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

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

  2. Find the arc length of [ 0 ,  0 ,  4t ] from $t= 1$ to $t=u$.

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

  3. Find the curvature of $\displaystyle \left[ t^3 , 1 , t^2 \right] $ at a general point.

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

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

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

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

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

  6. Find the acceleration for the position $\displaystyle \left[ t^4 , 1 , t^3 \right] $ at a general point.

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

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

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

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

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

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

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

  10. Reparametrize $\displaystyle \left[ t , 3\,\cos t , 3\,\sin t \right] $ with respect to arc length from [ 0 ,  3 ,  0 ]

    $\displaystyle \left[ \frac{s}{\sqrt{17}} , 4\,\cos \left(\frac{s}{\sqrt{17}} \right) , 4\,\sin \left(\frac{s}{\sqrt{17}}\right) \right] $ $\displaystyle \left[ \frac{s}{\sqrt{10}} , 3\,\cos \left(\frac{s}{\sqrt{10}} \right) , 3\,\sin \left(\frac{s}{\sqrt{10}}\right) \right] $ $\displaystyle \left[ \frac{s}{\sqrt{17}} , 3\,\cos \left(\frac{s}{\sqrt{17}} \right) , 3\,\sin \left(\frac{s}{\sqrt{17}}\right) \right] $ $\displaystyle \left[ \frac{s}{\sqrt{10}} , 4\,\cos \left(\frac{s}{\sqrt{10}} \right) , 4\,\sin \left(\frac{s}{\sqrt{10}}\right) \right] $


Department of Mathematics
Last modified: 2025-08-30