NCERT solution Class 12 application of derivatives

NCERT Solutions

Class 12 Maths

Application of Derivatives

NCERT Questions

Ex.6.1 Q.1
Ex.6.1 Q.2
Ex.6.1 Q.3
Ex.6.1 Q.4
Ex.6.1 Q.5
Ex.6.1 Q.6
Ex.6.1 Q.7
Ex.6.1 Q.8
Ex.6.1 Q.9
Ex.6.1 Q.10
Ex.6.1 Q.11
Ex.6.1 Q.12
Ex.6.1 Q.13
Ex.6.1 Q.14
Ex.6.1 Q.15
Ex.6.1 Q.16
Ex.6.1 Q.17
Ex.6.1 Q.18
Ex.6.2 Q.1
Ex.6.1 Q.2
Ex.6.2 Q.3
Ex.6.2 Q.4
Ex.6.2 Q.5
Ex.6.2 Q.6
Ex.6.2 Q.7
Ex.6.2 Q.8
Ex.6.2 Q.9
Ex.6.2 Q.10
Ex.6.2 Q.11
Ex.6.2 Q.12
Ex.6.2 Q.13
Ex.6.2 Q.14
Ex.6.2 Q.15
Ex.6.2 Q.16
Ex.6.2 Q.17
Ex.6.2 Q.18
Ex.6.2 Q.19
Ex.6.3 Q.1
Ex.6.3 Q.2
Ex.6.3 Q.3
Ex.6.3 Q.4
Ex.6.3 Q.5
Ex.6.3 Q.6
Ex.6.3 Q.7
Ex.6.3 Q.8
Ex.6.3 Q.9
Ex.6.3 Q.10
Ex.6.3 Q.11
Ex.6.3 Q.11
Ex.6.3 Q.13
Ex.6.3 Q.14
Ex.6.3 Q.15
Ex.6.3 Q.16
Ex.6.3 Q.17
Ex.6.3 Q.18
Ex.6.3 Q.19
Ex.6.3 Q.20
Ex.6.3 Q.21
Ex.6.3 Q.22
Ex.6.3 Q.23
Ex.6.3 Q.24
Ex.6.3 Q.25
Ex.6.3 Q.26
Ex.6.3 Q.27
Ex.6.4 Q.1
Ex.6.4 Q.2
Ex.6.4 Q.3
Ex.6.4 Q.4
Ex.6.4 Q.5
Ex.6.4 Q.6
Ex.6.4 Q.7
Ex.6.4 Q.8
Ex.6.4 Q.9
Ex.6.5 Q.1
Ex.6.5 Q.2
Ex.6.5 Q.3
Ex.6.5 Q.4
Ex.6.5 Q.5
Ex.6.5 Q.6
Ex.6.5 Q.7
Ex.6.5 Q.8
Ex.6.5 Q.9
Ex.6.5 Q.10
Ex.6.5 Q.11
Ex.6.5 Q.12
Ex.6.5 Q.13
Ex.6.5 Q.14
Ex.6.5 Q.15
Ex.6.5 Q.16
Ex.6.5 Q.17
Ex.6.5 Q.18
Ex.3.5 Q.19
Ex.6.5 Q.20
Ex.6.5 Q.21
Ex.6.5 Q.22
Ex.6.5 Q.23
Ex.6.5 Q.24
Ex.6.5 Q.25
Ex.6.5 Q.26
Ex.6.5 Q.27
Ex.6.5 Q.28
Ex.6.5 Q.29
Ex.Misc.Q.1
Ex.Misc.Q.2
Ex.Misc.Q.3
Ex.Misc.Q.4
Ex.Misc.Q.5
Ex.Misc.Q.6
Ex.Misc.Q.7
Ex.Misc.Q.8
Ex.Misc.Q.9
Ex.Misc.Q.10
Ex.Misc.Q.11
Ex.Misc.Q.12
Ex.Misc.Q.13
Ex.Misc.Q.14
Ex.Misc.Q.15
Ex.Misc.Q.16
Ex.Misc.Q.17
Ex.Misc.Q.18
Ex.Misc.Q.19
Ex.Misc.Q.20
Ex.Misc.Q.21
Ex.Misc.Q.22
Ex.Misc.Q.23
Ex.Misc.Q.24
Ex.2.2 Q.21

Ex.6.3 Q.14

Find the equations of the tangent and normal to the given curves at the indicated points:

(i) y = x⁴ − 6x³ + 13x² − 10x + 5 at (0, 5)

(ii) y = x⁴ − 6x³ + 13x² − 10x + 5 at (1, 3)

(iii) y = x³ at (1, 1)

(iv) y = x² at (0, 0)

(v) x = cos t, y = sin t at t = π/4

(i) The equation of the curve is y = x⁴ − 6x³ + 13x² − 10x + 5

On differentiating with respect to x, we get:

= 4x³ − 18x² + 26x – 10

= 4 (0)³ – 18 (0)² + 26 (0) – 10

⟹ = -10

Thus, the slope of the tangent at (0, 5) is −10. The equation of the tangent is given as:

y − 5 = − 10(x − 0)

⟹ y − 5 = − 10x

⟹ 10x + y = 5

The slope of the normal at (0, 5) = at (0, 5) = -1/(-10) = 1/10

Therefore, the equation of the normal at (0, 5) is given as:

y − 5 = (x − 0)

⟹ 10y − 50 = x

⟹ x - 10y + 50 = 0

(ii) The equation of the curve is y = x⁴ − 6x³ + 13x² − 10x + 5

On differentiating with respect to x, we get:

= 4x³ − 18x² + 26x – 10

= 4 * 1³ – 18 * 1² + 26 * 1 – 10

⟹ = 4 – 18 + 26 – 10

⟹ = 30 – 28

⟹ = 2

Thus, the slope of the tangent at (1, 3) is 2. The equation of the tangent is given as:

y − 3 = 2(x − 1)

⟹ y − 3 = 2x - 2

⟹ y = 2x + 1

The slope of the normal at (1, 3) = at (1, 3) =

Therefore, the equation of the normal at (1, 3) is given as:

y − 3 = (x − 1)

⟹ 2y − 6 = -x + 1

⟹ x + 2y - 7 = 0

(iii) The equation of the curve is y = x³

On differentiating with respect to x, we get:

= 3x²

= 3 (1)²= 3

Thus, the slope of the tangent at (1, 1) is 3 and the equation of the tangent is given as:

y – 1 = 3(x – 1)

⟹ y – 1 = 3x – 3

⟹ y = 3x - 2

The slope of the normal at (1, 1) = -1/Slope of the tangent at (1, 3) =

Therefore, the equation of the normal at (1, 1) is given as:

y – 1 = (x – 1)

⟹ 3y – 3 = -x + 1

⟹ x + 3y – 4 = 0

(iv) The equation of the curve is y = x²

On differentiating with respect to x, we get:

= 2x

= 2 (0)= 0

Thus, the slope of the tangent at (0, 0) is 0 and the equation of the tangent is given as:

y – 0 = 0(x – 0)

⟹ y = 0

The slope of the normal at (0, 0) = -1/Slope of the tangent at (0, 0) = -1/0

Which is not defined.

Therefore, the equation of the normal at (x₀, y₀) = (0, 0) is given as:

x = x₀ = 0

(v) The equation of the curve is x = cos t, y = sin t

So, and

Now,

Thus, the slope of the tangent at is -1

When ,

Thus, the equation of the tangent at i.e. is:

The slope of the normal at t at

Thus, the equation of the normal at is

⟹

Video solution:

Application of Derivatives Class 12 Maths NCERT Chapter 6 NCERT Solutions

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