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Draw the parametric curve for the subsequent set of parametric equations.
X = t2+t
Y=2t-1
-1 < t < 1
Solution
Note that the only dissimilarity here is the existence of the limits on t. All these types of limits do is tell us that we cannot take any type of value of t outside of this range. Hence, the parametric curve will just only be a portion of the curve. Now here is the parametric curve for this instance.
Note: that with this diagram we started and stopped the diagram right on the points originating from the end points of the range of t's. In difference this with the diagram in the preceding example in which we had a portion of the diagram to the right of the "start" and "end" points that we computed.
In this case the curve begins at t = -1 and ends at t = 1, while in the preceding example the curve didn't actually start at the right most points that we calculated. We require to be clear in our diagrams if the curve starts/ends right at a point, or if that point was just simply the first or last one that we calculated.
sin(2x+x)=sin2x.cosx+cos2x.sinx =2sinxcosx.cosx+(-2sin^2x)sinx =2sinxcos^2+sinx-2sin^3x =sinx(2cos^2x+1)-2sin^3x =sinx(2-2sin^2x+1)-2sin^3
4into 5 is;
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