__Geometrical interpretation of derivative :__

Consider a function y = f(x) and points P(x_{1}, y_{1}) and Q(x_{2}, y_{2}) on it. As x_{1} changes to x_{2}, y_{1} becomes y_{2}. 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 x_{2} → x_{1} 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 (x_{1}, y_{1}) be any point on the curve, that means, y_{1} = f(x_{1}).

Now the slope of the tangent that can be drawn to the curve at (x_{1}, y_{1}) will be

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

Thus the equation of the tangent at (x_{1}, y_{1}) would be,

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

Similarly, the equation of the normal at (x_{1}, y_{1}) 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 = x_{1}

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

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

Illustration : Tangent at point P_{1} (other then (0, 0)) on the curve y = x^{3} meets the curve again at P_{2}. The tangent at P_{2} meets the curve at P_{3} and so on. Show that the abscissa of P_{1}, P_{2}, P_{3} ,….. P_{n}, 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 P_{1} on y = x^{3} be (h, h^{3})

tangent at P_{1} is y – h^{3} = 3h^{2} (x − h), it meets y = x^{3} at P_{2}

x^{3} – h^{3} = 3h^{2} (x – h)

x^{2} + xh + h^{2} − 3h^{2} = 0

x^{2} + xh – 2h^{2} = 0

(x – h) (x + 2h) = 0

x = – 2h for P_{2} as x = h is for point P_{1}

P_{2} is (-2h, – 8h^{3})

tangent at P_{2} is y + 8h^{3} = 3(2h)^{2} (x + 2h),

it meets y = x^{3} at P_{3} ,

x^{3} + 8h^{3} =12h^{2}(x + 2h)

x^{2} – 2hx – 8h^{2} = 0

(x – 4h) (x + 2h) = 0

x = 4h for P_{3}

P_{3} is (4h, 64h^{3})

continuing like this, we get x = -8h for P_{4} etc,

hence the abscissae of P_{1}, P_{2} , P_{3} …. are h , -2h , 4h, -8h , … which are in G.P.

Δ_{1} = ΔP_{1}P_{2}P_{3}

Δ_{2} = ΔP_{2}P_{3}P_{4}

$ \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 P_{1} and P_{2} are the lengths of perpendiculars from the origin to the tangent and to the normal to the curve x^{2/3} + y^{2/3} = a^{2/3} 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 $

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

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

### Also Read :

→ Equation of Tangent & Normal → Length of Tangent,Normal ,Sub-tangent & Sub-normal → Angle of intersection of two Curves |