QBManager

Application of Derivatives

85791 If $θ$ is an angle between the curves $x^{2} + 4 y = 0$ and $x y = 2$ , then $\tan θ =$

1 -1

2

\frac{1}{3}

3

\frac{1}{2}

4 3

Explanation:

(D) : Given that,
The curves $x^{2} + 4 y = 0$
And $x y = 2$
From $x^{2} + 4 y = 0$
$4 y = - x^{2}$
$y = \frac{- x^{2}}{4}$
Putting the value of $y$ in equation (ii), we get-
$x (\frac{- x^{2}}{4}) = 2$
$x^{3} = - 8$
$x = - 2$
Substitute value of $x$ in $x y = 2$ , we get -
$(- 2) y = 2$
$y = - 1$
So, the point of intersection is -
$(x, y) = (- 2, - 1)$
Again, differentiate $x y = 2$ w. r. t. $x$ both sides-
$\frac{x d y}{d x} + y = 0 \Rightarrow \frac{d y}{d x} = \frac{- y}{x}$
Then, $m_{1} = \frac{dy}{dx} = \frac{- y}{x} \Rightarrow m_{1} = \frac{- 1}{2}$
And, differentiate $x^{2} + 4 y = 0$ w.r.t $x$ both sides-
$2 x + 4 \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = \frac{- 2 x}{4}$
$\frac{d y}{d x} = \frac{- x}{2}$
Then, $m_{2} = \frac{dy}{dx} = \frac{- x}{2} \Rightarrow m_{2} = + \frac{2}{2}$
$m_{2} = 1 ∵ \tan θ = | \frac{(m_{1} - m_{2})}{1 + m_{1} m_{2}} |$
$∴ \tan θ = | \frac{\frac{- 1}{2} - 1}{1 - \frac{1}{2}} | = | \frac{\frac{- 3}{2}}{\frac{1}{2}} |$
$\tan θ = | - 3 |$
$\tan θ = 3$

Shift-I

Application of Derivatives

85792 The angle between the curves $y^{2} = 8 (x + 4)$ and $y^{2} = 24 (4 - x)$ is

1

\tan^{- 1} (\frac{1}{6})

2

\tan^{- 1} (3)

3

\frac{π}{2}

4

\frac{π}{4}

Explanation:

(C) : Given,
The curves $y^{2} = 8 (x + 4)$ and $y^{2} = 24 (4 - x)$
Then, $8 (x + 4) = 24 (4 - x)$
$(x + 4) = 3 (4 - x)$
$x + 4 = 12 - 3 x$
$4 x = 12 - 4$
$4 x = 8$
$x = 2$
Put, $x = 2$ in $y^{2} = 8 (x + 4)$ , we get -
$y^{2} = 8 (2 + 4)$
$y^{2} = 8 \times 6$
$y^{2} = 48$
$y^{2} = 16 \times 3$
$y = \sqrt{16 \times 3}$
$y = \pm 4 \sqrt{3}$
Then, intersection points $(2, \pm 4 \sqrt{3})$
$y^{2} = 8 (x + 4)$
$2 \frac{d y}{d x} \cdot y = 8$
$\frac{d y}{d x} = m_{1} = \frac{8}{2 y}$
$m_{1} = \frac{8}{2 \times (\pm 4 \sqrt{3})} = \frac{1}{\pm \sqrt{3}} = \frac{1}{\sqrt{3}}$
$y^{2} = 24 (4 - x)$
And,
$2 y \frac{d y}{d x} = - 24$
$\frac{d y}{d x} = m_{2} = \frac{- 24}{2 y} \Rightarrow \frac{d y}{d x} = \frac{- 24}{2 \times (\pm 4 \sqrt{3})}$
$\frac{d y}{d x} = \pm \sqrt{3} = - \sqrt{3} = m_{2}$
Therefore, $m_{1} m_{2}$
$\frac{1}{\sqrt{3}} \times - \sqrt{3} = - 1$
$θ = 90^{\circ}$
So, angle between curves is $\frac{π}{2}$ .

