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Here's the set up of the problem:

Let $r(t)=<at, bt, ct>$ be a vector valued function. This is the equation of a line going through the origin of $\mathbb{R}^3$.

The problem (taken from a calculus book) asks to show that the angle between $r(t)$ and $r'(t)$ is constant.

I think that the problem as stated in the book is false: since $r'(t)= <a, b, c>$ then using the dot product we get that $cos(\alpha)= \frac{t}{|t|}$, after some computations, where $\alpha$ is the angle between $r$ and $r'$ at a given t. This angle is not constant since if $t$ is positive then the angle is 0 and if t is negative the angle will be $\pi$.

can anyone confirm or point my possible mistakes. Thanks.

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I suspect that the book intended to write $r(t')$ instead of $r'(t)$. Double check it. –  Emanuele Paolini Feb 12 '13 at 9:09

1 Answer 1

The angle is indeed zero. Geometrically the tangent of a straight line coincides with the line at any point. You can show that by computing the angles between the x-axis and your function and between the x-axis and your function's tangent. Note that the tangent always points in the direction of the depiction of your parametric interval. Thus if you are going in the negative direction, your tangent will point the opposite way, leaving the angle constant.

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