# Geometrical interpretation of derivative : Equation of Tangent & Normal

##### Geometrical interpretation of derivative :

Consider a function y = f(x) and points P(x1, y1) and Q(x2, y2) on it. As x1 changes to x2, y1 becomes y2. Average rate of change will be given by

$\displaystyle \frac{\Delta y}{\Delta x}= \frac{y_2 -y_1}{x_2-x_1}$

(which is clearly the slope of line PQ).

As Q → P i.e. as x2 → x1 or Δx → 0,

the average rate of change Δy/Δx becomes the instantaneous rate of change represented by dy/dx and thus dy/dx represent the slope of the tangent at P.

### Equation of Tangent and Normal

The derivative of a function y = f(x) represents the slope of the tangent to the curve at the general point (x, y).

Let y = f (x) be the given curve. We already know that dy/dx at any point lying on the curve would give us the slope of the tangent that can be drawn at that point.

Let (x1, y1) be any point on the curve, that means, y1 = f(x1).

Now the slope of the tangent that can be drawn to the curve at (x1, y1) will be

$\displaystyle (\frac{dy}{dx})_{x_1,y_1}$

Thus the equation of the tangent at (x1, y1) would be,

$\displaystyle y-y_1 = (\frac{dy}{dx})_{x_1,y_1} (x-x_1)$

Similarly, the equation of the normal at (x1, y1) would be

$\displaystyle y-y_1 = -\frac{1}{(\frac{dy}{dx})_{x_1,y_1}} (x-x_1)$

provided that $\displaystyle (\frac{dy}{dx})_{(x_1,y_1)} \ne 0$

If $\displaystyle (\frac{dy}{dx})_{(x_1,y_1)} = 0$ ; then the equation of the normal would be x = x1

Note: If x = g (t) , y = h (t)

then ,  $\displaystyle \frac{dy}{dx} = \frac{dh/dt}{dg/dt}$

Illustration : Tangent at point P1 (other then (0, 0)) on the curve y = x3 meets the curve again at P2. The tangent at P2 meets the curve at P3 and so on. Show that the abscissa of P1, P2, P3 ,….. Pn, form a G.P. also, find the ratio

$\displaystyle \frac{area \Delta P_1 P_2 P_3}{area \Delta P_2 P_3 P_4 }$

Solution: Let a point P1 on y = x3 be (h, h3)

tangent at P1 is y – h3 = 3h2 (x − h), it meets y = x3 at P2

x3 – h3 = 3h2 (x – h)

x2 + xh + h2 − 3h2 = 0

x2 + xh – 2h2 = 0

(x – h) (x + 2h) = 0

x = – 2h for P2 as x = h is for point P1

P2 is (-2h, – 8h3)

tangent at P2 is y + 8h3 = 3(2h)2 (x + 2h),

it meets y = x3 at P3 ,

x3 + 8h3 =12h2(x + 2h)

x2 – 2hx – 8h2 = 0

(x – 4h) (x + 2h) = 0

x = 4h for P3

P3 is (4h, 64h3)

continuing like this, we get x = -8h for P4 etc,

hence the abscissae of P1, P2 , P3 …. are h , -2h , 4h, -8h , … which are in G.P.

Δ1 = ΔP1P2P3

Δ2 = ΔP2P3P4

$\large \frac{\Delta_1}{\Delta_2} = \frac{1}{2}\left| \begin{array}{ccc} h & h^3 & 1 \\ -2h & -8h^3 & 1 \\ 4h & 64h^3 & 1 \end{array} \right| \div \frac{1}{2}\left| \begin{array}{ccc} -2h & -8h^3 & 1 \\ 4h & 64h^3 & 1 \\ -8h & -512h^3 & 1 \end{array} \right|$

$\displaystyle \frac{\Delta_1}{\Delta_2} = \frac{1}{16}$

Illustration : If P1 and P2 are the lengths of perpendiculars from the origin to the tangent and to the normal to the curve x2/3 + y2/3 = a2/3 respectively, prove that
$\displaystyle P_1^2 + \frac{P_2^2}{4}$ is a constant.

Solution : Parametric form of curve can some times be tried to solve the problems.

$\large x = a cos^3\theta \; , y = a sin^3 \theta$

$\large \frac{dy}{dx} = – \frac{3 sin^2 \theta cos\theta}{3cos^2 \theta sin\theta} = – tan\theta$

Tangent : $\large \frac{y-a sin^3 \theta}{x- a cos^3 \theta} = – tan\theta$

Normal : $\large \frac{y – a sin^3 \theta}{x – a cos^3 \theta} = cot\theta$

$\large P_1^2 = \frac{(a sin^3 \theta + a cos^3 \theta tan\theta)^2}{1 + tan^2 \theta}$

$\large P_1^2 = \frac{a^2 sin^2 \theta}{sec^2 \theta}$

$\large P_1^2 = a^2 sin^2 \theta cos^2 \theta$

$\large P_1^2 = \frac{a^2}{4} sin^2 2\theta$

$\large P_2^2 = \frac{(a sin^3 \theta – a cos^3 \theta cot\theta)^2}{1 + cot^2 \theta}$

$\large P_2^2 = a^2 (sin^4 \theta – cos^4 \theta)^2$

$\large = a^2 cos^2 2\theta$

$\large P_1^2 + \frac{P_2^2}{4} = \frac{a^2(sin^2 2\theta + cos^2 2\theta)}{4} = \frac{a^2}{4}$