### 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 $

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

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

### Length of Tangent , Normal , Sub-tangent , Sub-normal

Let P (x, y) be any point on y = f (x). Let the tangent drawn at ‘ P ‘ meets the x-axis at ‘ T ‘, and normal drawn at ‘ P ‘ meets the x-axis at ‘ N ‘.

PT is called length of the tangent and PN is called the length of the normal.

If ‘ P_{1} ‘ be the projection of the point P on the x-axis then TP_{1} is called the sub-tangent (projection of line segment PT on the x-axis) and NP_{1} is called the sub normal (projection of line segment PN on the x-axis).

Let ∠PTN = θ ⇒ ∠P_{1}PN = θ

We have tanθ = dy/dx and PP_{1} = |y|

Now, PT= |y cosec θ|

or, $ \displaystyle PT = |y| \sqrt{1+ cot^2 \theta} $

$ \displaystyle PT = |y| \sqrt{1 + (\frac{dx}{dy})^2} $

Hence length of the tangent $ \displaystyle PT = |y| \sqrt{1 + (\frac{dx}{dy})^2} $

Now, PN = |y sec θ|

$ \displaystyle PN = |y| \sqrt{1 + tan^2 \theta} $

$ \displaystyle PN = |y| \sqrt{1 + (\frac{dy}{dx})^2} $

Length of the normal, $ \displaystyle PN = |y| \sqrt{1 + (\frac{dy}{dx})^2} $

Now, TP_{1} = |y cot θ|

$ \displaystyle TP_1 = |y \frac{dx}{dy}| $

⇒ Sub-tangent $ \displaystyle TP_1 = |y \frac{dx}{dy}| $

Finally, NP_{1} = |y tan θ|

$ \displaystyle NP_1 = |y \frac{dy}{dx}| $

⇒ Sub-normal, $ \displaystyle NP_1 = |y \frac{dy}{dx}| $

### Angle of Intersection of Two Curves

Let y = f (x) and y = g (x) be two given intersecting curves. Angle of intersection of these curves is defined as the acute angle between the tangents that can be drawn to the given curves at the point of intersection.

Let (x_{1}, y_{1}) be the point of intersection two given intersecting curves .

Slope of the tangent drawn to the curve y = f (x) at (x_{1}, y_{1})

i.e. $ \displaystyle m_1 = \frac{d f(x)}{dx}_{(x_1 , y_1)} $

Similarly, Slope of the tangent drawn to the curve y = g(x) at (x_{1}, y_{1})

i.e. $ \displaystyle m_2 = \frac{d g(x)}{dx}_{(x_1 , y_1)} $

If ‘ α ‘ be the angle (acute) of intersection, then

$ \displaystyle tan\alpha = |\frac{m_1 – m_2}{1 + m_1 m_2}| $

(i) If α = 0, then m_{1} = m_{2} .

Thus the given curves would touch each other at the point (x_{1}, y_{1})

(ii) If α = π/2 , then m_{1} m_{2} = -1.

Thus the given curves would meet at right angle at the point (x_{1}, y_{1}) (or curves cut orthogonally at the point (x_{1}, y_{1}))

### Also Read :

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