1. Find the arc length of $\displaystyle \left[ 2\,\cos t , 4\,t , 2\,\sin t \right] $ from $t= 1$ to $t=u$.

    $\displaystyle \sqrt{10}\,\left(u-1\right)$ 5(u − 1) $\displaystyle 2\,\sqrt{5}\,\left(u-1\right)$ $\displaystyle \sqrt{5}\,\left(u-1\right)$

  2. Reparametrize [ 2t ,  0 ,  0 ] with respect to arc length from [ 0 ,  0 ,  0 ]

    $\displaystyle \left[ \frac{2\,s}{\sqrt{13}} , 3\,\sin \left(\frac{s}{\sqrt{13}} \right) , 3\,\cos \left(\frac{s}{\sqrt{13}}\right) \right] $ [ s ,  0 ,  0 ] $\displaystyle \left[ s , 3\,\sin \left(\frac{s}{2}\right) , 3\,\cos \left(\frac{s }{2}\right) \right] $ $\displaystyle \left[ \frac{2\,s}{\sqrt{13}} , 0 , 0 \right] $

  3. Find the unit tangent vector for $\displaystyle \left[ 3\,\sin t , 3\,\cos t , t \right] $ at a general point.

    $\displaystyle \left[ \frac{3\,\cos t}{\sqrt{10}} , -\frac{3\,\sin t}{\sqrt{10}} , \frac{1}{\sqrt{10}} \right] $ $\displaystyle \left[ \frac{2\,\cos t}{\sqrt{5}} , -\frac{2\,\sin t}{\sqrt{5}} , \frac{1}{\sqrt{5}} \right] $ $\displaystyle \left[ \frac{4\,\cos t}{\sqrt{17}} , -\frac{4\,\sin t}{\sqrt{17}} , \frac{1}{\sqrt{17}} \right] $ $\displaystyle \left[ \frac{3\,\cos t}{\sqrt{13}} , -\frac{3\,\sin t}{\sqrt{13}} , \frac{2}{\sqrt{13}} \right] $

  4. Find the curvature of $\displaystyle \left[ 2\,\cos t , 4\,t , 2\,\sin t \right] $ at a general point.

    $\displaystyle \frac{1}{2}$ $\displaystyle \frac{3}{25}$ $\displaystyle \frac{1}{10}$ $\displaystyle \frac{1}{3}$

  5. Find the normal vector for $\displaystyle \left[ -3\,t , -\sin t , -\cos t \right] $ at a general point.

    $\displaystyle \left[ 0 , -\sin t , -\cos t \right] $ $\displaystyle \left[ 0 , -\sin t , \cos t \right] $ $\displaystyle \left[ 0 , \sin t , -\cos t \right] $ $\displaystyle \left[ 0 , \sin t , \cos t \right] $

  6. Find the binormal vector for $\displaystyle \left[ 2\,\sin t , 4\,t , 2\,\cos t \right] $ at a general point.

    $\displaystyle \left[ -\frac{\cos t}{\sqrt{5}} , \frac{2}{\sqrt{5}} , \frac{\sin t }{\sqrt{5}} \right] $ $\displaystyle \left[ -\frac{4\,\cos t}{\sqrt{10}} , \frac{3}{\sqrt{10}} , \frac{4 \,\sin t}{\sqrt{10}} \right] $ $\displaystyle \left[ -\frac{4\,\cos t}{5} , \frac{3}{5} , \frac{4\,\sin t}{5} \right] $ $\displaystyle \left[ -\frac{2\,\cos t}{\sqrt{5}} , \frac{1}{\sqrt{5}} , \frac{2\, \sin t}{\sqrt{5}} \right] $

  7. Find the tangential component of acceleration for the position $\displaystyle \left[ 2\,\sin t , t^2 , 2\,\cos t \right] $ at a general point.

