Physics |
Chemistry |
Biology |
Mathematics |

Science > Mathematics > Applications of Derivatives > You are Here |

#### Example – 12:

- Find the equation of the tangent and normal to the curve x = sin θ and y = cos 2θ at θ = π/6
**Solution:**

The equation of the curve is x = sin θ and y = cos 2θ ………. (1)

Parameter θ = π/6

∴ x = sin π/6 = 1/2 and y = cos 2(π/6) = cos π/3 = ½

Therefore the point on the curve is P(1/2, 1/2)

Differentiating equation (1) w.r.t. x

The slope of the tangent at P(1/2, 1/2) is -2 and that of normal is 1/2.

**Equation of tangent:**

Its slope = m = -2

It passes through P(1/2, 1/2) ≡ P(x_{1}, y_{1})

By slope point form

y – y_{1} = m(x – x_{1})

∴ y – 1/2 = -2(x – 1/2)

∴ y – 1/2 = -2x + 1

∴ 2y – 1 = -4x + 2

∴ 4x + 2y – 3 = 0

**Equation of normal:**

By slope point form

y – y_{1} = m(x – x_{1})

∴ y – 1/2 = (1/2)(x – 1/2)

∴ 2y – 1 = x – 1/2

∴ 4y – 2 = 2x – 1

∴ 2x – 4y + 1= 0

**Ans:** Equation of tangent is 4x + 2y – 3 = 0 and that of normal is 2x – 4y + 1= 0

**Example – 13:**

- Find the equation of the tangent and normal to the curve x = acos
^{3}θ and y = asin^{3}θ at θ = π/4 **Solution:**

Equation of curve is x = acos^{3}θ and y = asin^{3}θ ………. (1)

Parameter θ = π/4

∴ x = acos^{3} (π/4) = a(1/√2)^{3} = a (1/2√2) = a/2√2

∴ y = asin^{3} (π/4) = a(1/√2)^{3} = a (1/2√2) = a/2√2

Therefore the point on the curve is P(a/2√2, a/2√2)

Differentiating equation (1) w.r.t. x

The slope of the tangent at P(a/2√2, a/2√2) is – 1 and that of normal is 1.

**Equation of tangent:**

Its slope = m = – 1

It passes through P(a/2√2, a/2√2) ≡ P(x_{1}, y_{1})

By slope point form

y – y_{1} = m(x – x_{1})

∴ y – a/2√2 = -1(x – a/2√2)

∴ 2√2 y – a = – 2√2 x + a

∴ 2√2 x + 2√2 y – 2a = 0

∴ √2 x + √2 y –a = 0

**Equation of normal:**

By slope point form

y – y_{1} = m(x – x_{1})

∴ y – a/2√2 = 1(x – a/2√2)

∴ y – a/2√2 = x – a/2√2

∴ 2√2 y – a = 2√2 x – a

∴ 2√2 x – 2√2 y = 0

∴ x – y = 0

**Ans:** Equation of tangent is √2 x + √2 y – a = 0 and that of normal is x – y = 0

#### Example – 14:

- Find the equation of the tangent and normal to the curve x = a sec θ and y = a tan θ at θ = π/6
**Solution:**

Equation of curve is x = a sec θ and y = a tan θ ………. (1)

Parameter θ = π/6

∴ x = a sec (π/6) = 2a/√3 and y = a tan (π/6) = a/√3

Therefore the point on the curve is P(2a/√3, a/√3)

Differentiating equation (1) w.r.t. x

The slope of the tangent at P(2a/√3, a/√3) is 2 and that of normal is – 1/2.

