Respuesta :
Given:- [tex]r(t)=< at^2+1,t>[/tex] ; [tex]-\infty < t< \infty[/tex] , where a is any positive real number.
Consider the helix parabolic equation :
[tex]r(t)=< at^2+1,t>[/tex]
now, take the derivatives we get;
[tex]r{}'(t)=<2at,1>[/tex]
As, we know that two vectors are orthogonal if their dot product is zero.
Here, [tex]r(t) and r{}'(t)[/tex] are orthogonal i.e, [tex]r\cdot r{}'=0[/tex]
Therefore, we have ,
[tex]< at^2+1,t>\cdot < 2at,1>=0[/tex]
[tex]< at^2+1,t>\cdot < 2at,1>=<at^2+1\cdot\left(2at\right), t\cdot \left(1)>[/tex]
[tex]=2a^2t^3+2at+t[/tex]
[tex]2a^2t^3+2at+t=0[/tex]
take t common in above equation we get,
[tex]t\cdot \left (2a^2t^2+2a+1\right )=0[/tex]
⇒[tex]t=0[/tex] or [tex]2a^2t^2+2a+1=0[/tex]
To find the solution for t;
take [tex]2a^2t^2+2a+1=0[/tex]
The number[tex]D = b^2 -4ac[/tex] determined from the coefficients of the equation [tex]ax^2 + bx + c = 0.[/tex]
The determinant [tex]D=0-4(2a^2)(2a+1)[/tex][tex]=-8a^2\cdot(2a+1)[/tex]
Since, for any positive value of a determinant is negative.
Therefore, there is no solution.
The only solution, we have t=0.
Hence, we have only one points on the parabola [tex]r(t)=< at^2+1,t>[/tex] i.e <1,0>
