Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 7 Applications of Differential Calculus Ex 7.2 Textbook Questions and Answers, Notes. BANDHANBNK Pivot Point Calculator

## Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.2

Question 1.

Find the slope of the tangent to the following curves at the respective given points.

(i) y = x^{4} + 2x² – x at x = 1

(ii) x = a cos³ t, y = b sin³ t at t = \(\frac { π }{ 2 }\)

Solution:

(i) y = x^{4} + 2x² – x

Differentiating w.r.t. ‘x’

\(\frac { dy }{ dx }\) = 4x³ + 4x – 1

Slope of the tangent (\(\frac { dy }{ dx }\))_{(x=1)}

= 4(1)³ + 4(1) – 1

= 4 + 4 – 1 = 7

(ii) x = a cos³ t, y = b sin³ t

Differenriating w.r.t. ‘t’

\(\frac { dx }{ dt }\) = – 3a cos² t sin t

\(\frac { dy }{ dt }\) = 3b sin² t sin t

Question 2.

Find the point on the curve y = x² – 5x + 4 at which the tangent is parallel to the line 3x + y = 7.

Solution:

y = x² – 5x + 4

Differentiating w.r.t. ‘x’

Slope of the tangent \(\frac { dy }{ dx }\) = 2x – 5

Given line 3x + y = 7

Slope of the line = –\(\frac { 3 }{ 1 }\) = -3

Since the tangent is parallel to the line, their slopes are equal.

∴ \(\frac { dy }{ dx }\) = -3

⇒ 2x – 5 = -3

2x = 2

x = 1

When x = 1, y = (1)² – 5 (1) + 4 = 0

∴ Point on the curve is (1, 0).

Question 3.

Find the points on curve y = x³ – 6x² + x + 3 where the normal is parallel to the line x + y = 1729.

Solution:

y = x³ – 6x² + x+ 3

Differentiating w.r.t. ‘x’

Slope of the tangent \(\frac { dy }{ dx }\) = 3x² – 12x + 1

Slope of the normal = \(\frac { 1 }{ 3x^2 – 12x + 1 }\)

Given line is x + y = 1729

Slope of the line is – 1

Since the normal is parallel to the line, their slopes are equal.

\(\frac { 1 }{ 3x^2 – 12x + 1 }\) = -1

3x² – 12x + 1 = 1

3x² – 12x =0

3x(x – 4) = 0

x = 0, 4

When x = 0, y = (0)³ – 6(0)² + 0 + 3 = 3

When x = 4, y = (4)³ – 6(4)² + 4 + 3

= 64 – 96 + 4 + 3 = -25

∴ The points on the curve are (0, 3) and (4, -25).

Question 4.

Find the points on the curve y² – 4xy = x² + 5 for which the tangent is horizontal.

Solution:

y² – 4xy = x² + 5 ………… (1)

Differentiating w.r.t. ‘x’

When the tangent is horizontal(Parallel to X-axis) then slope of the tangent is zero.

\(\frac { dy }{ dx }\) = 0 ⇒ \(\frac { x+2y }{ y-2x }\) = 0

⇒ x + 2y = 0

x = -2y

Substituting in (1)

y² – 4 (-2y) y = (-2y)² + 5

y² + 8y² = 4y² + 5

5y² = 5 ⇒ y² = 1

y = ±1

When y = 1, x = -2

When y = – 1, x = 2

∴ The points on the curve are (- 2, 1) and (2, -1).

Question 5.

Find the tangent and normal to the following curves at the given points on the curve.

(i) y = x² – x^{4} at (1, 0)

(ii) y = x^{4} + 2e^{x} at (0, 2)

(iii) y = x sin x at (\(\frac { π }{ 2 }\), \(\frac { π }{ 2 }\))

(iv) x = cos t, y = 2 sin² t at t = \(\frac { π }{ 3 }\)

Solution:

(i) y = x² – x^{4} at (1, 0)

Differentiating w.r.t. ‘x’

\(\frac { dx }{ dy }\) = 2x – 4x³

Slope of the tangent ‘m’ = (\(\frac { dx }{ dy }\))_{(1, 0)}

= 2 (1) – 4 (1)³ = -2

Slope of the normal –\(\frac { 1 }{ m }\) = \(\frac { -1 }{ -2 }\) = \(\frac { 1 }{ 2 }\)

Equation of tangent is

y – y_{1} = m (x – x_{1})

y – 0 = – 2 (x – 1)

y = -2x + 2

2x + y – 2 = 0

Equation of Normal is

y – y_{1} = –\(\frac { 1 }{ m }\)(x – x_{1})

y – 0 = \(\frac { 1 }{ 2 }\)(x – 1)

