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

    $\displaystyle \frac{2}{\sqrt{4\,t^2+1}}$ 0 0 0

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

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

  3. Find the binormal vector for $\displaystyle \left[ 2\,t , 3\,\cos t , 3\,\sin t \right] $ at a general point.

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

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

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

  5. Find the normal vector for $\displaystyle \left[ -\sin t , 4\,t , -\cos t \right] $ at a general point.

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

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

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

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

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

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

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

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

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

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

    [ 0 ,  s ,  0 ] $\displaystyle \left[ 0 , \frac{2\,s}{\sqrt{5}} , 0 \right] $ $\displaystyle \left[ \sin \left(\frac{s}{\sqrt{5}}\right) , \frac{2\,s}{\sqrt{5}} , \cos \left(\frac{s}{\sqrt{5}}\right) \right] $ $\displaystyle \left[ \sin \left(\frac{s}{2}\right) , s , \cos \left(\frac{s}{2} \right) \right] $


Department of Mathematics
Last modified: 2025-02-21