Find the curvature of r(t) = 7t, t2, t3 at the point (7, 1, 1).

This question aims to find the curvature of the given equation for the points (7,1,1).This question uses the concept of calculus and curvature. Curvature is used for graphs which tells us how sharply a graph bends. Mathematically it is represented as:

$K \space= \space || \space \frac{dT}{ds} \space ||$

We are given  the equation:

$r(t)\space = \space <\space7t,\space t^2, \space t^3 \space >$

We have to find the curvature of the given equation at point $(7,1,1)$.

We have to use the concept of curvature to find the curvature for the given points.

$r(t) \space = \space < \space 7t,t^2,t^3 \space >$

The first derivative results in:

$\gamma'(t) \space = \space < \space 7,2t,3t^2 \space >$

And the second derivative results in :

$\gamma”(t) \space = \space < \space 0,2,6t \space >$

Thus:

$\gamma'(t) \space \times \space \gamma”(t)\space = \begin{bmatrix} \hat{i} & \hat{j} & \hat{k} \\ 7 & 2t & 3t^2 \\ 0 & 2 & 6t \end{bmatrix} \space$

The cross product results in:

$(\space 12t^2 \space – \space 6t^2)\hat{i} \space – \space (\space 42t \space – \space 0)\hat{j} \space + \space (\space 14 \space – \space 0)\hat{k}$

$(\space 6t^2)\hat{i} \space – \space (\space 42t )\hat{j} \space + \space (\space 14 \space )\hat{k}$

$| \space \gamma'(1) \space \times \gamma”(1) \space| = \sqrt{(6t^2)^2 \space + \space (-42t)^2 \space + \space (14)^2}$

By putting $t=1$,we get:

$=\sqrt{36 \space + \space 1764 \space + \space 196}$

$\sqrt{1996}$

$| \space \gamma'(1) \space| = \sqrt{(7)^2 \space + \space (2)^2 \space + \space (3)^2}$

$\sqrt{45 \space + \space 4 \space + \space 9 }$

$\sqrt{62}$

so $K$ = 0.091515

The curvature of the given equation for the given point $(7,1,1)$ is $0.091515$.

Example

Calculate the curvature for the equation given below at point (7,1,1).

$r(t)\space = \space <\space7t,\space 2t^2, \space 3t^3 \space >$

We have to find the curvature of the given equation at point $(7,1,1)$.

We have to use the concept of curvature to find the curvature for the given points.

$r(t) \space = \space < \space 7t,2t^2,3t^3 \space >$

The first derivative of the given equation results in:

$\gamma'(t) \space = \space < \space 7,4t,9t^2 \space >$

And the second derivative of the given equation results in :

$\gamma”(t) \space = \space < \space 0,4,18t \space >$

Thus:

$\gamma'(t) \space \times \space \gamma”(t)\space = \begin{bmatrix} \hat{i} & \hat{j} & \hat{k} \\ 7 & 4t & 9t^2 \\ 0 & 4 & 18t \end{bmatrix} \space$

The cross product results in:

$(\space 6t^2)\hat{i} \space – \space (\space 42t )\hat{j} \space + \space (\space 14 \space )\hat{k}$

$| \space \gamma'(1) \space \times \gamma”(1) \space| = \sqrt{(36t^2)^2 \space + \space (-126t)^2 \space + \space (28)^2}$

By putting $t=1$,we get:

$=\sqrt{1296 \space + \space 15876 \space + \space 784}$

$\sqrt{17956}$

Now:

$| \space \gamma'(1) \space| = \sqrt{(7)^2 \space + \space (4)^2 \space + \space (9)^2}$

$\sqrt{49 \space + \space 16 \space + \space 81 }$

$\sqrt{146}$

so $K$ = $\frac{17956}{146^{\frac{2}{3}}}$

Hence it is calculated that the curvature for the given equation at a given point is $\frac{17956}{146^{\frac{2}{3}}}$.