$-\operatorname{Giren} x(t)=\frac{t^{4}-t^{3}-4 t}{2 t^{3}-2 t^{2}+t+4} \quad y(t)=2 t^{3}+t^{2}-t$ F. Find the tangent to the curve when $t=2$-give ans is $y=m \times+c$ $y=m x t c$ find cordinates of stationary points
Solution
Step 1 ：Given the function \(y(t) = 2t^3 + t^2 - t\), we first need to find the derivative of the function to find the slope of the tangent at \(t=2\).
Step 2 ：The derivative of \(y(t)\) is \(y'(t) = 6t^2 + 2t - 1\).
Step 3 ：Substituting \(t=2\) into the derivative, we get the slope of the tangent at \(t=2\), which is \(y'(2) = 27\).
Step 4 ：Substituting \(t=2\) into the original function, we get the y-coordinate of the point at \(t=2\), which is \(y(2) = 18\).
Step 5 ：Using the point-slope form of the line equation \(y - y1 = m(x - x1)\), where \(m\) is the slope and \((x1, y1)\) are the coordinates of a point on the line, we can find the equation of the tangent line. In this case, \(x1 = 2\), \(y1 = 18\), and \(m = 27\).
Step 6 ：Substituting these values into the point-slope form, we get the equation of the tangent line as \(y = 27x - 36\).
Step 7 ：Next, we need to find the stationary points of the function. Stationary points occur where the derivative of the function is zero. So, we need to find the values of \(t\) for which the derivative of \(y(t)\) is zero.
Step 8 ：Solving the equation \(y'(t) = 0\), we get two solutions for \(t\), which are \(t = -\frac{1}{6} + \frac{\sqrt{7}}{6}\) and \(t = -\frac{\sqrt{7}}{6} - \frac{1}{6}\).
Step 9 ：Substituting these values of \(t\) into the original function, we get the y-coordinates of the stationary points, which are \(y(-\frac{1}{6} + \frac{\sqrt{7}}{6}) = -\frac{\sqrt{7}}{6} + 2(-\frac{1}{6} + \frac{\sqrt{7}}{6})^3 + (-\frac{1}{6} + \frac{\sqrt{7}}{6})^2 + \frac{1}{6}\) and \(y(-\frac{\sqrt{7}}{6} - \frac{1}{6}) = 2(-\frac{\sqrt{7}}{6} - \frac{1}{6})^3 + \frac{1}{6} + (-\frac{\sqrt{7}}{6} - \frac{1}{6})^2 + \frac{\sqrt{7}}{6}\).
Step 10 ：\(\boxed{\text{Final Answer: The equation of the tangent to the curve when } t=2 \text{ is } y = 27x - 36. \text{ The coordinates of the stationary points are } (-\frac{1}{6} + \frac{\sqrt{7}}{6}, -\frac{\sqrt{7}}{6} + 2(-\frac{1}{6} + \frac{\sqrt{7}}{6})^3 + (-\frac{1}{6} + \frac{\sqrt{7}}{6})^2 + \frac{1}{6}) \text{ and } (-\frac{\sqrt{7}}{6} - \frac{1}{6}, 2(-\frac{\sqrt{7}}{6} - \frac{1}{6})^3 + \frac{1}{6} + (-\frac{\sqrt{7}}{6} - \frac{1}{6})^2 + \frac{\sqrt{7}}{6}).}\)
