Geometrical interpretation of derivative

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

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

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

Next Page → Length of Tangent , Normal , Sub-tangent and Sub-normal

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