Problem 23 - Essential Calculus

Question

Find the equations of the tangent lines at the point where the curve crosses itself. $$ x=t^{2}-t, \quad y=t^{3}-3 t-1 $$

Step-by-Step Solution

Answer

The equations of the tangent lines at the points where the curve crosses itself are \( y = x - 3 \) and \( y = -2x - 1 \) respectively.

Step 1: Find the Intersection Point

To locate the intersection point, set \( x(t) = t^{2}-t \) and \( y(t) = t^{3}-3t-1 \) then solve for common solutions. After solving, the intersection point comes out to be \( t = 1, -1 \). Since the solution has two points, the curve crosses itself.

Step 2: Find the Derivative

To find the tangent lines, we need to first find the slope of the curve at the intersection point. This can be calculated by the derivative \( y'(t) / x'(t) \). So, calculate \( x'(t) = 2t - 1 \) and \( y'(t) = 3t^{2} - 3 \).

Step 3: Calculate the Slope

After calculating \( x'(t) \) and \( y'(t) \), we insert \( t = 1, -1 \) into \( y'(t) / x'(t) \) to calculate the slope. After calculating, the slopes at \( t = 1, -1 \) are 0 and -2 respectively.

Step 4: Write the Equation for Tangents

Now, knowing the slope and the point at which the tangents intersect, we obtain the equations of the tangents using the following form: \( y - y_1 = m(x - x_1) \). Thus, this gives us \( y = x - 3 \) and \( y = -2x - 1 \), which are the equations of the tangents at \( t = 1, -1 \) respectively.