Chapter 6: Application of Derivatives

Let’s consider the area of the first square A₁ = x²

And, area of the second square be A₂ = y²

Now, A₁ = x² and A₂ = y² = (x – x²)²

Differentiating both A₁ and A₂ w.r.t. t, we get

Therefore, the rate of change of area of the second square with respect to first is 2x² – 3x + 1.

It’s seen that the given curves are equation of two circles.

2x = y² ….. (1) and

2xy = k ….. (2)

We know that, two circles intersect orthogonally if the angle between the tangents drawn to the two circles at the point of their intersection is 90^o.

Now, differentiating equations (1) and (2) w.r.t. t, we get

Now, solving equations (1) and (2), we have

y = k/2x [From (2)]

Putting the value of y in equation (1),

2x = (k/2x)² ⇒ 2x = k²/4x²

8x³ = k²

8(1)³ = k²

k² = 8

Therefore, the required condition is k² = 8.

Given curves are equations of two circles,

xy = 4 ….. (i) and

x² + y² = 8 …. (ii)

Different equation (i) w.r.t., x

Therefore, the point of contact of the two circles are (2, 2) and (-2, 2).

Equation of the curve is given by, √x + √y = 4

Now, let (x₁, y₁) be he required point on the curve

So, √x₁ + √y₁ = 4

On differentiating on both the sides w.r.t. x₁, we get

On squaring on both the sides, we get

y₁/x₁ = 1 ⇒ y₁ = x₁

Now, putting the value of y1 in the given equation of the curve.

√x₁ + √y₁ = 4

√x₁ + √x₁ = 4 ⇒ 2√x₁ = 4 ⇒ √x₁ = 2 ⇒ x₁ = 4

As y₁ = x₁

⇒ y₁ = 4

Therefore, the required point is (4, 4).

The given curves are y = 4 – x² …. (i) and y = x² …… (ii)

And, we know that the angle of intersection of two curves is equal to the angle between the tangents drawn to the curves at their point of intersection.

Now, differentiating equations (i) and (ii) w.r.t x, we have

dy/dx = -2x ⇒ m₁ = -2x

m₁ is the slope of the tangent to the curve (i).

and dy/dx = 2x ⇒ m₂ = 2x

m₂ is the slope of the tangent to the curve (ii).

So, m₁ = -2x and m₂ = 2x

On solving equation (i) and (ii), we get

4 – x² = x² ⇒ 2x² = 4 ⇒ x² = 2 ⇒ x = ±√2

So, m₁ = -2x = -2√2 and m₂ = 2x = 2√2

Let θ be the angle of intersection of two curves

Given curve equations are: y² = 4x …. (1) and x² + y² – 6x + 1 = 0 ….. (2)

Now, differentiating (i) w.r.t. x, we get

2y.(dy/dx) = 4 ⇒ dy/dx = 2/y

Slope of tangent at (1, 2), m₁ = 2/2 = 1

Differentiating (ii) w.r.t. x, we get

2x + 2y.(dy/dx) – 6 = 0

2y. dy/dx = 6 – 2x ⇒ dy/dx = (6 – 2x)/ 2y

Hence, the slope of the tangent at the same point (1, 2)

⇒ m₂ = (6 – 2 x 1)/ (2 x 2) = 4/4 = 1

It’s seen that m₁ = m₂ = 1 at the point (1, 2).

Therefore, the given circles touch each other at the same point (1, 2).

Given curve, 3x² – y² = 8

Differentiating both sides w.r.t. x, we get

6x – 2y. dy/dx = 0 ⇒ -2y(dy/dx) = -6x ⇒ dy/dx = 3x/y

So, slope of the tangent to the given curve = 3x/y

Thus, the normal to the curve = -1/(3x/y) = -y/3x

Now, differentiating both sides of the given line x + 3y = 4, we have

1 + 3.(dy/dx) = 0

dy/dx = -1/3

As the normal to the curve is parallel to the given line x + 3y = 4

We have, -y/3x = -1/3 ⇒ y = x

On putting the value of y in 3x² – y² = 8, we get

3x² – x² = 8

2x² = 8 ⇒ x² = 4 ⇒ x = ±2

So, y = ±2

Thus, the points on the curve are (2, 2) and (-2, -2).

Now, the equation of the normal to the curve at (2, 2) is given by

Therefore, the required equations are x + 3y = 8 and x + 3y = -8.

Given, the equation of the curve is x² + y² – 2x – 4y + 1 = 0 ….. (i)

Differentiating both the sides w.r.t. x, we get

Now putting the value of y in equation (i), we get

x² + (2)² – 2x – 8 + 1 = 0

x² – 2x + 4 – 8 + 1 = 0

x² – 2x – 3 = 0 ⇒ x² – 3x + x – 3 = 0

x(x – 3) + 1(x – 3) = 0 ⇒ (x – 3)(x + 1) = 0

x = -1 or 3

Therefore, the required points are (-1, 2) and (3, 2).

Given curve equation, y = b . e^-x/a and line equation x/a + y/b = 1

Now, let the coordinates of the point where the curve intersects the y-axis be (0, y₁)

Now differentiating y = b . e^-x/a both sides w.r.t. x, we get

Therefore, the equation of the curve intersects at (0, b) which is on the y-axis.

Given,

f (x) = 2x + cot^-1 x + log [√(1 + x²) – x]

Differentiating both sides w.r.t. x, we get

On squaring both the sides, we get

4x⁴ + 1 + 4x² ≥ 1 + x²

4x⁴ + 4x² – x² ≥ 0

4x⁴ + 3x² ≥ 0

x²(4x² + 3) ≥ 0

The above is true for any value of x ∈ R.

Therefore, the given function is an increasing function over R.