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

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+2e^x at (0, 2)

Solution :

dy/dx  =  4x³ + 2e^x

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 :

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

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

Apart from the stuff given above,  if you need any other stuff in math, please use our google custom search here.

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.