r}}, \ddot{\vec{r}}, \overset{⃛}{\vec{r}}]}{| \dot{\vec{r}} \times \ddot{\vec{r}} |^{2}}$
  or $τ = - \frac{a^{2} b}{(a \sqrt{a^{2} + b^{2}})^{2}}$
  or $τ = - \frac{b}{a^{2} + b^{2}}$
  This completes the solution.
6. $x = a c o s θ, y = a s i n θ, z = a θ t a n α$
  
  Show/Hide 👉 Click Here
  
  Solution
  Given space curve is
  $\vec{r} = (a \cos θ, a \sin θ, a θ \tan α)$
  By successive differentiation w. r. to. $θ$ , we get
  $\dot{\vec{r}} = (- a \sin θ, a \cos θ, a \tan α)$
  $\ddot{\vec{r}} = (- a \cos θ, - a \sin θ, 0)$
  $\overset{⃛}{\vec{r}} = (a \sin θ, - a \cos θ, 0)$
  Since, curvature and torsion of space curve is given by the formula
  $κ = \frac{| \dot{\vec{r}} \times \ddot{\vec{r}} |}{| \dot{\vec{r}} |^{3}}, τ = \frac{[\dot{\vec{r}}, \ddot{\vec{r}}, \overset{⃛}{\vec{r}}]}{| \dot{\vec{r}} \times \ddot{\vec{r}} |^{2}}$
  Thus, taking magnitude of (i), we get
  $| \dot{\vec{r}} | = \sqrt{(- a \sin θ)^{2} + (a \cos θ)^{2} + (a \tan α)^{2}}$
  or $| \dot{\vec{r}} | = \sqrt{a^{2} (\sin^{2} θ + \cos^{2} θ) + a^{2} \tan^{2} α} = a \sqrt{1 + \tan^{2} α} = a \sec α$
  Taking cross product between (ii) and (iii), we get
  $\dot{\vec{r}} \times \ddot{\vec{r}} = [\begin{matrix} \vec{i} & \vec{j} & \vec{k} \\ - a \sin θ & a \cos θ & a \tan α \\ - a \cos θ & - a \sin θ & 0 \end{matrix}]$
  or $\dot{\vec{r}} \times \ddot{\vec{r}} = [\begin{matrix} a^{2} \tan α \sin θ & a^{2} \tan α \cos θ & a^{2} \end{matrix}]$
  or $\dot{\vec{r}} \times \ddot{\vec{r}} = (a^{2} \tan α \sin θ, a^{2} \tan α \cos θ, a^{2})$
  Taking magnitude, we get
  $| \dot{\vec{r}} \times \ddot{\vec{r}} | = a^{2} \sqrt{\tan^{2} α (\sin^{2} θ + \cos^{2} θ) + 1} = a^{2} \sqrt{\tan^{2} α + 1} = a^{2} \sec α$
  Operating scalar triple product among (i) ,(ii) (iii), we get
  $[\dot{\vec{r}}, \ddot{\vec{r}}, \overset{⃛}{\vec{r}}] = | \begin{matrix} - a \sin θ & a \cos θ & a \tan α \\ - a \cos θ & - a \sin θ & 0 \\ a \sin θ & - a \cos θ & 0 \end{matrix} |$
  or $[\dot{\vec{r}}, \ddot{\vec{r}}, \overset{⃛}{\vec{r}}] = a^{3} \tan α | \begin{matrix} - \sin θ & \cos θ & 1 \\ - \cos θ & - \sin θ & 0 \\ \sin θ & - \cos θ & 0 \end{matrix} |$
  or $[\dot{\vec{r}}, \ddot{\vec{r}}, \overset{⃛}{\vec{r}}] = a^{3} \tan α (\sin^{2} θ + \cos^{2} θ)$
  or $[\dot{\vec{r}}, \ddot{\vec{r}}, \overset{⃛}{\vec{r}}] = a^{3} \tan α = a^{3} \tan α$
  Hence, curvature and torsion of the curve are
  $κ = \frac{| \dot{\vec{r}} \times \ddot{\vec{r}} |}{| \dot{\vec{r}} |^{3}}$
  or $κ = \frac{a^{2} \sec α}{(a \sec α)^{3}}$
  or $κ = \frac{1}{a \sec^{2} α} = \frac{\cos^{2} α}{a}$
  and
  $τ = \frac{[\dot{\vec{r}}, \ddot{\vec{r}}, \overset{⃛}{\vec{r}}]}{| \dot{\vec{r}} \times \ddot{\vec{r}} |^{2}}$
  or $τ = \frac{a^{3} \tan α}{(a^{2} \sec α)^{2}}$
  or $τ = \frac{a^{3} \tan α}{a^{4} \sec^{2} α} = \frac{\tan α}{a \sec^{2} α} = \frac{\sin α \cos α}{a}$
  This completes the solution.
7. $x = a \sqrt{6} t^{3}, y = a (1 + 3 t^{2}), z = a \sqrt{6} t$
  
