Analyzing Tangents and Normal Equations

“Tangent” means “touch”. In geometry, the straight line which just touches the curve is called the line of a tangent. There are many tangents to one circle. Tangents are perpendicular to the radius of the circle. Considering a real-world example, while riding a bicycle the wheel makes a tangent at every point on the circumference when it touches the road. Geometrically, a normal is some object which is perpendicular to a given object such as a line, ray, or vector. The line perpendicular to a tangent is the normal of a curve creating a point where the normal lines intersect the curves. Therefore the products of the slope of the tangent and normal come up to be -1.

Equation of Tangent and Normal Equation 

The straight line touching the curve at the point is the tangent at a point on the curve, the slope of which is the derivative of the curve at the same point. Through the reference of the definition, we can work out to find the tangent equations to curves at a given point. If a given function is y=f(a), then the tangent equation  to the curve which is a=a0 can further be found in the ways following:

(a)Firstly finding out the derivative of the curve at the certain point a=a0.And for doing so one needs to find out the data of a=a0. Let’s now call out this value to be m, in analyzing the slope of the straight line.

(b)Secondly finding out the straight-line equation which passes through the point(a0,y0)with the slope denoted by m. This being quite straightforward, and found out as 

y−y0=m(a−a0)

So, now you have found the equation out of the tangent to the curve at the point given.

If the slope of a normal is denoted by n and the slope of the same tangent at that point is denoted by then we have already proved that m x n = -1. Therefore the steps for finding normal to the given curve of y = f(a)at point a = a0

Now you have to find out the derivative of the curve at the point a = a0. This first step is the same method while finding the equation of tangent with the curve that is m=dyda[a = a0

The following step is to find the slope of the normal line;

As we know normal is perpendicular to tangent we have n = -1/m.

Now finding the equation of the straight line which is passing through the point (ao,f(ao)) is with slope n the equation coming out is

y-f(a0)=-1(a-a0)/F(a0)

Both the tangents and normal to the curve go hand in hand which is quite noticeable. The tangents and the normal are very easily derivable from each other.

Conclusion

The tangent line linear approximation concept is followed from the tangent line equation. Which means that the equation of the line of tangent of a function Y = f(a) on point (a0 ,y0) can be utilized  to find the value of the function of any close point to (a0y0) let us understand this from an example:

By using that approximation of tangents line we need to find out the approximate value of ∛ 8.1.

Practically ∛ 8 is 3 and 8.1 lies very close to 8.

Assuming the function to be f(a)= ∛a and the point of tangent to be a= 8.

Derivating the function of f ‘(a) = (1/3) a-2/3

m(slope) = (f ‘(a))₍₈, ₂₎ = (1/3) (21)-2/3 = (1/3) (1/4) = 1/12

the tangent line equation is  y – y₀ = m (a – a₀)

y – 2 = (1/12) (a- 8)

y = a/12 – 2/3 + 2

y = a/12 + 4/3

on Substituting y = f(a) here,

f(a) = a/12 + 4/3

by substituting the value of  x = 8.1 in the equation ,we can calculate approximate value of ∛8.1.

f(8.1) = (8.1)/12 + 4/3

∛8.1 =2.008