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Each equation gives information about the body's location on each axis in Cartesian coordinate system (A is some constant and $t$ is time). We know that $\sin^2(x)+\cos^2(x)=1$ (Pythagora's theorem applied to unit circle which gives us the radius of unit circle). This is the answer given for this problem:

Each velocity component is given by $$v_x=\frac{\mathrm dx}{\mathrm dt}=-\sin(t)$$ $$v_y=\frac{\mathrm dy}{\mathrm dt}=\cos(t)$$ $$v_z=\frac{\mathrm dz}{\mathrm dt}=A$$ so the body's speed is
$$v=\sqrt{v_x^2+v_y^2+v_z^2}$$
Since $v$ is constant, $a_\mathrm t=\frac{\mathrm dv}{\mathrm dt}=0$ (I guess this is tangential acceleration while $a_\mathrm n$is normal acceleration), so radius of the curve is
$$R=\frac{v^2}{a_\mathrm n}=\frac{v^2}{a}=\frac{v^2}{\sqrt{a_x^2+a_y^2+a_z^2}}$$

Accelerations are just derivations of each velocity component.

The answer isn't one, how is that possible? It would be one if there was no vertical movement so how does that affect unit circle?

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closed as off-topic by Aaron Stevens, Jon Custer, Kyle Kanos, Qmechanic Jul 24 at 7:57

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  • $\begingroup$ Are you asking why the answer does not reduce to 1 when $A=0$? $\endgroup$ – garyp Jul 21 at 8:45
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    $\begingroup$ I haven't dug into this deeply, but since the trajectory is not a circle, I would not be surprised to find that the radius of curvature is not 1. $\endgroup$ – garyp Jul 21 at 8:50
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    $\begingroup$ Because it's not a circle - it's a helix. But if you calculate $x(t)^2+y(t)^2=1$ it is a circle and it has a circumference in the x-y plane - it's the projection of the helix onto the x-y plane. A circle is 2 dimensional object . A helix is a 3 dimensional object. It's z(t) component which turns the circle into a helix - or a spiral. $\endgroup$ – Cinaed Simson Jul 21 at 8:59
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    $\begingroup$ $1+A^2$ is the square of the distance the body travels per second $\endgroup$ – Hagen von Eitzen Jul 21 at 9:34
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    $\begingroup$ Sorry I wasn't being too technical. I just mean that it tells you the local radius at some point on the curve. You haven't done anything yet to look at projection. And I made a typo and meant to say as $A$ increases the curvature decreases. $\endgroup$ – Aaron Stevens Jul 21 at 11:59
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Are you asking about circumference or radius? The title says circumference but then the text of the question says radius.

Your question has the word "circle" in brackets. That is wrong. The curve described by $\{x=\sin(t),\;y=\cos(t),\;z=A t\}$ is a helix not a circle. We can't say what circumference it has until we have first agreed a definition of circumference for a curve which does not close on itself.

If you want the circumference of the projection of this curve onto the xy plane, then the answer is $2\pi$, because by using $\sin^2(t) + \cos^2(t)=1$ we find the radius (not the circumference) is 1.

The answer you quote is about the radius of any small section of the helix. The idea is that you can imagine a circle, tipped over in 3 dimensions, that lines up with some small section of the helical track. The radius of such a circle could be said to be the radius of that part of the track.

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  • $\begingroup$ I'm asking about radius (by definition, distance from point $(0,0)$ to the point in circle).Again, I implied the projection to the xy plane because it's obvious that the trajectory isn't a circle (which is why the given answer doesn't make sense).I don't understand the last part. $\endgroup$ – user3711671 Jul 21 at 9:41
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    $\begingroup$ @user3711671 I'm not sure but I think the problem you are looking at is to do with the concept of radius of curvature. You might find it helpful to look that one up. $\endgroup$ – Andrew Steane Jul 21 at 10:36
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$R$ isn't the radius, $R$ is the radius of curvature of the path

We use the term radius of curvature even when the motion isn't exactly in a circle. For any point on a curve, the radius of curvature is $1/\kappa$ [where $\kappa$ is the curvature of the path].

In other words, the radius of curvature is the radius of a circle with the same instantaneous curvature as the curve.

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If the object is moving with constant speed along a curved path, then $dv/dt=0$, so there is no tangential component of acceleration. The acceleration vector $\mathbf{a}(t)=\kappa(t)v^2(t)\mathbf{N}(t)$ lies in the normal direction. The magnitude of the acceleration is often written as $v^2/R$, where $R$ is the radius of curvature.

(emphasis mine)

Credit

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