  Show/Hide 👉 Click Here
  
  Solution
  Given space curve is
  $\vec{r} = (a \sqrt{6} t^{3}, a (1 + 3 t^{2}), a \sqrt{6} t)$
  By successive differentiation w. r. to. t, we get
  $\dot{\vec{r}} = (3 a \sqrt{6} t^{2}, 6 a t, a \sqrt{6})$
  $\ddot{\vec{r}} = (6 a \sqrt{6} t, 6 a, 0)$
  $\overset{⃛}{\vec{r}} = (6 a \sqrt{6}, 0, 0)$
  Since, curvature and torsion of space curve is given by the formula
  $κ = \frac{| \dot{\vec{r}} \times \ddot{\vec{r}} |}{| \dot{\vec{r}} |^{3}}, τ = \frac{[\dot{\vec{r}}, \ddot{\vec{r}}, \overset{⃛}{\vec{r}}]}{| \dot{\vec{r}} \times \ddot{\vec{r}} |^{2}}$
  Thus, taking magnitude of (i), we get
  $| \dot{\vec{r}} | = \sqrt{(3 a \sqrt{6} t^{2})^{2} + (6 a t)^{2} + (a \sqrt{6})^{2}}$
  or $| \dot{\vec{r}} | = a \sqrt{6} (1 + 3 t^{2})$
  Taking cross product between (ii) and (iii), we get
  $\dot{\vec{r}} \times \ddot{\vec{r}} = [\begin{matrix} \vec{i} & \vec{j} & \vec{k} \\ a \sqrt{6} t^{3} & a (1 + 3 t^{2}) & a \sqrt{6} t \\ 3 a \sqrt{6} t^{2} & 6 a t & a \sqrt{6} \\ 6 a \sqrt{6} t & 6 a & 0 \end{matrix}]$
  or $\dot{\vec{r}} \times \ddot{\vec{r}} = 6 a^{2} \sqrt{6} [\begin{matrix} \vec{i} & \vec{j} & \vec{k} \\ t^{3} & 1 + 3 t^{2} & t \\ 3 t^{2} & 6 & 1 \\ 1 & 0 & 0 \end{matrix}]$
  or $\dot{\vec{r}} \times \ddot{\vec{r}} = 6 a^{2} \sqrt{6} (0, - 3 t^{3}, 3 t^{4} + 1)$
  Taking magnitude, we get
  $| \dot{\vec{r}} \times \ddot{\vec{r}} | = 6 a^{2} \sqrt{6} (1 + 3 t^{2})$
  Operating scalar triple product among (i) ,(ii) (iii), we get
  $[\dot{\vec{r}}, \ddot{\vec{r}}, \overset{⃛}{\vec{r}}] = | \begin{matrix} a \sqrt{6} t^{3} & a (1 + 3 t^{2}) & a \sqrt{6} t \\ 3 a \sqrt{6} t^{2} & 6 a t & a \sqrt{6} \\ 6 a \sqrt{6} t & 6 a & 0 \end{matrix} |$
  or $[\dot{\vec{r}}, \ddot{\vec{r}}, \overset{⃛}{\vec{r}}] = 36 a^{3} \sqrt{6} | \begin{matrix} t^{3} & 1 + 3 t^{2} & t \\ 3 t^{2} & 6 & 1 \\ 1 & 0 & 0 \end{matrix} |$
  or $[\dot{\vec{r}}, \ddot{\vec{r}}, \overset{⃛}{\vec{r}}] = 36 a^{3} \sqrt{6} (- 1 - 9 t^{2})$
  or $[\dot{\vec{r}}, \ddot{\vec{r}}, \overset{⃛}{\vec{r}}] = - 216 a^{3}$
  Hence, curvature and torsion of the curve are
  $κ = \frac{| \dot{\vec{r}} \times \ddot{\vec{r}} |}{| \dot{\vec{r}} |^{3}}$
  or $κ = \frac{6 a^{2} \sqrt{6} (1 + 3 t^{2})}{a^{3} {\sqrt{6}}^{3} (1 + 3 t^{2})^{3}}$
  or $κ = \frac{1}{a (1 + 3 t^{2})^{2}}$
  and
  $τ = \frac{[\dot{\vec{r}}, \ddot{\vec{r}}, \overset{⃛}{\vec{r}}]}{| \dot{\vec{r}} \times \ddot{\vec{r}} |^{2}}$
  or $τ = \frac{- 216 a^{3}}{6^{2} a^{4} {\sqrt{6}}^{2} (1 + 3 t^{2})^{2}}$
  or $τ = - \frac{1}{a (1 + 3 t^{2})^{2}}$
  This completes the solution.
8. $x = 3 t, y = 3 t^{2}, z = 2 t^{3}$
  
