can someone help please: Find what kind of curves are given by the following representations and draw the curves: 1. r(t) = (2t-5, -3t+1, 4) 2. r(t) = (0, -cos(t), 3sin(t))
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Originally Posted by sonia1 can someone help please: Find what kind of curves are given by the following representations and draw the curves: 1. r(t) = (2t-5, -3t+1, 4) 2. r(t) = (0, -cos(t), 3sin(t)) The first is a straight line parallel to the xy plane moved 4 units in the z direction and the second, an ellipse in the yz plane.
how did you find that out, could u please show me
Originally Posted by sonia1 how did you find that out, could u please show me Sure. In the first case z isn't changing, x and y are. Eliminating t gives a straight line. In the second, , it's not changing and Eliminating t using the fact that gives , the formula for an ellipse in the yz plane. Hope that helps.
yes thanks alot, i'm also stuck on a different question on the same topic wonder if you can help: find a parametric representation of the curve : x^2 + y^2= 36 and z= (1/pi) arctan(x/y) i.e. find a representation in the form x=x(t), y=y(t), z=z(t)
Originally Posted by sonia1 yes thanks alot, i'm also stuck on a different question on the same topic wonder if you can help: find a parametric representation of the curve : x^2 + y^2= 36 and z= (1/pi) arctan(x/y) i.e. find a representation in the form x=x(t), y=y(t), z=z(t) You'll notice with the first equation and the identity then so comparing gives . Now sub. these into z and simplify.
thanks..i simplified it and i get z= (1/pi) arc 6tan^2 t, is that correct
regarding the before question, could u please show me a drawing of what the ellipse actually looks like.
Originally Posted by sonia1 thanks..i simplified it and i get z= (1/pi) arc 6tan^2 t, is that correct Not but
r u sure it should be tan^2 t because (6sint/6cost) =tan t and then we get 1/pi arc tan^2t right
Here's a picture of the ellipse.
thankz for your help, did u get tan^2 t then not tant
Originally Posted by sonia1 thankz for your help, did u get tan^2 t then not tant From your earlier post, if then
thankz for your help!
Originally Posted by sonia1 thankz for your help! You can thank danny (and add to his reputation) by clicking the Thanks button (bottom right).
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