For a curve given by y = f(x), \( \frac{dy}{dx} \) is the gradient function. If given the coordinates of a point as on the curve, The value of \( \frac{dy}{dx} \) and the equation of the tangent at the point can be found.
Note: if two lines with gradients m1, m2 are perpendicular, then m1, m2 = -1
Hence the gradient of the line perpendicular to the tangent, the normal to the curve is \( \scriptsize m_2 = \normalsize \frac{-1}{m} \)
m2 = \( \frac {-1}{\frac{dy}{dx}} \)
Examples:
1. Find the equation of the tangent and normal to the curve y = (1 – x)(2 + x), at the point where x = 2.
Solution:
y = (1 – x)(2 + x)
y = 2 – x – x2
\( \frac{dy}{dx} \) = -1 -2x
x = 2
y = 2 – 2 – (22)
y = -4
Gradient of the tangent at the point (2, -4)
\( \frac {dy}{dx} \scriptsize = -1 \; – \; 2x \\ \scriptsize = \; -1 \; – \; 2(2) \\ \scriptsize = 5 \)y + 4 = -5(x – 2)
So its equation is y + 5x = 6
Gradient of the normal at the point = \( \frac{1}{5} \) and its equation is
y + 4 = \( \frac {1}{5} \scriptsize(x \; – \; 2) \\ \scriptsize 5y = x \; – \; 22\)
Responses