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Chapter 8: Problem 28

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

Short Answer

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

Step by step solution

01

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 \(3t^2 - 3\).
02

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 \((3t^2 - 3) / (2t - 1)\).
03

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.
04

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.
05

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.

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