Question

Determine any differences between the curves of the parametric equations. Are the graphs the same? Are the orientations the same? Are the curves smooth? (a) \(x=t+1, y=t^{3}\) (b) \(x=-t+1, y=(-t)^{3}\)

Short answer

Both the curves have a cubic shape and are smooth, however the orientations are different. The parametric equations of set (b) produce a curve that is a mirror image of the curve produced by set (a) reflected over the y-axis.

Step-by-step solution

Sketch the curves

To visualize anything in mathematics, it's beneficial to graph it. Therefore, both parametric equations will be graphed.\n(a) The equation set is \(x=t+1, y=t^{3}\). As t varies from -∞ to ∞, x will vary from -∞ to ∞ and y will also vary from -∞ to ∞. So, the plot of this parametric set will be a cubic curve.\n(b) The equation set is \(x=-t+1, y=(-t)^{3}\). As t ranges from -∞ to ∞, x varies from ∞ to -∞ (opposite of (a)) and y will vary from -∞ to ∞ (which is same as (a)). Thus, the plot of this set will also be a cubic curve, but will be a mirrored version about y-axis of (a).

Compare shape and orientation of the curves

From the sketches made in step 1, it can be observed that both curves have a cubic shape, however, their orientations are different. For set (a), as t increases, x also increases, while for set (b), as t increases, x decreases. Hence, the two cubic graphs are mirror images of each other about the y-axis. Therefore, while the curves are the same in terms of shape, their orientations are different.

Check for smoothness

A curve is smooth if its derivative is continuous. For a parametrically defined curve described by \(x=f(t), y=g(t)\), the derivative \(dy/dx\) exists and is continuous wherever \(df(t)/dt\) and \(dg(t)/dt\) are continuous. In this case, he functions in both sets (a) and (b) are continuous for all t, and thus their derivatives are also continuous. Hence, both the curves are smooth