Find the equation for the line tangent to the parametric curve: \[ \begin{array}{l} x=t^{3}-16 t \\ y=16 t^{2}-t^{4} \end{array} \] at the points where $t=4$ and $t=-4$. For $t=4$, the tangent line (in form $y=m x+b$ ) is \[ y= \] For $t=-4$, the tangent line is \[ y= \]
Solution
Step 1 ：First, we need to find the derivative of the parametric curve. The derivative of a parametric curve is given by \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt}\).
Step 2 ：Given the parametric equations \(x = t^3 - 16t\) and \(y = 16t^2 - t^4\), we can find \(\frac{dx}{dt} = 3t^2 - 16\) and \(\frac{dy}{dt} = -4t^3 + 32t\).
Step 3 ：Substituting these into the equation for \(\frac{dy}{dx}\), we get \(\frac{dy}{dx} = \frac{-4t^3 + 32t}{3t^2 - 16}\).
Step 4 ：We can find the slope of the tangent line at the given points by substituting the values of t into the derivative. For \(t = 4\), the slope is -4. For \(t = -4\), the slope is 4.
Step 5 ：Next, we need to find the coordinates of the points on the curve where \(t = 4\) and \(t = -4\). Substituting these values into the equations for x and y, we find that both points are (0, 0).
Step 6 ：Finally, we can use the point-slope form of the equation of a line, \(y - y1 = m(x - x1)\), where m is the slope of the line and \((x1, y1)\) is a point on the line, to find the equation of the tangent line.
Step 7 ：For \(t = 4\), the equation of the tangent line is \(y = -4x\).
Step 8 ：For \(t = -4\), the equation of the tangent line is \(y = 4x\).
Step 9 ：\(\boxed{\text{Final Answer: For } t=4, \text{ the tangent line is } y = -4x. \text{ For } t=-4, \text{ the tangent line is } y = 4x.}\)
