Let's solve each part step by step for the parametric curve: \[ x(t) = t^2 - 2t, \quad y(t) = t^3 + t^2 - 6t, \quad -\infty < t < \infty \] --- ### **(a) Write the equation for the curve in \(x\) and \(y\) (eliminate \(t\))** Given: \[ x = t^2 - 2t \implies t^2 - 2t - x = 0 \] This is a quadratic in \(t\): \[ t = 1 \pm \sqrt{1 + x} \] Now, plug \(t\) into \(y\): Let’s use \(y = t^3 + t^2 - 6t\). Substitute \(t = 1 \pm \sqrt{1 + x}\): \[ y = (t)^3 + (t)^2 - 6(t) \] So, the equation in \(x\) and \(y\) is: \[ y = (t^3 + t^2 - 6t) \quad \text{where} \quad t = 1 \pm \sqrt{1 + x} \] Or more explicitly: \[ y = (1 \pm \sqrt{1 + x})^3 + (1 \pm \sqrt{1 + x})^2 - 6(1 \pm \sqrt{1 + x}) \] --- ### **(b) Find all critical points and indicate if the tangent is vertical or horizontal** Critical points occur when either \( \frac{dx}{dt} = 0 \) (vertical tangent) or \( \frac{dy}{dt} = 0 \) (horizontal tangent). #### **Step 1: Compute derivatives** \[ \frac{dx}{dt} = 2t - 2 \] \[ \frac{dy}{dt} = 3t^2 + 2t - 6 \] #### **Step 2: Horizontal Tangent (\(dy/dt = 0\))** Set \(3t^2 + 2t - 6 = 0\): \[ 3t^2 + 2t - 6 = 0 \] Use quadratic formula: \[ t = \frac{-2 \pm \sqrt{4 + 72}}{6} = \frac{-2 \pm \sqrt{76}}{6} = \frac{-2 \pm 2\sqrt{19}}{6} = \frac{-1 \pm \sqrt{19}}{3} \] So, two values: \[ t_1 = \frac{-1 + \sqrt{19}}{3}, \quad t_2 = \frac{-1 - \sqrt{19}}{3} \] Plug into \(x(t)\) and \(y(t)\) for coordinates. #### **Step 3: Vertical Tangent (\(dx/dt = 0\))** Set \(2t - 2 = 0 \implies t = 1\) Find point: \[ x(1) = 1^2 - 2\times1 = 1 - 2 = -1 \] \[ y(1) = 1^3 + 1^2 - 6\times1 = 1 + 1 - 6 = -4 \] So, the critical point for vertical tangent is \((1, (-1, -4))\). #### **Summary of critical points:** - Horizontal: \(t_1, t_2\) as above, coordinates \( (x(t_1), y(t_1)) \), \( (x(t_2), y(t_2)) \). - Vertical: \(t = 1\), coordinate \((-1,-4)\). --- ### **(c) Where does the concavity change? Where is the curve concave up?** We need to consider the second derivative of \(y\) with respect to \(x\): \[ \frac{d^2y}{dx^2} = \frac{d}{dt}\left(\frac{dy}{dx}\right)\Big/\frac{dx}{dt} \] First, \[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{3t^2 + 2t - 6}{2t - 2} \] Now, \[ \frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{(6t + 2)(2t - 2) - (3t^2 + 2t - 6) \cdot 2}{(2t - 2)^2} \] Set numerator to zero to find inflection points: \[ (6t + 2)(2t - 2) - 2(3t^2 + 2t - 6) = 0 \] \[ (12t^2 - 12t + 4t - 4) - (6t^2 + 4t - 12) = 0 \] \[ (12t^2 - 8t - 4) - (6t^2 + 4t - 12) = 0 \] \[ 12t^2 - 8t - 4 - 6t^2 - 4t + 12 = 0 \] \[ (12t^2 - 6t^2) + (-8t - 4t) + (-4 + 12) = 0 \] \[ 6t^2 - 12t + 8 = 0 \] \[ 3t^2 - 6t + 4 = 0 \] \[ t = \frac{6 \pm \sqrt{36 - 48}}{6} = \frac{6 \pm \sqrt{-12}}{6} \] \[ t = \frac{6 \pm 2i\sqrt{3}}{6} = 1 \pm \frac{i\sqrt{3}}{3} \] No **real** solutions, so the curve never changes concavity; the sign of the second derivative is always the same. Check concavity at a test value (say, \(t = 0\)): Plug \(t = 0\) into numerator: \[ (6(0) + 2)(2(0) - 2) - 2(3(0)^2 + 2(0) - 6) = (2)(-2) - 2(-6) = -4 + 12 = 8 \] Since denominator \((2t-2)^2 > 0\) always, and numerator is positive, the curve is always **concave up**. --- ### **(d) Self-intersection: find \(t\), \((x, y)\), and tangent lines** Self-intersection: \((x(t_1), y(t_1)) = (x(t_2), y(t_2)), t_1 \neq t_2\). Set \(x(t_1) = x(t_2)\): \[ t_1^2 - 2t_1 = t_2^2 - 2t_2 \] \[ t_1^2 - t_2^2 - 2t_1 + 2t_2 = 0 \] \[ (t_1 - t_2)(t_1 + t_2) - 2(t_1 - t_2) = 0 \] \[ (t_1 - t_2)[(t_1 + t_2) - 2] = 0 \] Since \(t_1 \neq t_2\), \(t_1 + t_2 = 2\). Now, set \(y(t_1) = y(t_2)\): \[ t_1^3 + t_1^2 - 6t_1 = t_2^3 + t_2^2 - 6t_2 \] \[ (t_1^3 - t_2^3) + (t_1^2 - t_2^2) - 6(t_1 - t_2) = 0 \] \[ (t_1 - t_2)[t_1^2 + t_1 t_2 + t_2^2 + t_1 + t_2 - 6] = 0 \] Again, \(t_1 \neq t_2\), so: \[ t_1^2 + t_1 t_2 + t_2^2 + t_1 + t_2 - 6 = 0 \] But \(t_1 + t_2 = 2\), let \(s = t_1 + t_2 = 2\), \(p = t_1 t_2\): \[ t_1^2 + t_1 t_2 + t_2^2 = (t_1 + t_2)^2 - t_1 t_2 = s^2 - p \] So: \[ s^2 - p + p + s - 6 = 0 \] \[ s^2 + s - 6 = 0 \] With \(s = 2\): \[ 2^2 + 2 - 6 = 4 + 2 - 6 = 0 \] So any two values \(t_1, t_2\) with \(t_1 + t_2 = 2\) and \(t_1 \neq t_2\) give a self-intersection. Pick \(t_1 = a\), \(t_2 = 2 - a\), \(a \neq 1\). Let’s pick \(t_1 = 0, t_2 = 2\): \[ x(0) = 0^2 - 2(0) = 0 \] \[ y(0) = 0^3 + 0^2 - 6(0) = 0 \] \[ x(2) = 2^2 - 2 \times 2 = 4 - 4 = 0 \] \[ y(2) = 8 + 4 - 12 = 0 \] So, both points are \((0, 0)\). The curve passes through \((0, 0)\) at \(t = 0\) and \(t = 2\). Find tangent lines at \(t = 0\) and \(t = 2\): - At \(t = 0\): \[ \frac{dx}{dt} = 2t-2 = -2, \quad \frac{dy}{dt} = 3t^2 + 2t - 6 = -6 \] Slope: \(\frac{dy}{dx} = \frac{-6}{-2} = 3\) Equation: \(y = 3x\) at \((0, 0)\) - At \(t = 2\): \[ \frac{dx}{dt} = 2(2) - 2 = 2, \quad \frac{dy}{dt} = 3(4) + 4 - 6 = 12 + 4 - 6 = 10 \] Slope: \(\frac{dy}{dx} = \frac{10}{2} = 5\) Equation: \(y = 5x\) at \((0, 0)\) --- ### **(e) Set up the arc length integral between self-intersection points** Arc length from \(t = 0\) to \(t = 2\): \[ L = \int_{0}^{2} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt \] Where: \[ \frac{dx}{dt} = 2t - 2 \] \[ \frac{dy}{dt} = 3t^2 + 2t - 6 \] So, \[ L = \int_{0}^{2} \sqrt{(2t - 2)^2 + (3t^2 + 2t - 6)^2} \, dt \] --- ## **Final Answers (Summary):** **(a)** \(\boxed{y = (1 \pm \sqrt{1 + x})^3 + (1 \pm \sqrt{1 + x})^2 - 6(1 \pm \sqrt{1 + x})}\) **(b)**  - Horizontal tangents at \(t = \frac{-1 \pm \sqrt{19}}{3}\), coordinates \((x(t), y(t))\). - Vertical tangent at \(t = 1\), coordinate \((-1, -4)\). **(c)** Curve is always **concave up**; concavity never changes. **(d)** Self-intersection at \(t = 0\) and \(t = 2\) at point \((0, 0)\). - Tangent at \(t = 0\): \(y = 3x\) - Tangent at \(t = 2\): \(y = 5x\) **(e)** Arc length between first and second pass through \((0, 0)\): \[ \boxed{L = \int_0^2 \sqrt{(2t - 2)^2 + (3t^2 + 2t - 6)^2} \, dt} \]