**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.