Question

Find equations for the lines that are tangent and normal to the graph of y=\sin x+3\( at \)x=\pi

Short answer

The tangent line to the function \(y = \sin(x) + 3\) at \(x = \pi\) is \(y = -1(x - \pi) + 3\) and the normal line is \(y = (x - \pi) + 3.\)

Step-by-step solution

Calculate the derivative of the function

The function given is \(y=\sin x+3\) and its derivative is \( y'=\cos(x)\). We take the derivative of the function because the slope of the tangent line at any given point of a function is equal to the derivative of that function at that same point.

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

We substitute \(x=\pi\) into the derivative to find the slope of the tangent line. The cosine of \(\pi\) is \(-1\), therefore slope of the tangent line, \(m_t= -1\). Since the slope of the normal line is the negative reciprocal of the slope of the tangent line, slope of the normal line, \(m_n=-1/-1\)=1.

Find the y-coordinates for the point

Substitute \(x=\pi\) into the function \(y=\sin x+3\) to get \(y= \sin(\pi)+3=3\). So the point of tangency is \((\pi,3)\). This is the point through which both the tangent and normal lines pass.

Write out the equations of the tangent and normal lines

The equation of a line that passes through a point \((x_1, y_1)\) is defined by \(y-y_1=m(x-x_1)\). Using the slope of the tangent line \(m_t=-1\) and normal line \(m_n=1\), with the point of tangency \((\pi,3)\), we get the equations: Tangent line: \(y-3=-1(x-\pi)\) and Normal line: \(y-3=(x-\pi)\).