next up previous
Next: この文書について...

Example Find the tangent line of the following curve at the point corresponding to $x = a$.

$\displaystyle \displaystyle{f(x) = x^{2} - 5x + 3, \ a = 2} $

Answer The slope of the tangent line at $x = a$ is given by $f'(a)$. So,

$\displaystyle f'(2) = 2*2 - 5 = -1$

We know every straight line can be determined by knowing the slope $m$ and the point $(a,f(a))$ goes through. Then the equation of the tangent line is given by $y = m(x-a) + f(a)$．Now we find the point where the tangent line goes through.

$\displaystyle f(2) = 4 -10+3 = -3$

Thus the equation of the tangent line is

$\displaystyle y = -(x-2) -3 = -x -1$

To find the normal line, it is enough to know that the slope $m$ of the tangent line and the slope $m'$ of the normal line, $mm' = -1$. To see this, consider the vector $(1,m)$ on the tangent line and the vector $(1,m')$ on the normal line. Then these vector are orthogonal which implies $(1,m)\cdot (1,m') = 1 + mm' = 0$. Thus，$mm' = -1$. Therefore, the equation of normal line is

$\displaystyle y = (x-2)-3 = x-5$