Introduction

Compared to a straight line, the slope of a curve changes as you walk along with the graph. Calculus teaches students that each point on this graph can be defined by a slope, often known as an “instantaneous rate of change.” The tangent line is a straight line with that slope passing through that spot on the graph. This article will delve into the concept of tangent and equation of the tangent in depth. 

What is a tangent?

The term “tangent” is derived from the Latin word “tangere” (which means “to touch”), which was coined by a Danish mathematician named “Thomas Fineko” in the year 1583. 

One of the most prominent applications of differentiation is the “tangent line.” At one point on the curve, the tangent line comes into contact with the curve. The “point of tangency” is where the tangent is drawn.

A line that simply touches the curve (function) at a specific location is called a tangent line of a curve. In calculus, the tangent line can touch the curve at any point(s) and cross the graph at any point(s). 

It is not a tangent curve line at each of the two points if a line travels through two points of the curve but does not contact the curve at each point. The line is known as a secant line in this situation.

Real-life examples of tangent 

• If we drive around a corner and hit something slick on the road (oil, ice, water, or loose gravel) and our automobile begins to skid, it will continue tangent to the curve direction.
• As a cycle travels down a road, the road becomes the tangent at each location where the wheels roll.
• A stone will follow a circular path if connected to one end of a string and rotated from the other end. If we halt the motion abruptly, the stone will move in a tangential direction to the circular motion.

Application of derivatives in the equation of the tangents

The derivatives can be helpful in many ways related to functions, such as: 

• calculate the equations of the tangent and normal to a curve at a point, thereby figuring out the tangent of a circle formula, 
• locate turning points on a function’s graph, which will assist us in locating points where the function’s highest or smallest value (locally) occurs
• determine the rate of change of numbers

The slope of the line tangent to the curve at that point is a derivative at that point in a curve. 

Furthermore, derivatives and second derivatives can aid in the understanding of the function’s form (whether they are concave upward or downward). If you have a rudimentary conceptual understanding of derivatives, you can use it to discover crucial points, extreme points, inflection points, and even graph functions.

Equation of a tangent to a circle

A straight line passing through a point (x1, y1) and having  gradient m is given by the equation:

                                             (y − y1)/(x − x1) = m 

Where, 

x,y = equation variable

x1, y1 = coordinate on a graph where the equation of a tangent is required 

m = gradient (slope) of the tangent 

Procedure to find out the equation of tangent to a circle y = f(x) at a given point P (x, y1) :

(i) find the (dy/dx) from the given equation y = f(x)

(ii) find the value of (dy/dx) at the given point P (x1, y1), and then let m= (dy/dx) at x=x1 and y=y1

(iii) the tangent of a circle formula becomes: 

                                         (y − y1)/ (x − x1)   = m 

Remark: If (dy/dx) at x=x1 and y=y1 does not exist, then the tangent at P is parallel to the y-axis, and its equation is x = x1. 

Examples 

• Find the slope of a tangent to the curve y= x3-x+1 at the point where x- coordinate is 2. 

Solution: 

The given curve is y= x3-x+1……. (eq 1)

Differentiating the equation with respect to x, 

(dy/dx)= d/dx (x3-x+1)

           = 3x2– 1……………………… (eq 2)

The slope of the tangent at the given point where x- coordinate is 2:

(dy/dx) at x=2 => 3. (2)2– 1 ( putting the value of the x- coordinate in the eqn 2)

                       =>11

Ans: The slope of the tangent to the curve y= x3-x+1 at x-coordinate 2 is 11. 

• The slope of the curve 2y2= ax2++b at ( 1, -1) is -1. Find a and b. 

Solution: 

The given curve is 2y2= ax2++b………….. (eq 1)

On differentiating the given equation:

                    4y (dy/dx)= 2ax…………… (eq 2)

At x=1, y= -1, and m (dy/dx)= -1, replace the value in eq 2, 

4 (-1) (-1)= 2 a (1)

 => a= 2     ………………………………… (eq 3)

Putting the value of ‘a’ in eq 1 

2 (-1)2 = 2 ( 1)2 +b 

2= 2+b

b= 0

Ans: The value of a and b in the given equation is 2 and 0, respectively. 

1. What is the tangent line equation of the curve x = cos a and y = sin(a) at a= π/2?
   Solution:
   The tangent passes through the point (x1, y1) which is given by,

(x1, y1) = (cos π/2, sin π/2) = (0, 1)

The curve is identified with parametric equations.

So dy/dx = (dy/dt) / (dx/dt) = (cos a) / (- sin a).

The slope of the tangent is, m = (dy/dx)ₜ ₌ π/₂ = (cos π/2) / (-sin π/2) = 0/(-1) = 0

The equation of the tangent line is,
y – y1 = m (x – x1)
y – 1 = 0 (x – 0)
y – 1 = 0
y = 1
Answer: y = 1 is the equation of the tangent line.

Conclusion

A tangent to a circle with coordinates at point P is a straight line intersecting the circle at P. The tangent is perpendicular to the radius that connects the circle’s center to the point P. 

Because the equation of the tangent  will be of the form y=mx+c as it is a straight line,. Thus the equation can be formed using the perpendicular gradient of the tangent and the point of intersection of the tangent and the circle, which will lead to an equation of the tangent on the particular point of intersection on the circle.