Problem 28 - Essential Calculus

Question

Find all points (if any) of horizontal and vertical tangency to the curve. Use a graphing utility to confirm your results. $$ x=t^{2}-t+2, \quad y=t^{3}-3 t $$

Step-by-Step Solution

Answer

The points of horizontal tangency are (-1,2) and (1, -2). The point of vertical tangency is (1.75, -0.125).

Step 1: Compute the Derivatives dx/dt and dy/dt

From \(x = t^2 - t + 2\), the derivative dx/dt can be computed as \(2t - 1\). Similarly, from \(y = t^3 - 3t\), the derivative dy/dt is obtained as \(3t^2 - 3\).

Step 2: Compute the Ratio dy/dx

The derivative dy/dx is actually a ratio of dy/dt to dx/dt. Plug the obtained values from Step 1 to get \((3t^2 - 3) / (2t - 1)\).

Step 3: Find the Values of t for Horizontal Tangency

For horizontal tangency, dy/dx should be equal to zero. So set \((3t^2 - 3) / (2t - 1) = 0\) and solve for t to get \( t = sqrt(1)\) and \(t = -sqrt(1)\). These are the parameter values at which horizontal tangency occurs. Substitute these values in the original parametric equations to get the corresponding (x, y) coordinates.

Step 4: Find the Values of t for Vertical Tangency

For vertical tangency, dx/dt should be equal to zero. Thus, set \(2t - 1 = 0\) and solve for t. You will get \(t = 1/2\). This is the parameter value at which vertical tangency occurs. Substitute this value in the original parametric equations to get the corresponding (x, y) coordinates.

Step 5: Graphing the Function to Confirm the Results

You can graph the functions using a graphic calculator or software by plotting the original parametric equations \(x = t^2 - t + 2\) and \(y = t^3 - 3t\). The points of tangency found in the previous steps should align with the horizontal and vertical tangents noticeable on the graph.