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

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

  2. Find the curvature of $\displaystyle \left[ t^2 , t , t^3 \right] $ at a general point.

    $\displaystyle \frac{6\,\left\vert t\right\vert }{\left(9\,t^4+1\right)^{\frac{3}{2}}}$ $\displaystyle \frac{3\,2^{\frac{3}{2}}\,\left\vert t\right\vert }{\left(18\,t^4+1\right) ^{\frac{3}{2}}}$ 0 $\displaystyle \frac{\sqrt{36\,t^4+36\,t^2+4}}{\left(9\,t^4+4\,t^2+1\right)^{ \frac{3}{2}}}$

  3. Find the unit tangent vector for $\displaystyle \left[ t^2 , t^3 , t^3 \right] $ at a general point.

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

  4. Find the acceleration for the position $\displaystyle \left[ t^4 , t^2 , t^3 \right] $ at a general point.

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

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

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

  6. Reparametrize $\displaystyle \left[ 2\,t , 4\,\sin t , 4\,\cos t \right] $ with respect to arc length from [ 0 ,  0 ,  4 ]

    $\displaystyle \left[ \frac{2\,s}{\sqrt{13}} , 4\,\sin \left(\frac{s}{\sqrt{13}} \right) , 4\,\cos \left(\frac{s}{\sqrt{13}}\right) \right] $ $\displaystyle \left[ \frac{2\,s}{\sqrt{13}} , 3\,\sin \left(\frac{s}{\sqrt{13}} \right) , 3\,\cos \left(\frac{s}{\sqrt{13}}\right) \right] $ $\displaystyle \left[ \frac{s}{\sqrt{5}} , 3\,\sin \left(\frac{s}{2\,\sqrt{5}} \right) , 3\,\cos \left(\frac{s}{2\,\sqrt{5}}\right) \right] $ $\displaystyle \left[ \frac{s}{\sqrt{5}} , 4\,\sin \left(\frac{s}{2\,\sqrt{5}} \right) , 4\,\cos \left(\frac{s}{2\,\sqrt{5}}\right) \right] $

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

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

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

    $\displaystyle \frac{1}{2}$ $\displaystyle \frac{1}{4}$ $\displaystyle \frac{2}{5}$ $\displaystyle \frac{4}{17}$

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

    $\displaystyle \frac{2^{\frac{3}{2}}}{\sqrt{4\,t^2+2}}$ $\displaystyle \frac{2^{\frac{3}{2}}}{\sqrt{8\,t^2+1}}$ $\displaystyle \frac{\sqrt{36\,t^4+36\,t^2+4}}{\sqrt{9\,t^4+4\,t^2+1}}$ 0

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

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


Department of Mathematics
Last modified: 2026-06-13