Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Show that the tangent lines to the regular parametrized curve $$ \alpha \left( t \right) = \left( {3t,3t^2 ,2t^3 } \right) $$ make a constant angle with the line $y=0$ , $z=x$

First of all, the derivate of that curve is $$ \left( {3,6t,6t^2 } \right) $$ So in an arbitrary point of the curve at $ t=$$ \varphi $$ _0 $ the tangent line is $$ \eqalign{ & \left( {3\varphi _0 ,3\varphi _0 ^2 ,2\varphi _0 ^3 } \right) + t\left( {3,6\varphi _0 ,6\varphi _0 ^2 } \right) \cr & = \left( {3\varphi _0 + 3\,t\,\,,\,\,\,3\varphi _0 ^2 + 6\,t\,\varphi _0 ,\,\,\,2\varphi _0 ^3 + 2\,t\,\varphi _0 ^2 } \right) \cr} $$ The other line is $ (u,0,u) $ but the dot product betweem this two lines is not constant, what is bad?

share|improve this question
add comment

1 Answer

up vote 2 down vote accepted

The vectors we want to take the dot product of are $$\mathbf{a}=(u,0,u),$$ $$\mathbf{b}=(3,6t,6t^2).$$ (As Jim points out below, we want to take the dot product of the direction vectors of the lines; what I wrote before was the dot product of $(u,0,u)$ and the position vector of a point on the tangent line). Recall that $$\mathbf{a}\cdot\mathbf{b}=||\mathbf{a}||\,\,||\mathbf{b}||\cos(\theta).$$ The fact that the dot product you're getting isn't constant is due to factor of $||\mathbf{a}||\,\,||\mathbf{b}||$ that you are forgetting to compensate for (or at least, I assume this is the problem that's occuring). Once you divide your answer by $||(u,0,u)||=u\sqrt{2}$ and $\left|\left|\left( {3\,\,,\,6\,t\,,6\,t ^2 } \right)\right|\right|=\sqrt{9+36t^2+36t^4}$, it should be constant, i.e. not have any $t$'s or $u$'s.

share|improve this answer
    
But you shouldn't add $(3\phi_o,3\phi_0^2,2\phi_0^3)$ since that's not part of the direction vector of the tangent line. –  Grumpy Parsnip Aug 9 '11 at 20:37
    
@Jim: Ach, of course. I'll edit. –  Zev Chonoles Aug 9 '11 at 20:39
    
Zev, how did you get that it has a constant angle from your answer? I get how you got $a$ and $b$ but I am not sure how to show it forms a constant angle? –  Lays Aug 18 '13 at 8:23
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.