**Equation of tangent:**

Its slope = m = 2

It passes through P(2a/√3, a/√3) ≡ P(x_{1}, y_{1})

By slope point form

y – y_{1} = m(x – x_{1})

∴ y – a/√3 = 2(x – 2a/√3)

∴ y – a/√3 = 2x – 4a/√3

∴ √3 y – a = 2 √3 x – 4a

∴ 2√3 x – √3 y – 3a = 0

∴ 2x – y – √3a = 0

**Equation of normal:**

By slope point form

y – y_{1} = m(x – x_{1})

∴ y – a/√3 = (-1/2)(x – 2a/√3)

∴ 2y – 2a/√3= – x + 2a/√3

∴ 2√3 y – 2a = – √3 x + 2a

∴ √3 x + 2√3 y – 4a = 0

**Ans:** Equation of tangent is 2x – y – √3 a = 0 and that of normal is √3 x + 2√3 y – 4a = 0

#### Example – 15:

- Find the equation of the tangent to the curve x = a(θ + sin θ) and y = a(1 + cos θ) at θ = π/2
**Solution:**

Equation of curve is x = a(θ + sin θ) and y = a(1 + cos θ) ………. (1)

Parameter θ = π/6

∴ x = a(θ + sin θ) = a(π/2 + sin π/2)=a(π/2 + 1) and

∴ y = a(1 + cos θ) = a(1 + cos π/2) = a(1 + 0) = a

Therefore the point on the curve is P(a(π/2 + 1), a)

Differentiating equation (1) w.r.t. x

The slope of the tangent at P(a(π/2 + 1), a) is -1 and that of normal is 1.

**Equation of tangent:**

Its slope = m = -1

It passes through P(a(π/2 + 1), a) ≡ P(x_{1}, y_{1})

By slope point form

y – y_{1} = m(x – x_{1})

∴ y – a = -1(x – a(π/2 + 1))

∴ y – a = – x + a(π/2) + a

∴ x + y – a – a(π/2) – a = 0

∴ x – y – a(π/2) – 2a = 0

Ans: Equation of tangent is x – y – a(π/2) – 2a = 0

#### Example – 16:

- Find the equation of tangent and normal to the curve x = 1/t and y = t – 1/t at t = 2
**Solution:**

Equation of curve is x = 1/t and y = t – 1/t ………. (1)

Parameter t = 2

∴ x = 1/2 and y = 2 – 1/2 = 3/2

Therefore the point on the curve is P(1/2, 3/2)

Differentiating equation (1) w.r.t. x

The slope of the tangent at P(1/2, 3/2) is -5 and that of normal is 1/5.

**Equation of tangent:**

Its slope = m = – 5

It passes through P(1/2, 3/2) ≡ P(x_{1}, y_{1})

By slope point form

y – y_{1} = m(x – x_{1})

∴ y – 3/2 = – 5(x – 1/2)

∴ y – 3/2 = – 5x + 5/2

∴ 5x + y – 3/2 – 5/2 = 0

∴ 5x + y – 4 = 0

**Equation of normal:**

By slope point form

y – y_{1} = m(x – x_{1})

∴ y – 3/2 = 1/5(x – 1/2)

∴ 5y – 15/2 = x – 1/2

∴ x – 5y – 1/2 + 15/2 = 0

∴ x – 5y + 7 = 0

**Ans:** Equation of tangent is 5x + y – 4 = 0 and that of normal is x – 5y + 7 = 0

#### Example – 17:

- Show that the tangent to the curve 8y = (x -2)
^{2}at the point (-6, 8) is parallel to the tangent to the curve y = x + 3/x at point (1, 4) **Solution:**

Consider the first curve 8y = (x -2)^{2}

Differentiating both sides w.r.t. x

8 = 2(x – 2) (1)

= (x – 2)/4

∴ (dy/dx)_{at P(-6, 8)} = (- 6 – 2)/4 = -8/4 = -2

Thus slope of tangent to curve 8y = (x -2)^{2} at the point (-6, 8) is – 2 ………….. (1)

Consider the second curve y = x + 3/x

Differentiating both sides w.r.t. x

= 1 – 3/x^{2}

∴ (dy/dx)_{at Q(1, 4)} = 1- 3/1^{2} = 1- 3 = -2

Thus slope of tangent to curve y = x + 3/x at the point (1, 4) is – 2 ………….. (2)

From statements (1) and (2) we can see that the two slopes are equal.