2y = x- 1

x – 2y – 1 = 0

(ii) y = x^{4} + 2e^{x} at (0, 2)

Differentiating w.r.t. ‘x’

\(\frac { dy }{ dx }\) = 4x^{3} + 2e^{x}

Slope of the tangent ‘m’

(\(\frac { dy }{ dx }\))_{(0, 2)} = 4(0)³ + 2e^{0} = 2

Slope of the Normal –\(\frac { 1 }{ m }\) =-\(\frac { 1 }{ 2 }\)

Equation of tangent is

y – y_{1} = m(x – x_{1})

⇒ y – 2 = 2(x – 0)

⇒ y – 2 = 2x

⇒ 2x – y + 2 = 0

Equation of Normal is

y – y_{1} = –\(\frac { 1 }{ m }\) (x – x_{1})

y – 2 = –\(\frac { 1 }{ 2 }\)(x – 0)

2y – 4 = -x

x + 2y – 4 = 0

(iii) y = x sin x at (\(\frac { π }{ 2 }\), \(\frac { π }{ 2 }\))

Differentiating w.r.t. ‘x’

\(\frac { dy }{ dx }\) = x cos x + sin x

Slope of the tangent ‘m’ = (\(\frac { dy }{ dx }\))_{(π/2, π/2)}

= \(\frac { π }{ 2 }\) cos \(\frac { π }{ 2 }\) + sin \(\frac { π }{ 2 }\) = 1

Slope of the Normal –\(\frac { 1 }{ m }\) = -1

Equation of tangent is

y – y_{1} = m(x – x_{1})

⇒ y – \(\frac { π }{ 2 }\) = 1 (x – \(\frac { π }{ 2 }\))

⇒ x – y = 0

Equation of Normal is

y – y_{1} = –\(\frac { 1 }{ m }\)(x – x_{1})

⇒ y – \(\frac { π }{ 2 }\) = -1(x – \(\frac { π }{ 2 }\))

⇒ y – \(\frac { π }{ 2 }\) = -x + \(\frac { π }{ 2 }\)

⇒ x + y – π = 0

(iv) x = cos t, y = 2 sin² t at t = \(\frac { π }{ 2 }\)

at t = \(\frac { π }{ 3 }\), x = cos \(\frac { π }{ 3 }\) = \(\frac { 1 }{ 2 }\)

at t = \(\frac { π }{ 3 }\), y = 2 sin² \(\frac { π }{ 3 }\) = 2(\(\frac { 3 }{ 4}\)) = \(\frac { 3 }{ 2 }\)

Point is (\(\frac { 1 }{ 2 }\), \(\frac { 3 }{ 2 }\))

Now x = cos t y = 2 sin² t

Differentiating w.r.t. ‘t’,

\(\frac { dx }{ dt }\) = -sin t; \(\frac { dy }{ dt }\) = 4 sin t cos t

Slope of the tangent

Slope of the Normal –\(\frac { 1 }{ m }\) = \(\frac { 1 }{ 2 }\)

Equation of tangent is

y – y_{1} = m(x – x_{1})

⇒ y – \(\frac { 3 }{ 2 }\) = -2(x – \(\frac { 1 }{ 2 }\))

⇒ 2y – 3 = – 4x + 2

⇒ 4x + 2y – 5 = 0

Equation of Normal is

y – y_{1} = –\(\frac { 1 }{ m }\)(x – x_{1})

⇒ y – \(\frac { 3 }{ 2 }\) = \(\frac { 1 }{ 2 }\)(x – \(\frac { 1 }{ 2 }\))

⇒ 2 (2y – 3) = 2x – 1

⇒ 4y – 6 = 2x – 1

⇒ 2x – 4y + 5 = 0

Question 6.

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

Solution:

Curve is y = 1 + x³

Differentiating w.r.t ‘x’,

Slope of the tangent ‘m’ = \(\frac { dy }{ dx }\) = 3x²

Given line is x + 12y = 12

Slope of the line is –\(\frac { 1 }{ 12 }\)

Since the tangent is orthogonal with the line, the slope of the tangent is 12.

