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

Related Questions

Show that the tangents to the curve y = $$2x^3 – 3$$ at the points where x =2 and x = -2 are parallel.

Find the slope of normal to the curve x = 1 – $$a sin\theta$$, y = $$b cos^2\theta$$ at $$\theta$$ = $$\pi\over 2$$.

Find the slope of the normal to the curve x = $$a cos^3\theta$$, y = $$a sin^3\theta$$ at $$\theta$$ = $$\pi\over 4$$.