Geometrical Applications of Derivatives

GEOMETRICAL MEANING OF DERIVATIVE AT A POINT

The derivative $\left[f^{\prime }\left(x\right)or\frac{dy}{dx}\right]$ of the function y = f(x) at the point P(x, y) (when exists) is equal to the slope (or gradient) of the tangent line to the curve y = f(x) at P(x, y).

Slope of tangent to the curve y = f(x) at the point (x, y) is m = $\tan \theta =$ ${\left[\frac{dy}{dx}\right]}_{\left(x,y\right)}$

EQUATION OF TANGENT

The equation of tangent to the curve y = f(x) at the point P(x₁, y₁) is given by $y-y_1={\left[\frac{dy}{dx}\right]}_{(x_1,y_1)}(x-x_1)$.

Notes:

(i) If $\frac{dy}{dx}$ = 0 then the tangent to curve y = f(x) at the point (x, y) is parallel to the x-axis.

(ii) If $\frac{dy}{dx}$ $\to$ $\infty$or $\frac{dx}{dy}$ = 0, then the tangent to the curve y = f(x) at the point (x, y) is parallel to the y-axis.

(iii) If $\frac{dy}{dx}$ = $\tan \theta >0$ then the tangent to the curve y = f(x) at the point (x, y) makes an acute angle with positive x-axis and vice versa.

EQUATION OF NORMAL

The normal to the curve at the point P(x₁, y₁) is a line perpendicular to the tangent at the point P(x₁, y₁) and passing through it. The angle between a tangent and a normal at a point is always 90⁰. The equation of the normal to the curve y=f(x) at a given point P(x₁, y₁)is given by(x - x₁) + (y - y₁) $\left[\frac{dy}{dx}\right]$${}_{\left(x_1,y_1\right)}$ = 0.EQUATION OF TANGENT AND NORMAL IF EQUATION OF THE CURVE IS GIVEN IN PARAMETRIC FORMIf the equation of the curve is in the parametric form x = f(t) and y = g(t), then the equations of the tangent and the normal are $\text{y}-g(t)=\frac{gapos(t)}{fapos(t)}(x-f(t)$) and $fapos(t)\left[x-f(t)\right]+gapos(t)\left[y-g(t)\right]=0$ respectivelyIllustration 1 If at each point of the curve y = x³ – ax² + x + 1 the tangents is inclined at an acute angle with the positive direction of the x-axis, then find the interval in which a lies.Key concept: since the tangent is always inclined at an acute angle with the x-axis, hence$\frac{dy}{dx}>0$ and the use the conceptof quadratic equation that ax² + bx + c > 0 for all x $\in$R ifa > 0 and D < 0 Solution y = x³ – ax² + x + 1 and the tangent is inclined at an acute angle with the positive direction of x-axis,$\Rightarrow \frac{dy}{dx}>0$$\therefore$ 3x² – 2ax + 1 >0, for all x $\in$ R$\Rightarrow$(2a)² – 4 (3)(1) <0$\Rightarrow$ 4(a² ;– 3)< 0$\Rightarrow \left(a-\sqrt{3}\right)\left(a+\sqrt{3}\right)<0$$\therefore$$-\sqrt{3}<a<\sqrt{3}$

ANGLE OF INTERSECTION OF TWO CURVES

Let C₁: y= f(x) and C₂: y = g(x) be two curves. If two curves intersect at point P (x₁, y₁). Then the angle of intersection of two curves is defined as the angles between the tangents at their intersection. And is given by

$\tan \theta =\frac{f^{\prime }(x_1)-g^{\prime }(x)}{1+f^{\prime }(x_1)g^{\prime }(x_1)}$

Note:

(i) If the curves touch at P, then q = 0 so that f ' (x₁) = g ' (x₁)

(ii) If the angle $\theta ={90}^{\circ }\left(\frac{{\pi }^c}{2},rightangle\right)$i.e. Two tangents are perpendicular to each other then the curves are said to cut orthogonally, then f '(x₁). g '(x₁) =- 1.

Illustration 2: If the curves x²-4y²+c=0 and y² = 4x intersect orthogonally then find the range of c.

Solution: Curves will intersect if x² – 16 x + c = 0 has real roots. Thus c $⩽$ 64.

For x² – 4y² + c = 0, $\frac{dy}{dx}=\frac{x}{4y}$

For y² = 4x, $\frac{dy}{dx}=\frac{2}{y}$

If curves intersect orthogonally then $\frac{x}{4y}.\frac{2}{y}=-1\Rightarrow x=-2y^2=-8x$

$\Rightarrow x=0\Rightarrow y=0$. But if y = 0, slope of both curves undefined.