Question

How do you find the curvature $\kappa $of the curve $r\left( t \right) = ti + {t^2}j + {t^2}k$?

Answer 1

Hint: We will solve this question using the formula for curvature of the curve.
2. To solve using that formula we will also use the first and the second derivative of the function.
3. After that find the cross product of the derivatives and substitute them into the formula to reach the result.

Formula used:
Formula for curvature of the curve: $\kappa = \dfrac{{\left\| {r'\left( t \right) \times r''\left( t \right)} \right\|}}{{{{\left\| {r'\left( t \right)} \right\|}^3}}}$.

Complete Step by Step Solution:
 Firstly, to compute the arc length of the curvature of the curve of the form $\vec r\left( t \right) = a\vec i + b\vec j + c\vec k$.
We use the formula$\kappa = \dfrac{{\left\| {r'\left( t \right) \times r''\left( t \right)} \right\|}}{{{{\left\| {r'\left( t \right)} \right\|}^3}}}$.
Where to find the norm length of the vector is given by: $\vec r'\left( t \right)$is: $\left\| {\vec r'\left( t \right)} \right\| = \sqrt {{a^2} + {b^2} + {c^2}} $.
We will find the first derivative$\vec r'\left( t \right)$:
$ \Rightarrow r\left( t \right) = ti + {t^2}j + {t^2}k$
$ \Rightarrow \vec r'\left( t \right) = i + 2tj + 2tk$
Find the norm length of the vector $\vec r'\left( t \right)$
$ \Rightarrow \left\| {\vec r'\left( t \right)} \right\| = \sqrt {{1^2} + {{(2t)}^2} + {{(2t)}^2}} $
On simplify we get,
$ \Rightarrow \left\| {\vec r'\left( t \right)} \right\| = \sqrt {1 + 4{t^2} + 4{t^2}} $
Let us add the term and we get
$ \Rightarrow \left\| {\vec r'\left( t \right)} \right\| = \sqrt {1 + 8{t^2}} $ …… let it be eq. (1)
Now, we find the second derivative $\vec r''\left( t \right)$:
$ \Rightarrow \vec r''\left( t \right) = 2j + 2k$
Find the norm length of the vector $\vec r''\left( t \right)$:
$ \Rightarrow \left\| {\vec r''\left( t \right)} \right\| = \sqrt {{0^2} + {{(2)}^2} + {{(2)}^2}} $
On simplify the term and we get,
$ \Rightarrow \left\| {\vec r''\left( t \right)} \right\| = \sqrt {4 + 4} $
Let us add the term and we get
$ \Rightarrow \left\| {\vec r''\left( t \right)} \right\| = \sqrt 8 $
Now, to solve further we need to know the concept of the cross product. Then, cross product also looks like $3 \times 3$ determinant.
$\vec a \times \vec b = \left| {\begin{array}{*{20}{c}}
  i&j&k \\
  {{a_1}}&{{a_2}}&{{a_3}} \\
  {{b_1}}&{{b_2}}&{{b_3}}
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
  {{a_2}}&{{a_3}} \\
  {{b_2}}&{{b_3}}
\end{array}} \right|i - \left| {\begin{array}{*{20}{c}}
  {{a_1}}&{{a_3}} \\
  {{b_1}}&{{b_3}}
\end{array}} \right|j + \left| {\begin{array}{*{20}{c}}
  {{a_1}}&{{a_2}} \\
  {{b_1}}&{{b_2}}
\end{array}} \right|k$
$ \Rightarrow \vec a \times \vec b = \left( {{a_2}{b_3} - {b_2}{a_3}} \right)i - \left( {{a_1}{b_3} - {b_1}{a_3}} \right)j + \left( {{a_1}{b_2} - {b_1}{a_2}} \right)k$
 This is the compact way to remember how to compute the cross product.
Now we find the cross product of $\vec r'\left( t \right)$and $\vec r''\left( t \right)$using the above formula:
$\vec r'(t) \times \vec r''(t) = \left| {\begin{array}{*{20}{c}}
  i&j&k \\
  1&{2t}&{2t} \\
  0&2&2
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
  {2t}&{2t} \\
  2&2
\end{array}} \right|i - \left| {\begin{array}{*{20}{c}}
  1&{2t} \\
  0&2
\end{array}} \right|j + \left| {\begin{array}{*{20}{c}}
  1&{2t} \\
  0&2
\end{array}} \right|k$
$ \Rightarrow \left( {2t \times 2 - 2 \times 2t} \right)i - \left( {2 \times 1 - 0 \times 2t} \right)j + \left( {2 \times 1 - 0 \times 2t} \right)k$
On simplify we get,
$ \Rightarrow \left( {4t - 4t} \right)i - \left( {2 - 0} \right)j + \left( {2 - 0} \right)k$
On subtract the term and we get,
$ \Rightarrow 0i - 2j + 2k$
Find the norm length of the vector $\vec r'(t) \times \vec r''(t)$:
$ \Rightarrow \left\| {\vec r'(t) \times \vec r''(t)} \right\| = \sqrt {{0^2} + {{( - 2)}^2} + {2^2}} $
On squaring we get,
$ \Rightarrow \left\| {\vec r'(t) \times \vec r''(t)} \right\| = \sqrt {0 + 4 + 4} $
On adding the term and we get,
$ \Rightarrow \left\| {\vec r'(t) \times \vec r''(t)} \right\| = \sqrt 8 $ …….. let it be eq. (2)
Now, applying the final formula to find the curvature of the curve:
$ \Rightarrow \kappa = \dfrac{{\left\| {r'\left( t \right) \times r''\left( t \right)} \right\|}}{{{{\left\| {r'\left( t \right)} \right\|}^3}}}$
Substituting values from equation 1 and 2 in the above formula will give:
$ \Rightarrow \kappa = \dfrac{{\sqrt 8 }}{{{{\left( {\sqrt {1 + 8{t^2}} } \right)}^3}}}$

So, the curvature of the curve is: $\kappa = \dfrac{{\sqrt 8 }}{{{{\left( {1 + 8{t^2}} \right)}^{3/2}}}}$.

Note: Alternative formulas for curvature:
If C is a smooth curve given by r(t), then the curvature $\kappa $ at t is given by: $\kappa = \dfrac{{\left\| {T'\left( t \right)} \right\|}}{{\left\| {r'\left( t \right)} \right\|}}$.
If C is a three-dimensional curve given by r(t), then the curvature can be given by the formula: $\kappa = \dfrac{{\left\| {r'\left( t \right) \times r''\left( t \right)} \right\|}}{{{{\left\| {r'\left( t \right)} \right\|}^3}}}$
If C is the graph of a function $y = f(x)$and body $y'$ and $y''$ exist, then the curvature $\kappa $ at point $(x,y)$ is given by: $\kappa = \dfrac{{\left| {y''} \right|}}{{{{\left[ {1 + {{\left( {y'} \right)}^2}} \right]}^{3/2}}}}$.