Hence the two tangents are parallel to each other (proved as required)

#### Example – 18:

- Show that the tangents to the curve x
^{2}= 2y and 6y = 5 – 2x^{3}at x = 1 are perpendicular to each other. **Solution:**

Consider the first curve x^{2} = 2y

Differentiating both sides w.r.t. x

2x = 2

∴ = x

∴ (dy/dx)_{at x = 1} = 1

Thus slope of tangent to curve x^{2} = 2y at x = 1 is 1 ………….. (1)

Consider the second curve 6y = 5 – 2x^{3}

Differentiating both sides w.r.t. x

6 = -6x^{2}

∴ = -x^{2}

∴ (dy/dx)_{at x = 1} = -(1)^{2} = -1

Thus slope of tangent to curve 6y = 5 – 2x^{3} at x = 1 is -1 ………….. (2)

From statements (1) and (2) we can see that the product of the two slopes is -1.

Hence the two tangents are perpendicular to each other (proved as required)

#### Example – 19:

- Find the points on the curve y = 3x
^{2}– 9x + 8 where the tangent is parallel to x- axis **Solution:**

The tangent is parallel to x-axis hence its slope m = (dy/dx) = 0

The equation of the curve is y = 3x^{2} – 9x + 8 …………… (1)

Differentiating both sides w.r.t. x

= 6x – 9

Thus 6x – 9 = 0

∴ 6x = 9

∴ x = 3/2

Substituting in equation (1)

y = 3(3/2)^{2} – 9(3/2) + 8= 3(9/4) – 27/2 + 8

∴ y = 27/4 -27/2 + 8 = -27/4 + 8 = 5/4

The point on the curve is (3/2, 5/4)

#### Example – 20:

- Find the points on the curve y = 3x
^{2}– 9x + 8 where the tangent makes an angle 45^{o}with the x- axis **Solution:**

The tangent makes an angle 45^{o} with the x- axis

Slope of tangent = m = tan 45^{o} = 1 = (dy/dx)

The equation of the curve is y = 3x^{2} – 9x + 8 …………… (1)

Differentiating both sides w.r.t. x

= 6x – 9

Thus 6x – 9 = 1

∴ 6x = 10

∴ x = 5/3

Substituting in equation (1)

y = 3(5/3)^{2} – 9(5/3) + 8= 3(25/9) – 15 + 8

∴ y = 25/3 – 15 + 8 = 25/3 – 7 = 4/3

The point on the curve is (5/3, 4/3)

#### Example – 21:

- Find the point on the curve y = √x – 3 where the tangent is perpendicular to the line 6x + 3y – 5 = 0
**Solution:**

Equation of given line 6x + 3y – 5 = 0

Slope of given line = – (coefficient of x/coefficient of y) = – (6/3) = – 2

As the tangent is perpendicular to given line

Slope of tangent = 1/2

Equation of curve is y = √x – 3

Differentiating both sides w.r.t. x

squaring both sides

1 = x – 3

∴ x = 4

Substituting in equation (1)

y = √4 – 3 = √1 = ±1

The points on the curve is (4, 1) and (4, -1)

#### Example – 22:

- Find the coordinates of the point on the curve y = x – 4/x, where the tangent at that point is parallel to the line y = 2x.
**Solution:**

Equation of given line y = 2x

Slope of given line = 2

Now the tangent is parallel to the given line

Slope of tangent = m = 2 = (dy/dx)

Equation of the curve is y = x – 4/x

Differentiating both sides w.r.t.x

= 1 – 4(- 1/x^{2}) = 1 + 4/x^{2}

∴ 1 + 4/x^{2} = 2

∴ 4/x^{2} = 1

∴ x^{2} = 4

∴ x = ± 2

When x = 2

y = 2 – 4/2 = 2 – 2 = 0

Hence the point is (2, 0)

When x = -2

y = – 2 – 4/(-2) = – 2 + 2 = 0

Hence the point is (- 2, 0)

The required points are (2, 0) and (-2, 0)

#### Example – 23:

- If the line x + y = 0 touches the curve 2y
^{2}= ax^{2}+ b at (1, -1) find a and b **Solution:**

