According to Leibniz, tangent is the line through a pair of very close points on the curve.

**Equations of
Tangent and Normal**

According
to Leibniz, tangent is the line through a pair of very close points on the
curve.

** **

The tangent line (or simply tangent) to a plane curve at a given
point is the straight line that **just touches** the curve at that
point.

The normal at a point on the curve is the straight line which is
perpendicular to the tangent at that point.

The tangent and the normal of a curve at a point are illustrated
in Fig. 7.7.

Consider
the given curve *y* =
*f* ( *x*) .

The
equation of the tangent to the curve at the point, say (*a* , *b*) , is given by

In order
to get the equation of the normal to the same curve at the same point, we
observe that normal is perpendicular to the tangent at the point. Therefore,
the slope of the normal at (*a* , *b*) is the negative of the reciprocal of
the slope of the tangent which is âˆ’( 1 /*
dy*/* dx)*_{( a , b)}.

Hence,
the equation of the normal is ,

**Remark**

(i) If
the tangent to a curve is horizontal at a point, then the derivative at that
point is 0. Hence, at that point ( *x*_{1}
, *y*_{1} )
the equation of the tangent is *y* =
*y*_{1} and equation of the
normal is *x* =
*x*_{1} .

(ii) If
the tangent to a curve is vertical at a point, then the derivative exists and
infinite ( âˆž) at the point. Hence, at that point (
*x*_{1} , *y*_{1} ) the equation of the tangent is *x* = *x*_{1}
and the equation of the normal is *y* =
*y*_{1} .

** **

**Example 7.11**

Find the
equations of tangent and normal to the curve *y *=* x*^{2}* *+* *3*x *âˆ’* *2*
*at the point* *(1, 2)* *

**Solution**

We have,
*dy*/*dx* = 2*x +* 3 . Hence at
(1, 2), *dy/dx* = 5 .

Therefore,
the required equation of tangent is

( *y* âˆ’ 2) = 5(*x* âˆ’ 1) â‡’
5*x* âˆ’ *y*
âˆ’
3 =
0 .

The
slope of the normal at the point ( 1, 2) is â€“ 1/5.

Therefore,
the required equation of normal is

(* y *âˆ’
2) = âˆ’ 1/5 (*x* âˆ’1) â‡’* x *+ 5* y *âˆ’ 11 = 0 .

** **

Find the
points on the curve *y* =
*x*^{3} âˆ’
3*x*^{2} +
*x* âˆ’ 2 at which the tangent is parallel
to the line *y *=* x *.

**Solution**

The
slope of the line *y *=* x *is 1. The tangent to the given curve
will be parallel to the line, if the slope of the tangent to the curve at a
point is also 1. Hence,

*dy/dx* = 3x^{2} âˆ’ 6x +1 = 1

which
gives 3x^{2} âˆ’ 6x = 0 .

Hence,* x *= 0 and* x *= 2.

Therefore,
at (0, â€“2) and (2, â€“4) the tangent is parallel to the line *y* = *x* .

** **

Find the
equation of the tangent and normal at any point to the Lissajous curve given by
*x *=* *2 cos 3*t *and* y *=* *3 sin 2*t*,* t *âˆˆ* ***R**.

**Solution**

Observe
that the given curve is neither a circle nor an ellipse. For your reference the
curve is shown in Fig. 7.9.

Therefore,
the tangent at any point is

* y*
âˆ’ 3sin 2*t* = âˆ’ cos 2t/ sin 3t (*x *âˆ’
2 cos 3t)

That is,* x *cos 2t +* y *sin 3t = 3sin 2t sin 3t + 2 cos 2t cos 3t .

The
slope of the normal is the negative of the reciprocal of the tangent which in
this case is sin3*t / *cos2*t* . Hence, the equation of the normal is

*y *âˆ’* *3sin 2*t *= [sin 3*t /* cos 2*t*] (*x *âˆ’* *2 cos 3*t*)* *.

That is,
*x* sin 3*t* âˆ’ *y* cos 2*t* = 2 sin 3*t* cos 3*t* âˆ’
3sin 2*t* cos 2*t* = sin 6*t* âˆ’
3/2 sin 4*t* .

Tags : Differential Calculus | Mathematics , 12th Maths : UNIT 7 : Applications of Differential Calculus

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12th Maths : UNIT 7 : Applications of Differential Calculus : Equations of Tangent and Normal | Differential Calculus | Mathematics

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