FIND THE EQUATION OF THE TANGENT AND NORMAL TO THE GIVEN CURVE

Tangent :

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.

Equation of tangent :

(y-y1)  =  m(x-x1)

Normal :

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 the adjoining figure.

Equation of normal :

(y-y1)  =  (-1/m)(x-x1)

Find the tangent and normal to the following curves at the given points on the curve.

(i)  y  =  x2 - x4 at (1, 0)

Solution :

y  =  x2 - x4 

Differentiating with respect to x, we get

dy/dx  =  2x-4x3

Slope at (1, 0)

dy/dx  =  2(1)-4(1)3

dy/dx  =  -2

Slope of tangent  =  -2

Slope of normal  =  1/2

Equation of tangent :

(y-y1)  =  m(x-x1)

(y-0)  =  -2(x-1)

y  =  -2x+2

2x+y-2  =  0

So, the equation of tangent is 2x+y-2  =  0.

Equation of normal :

(y-y1)  =  (-1/m)(x-x1)

(y-0)  =  (1/2)(x-1)

2y  =  x-1

x-2y-1  =  0

So, equation of normal is x-2y-1  =  0.

(ii)  y  =  x4+2ex at (0, 2)

Solution :

dy/dx  =  4x3 + 2ex

Slope at (0, 2)

dy/dx  =  4(0)3 + 2e0

dy/dx  =  2

Slope of the tangent  =  2

Slope of the normal  =  -1/2

Equation of tangent :

(y-2)  =  2(x-0)

y-2  =  2x

2x-y+2  =  0

Equation of normal :

(y-2)  =  (-1/2)(x-0)

2(y-2)  =  -x

2y-4  =  -x

x+2y-4  =  0

So, equation of tangent is 2x-y+2  =  0 and equation of normal is x+2y-4  =  0.

(iii)  y  =  x sin x at (π/2, π/2)

Solution :

y  =  x sin x

dy/dx  =  x cos x + sin x (1)

dy/dx  =  x cos x + sin x

dy/dx at (π/2, π/2)

dy/dx  =  π/2 cos π/2 + sin π/2

dy/dx  =  π/2 (0) + 1

dy/dx  =  1

Slope of tangent  =  1

Slope of normal  =  -1

Equation of tangent :

(y - π/2)  =  1(x - π/2)

y - π/2  =  x - π/2

x-y  =  0

Equation of normal :

(y - π/2)  =  -1(x - π/2)

y - π/2  =  -x + π/2

x+y-π  =  0

(iv)  x  =  cost, y  =  2 sin2t at t  =  π/3

Solution :

x  =  cost

dx/dt  =  -sin t

y  =  2 sin2t

dy/dt  =  4 sin t cost

dy/dx  =  (dy/dt) / (dx/dt)

dy/dx  =  4 sin t cost / (-sin t)

dy/dx  =  -4 cost

dy/dx at t  =  π/3  =  -4cos (π/3)

=  -4(1/2)

=  -2

Slope of tangent  =  -2

Slope of normal  =  -1/(-2)  ==>  1/2

x  =  cost

x  =  cos ( π/3)

x  =  1/2

y  =  2 sin2t

y  =  2 sin2(π/3)

y  =  2 (3/4)

y  =  3/2

Equation of tangent :

(y-(3/2))  =  -2(x-(1/2))

(2y-3)  =  -2(2x-1)

2y-3  =  -4x+2

4x+2y-5  =  0

Equation of normal :

(y-(3/2))  =  (1/2)(x-(1/2))

2(2y-3)  =  (2x-1)

4y-6  =  2x-1

2x-4y-1+6  =  0

2x-4y+5  =  0

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