EQUATION OF TANGENT TO THE CURVE AT THE GIVEN CONDITION

Problem 1 :

Find the equations of the tangent to the curve y = 1 + x³ for which the tangent is orthogonal with the line x + 12y = 12

Solution :

y = 1 + x³

dy/dx = 0 + 3x²

dy/dx = 3x²

x + 12y = 12

Slope :

12y = -x + 12

y = (-1/12)x + 1

Slope = -1/12

Slope of perpendicular line = 12

3x² = 12

x² = 4

x = 2 and -2

When x = 2, y = 1 + 2³ ==> 9

When x = -2, y = 1 + (-2)³ ==> -7

So, the required points are (2, 9) and (-2, -7).

Equation of tangent :

(2, 9) and slope = 12

y - y₁ = m(x - x₁)

y - 9 = 12(x - 2)

y - 9 = 12x - 24

12x - y = -9 + 24

12x - y = 15

(-2, -7) and slope = 12

y - y₁ = m(x - x₁)

y + 7 = 12(x + 2)

y + 7 = 12x + 24

12x - y = 7 - 24

12x - y = -17

Problem 2 :

Show that the two curves x² - y² = r² and xy = c² where c, r are constants cut orthogonally.

Solution :

x² - y² = r²

2x - 2y(dy/dx) = 0

2y(dy/dx) = 2x

(dy/dx) = 2x/2y

(dy/dx) = x/y -----(1)

xy = c²

x(dy/dx) + y(1) = 0

x(dy/dx) = -y

dy/dx = -y/x -----(2)

(1) x (2)

= (x/y) ( -y/x)

= -1

The product of the slopes is equal to -1. So, the curves are orthogonal.

Find the tangent and normal of the following curves at the given points on the curve :

Problem 3 :

y = x² - x⁴ at (1, 0)

Solution :

Given that, y = x² - x⁴

dy/dx = 2x - 4x³

Slope at (1, 0)

dy/dx = 2(1) - 4(1)³

= 2 -  4

Slope = -2

Equation of tangent :

(y - y₁) = m(x - x₁)

Point is at (1, 0)

y - 0 = -2(x - 1)

y = -2x + 2

2x + y = 2

Equation of normal :

(y - y₁) = -1/m(x - x₁)

Point is at (1, 0)

y - 0 = (1/2)(x - 1)

2y = x - 1

x - 2y = 1

Problem 4 :

y = x⁴ + 2e^x at (0, 2)

Solution :

y = x⁴ + 2e^x

dy/dx = 4x³ + 2e^x

Slope at (0, 2)

dy/dx = 4(0)³ + 2e⁰

= 0 + 2(1)

Slope of tangent = 2

Equation of tangent :

(y - y₁) = m(x - x₁)

Point is at (0, 2)

y - 2 = 2(x - 0)

y - 2 = 2x

2x - y = - 2

Equation of normal :

(y - y₁) = -1/m(x - x₁)

(y - 2) = (-1/2)(x - 0)

2(y - 2) = -x

2y - 4 = -x

x + 2y = 4

Problem 5 :

y = x sin x at (π/2, π/2)

Solution :

Using product rule, finding the derivative

dy/dx = x(cos x) + sin x (1)

= x cos x + sin x

Slope at  (π/2, π/2)

= π/2 cos (π/2) + sin (π/2)

= π/2 (0) + 1

Slope of tangent = 1

Slope of normal = -1

Equation of tangent :

(y - y₁) = m(x - x₁)

(y - π/2) = 1(x - π/2)

y = x - π/2 + π/2

y = x

Equation of normal:

(y - y₁) = -1/m(x - x₁)

(y - π/2) = -1(x - π/2)

y = -x + π/2 + π/2

y = -x + π

Problem 6 :

x = cos t, y = 2 sin² t at t = π/3

Solution :

Applying the value of t in the given function, we get the value of x and y.

dx/dt = -sin t, dy/dt = 4 sin t cost t

dy/dx = (dy/dt) / (dx/dt)

= (4 sin t cost t) / (-sin t)

dy/dx = -4 cos t 

Slope at t = π/3

dy/dx = -4 cos (π/3)

= - 4(1/2)

Slope of tangent = -2

Slope of normal = 2

Equation of tangent :

(y - y₁) = m(x - x₁)

x = 1/2 and y = 3/2

Slope = -2

(y - (3/2)) = -2 (x - (1/2))

2y - 3 = -2(2x - 1)

2y - 3 = -4x + 2

4x + 2y -3 - 2 = 0

4x + 2y -5 = 0

Equation of normal :

(y - y₁) = (-1/m)(x - x₁)

x = 1/2 and y = 3/2

Slope = 1/2

(y - (3/2)) = 1/2(x - (1/2))

2y - 3 = 1/2(x - 1)

4y - 6 = x - 1

x - 4y - 1 + 6 = 0

x - 4y + 5 = 0

Problem 7 :

Find the equations of the tangents to the curve

y = (x + 1) / (x - 1)

which are parallel to the line x + 2y = 6.

Solution :

Find the derivative of the given function to figure out slope.

x + 2y = 6

2y = -x + 6

y = (-1/2)x + 6/2

y = (-1/2)x + 3

Slope = -1/2 -----(2)

(1) = (2)

Applying x = 3, we get y = (3+1)/(3-1)

y = 4/2

y = 2

Applying x = -1, we get y = (-1+1)/(-1-1)

y = 0

So, the required points are (3, 2) and (-1, 0)

Equation of tangents :

(y - y₁) = m(x - x₁)

(3, 2) and slope = -1/2

(y - 2) = -1/2 (x - 3)

2(y - 2) = -1(x - 3)

2y - 4 = -x + 3

x + 2y - 4 - 3 = 0

x + 2y - 7 = 0

(-1, 0) and slope = -1/2

(y - 0) = -1/2 (x + 1)

2y = -x - 1

x + 2y = -1

So, equation of the tangent is x + 2y = -1.