11 A bug moves along the $x y$-plane, with motion described by the equations \[ x=t^{3}-3 t \text { and } y=2 t^{3}-9 t^{2}+12 t \] a. Does the bug ever come to a complete stop? If so, when? b. Does the bug ever move straight up or down? If so, when and where? c. Is the bug ever moving straight horizontally right or left? If so, when and where?

Step 1 ：First, we need to find the velocity of the bug. The velocity is given by the derivative of the position function. So we differentiate the given equations for x and y with respect to t.

Step 2 ：The derivative of \(x = t^{3} - 3t\) with respect to t is \(x' = 3t^{2} - 3\).

Step 3 ：The derivative of \(y = 2t^{3} - 9t^{2} + 12t\) with respect to t is \(y' = 6t^{2} - 18t + 12\).

Step 4 ：The bug comes to a complete stop when both \(x'\) and \(y'\) are zero. So we set these equal to zero and solve for t.

Step 5 ：Setting \(x' = 0\) gives \(3t^{2} - 3 = 0\), which simplifies to \(t^{2} = 1\). So \(t = 1\) or \(t = -1\).

Step 6 ：Setting \(y' = 0\) gives \(6t^{2} - 18t + 12 = 0\), which simplifies to \(t^{2} - 3t + 2 = 0\). This factors to \((t - 1)(t - 2) = 0\), so \(t = 1\) or \(t = 2\).

Step 7 ：Comparing the solutions, we see that the bug comes to a complete stop at \(t = 1\).

Step 8 ：The bug moves straight up or down when \(x' = 0\), which we found occurs at \(t = 1\) and \(t = -1\). We substitute these values into the original equations to find the corresponding y-coordinates.

Step 9 ：Substituting \(t = 1\) into \(x = t^{3} - 3t\) and \(y = 2t^{3} - 9t^{2} + 12t\) gives \(x = -2\) and \(y = 5\). So the bug moves straight up or down at the point \((-2, 5)\) when \(t = 1\).

Step 10 ：Substituting \(t = -1\) into \(x = t^{3} - 3t\) and \(y = 2t^{3} - 9t^{2} + 12t\) gives \(x = 2\) and \(y = -5\). So the bug moves straight up or down at the point \((2, -5)\) when \(t = -1\).

Step 11 ：The bug moves straight horizontally right or left when \(y' = 0\), which we found occurs at \(t = 1\) and \(t = 2\). We substitute these values into the original equations to find the corresponding x-coordinates.

Step 12 ：Substituting \(t = 1\) into \(x = t^{3} - 3t\) and \(y = 2t^{3} - 9t^{2} + 12t\) gives \(x = -2\) and \(y = 5\). So the bug moves straight horizontally right or left at the point \((-2, 5)\) when \(t = 1\).

Step 13 ：Substituting \(t = 2\) into \(x = t^{3} - 3t\) and \(y = 2t^{3} - 9t^{2} + 12t\) gives \(x = 2\) and \(y = 0\). So the bug moves straight horizontally right or left at the point \((2, 0)\) when \(t = 2\).

Step 14 ：Finally, we check our answers and find that they meet the requirements of the problem. The bug comes to a complete stop at \(t = 1\), moves straight up or down at the points \((-2, 5)\) and \((2, -5)\), and moves straight horizontally right or left at the points \((-2, 5)\) and \((2, 0)\).

Step 15 ：So the final answer is: the bug comes to a complete stop at \(t = 1\), moves straight up or down at the points \((-2, 5)\) and \((2, -5)\), and moves straight horizontally right or left at the points \((-2, 5)\) and \((2, 0)\). \(\boxed{t = 1, (-2, 5), (2, -5), (-2, 5), (2, 0)}\)