AP EAMCET-2019-22.04.2019

Application of Derivatives

85793 A ladder $5 m$ long is leaning against a wall. If the top of the ladder slides downwards at a rate of $10 cm . \sec^{- 1}$ , then the rate at which the angle between the floor and the ladder decreases, when the lower end of ladder is $2 m$ from the wall, is radian. sec

1

\frac{1}{10}

2

\frac{1}{20}

3 20

4 10

Explanation:

(B) : Given, length of the ladder $= 500 cm$ consider the horizontal length covered between the wall and the ladder be $x$ and vertical length covered between the wall and the ladder be $y$ .

And, let the angle between the floor and ladder be $θ$ .
Then, $\sin θ = \frac{y}{500}$
On differentiating w.r.t. t, we get-
$\cos θ \frac{d θ}{dt} = \frac{1}{500} \frac{dy}{dt}$
$∵$ Given that,
$\frac{dy}{dt} = - 10 cm / \sec .$
And also, $\cos θ = \frac{x}{500}$
When, $x = 200 cm, \cos θ = \frac{200}{500} = \frac{2}{5}$
Substituting equation (ii) and (iii) in equation (i), we get-
$\frac{2}{5} \frac{d θ}{dt} = \frac{1}{500} (- 10)$
$\frac{d θ}{dt} = \frac{- 1}{20} radian / second$
So, the angle between the floor and the ladder is decreasing at the rate of $\frac{1}{20}$ radian / second.

Shift-II

Application of Derivatives

85794 The angle between the curves $y^{2} = 4 ax$ and ay $=$

1

\tan^{- 1} \frac{3}{4}

2

\tan^{- 1} \frac{3}{5}

3

\tan^{- 1} \frac{4}{3}

4

\tan^{- 1} \frac{5}{3}

Explanation:

(B) : Given that, $y^{2} = 4 ax$
and, ay $= 2 x^{2}$
Solving equation (i) and equation (ii), we get -
$x = a and y = 2 a$
Differentiating equation (i) w.r.t.'x', we get -
$2 y \frac{d y}{d x} = 4 a$
Now, ${[\frac{dy}{dx}]}_{(a .2 a)} = \frac{4 a}{2 (2 a)} = 1$
$m_{1} = 1$
Now, differentiating equation (ii) w.r.t. 'x', we get -
$a \frac{dy}{dx} = 4 x$
${[\frac{dy}{dx}]}_{(a, 2 a)} = \frac{4 a}{a} = 4$
$m_{2} = 4$
Angle between curves is equal to angle between their tangents.
$\tan θ = | \frac{m_{2} - m_{1}}{1 + m_{1} m_{2}} |$
$\tan θ = | \frac{4 - 1}{1 + 4 \times (1)} |$
$\tan θ = \frac{3}{5}$
Or $θ = \tan^{- 1} (\frac{3}{5})$

COMEDK-2015

Application of Derivatives

85795 If the angle between the curves $y^{2} = 4 x$ and $y =$ $e^{- x / 2}$ is $θ$ , then ${cosec}^{2} (θ / 2) =$

1 2

2 3

3

\sqrt{3}

4

\sqrt{2}

Explanation:

(A) : $y^{2} = 4 x$
$\frac{dy}{dx} = \frac{2 y}{y}$
$m_{1} = {(\frac{dy}{dx})}_{(x_{1} y_{1})} = \frac{2}{y_{1}}$
Also, $y = e^{- \frac{x}{2}}$
$\frac{d y}{d x} = e^{- \frac{x}{2}} \times - \frac{1}{2}$
$m_{2} = {(\frac{d y}{d x})}_{(x_{1} y_{1})} = - \frac{1}{2} \times e^{- \frac{x_{1}}{2}} = \frac{- 1}{2} y_{1}$
$\tan θ = | \frac{m_{2} - m_{1}}{1 + m_{1} m_{2}} |$
$= | \frac{- \frac{y_{1}}{2} - \frac{2}{y_{1}}}{1 + (\frac{- y_{1}}{2} \times \frac{2}{y_{1}})} | = \infty$
$θ = \frac{π}{2} = 90^{\circ}$
Then ${cosec}^{2} (\frac{θ}{2}) = {cosc}^{2} (\frac{90^{\circ}}{2}) = {cosec}^{2} 45^{\circ} = 2$

AP EAMCET-2022-08.07.2022

Application of Derivatives

85791 If $θ$ is an angle between the curves $x^{2} + 4 y = 0$ and $x y = 2$ , then $\tan θ =$

1 -1

2

\frac{1}{3}

3

\frac{1}{2}

4 3

Explanation:

(D) : Given that,
The curves $x^{2} + 4 y = 0$
And $x y = 2$
From $x^{2} + 4 y = 0$
$4 y = - x^{2}$
$y = \frac{- x^{2}}{4}$
Putting the value of $y$ in equation (ii), we get-
$x (\frac{- x^{2}}{4}) = 2$
$x^{3} = - 8$
$x = - 2$
Substitute value of $x$ in $x y = 2$ , we get -
$(- 2) y = 2$
$y = - 1$
So, the point of intersection is -
$(x, y) = (- 2, - 1)$
Again, differentiate $x y = 2$ w. r. t. $x$ both sides-
$\frac{x d y}{d x} + y = 0 \Rightarrow \frac{d y}{d x} = \frac{- y}{x}$
Then, $m_{1} = \frac{dy}{dx} = \frac{- y}{x} \Rightarrow m_{1} = \frac{- 1}{2}$
And, differentiate $x^{2} + 4 y = 0$ w.r.t $x$ both sides-
$2 x + 4 \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = \frac{- 2 x}{4}$
$\frac{d y}{d x} = \frac{- x}{2}$
Then, $m_{2} = \frac{dy}{dx} = \frac{- x}{2} \Rightarrow m_{2} = + \frac{2}{2}$
$m_{2} = 1 ∵ \tan θ = | \frac{(m_{1} - m_{2})}{1 + m_{1} m_{2}} |$
$∴ \tan θ = | \frac{\frac{- 1}{2} - 1}{1 - \frac{1}{2}} | = | \frac{\frac{- 3}{2}}{\frac{1}{2}} |$
$\tan θ = | - 3 |$
$\tan θ = 3$

Shift-I

Application of Derivatives

85792 The angle between the curves $y^{2} = 8 (x + 4)$ and $y^{2} = 24 (4 - x)$ is

1

\tan^{- 1} (\frac{1}{6})

2

\tan^{- 1} (3)

3

\frac{π}{2}

4

\frac{π}{4}

Explanation:

(C) : Given,
The curves $y^{2} = 8 (x + 4)$ and $y^{2} = 24 (4 - x)$
Then, $8 (x + 4) = 24 (4 - x)$
$(x + 4) = 3 (4 - x)$
$x + 4 = 12 - 3 x$
$4 x = 12 - 4$
$4 x = 8$
$x = 2$
Put, $x = 2$ in $y^{2} = 8 (x + 4)$ , we get -
$y^{2} = 8 (2 + 4)$
$y^{2} = 8 \times 6$
$y^{2} = 48$
$y^{2} = 16 \times 3$
$y = \sqrt{16 \times 3}$
$y = \pm 4 \sqrt{3}$
Then, intersection points $(2, \pm 4 \sqrt{3})$
$y^{2} = 8 (x + 4)$
$2 \frac{d y}{d x} \cdot y = 8$
$\frac{d y}{d x} = m_{1} = \frac{8}{2 y}$
$m_{1} = \frac{8}{2 \times (\pm 4 \sqrt{3})} = \frac{1}{\pm \sqrt{3}} = \frac{1}{\sqrt{3}}$
$y^{2} = 24 (4 - x)$
And,
$2 y \frac{d y}{d x} = - 24$
$\frac{d y}{d x} = m_{2} = \frac{- 24}{2 y} \Rightarrow \frac{d y}{d x} = \frac{- 24}{2 \times (\pm 4 \sqrt{3})}$
$\frac{d y}{d x} = \pm \sqrt{3} = - \sqrt{3} = m_{2}$
Therefore, $m_{1} m_{2}$
$\frac{1}{\sqrt{3}} \times - \sqrt{3} = - 1$
$θ = 90^{\circ}$
So, angle between curves is $\frac{π}{2}$ .