Equation of tangent is x + y = 0

Slope of tangent = – (coefficient of x/coefficient of y) = – (1/1) = -1

Equation of curve is 2y^{2} = ax^{2} + b

Differentiating both sides w.r.t. x

4y = 2ax

∴ =ax/2y

∴ (dy/dx)_{at P(1, -1)} = a(1)/2(-1) = – a/2

– a/2 = -1

∴ a = 2

Point (1, -1) lies on the curve 2y^{2} = ax^{2} + b

2(-1)^{2} = 2(1)^{2} + b

∴ 2 = 2 + b

∴ b = 0

**Ans:** a = 2 and b = 0

#### Example – 24:

- For the curve y = 3x – 2x
^{2}, find the equation of the tangent whose slope is – 1. **Solution:**

Slope of tangent = -1

Equation of curve is y = 3x – 2x^{2} ………. (1)

Differentiating both sides w.r.t. x

= 3 – 4x

∴ 3 – 4x = -1

∴ 4 = 4x

∴ x = 1

Substituting in equation (1)

y = 3(1) – 2(1)^{2} = 3 – 2 = 1

The tangent touches the curve at (1, 1) ≡ P(x_{1}, y_{1}) and its slope = m = -1

By slope point form

y – y_{1} = m(x – x_{1})

∴ y – 1 = – 1(x – 1)

∴ y – 1 = – x + 1

∴ x + y – 2 = 0

**Ans:** The equation of tangent is x + y – 2 = 0

#### Example – 25:

- Find equations of tangents and normal to the curve y = 6 – x
^{2}, where the normal is parallel to the cline x – 4y + 3 = 0 **Solution:**

Equation of given line x – 4y + 3 = 0

Slope of given line = – (coefficient of x/coefficient of y) = – (1/-4) = 1/4

As the normal is perpendicular to given line

Slope of normal = 1/4

Slope of tangent = – 4

The equation of curve isy = 6 – x^{2}

Differentiating both sides w.r.t. x

= – 2x = – 4

x = 2

Substituting in equation of curve

y = 6 – 2^{2} = 6 – 4 = 2

The tangent touches the curve at (2, 2)

**Equation of tangent:**

Its slope = m = – 4

It passes through P(2, 2) ≡ P(x_{1}, y_{1})

By slope point form

y – y_{1} = m(x – x_{1})

∴ y – 2 = – 4(x – 2)

∴ y – 2 = – 4x + 8

∴ 4x + y – 10 = 0

**Equation of normal:**

Its slope = m = 1/4

It passes through P(2, 2) ≡ P(x_{1}, y_{1})

By slope point form

y – y_{1} = m(x – x_{1})

∴ y – 2 = 1/4(x – 2)

∴ 4y – 8 = x – 2

∴ x – 4y + 6 = 0

**Ans:** Equation of tangent is 4x + y – 10 = 0 and that of normal is x – 4y + 6 = 0

#### Example – 25:

- If the line y = 4x -5 touches the curve y
^{2}= ax^{3}+ b at the point (2, 3) show that 7a + 2b = 0 **Solution:**

Equation of tangent is 4x – y -5 = 0

Slope of the tangent = – (coefficient of x/coefficient of y) = – (4/-1) = 4 = (dy/dx)

Equation of curve is

y^{2} = ax^{3} + b

Differentiating both sides w.r.t. x

2y(dy/dx) = 3ax^{2}

∴ (dy/dx) = 3ax^{2}/2y

∴ (dy/dx) _{at(2, 3)} = 3a(2)^{2}/ 2(3)= 2a

Thus slope of tangent to the curve at (2, 3) is 2a

2a = 4

∴ a = 2

Now point (2, 3) lies on the curve y^{2} = ax^{3} + b

3^{2} = 2(2)^{3} + b

∴ 9 = 16 + b

∴ b = 9 – 16 = -7

To prove that 7a + 2b = 0

L.H.S. = 7a + 2b = 7(2) + 2(-7) = 14 – 14 = 0 = R.H.S.

∴ 7a + 2b = 0 (Proved as required)

Science > Mathematics > Applications of Derivatives > You are Here |

Physics |
Chemistry |
Biology |
Mathematics |