Question

In Exercises \(41-48,\) find the equation of the line tangent to the curve at the point defined by the given value of \(t\) . $$x=\sec ^{2} t-1, \quad y=\tan t, \quad t=-\pi / 4$$

Short answer

The equation of the tangent line to the curve at \(t = -\pi / 4\) is \(y = -x - 1\).

Step-by-step solution

Calculating derivatives

First, we need to calculate the derivatives of \(x(t)\) and \(y(t)\) with respect to \(t\). Using the chain rule and the fact that the derivative of \(\sec t\) is \(\sec t \tan t\), we find that \(dx/dt = 2 \sec t \sec t \tan t = 2 \sec^{2}t \tan t\) and \(dy/dt = \sec^{2}t\). To find \(dy/dx\), we need to form the ratio \(dy/dt\) divided by \(dx/dt\). Hence, \(dy/dx = dy/dt / dx/dt = \sec^{2}t / (2 \sec^{2}t \tan t)\).

Evaluate dy/dx at \(t = -\pi / 4\)

We substitute \(t = -\pi / 4\) into our expression for \(dy/dx\) to evaluate the slope of the tangent line. This gives \(dy/dx = 1 / (2 \tan(-\pi / 4)) = -1\).

Calculate the point of tangency

To find the point at which the tangent line touches the curve, we substitute \(t = -\pi / 4\) into \(x(t)\) and \(y(t)\). This gives us the coordinates of the point: \(x = \sec^{2}(-\pi / 4)-1 = 1-1 = 0\) and \(y = \tan(-\pi / 4) = -1\). So the point is (0, -1).

Find the equation of the tangent line

Finally, we can use the slope of the tangent line and the point of tangency to find the equation of the tangent line. From the slope-intercept form: \(y = mx + c\), we know \(m = -1\) (from Step 2), and the point (0, -1) lies on the line, so when \(x = 0, y = -1\). Thus, the equation of the tangent line is \(y = -x - 1\).