Differential Geometry of Curves and Surfaces - Chapter 1 Section 3 Exercise 1

Problem:

Solution:

The tangent line has direction $\vec{a} = (3, 4t, 8t^2)$
The given line has parametric form $(u, 0, u)$ so it's direction $\vec{b} (1, 0, 1)$

The angle between these two direction is $\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} = \frac{(3 + 8t^2)}{\sqrt{3^2 + (4t)^2 + (8t^2)^2}\sqrt{1^2 + 0^2 + 1^2}} = \frac{(3 + 8t^2)}{\sqrt{9 + 16t^2 + 64t^4}\sqrt{2}} = \frac{(3 + 8t^2)}{\sqrt{(3 + 8t^2)^2}\sqrt{2}} = \frac{1}{\sqrt{2}}$

So the angle is constant.