Answer:
[tex]k=\frac{10}{(100t^2+1)^{\frac{3}{2}}}[/tex]
Or
[tex]k=\frac{10\sqrt{100t^2+1}}{(100t^2+1)^{2}}[/tex]
Step-by-step explanation:
We want to compute the curvature of the parameterized curve, [tex]r(t)=\:<\:6+5t^2,t,0)\:>\:[/tex] using the alternative formula:
[tex]k=\frac{|a\times v|}{|v|^3}[/tex].
We first compute the required ingredients.
The velocity vector is [tex]v=r'(t)=<\:10t,1,0\:>[/tex]
The acceleration vector is given by [tex]a=r''(t)=<\:10,0,0\:>[/tex]
The magnitude of the velocity vector is [tex]|v|=\sqrt{(10t)^2+1^2+0^2}=\sqrt{100t^2+1}[/tex]
The cross product of the velocity vector and the acceleration vector is
[tex]a\times v=\left|\begin{array}{ccc}i&j&k\\10&0&0\\10t&1&0\end{array}\right|=10k[/tex]
We now substitute ingredients into the formula to get:
[tex]k=\frac{|10k|}{(\sqrt{100t^2+1})^3}[/tex].
[tex]k=\frac{10}{(100t^2+1)^{\frac{3}{2}}}[/tex]
Or
[tex]k=\frac{10\sqrt{100t^2+1}}{(100t^2+1)^{2}}[/tex]
