Question

Multiple Choice Which of the following is an equation of the normal line to y=\sin x+\cos x\( at \)x=\pi ? (A) \(y=-x+\pi-1\) (B) \(y=x-\pi-1\) (C)\(y=x-\pi+1\) (D) \(y=x+\pi+1\) (E) \(y=x+\pi-1\)

Short answer

The equation of the normal line to \(y=\sin x + \cos x\) at \(x=\pi\) is \(y = x - \pi + 1\).

Step-by-step solution

Find the derivative of the function

The derivative of the function \(y=\sin x + \cos x\) is \(y'=\cos x - \sin x\). This gives the slope of the tangent line at any point on the curve.

Evaluate the derivative at \(x=\pi\)

Substitute \(x=\pi\) into the derivative to find the slope of the tangent line at \(x=\pi\). This gives \(y'(\pi)=\cos \pi - \sin \pi = -1 - 0 = -1\).

Find the negative reciprocal of the slope

The slope of the normal line is the negative reciprocal of the slope of the tangent line. Therefore, the slope of the normal line is -(-1) = 1.

Use the point-slope form to find the equation of the normal line

The point-slope form of a line is \(y - y1 = m(x - x1)\), where \(m\) is the slope and \((x1, y1)\) is a point on the line. The point on the line is \(\pi, \sin \pi + \cos \pi = \pi, -1\). Hence, the equation of the normal line is \(y - (-1) = 1(x - \pi)\), which simplifies to \(y = x - \pi + 1\).