AP EAMCET-2019-22.04.2019

Application of Derivatives

85793 A ladder $5 m$ long is leaning against a wall. If the top of the ladder slides downwards at a rate of $10 cm . \sec^{- 1}$ , then the rate at which the angle between the floor and the ladder decreases, when the lower end of ladder is $2 m$ from the wall, is radian. sec

1

\frac{1}{10}

2

\frac{1}{20}

3 20

4 10

Explanation:

(B) : Given, length of the ladder $= 500 cm$ consider the horizontal length covered between the wall and the ladder be $x$ and vertical length covered between the wall and the ladder be $y$ .

And, let the angle between the floor and ladder be $θ$ .
Then, $\sin θ = \frac{y}{500}$
On differentiating w.r.t. t, we get-
$\cos θ \frac{d θ}{dt} = \frac{1}{500} \frac{dy}{dt}$
$∵$ Given that,
$\frac{dy}{dt} = - 10 cm / \sec .$
And also, $\cos θ = \frac{x}{500}$
When, $x = 200 cm, \cos θ = \frac{200}{500} = \frac{2}{5}$
Substituting equation (ii) and (iii) in equation (i), we get-
$\frac{2}{5} \frac{d θ}{dt} = \frac{1}{500} (- 10)$
$\frac{d θ}{dt} = \frac{- 1}{20} radian / second$
So, the angle between the floor and the ladder is decreasing at the rate of $\frac{1}{20}$ radian / second.

Shift-II

Application of Derivatives

85794 The angle between the curves $y^{2} = 4 ax$ and ay $=$

1

\tan^{- 1} \frac{3}{4}

2

\tan^{- 1} \frac{3}{5}

3

\tan^{- 1} \frac{4}{3}

4

\tan^{- 1} \frac{5}{3}

Explanation:

(B) : Given that, $y^{2} = 4 ax$
and, ay $= 2 x^{2}$
Solving equation (i) and equation (ii), we get -
$x = a and y = 2 a$
Differentiating equation (i) w.r.t.'x', we get -
$2 y \frac{d y}{d x} = 4 a$
Now, ${[\frac{dy}{dx}]}_{(a .2 a)} = \frac{4 a}{2 (2 a)} = 1$
$m_{1} = 1$
Now, differentiating equation (ii) w.r.t. 'x', we get -
$a \frac{dy}{dx} = 4 x$
${[\frac{dy}{dx}]}_{(a, 2 a)} = \frac{4 a}{a} = 4$
$m_{2} = 4$
Angle between curves is equal to angle between their tangents.
$\tan θ = | \frac{m_{2} - m_{1}}{1 + m_{1} m_{2}} |$
$\tan θ = | \frac{4 - 1}{1 + 4 \times (1)} |$
$\tan θ = \frac{3}{5}$
Or $θ = \tan^{- 1} (\frac{3}{5})$

COMEDK-2015

Application of Derivatives

85795 If the angle between the curves $y^{2} = 4 x$ and $y =$ $e^{- x / 2}$ is $θ$ , then ${cosec}^{2} (θ / 2) =$

1 2

2 3

3

\sqrt{3}

4

\sqrt{2}

Explanation:

