Find equations of the tangent line and normal line to the curve $y=14\cos x$ at the point $(\pi /3,7)$.
The derivative $y^{\prime }(x)=$
The slope of the tangent line is $m_1=$
The equation of the tangent line is $y=$
The slope of the normal line is $m_2=$
The equation of the normal line is $y=$

Step 1 ：Given the equation for y as a function of x, we have $y=14\cos (x)$.

Step 2 ：The derivative of y with respect to x is $dy/dx=-14\sin (x)$.

Step 3 ：The slope of the tangent line to the curve at a given point is equal to the derivative of y with respect to x, which is $dy/dx$.

Step 4 ：Evaluating this derivative at $x=\pi /3$, we find that $dy/dx=-14\sin (\pi /3)=-14\sqrt{3}/2=-7\sqrt{3}$, so the slope of the tangent line is $m_1=dy/dx=-7\sqrt{3}$.

Step 5 ：The coordinates of the point on the curve are $x=\pi /3$ and $y=7$.

Step 6 ：The equation of a line is given by $y=mx+c$, where m is the slope and c is the y-intercept. Substituting the slope and the coordinates of the point into this equation, we can solve for c to find that $c=y-mx=7-(-7\sqrt{3})(\pi /3)=7+7\pi \sqrt{3}/3$.

Step 7 ：So, the equation of the tangent line is $y=-7\sqrt{3}x+7+7\pi \sqrt{3}/3$.

Step 8 ：The slope of the normal line is the negative reciprocal of the slope of the tangent line, so $m_2=-1/m_1=1/(7\sqrt{3})=\sqrt{3}/21$.

Step 9 ：Substituting the slope of the normal line and the coordinates of the point into the equation for a line, we can solve for the y-intercept to find that $c=y-m_{2}x=7-(\sqrt{3}/21)(\pi /3)=7-\pi \sqrt{3}/63$.

Step 10 ：So, the equation of the normal line is $y=\sqrt{3}/21x+7-\pi \sqrt{3}/63$.

Step 11 ：$\boxed{{\displaystyle \text{Final Answer: The derivative }{y}^{\prime }(x)=-14\sin (x),\text{ the slope of the tangent line is }{m}_1=-7\sqrt{3},\text{ the equation of the tangent line is }y=-7\sqrt{3}x+7+7\pi \sqrt{3}/3,\text{ the slope of the normal line is }{m}_2=\sqrt{3}/21,\text{ and the equation of the normal line is }y=\sqrt{3}/21x+7-\pi \sqrt{3}/63}}$