    $\displaystyle \frac{16\,t}{\sqrt{16\,t^2+4}}$ $\displaystyle \frac{4\,t}{\sqrt{4\,t^2+4}}$ $\displaystyle \frac{16\,t}{\sqrt{16\,t^2+1}}$ $\displaystyle \frac{4\,t}{\sqrt{4\,t^2+1}}$

  8. Find the normal component of acceleration for the position $\displaystyle \left[ 2\,\cos t , 2\,\sin t , 2\,t^2 \right] $ at a general point.

    $\displaystyle \frac{\sqrt{16\,t^2+32}}{\sqrt{4\,t^2+4}}$ $\displaystyle \frac{\sqrt{64\,t^2+80}}{\sqrt{16\,t^2+4}}$ $\displaystyle \frac{\sqrt{16\,t^2+17}}{\sqrt{16\,t^2+1}}$ $\displaystyle \frac{\sqrt{4\,t^2+5}}{\sqrt{4\,t^2+1}}$

  9. Find the unit tangent vector for $\displaystyle \left[ t^4 , t^3 , 1 \right] $ at a general point.

    $\displaystyle \left[ \frac{3\,t^2}{\sqrt{18\,t^4+4\,t^2}} , \frac{3\,t^2}{\sqrt{ 18\,t^4+4\,t^2}} , \frac{2\,t}{\sqrt{18\,t^4+4\,t^2}} \right] $ $\displaystyle \left[ \frac{4\,t^3}{\sqrt{16\,t^6+9\,t^4}} , \frac{3\,t^2}{\sqrt{ 16\,t^6+9\,t^4}} , 0 \right] $ $\displaystyle \left[ \frac{1}{\sqrt{4\,t^2+2}} , \frac{2\,t}{\sqrt{4\,t^2+2}} , \frac{1}{\sqrt{4\,t^2+2}} \right] $ $\displaystyle \left[ \frac{2\,t}{\sqrt{9\,t^4+4\,t^2}} , \frac{3\,t^2}{\sqrt{9\,t ^4+4\,t^2}} , 0 \right] $

  10. Find the acceleration for the position [ t ,  1 ,  1 ] at a general point.

    [ 0 ,  0 ,  0 ] [ 6t ,  0 ,  2 ] $\displaystyle \left[ 0 , 0 , 12\,t^2 \right] $ $\displaystyle \left[ 0 , 2 , 12\,t^2 \right] $

  11. Find the tangential component of acceleration for the position $\displaystyle \left[ t , t^2 , t^3 \right] $ at a general point.

    $\displaystyle \frac{18\,t^3+4\,t}{\sqrt{9\,t^4+4\,t^2}}$ $\displaystyle \frac{18\,t^3+4\,t}{\sqrt{9\,t^4+4\,t^2+1}}$ $\displaystyle \frac{2\,\sqrt{3}\,t}{\left\vert t\right\vert }$ $\displaystyle \frac{18\,t^3+4\,t}{\sqrt{9\,t^4+4\,t^2}}$

  12. Find the normal component of acceleration for the position $\displaystyle \left[ t^2 , 1 , t^3 \right] $ at a general point.

    $\displaystyle \frac{6\,\left\vert t\right\vert }{\sqrt{9\,t^4+1}}$ 0 $\displaystyle \frac{6\,t^2}{\sqrt{9\,t^4+4\,t^2}}$ $\displaystyle \frac{6\,\left\vert t\right\vert }{\sqrt{9\,t^4+1}}$

  13. Find the curvature of $\displaystyle \left[ t , t^2 , 1 \right] $ at a general point.

    $\displaystyle \frac{6\,t^2}{\left(9\,t^4+4\,t^2\right)^{\frac{3}{2}}}$ $\displaystyle \frac{2}{\left(4\,t^2+1\right)^{\frac{3}{2}}}$ $\displaystyle \frac{3\,2^{\frac{3}{2}}\,t^2}{\left(18\,t^4+4\,t^2\right)^{\frac{3 }{2}}}$ $\displaystyle \frac{6\,t^2}{\left(9\,t^4+4\,t^2\right)^{\frac{3}{2}}}$


Department of Mathematics
Last modified: 2026-02-22