Here you will learn slopes of tangent and normal to the curve with examples.
Let’s begin –
Slopes of Tangent and Normal to the Curve
(a) Slopes of Tangent
Let y = f(x) be a continuous curve, and let \(P(x_1, y_1)\) be a point on it. Then,
\(({dy\over dx})_P\) is the tangent to the curve y = f(x) at point P.
i.e. \(({dy\over dx})_P\) = tan \(\psi\) = Slope of the tangent at P,
where \(\psi\) is the angle which the tangent at \(P(x_1, y_1)\) makes with the positive direction of x-axis.
If the tangent at P is parallel to x-axis, then
\(\psi\) = 0 \(\implies\) tan \(\psi\) = 0 \(\implies\) Slope = 0 \(\implies\) \(({dy\over dx})_P\) = 0
If the tangent at P is perpendicular to x-axis, or parallel to y-axis, then
\(\psi\) = \(\pi\over 2\) \(\implies\) cot \(\psi\) = 0 \(\implies\) \(1\over tan \psi\) = 0 \(\implies\) \(({dx\over dy})_P\) = 0
(b) Slopes of Normal
The normal to the curve at \(P(x_1, y_1)\) is a line perpendicular to the tangent at P and passing through P.
\(\therefore\) Slope of the normal at P = \(-1\over Slope of the tangent at P\) = \(-({dx\over dy})_P\)
Example : find the slopes of the tangent and the normal to the curve \(x^2 + 3y + y^2\) = 5 at (1, 1).
Solution : The equation of the curve is \(x^2 + 3y + y^2\) = 5
Differentiating with respect to x, we get
2x + 3\(dy\over dx\) + 2y\(dy\over dx\) = 0
\(\implies\) \(dy\over dx\) = \(-2x\over 2y + 3\)
\(\implies\) \(({dy\over dx})_{(1, 1)}\) = -(\(2\over 2 + 3\)) = -\(2\over 5\)
\(\therefore\) Slope of the tangent at (1, 1) = -\(2\over 5\)
and, Slope of normal at (1, 1) = \(-1\over slope of tangent at (1, 1)\) = \(5\over 2\)
