Proving Constant Angle of Tangent Lines to a Curve with y=0 and z=x

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In summary, the conversation discusses the problem of showing that the tangent lines to the curve \alpha (t)=(3t,2t^2,2t^3) make a constant angle with the line y=0 and z=x. The speaker has taken the derivative of the right-hand side of the equation cos(\gamma (t)) = \frac{\alpha '(t) \cdot v}{|v| |\alpha '(t)|} with respect to t, where v=(x,0,x), in order to show that if the derivative is zero, then the angle is constant. However, the speaker has not achieved a zero derivative and is seeking assistance in finding the error. Another speaker suggests that perhaps the original statement was meant
  • #1
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I have this question: Show that the tangent lines to the curve [tex]\alpha (t)= (3t,2t^2,2t^3)[/tex] make a constant angle with the line y=0 and z=x.

Now what I have done is, well obviously we have:
[tex] (1)cos(\gamma (t)) = \frac{\alpha '(t) \cdot v}{|v| |\alpha '(t)|}[/tex] So what I have done is to take the derivative of the RHS in (1) wrt t, where v=(x,0,x).
My reasoning is that if the derivative is zero then the angle is constant.

My problem is that I don't get zero, where did I get it wrong?
:confused::cry:

Thanks.
 
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  • #2
If we're supposed to show that the tangent vectors of [itex]\alpha[/itex] make a constant angle with the tangent vectors of that line, then it seems to me that what we have to show is that [tex]\frac{d}{dt}\big(\alpha'(t)\cdot(1,0,1)\big)=0[/tex] for all t. This is obviously not true, so I'm wondering if he might have meant something else. But I don't see how he could have meant something that makes the claim true.
 
  • #3
Well, I guess you also have a copy of the book right?

Anyway, here's a scan for the others.

question 1.
 

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  • #4
I don't have a copy of the book, but I don't mind downloading a pdf for purposes like this. I have skimmed the first few pages now. I didn't see any hints that he might have meant something different.
 
  • #5
For me 'copy' doesn't necessarily mean hard copy.

Thanks, anyway.
 

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