FIND THE EQUATIONS OF THE TANGENT AND NORMAL TO THE GIVEN CURVE AT THE INDICATED POINT

Problem 1 :

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

Problem 2 :

Find the tangent and normal to the curve x³ + y³ = 2xy at (1, 1).

Solution :

x³ + y³ = 2xy

Differentiate with respect to x, we get

3x² + 3xy²(dy/dx) = 2[x(dy/dx) + y(1)]

3xy²(dy/dx) - 2x(dy/dx) = 2y - 3x²

(dy/dx)(3xy² - 2x) = 2y - 3x²

(dy/dx) = (2y - 3x²) / (3xy² - 2x)

Slope of the tangent at (1, 1) :

dy/dx = (2 - 3) / (3 - 2)

= -1

Slope of normal :

= -1/m

= -1/(-1)

= 1

Equation of tangent :

y - y₁ = m(x - x₁)

y - 1 = -1(x - 1)

y - 1 = -x + 1

x + y - 1 - 1 = 0

x + y = 2

Equation of normal :

y - y₁ = -1/m(x - x₁)

y - 1 = 1(x - 1)

y - 1 = x - 1

x - y - 1 + 1 = 0

x - y = 0

Problem 3 :

Consider the curve defined by 2y³ + 6x² y - 12x² + 6y = 1

i) Find dy/dx

ii)  Write an equation of each horizontal tangent line to the curve.

iii) The line through the origin with slope -1 is tangent to the curve at point P. Find the x and y coordinate of P.

Solution :

2y³ + 6x² y - 12x² + 6y = 1

i)  

2(3y² dy/dx) + 6[2x y + x² (dy/dx)] - 12(2x) + 6(dy/dx) = 0

(6y² dy/dx) + 12x y + 6x² (dy/dx) - 24x + 6(dy/dx) = 0

dy/dx(6y²  + 6x² + 6) = - 12xy + 24x

dy/dx = (- 12xy + 24x) / (6y² + 6x² + 6)

dy/dx = 6(- 2xy + 4x) / 6(y² + x² + 1)

dy/dx = (- 2xy + 4x) / (y² + x² + 1)

ii)  Equation of horizontal tangent :

For horizontal tangent, the slope = 0

(- 2xy + 4x) / (y² + x² + 1) = 0

-2xy + 4x = 0

-2x(y - 2) = 0

x = 0 and y = 2

When x = 0,

2y³ + 6(0)² y - 12(0)² - 6y = 1

2y³ - 6y = 1

When y = 2

2(2)³ + 6x² (2) - 12x² + 6(2) = 1

16 + 12x² - 12x² + 12 = 1

There are no points on the curve where y = 2.

iii) The line passes through origin will be y = x

Since it has the slope -1, it must be falling line and the required line will be y = -x

(- 2xy + 4x) / (y² + x² + 1) = -1

Applying y = -x, we get

(2x² + 4x) / (x² + x² + 1) = -1

2x² + 4x = -x² - x² - 1

2x² + 4x = -2x² - 1

4x² + 4x + 1 = 0

(2x + 1)² = 0

2x + 1 = 0

2x = -1

and x = -1/2

y = 1/2

So, the required point is (-1/2, 1/2).

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