Question

Find equations for the lines that are tangent and normal to the graph of y=\sec x\( at \)x=\pi / 4

Short answer

The equations for the tangent and normal lines to the graph of \(y = sec x\) at \(x = \pi / 4\) are \(y - \sqrt{2} = \sqrt{2}(x - \pi / 4)\) and \(y - \sqrt{2} = -1 / \sqrt{2}(x - \pi / 4)\) respectively.

Step-by-step solution

Find the derivative of y = sec x

The derivative of \(y = \sec x\) is \(y' = \sec x \cdot \tan x\).

Substitute x = \(\pi / 4\)

Substitute \(x = \pi / 4\) into the derivative to find the slope of the tangent line: \(m_t = \sec(\pi / 4) \cdot \tan(\pi / 4) = \sqrt{2}\).

Find the y-coordinate

Substitute \(x = \pi / 4\) into the original equation to find the y-coordinate of the point of tangency: \(y = \sec(\pi / 4) = \sqrt{2}\). So, the point of tangency is \((\pi / 4, \sqrt{2})\).

Find equation for the tangent line

Use the point-slope formula to find an equation for the tangent line: \(y - \sqrt{2} = \sqrt{2}(x - \pi / 4)\).

Find the slope and equation for the normal line

The slope of the normal line is the negative reciprocal of the slope of the tangent line: \(m_n = -1 / \sqrt{2}\). Using the point-slope formula, the equation for the normal line is \(y - \sqrt{2} = -1 / \sqrt{2}(x - \pi / 4)\).