∴ \(\frac { dy }{ dx }\) = 12

i.e 3x² = 12

x² = 4

x = ±2

When x = 2, y = 1 + 8 = 9 ⇒ point is (2, 9)

When x = -2, y = 1 – 8 = -7 ⇒ point is (-2, -7)

Equation of tangent with slope 12 and at the j point (2, 9) is

y – 9 = 12 (x – 2)

y – 9 = 12x – 24

12x – y – 15 = 0

Equation of tangent with slope 12 and at the point (-2, -7) is

y + 7 = 12 (x + 2)

y + 7 = 12x + 24

12x – y + 17 = 0

Question 7.

Find the equations of the tangents to the curve y = –\(\frac { x+1 }{ x-1 }\) which are parallel to the line x + 2y = 6.

Solution:

Given line is x + 2y = 6

Slope of the line = –\(\frac { 1 }{ 2 }\)

Since the tangent is parallel to the line, then the slope of the tangent is –\(\frac { 1 }{ 2 }\)

∴ \(\frac { dy }{ dx }\) = \(\frac { 2 }{ (x-1)^2 }\) = –\(\frac { 1 }{ 2 }\)

(x – 1)² = 4

x – 1 = ±2

x = -1, 3

When x = – 1, y = 0 ⇒ point is (-1, 0)

When x = 3, y = 2 ⇒ point is (3, 2)

Equation of tangent with slope –\(\frac { 1 }{ 2 }\) and at the point (-1, 0) is

y – o = –\(\frac { 1 }{ 2 }\)(x + 1)

2y = -x – 1 ⇒ x + 2y + 1 = 0

Equation of tangent with slope –\(\frac { 1 }{ 2 }\) and at the point (3, 2) is 2

y – 2 = –\(\frac { 1 }{ 2 }\) (x – 3)

2y – 4 = -x + 3

x + 2y – 7 = 0.

Question 8.

Find the equation of tangent and normal to the curve given by x – 7 cos t andy = 2 sin t, t ∈ R at any point on the curve.

Solution:

x = 7 cos t and y = 2 sin t, t ∈ R

Differentiating w.r.t. ‘t’,

\(\frac { dx }{ dt }\) = -7 sin t and \(\frac { dy }{ dt }\) = 2 cos t

Slope of the tangent ‘m’

\(\frac { dy }{ dx }\) = \(\frac{\frac { dy }{ dt }}{\frac{ dx }{ dt }}\) = \(\frac { 2 cot t }{ -7 sin t }\)

Any point on the curve is (7 Cos t, 2 sin t)

Equation of tangent is y – y_{1} = m (x – x_{1})

y – 2 sint = –\(\frac { 2 cot t }{ 7 sin t }\) (x – 7 cos t)

7y sin t – 14 sin² t = -2x cos t + 14 cos² t

2x cos t + 7 y sin t – 14 (sin² t + cos² t) = 0

2x cos t + 7y sin t – 14 = 0

Now slope of normal is –\(\frac { 1 }{ 3 }\) = \(\frac { 7 sin t }{ 2 cos t }\)

Equation of normal is y – y_{1} = –\(\frac { 1 }{ m }\)(x – x_{1})

y – 2 sin t = \(\frac { 7 sin t }{ 2 cos t }\) (x – 7 cos t)

2y cos t – 4 sin t cos t = 7x sin t – 49 sin t cos t 7x sin t – 2y cos t – 45 sin t cos t = 0

Question 9.

Find the angle between the rectangular hyperbola xy = 2 and the parabola x² + 4y = 0

Solution:

Given curves are xy = 2 ……… (1)

x² + 4y = 0 ………. (2)

Now solving (1) and (2)

Substituting (1) in (2)

⇒ x² + 4(2/x) = 0

x³ + 8 = 0

x³ = -8

x = -2

Substituting in (1) ⇒ y = \(\frac { 2 }{ -2 }\) = -1

∴ Point of intersection of (1) and (2) is (-2, -1)

xy = 2 ⇒ y = \(\frac { 2 }{ x }\) ……….. (1)

Differentiating w.r.t. ‘x’

The angle between the curves

Question 10.

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

Solution:

Given curves are x² – y² = r² ……….. (1)

xy = c² …….. (2)

Let (x_{1}, y_{1}) be the point of intersection of the given curves.

(1) ⇒ x² – y² = r²

Differentiating w.r.t ‘x’,

2x – 2y \(\frac { dx }{ dy }\) = 0

\(\frac { dx }{ dy }\) = \(\frac { x }{ y }\)

now (\(\frac { dx }{ dy }\))(_{x1,y1}) = m_{1} = \(\frac { x_1 }{ y_1 }\)

(2) ⇒ xy = c²

Differentiating w.r.t ‘x’,

Hence, the given curves cut orthogonally.