  Show/Hide 👉 Click Here
  
  Solution
  Given space curve is
  $\vec{r} = (3 t, 3 t^{2}, 2 t^{3})$
  By successive differentiation w. r. to. t, we get
  $\dot{\vec{r}} = (3, 6 t, 6 t^{2})$
  $\ddot{\vec{r}} = (0, 6, 12 t)$
  $\overset{⃛}{\vec{r}} = (0, 0, 12)$
  Since, curvature and torsion of space curve is given by the formula
  $κ = \frac{| \dot{\vec{r}} \times \ddot{\vec{r}} |}{| \dot{\vec{r}} |^{3}}, τ = \frac{[\dot{\vec{r}}, \ddot{\vec{r}}, \overset{⃛}{\vec{r}}]}{| \dot{\vec{r}} \times \ddot{\vec{r}} |^{2}}$
  Thus, taking magnitude of (i), we get
  $| \dot{\vec{r}} | = 3 \sqrt{1 + (2 t)^{2} + (2 t^{2})^{2}}$
  or $| \dot{\vec{r}} | = 3 \sqrt{1 + 4 t^{2} + 4 t^{4}}$
  or $| \dot{\vec{r}} | = 3 (1 + 2 t^{2})$
  Taking cross product between (ii) and (iii), we get
  $\dot{\vec{r}} \times \ddot{\vec{r}} = [\begin{matrix} \vec{i} & \vec{j} & \vec{k} \\ 3 & 6 t & 6 t^{2} \\ 0 & 6 & 12 t \end{matrix}]$
  or $\dot{\vec{r}} \times \ddot{\vec{r}} = 18 (2 t^{2}, - 2 t, 1)$
  Taking magnitude, we get
  $| \dot{\vec{r}} \times \ddot{\vec{r}} | = 18 (1 + 2 t^{2})$
  Operating scalar triple product among (i) ,(ii) (iii), we get
  $[\dot{\vec{r}}, \ddot{\vec{r}}, \overset{⃛}{\vec{r}}] = | \begin{matrix} 3 & 6 t & 6 t^{2} \\ 0 & 6 & 12 t \\ 0 & 0 & 12 \end{matrix} |$
  or $[\dot{\vec{r}}, \ddot{\vec{r}}, \overset{⃛}{\vec{r}}] = 216$
  Hence, curvature and torsion of the curve are
  $κ = \frac{| \dot{\vec{r}} \times \ddot{\vec{r}} |}{| \dot{\vec{r}} |^{3}}$
  or $κ = \frac{18 (1 + 2 t^{2})}{3^{2} (1 + 2 t^{2})^{2}}$
  or $κ = \frac{2}{3 (1 + 2 t^{2})^{2}}$
  and
  $τ = \frac{[\dot{\vec{r}}, \ddot{\vec{r}}, \overset{⃛}{\vec{r}}]}{| \dot{\vec{r}} \times \ddot{\vec{r}} |^{2}}$
  or $τ = \frac{216}{18^{2} (1 + 2 t^{2})^{2}}$
  or $τ = \frac{2}{3 (1 + 2 t^{2})^{2}}$
  This completes the solution.
For a space curve $\vec{r} = \vec{r} (s)$ show that
1. $[{\vec{t}}^{'}, {\vec{t}}^{″}, {\vec{t}}^{‴}] = κ^{5} \frac{d}{d s} (\frac{τ}{κ})$
  
