# Applications of differentiation – Errors and Tangent

## Applications of differentiation – Errors and Tangent

Errors Definition: let y = f(x) be a function defined in neighborhood of a point x. Let δx be a small change in x and δy be the corresponding change in y.

If δx is considered change in y.

i) δy is called the absolute error or error in y

ii) δy /y is called the relative error in y.

iii) $$\frac{\delta y\text{ }}{y}\times 100$$ is called the percentage error in y corresponding to the error δx in x

Example: Find $$\Delta y$$ , dy .  $$y={{e}^{x}},x=0,\Delta x=0.1$$

Solution: We know that

$$\Delta y=f(x+\Delta x)-f(x)$$

Given that x = 0 , $$\Delta x=0.1$$

$$\Delta y={{e}^{(\Delta x+x)}}-{{e}^{(x)}}$$

$$\Delta y={{e}^{(0.1+0)}}-{{e}^{(0)}}$$

$$\Delta y={{e}^{(0.1)}}-1$$ … (1)

$$y={{e}^{x}}$$

Above equation differentiation with respect to the x

$$\frac{dy}{dx}\text{ }=\Delta x.\text{ }{{e}^{x}}=(0.1){{e}^{0}}=0.1$$

Motion in a Straight Line: If x and v denote the displacement and velocity of a particle at any instant t, then velocity is given by

$$v\,=\,\frac{dx}{dt}$$.

and  $$a\,=\,\frac{dv}{dt}\,$$

We know that $$v\,=\,\frac{dx}{dt}$$

$$=\,v\frac{dv}{dt}\,=\,\frac{{{d}^{2}}x}{d{{t}^{2}}}$$

Where,

a is acceleration of particle.

If the sign of acceleration is opposite to that of velocity, then the acceleration is called retardation which means decrease in magnitude of the velocity.

Tangent and Normal: Tangent: The tangent to a curve at a point P on it is defined as the limiting position of the secant PQ as the point Q approaches the point P provided such a limiting position exists.

Slope of Tangent: Let y = f (x) be a continuous curve and let P (x₁, y₁) be the point on it. Then $${{\left( \frac{dy}{dx} \right)}_{\left( {{x}_{1}},\,{{y}_{1}} \right)}}$$ is the slope of tangent to the curve y = f (x) at a point P.

i.e., $${{\left( \frac{dy}{dx} \right)}_{P}}\,=\,\tan \theta$$

= Slope of tangent at P.

Where,

θ is the angle which the tangent at P (x₁, y₁) makes with the positive direction of x – axis.

Note:

1. If tangent is parallel to x – axis, then θ = 0°

$$\tan \theta \,=\,\left( \frac{dy}{dx} \right)$$

$$\tan 0\,=\,\left( \frac{dy}{dx} \right)=0$$

2. If tangent is perpendicular to x – axis (or parallel to y – axis), then

θ = 90°

$$\tan \theta \,=\,\left( \frac{dy}{dx} \right)$$

⇒ tanθ = ∞

or cot θ = 0

$${{\left( \frac{dy}{dx} \right)}_{\left( {{x}_{1}},\,{{y}_{1}} \right)}}\,=\,0$$

Example: x²+ y² + 7 = 0 curve and A (1, 2) be the point on it. Find the tangent of the curve.

Solution: Given that x²+ y² + 7 = 0 curve and A (1, 2)

Above equation differentiation with respect to the x

2x+ 2y. dy/dx + 0 = 0

dy/dx = -2x/2y

= -x/y at A(1, 2)

$${{\left( \frac{dy}{dx} \right)}_{\left( 1,\,2 \right)}}\,=\,\frac{-x}{y}$$

= -1/2.