How do I sketch the curve represented by the vector valued function:
r(t) = (sin t)i + t j + (cos t) k
As a two dimensional projection in terms of x, y. y, z. x, z. and then as a three dimensional projection
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If your function was just r(t) = (sin t)i + (cos t)k,
then that turns out to just keep on tracing the same circle over and over again on the x-z plane. It is 2pi periodic, and it will trace the same points over and over again.
But when you add in the extra term t*j to get r(t) = (sin t)i + t j + (cos t) k, that's not longer a circle in the x-z plane. That's a spiral who's axis is the y-axis. So you have a circular spiral of the same radius centered on the y-axis.
The projection of it onto the x-z plane would be the same circle we began with (since that projection is what happens when the j component is 0.)
The projection of it onto the y-z plane would be the cosine curve, since
our projection would be p(t) = tj + (cos t)k (since that projection is what happens when the i component is 0.)
The projection of it onto the x-y plane would be the sine curve, since
our projection would be p(t) = (sin t)i + tj (since that projection is what happens when the k component is 0.)