# Thread: The two curves make a constant angle?

1. ## The two curves make a constant angle?

How to show that the curve $\alpha(t)=(3, 4t, 6t^2)$ makes a constant angle with the line $x=z, y=0$?

2. The angle between the two curves would be the angle between their tangent vectors, right? If so, is the statement of the problem really true?

The line can be parametrized by $(x,y,z) = t(1,0,1)$, so you see that its tangent will point in the direction of $(1,0,1)$. The derivative of the curve is non-constant. What am I misunderstanding?

3. Originally Posted by Hinatico
How to show that the curve $\alpha(t)=(3, 4t, 6t^2)$ makes a constant angle with the line $x=z, y=0$?

I'm not sure I understand the question: that curve and that line do not intersect each other so how can you begin to evaluate the angle between them...??

Tonio

4. As Tonio said, the curve and line do not intersect so it really makes no sense to talk about the "angle they make". Even if we imagine one or the other being "shifted", without changing orientation, so they do intersect, the angle between them is NOT constant:

At any t, the tangent vector to the first curve is < 0, 4, 12t> and the tangent vector to the second curve is <1, 0, 1>. Their dot product is $<0, 4, 12t>\cdot<1, 0, 1>= 12t$ while their lengths are $\sqrt{16+ 144t^2}= 2\sqrt{1+ 9t^2}$ and $\sqrt{2}$. If $\theta$ is the angle between the vectors, $cos(\theta)= \frac{12t}{2\sqrt{2+ 18t^2}}$ which is NOT a constant so the angle is not a constant.

5. Originally Posted by tonio
I'm not sure I understand the question: that curve and that line do not intersect each other so how can you begin to evaluate the angle between them...??

Tonio
It is actually taken from an exercise book and I don't understand it either. But thanks for all your help!

6. Originally Posted by Hinatico
It is actually taken from an exercise book and I don't understand it either. But thanks for all your help!
What's that book's name and author, and in what page or section is that problem?

Tonio