Slopes of Tangent and Normal to the Curve

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) = \(-1\over slope of tangent at (1, 1)\) = \(5\over 2\) 

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