  Show/Hide 👉 Click Here
  
  Solution
  We know that:
  $\vec{t} = \frac{d \vec{r}}{d s}$
  By successive differentiation w.r.t. $s$ , we get:
  ${\vec{t}}^{'} = κ \vec{n}$
  and
  ${\vec{t}}^{″} = κ^{'} \vec{n} + κ \frac{d \vec{n}}{d s}$
  ${\vec{t}}^{″} = κ^{'} \vec{n} + κ (τ \vec{b} - κ \vec{t})$
  ${\vec{t}}^{″} = - κ^{2} \vec{t} + κ^{'} \vec{n} + κ τ \vec{b}$
  and
  ${\vec{t}}^{‴} = \frac{d}{d s} (- κ^{2} \vec{t} + κ^{'} \vec{n} + κ τ \vec{b})$
  ${\vec{t}}^{‴} = - 2 κ κ^{'} \vec{t} - κ^{3} \vec{n} + κ^{″} \vec{n} + κ^{'} (τ \vec{b} - κ \vec{t}) + κ^{'} τ \vec{b} + κ τ^{'} \vec{b} - κ τ^{2} \vec{n}$
  ${\vec{t}}^{‴} = (κ^{″} - κ^{3} - κ τ^{2}) \vec{n} + (2 κ^{'} τ + κ τ^{'}) \vec{b} - 3 κ κ^{'} \vec{t}$
  Next,
  $[{\vec{t}}^{'}, {\vec{t}}^{″}, {\vec{t}}^{‴}] = κ | \begin{matrix} 0 & 1 & 0 \\ - κ^{2} & κ^{'} & κ τ \\ - 3 κ κ^{'} & (κ^{″} - κ^{3} - κ τ^{2}) & (2 κ^{'} τ + κ τ^{'}) \end{matrix} |$
  This simplifies to:
  $[{\vec{t}}^{'}, {\vec{t}}^{″}, {\vec{t}}^{‴}] = κ^{5} \frac{d}{d s} (\frac{τ}{κ})$
  This completes the proof.
2. $[{\vec{b}}^{'}, {\vec{b}}^{″}, {\vec{b}}^{‴}] = τ^{5} \frac{d}{d s} (\frac{κ}{τ})$

Curvature and Torsion

Curvature

Expression of Curvature

Torsion

Expression of Torsion

Expression of Screw-Curvature:

Serret-frenet formula.

Theorems

Exercise

No comments:

Post a Comment

Follow Me

SEARCH

LATEST

SECTIONS

Pageviews

Popular

Archive

Courses

Categories

Comments