(A) : $y^{2} = 4 x$
$\frac{dy}{dx} = \frac{2 y}{y}$
$m_{1} = {(\frac{dy}{dx})}_{(x_{1} y_{1})} = \frac{2}{y_{1}}$
Also, $y = e^{- \frac{x}{2}}$
$\frac{d y}{d x} = e^{- \frac{x}{2}} \times - \frac{1}{2}$
$m_{2} = {(\frac{d y}{d x})}_{(x_{1} y_{1})} = - \frac{1}{2} \times e^{- \frac{x_{1}}{2}} = \frac{- 1}{2} y_{1}$
$\tan θ = | \frac{m_{2} - m_{1}}{1 + m_{1} m_{2}} |$
$= | \frac{- \frac{y_{1}}{2} - \frac{2}{y_{1}}}{1 + (\frac{- y_{1}}{2} \times \frac{2}{y_{1}})} | = \infty$
$θ = \frac{π}{2} = 90^{\circ}$
Then ${cosec}^{2} (\frac{θ}{2}) = {cosc}^{2} (\frac{90^{\circ}}{2}) = {cosec}^{2} 45^{\circ} = 2$

AP EAMCET-2022-08.07.2022

Application of Derivatives

85791 If $θ$ is an angle between the curves $x^{2} + 4 y = 0$ and $x y = 2$ , then $\tan θ =$

1 -1

2

\frac{1}{3}

3

\frac{1}{2}

4 3

Explanation:

(D) : Given that,
The curves $x^{2} + 4 y = 0$
And $x y = 2$
From $x^{2} + 4 y = 0$
$4 y = - x^{2}$
$y = \frac{- x^{2}}{4}$
Putting the value of $y$ in equation (ii), we get-
$x (\frac{- x^{2}}{4}) = 2$
$x^{3} = - 8$
$x = - 2$
Substitute value of $x$ in $x y = 2$ , we get -
$(- 2) y = 2$
$y = - 1$
So, the point of intersection is -
$(x, y) = (- 2, - 1)$
Again, differentiate $x y = 2$ w. r. t. $x$ both sides-
$\frac{x d y}{d x} + y = 0 \Rightarrow \frac{d y}{d x} = \frac{- y}{x}$
Then, $m_{1} = \frac{dy}{dx} = \frac{- y}{x} \Rightarrow m_{1} = \frac{- 1}{2}$
And, differentiate $x^{2} + 4 y = 0$ w.r.t $x$ both sides-
$2 x + 4 \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = \frac{- 2 x}{4}$
$\frac{d y}{d x} = \frac{- x}{2}$
Then, $m_{2} = \frac{dy}{dx} = \frac{- x}{2} \Rightarrow m_{2} = + \frac{2}{2}$
$m_{2} = 1 ∵ \tan θ = | \frac{(m_{1} - m_{2})}{1 + m_{1} m_{2}} |$
$∴ \tan θ = | \frac{\frac{- 1}{2} - 1}{1 - \frac{1}{2}} | = | \frac{\frac{- 3}{2}}{\frac{1}{2}} |$
$\tan θ = | - 3 |$
$\tan θ = 3$

Shift-I

Application of Derivatives

85792 The angle between the curves $y^{2} = 8 (x + 4)$ and $y^{2} = 24 (4 - x)$ is

1

\tan^{- 1} (\frac{1}{6})

2

\tan^{- 1} (3)

3

\frac{π}{2}

4

\frac{π}{4}

Explanation:

(C) : Given,
The curves $y^{2} = 8 (x + 4)$ and $y^{2} = 24 (4 - x)$
Then, $8 (x + 4) = 24 (4 - x)$
$(x + 4) = 3 (4 - x)$
$x + 4 = 12 - 3 x$
$4 x = 12 - 4$
$4 x = 8$
$x = 2$
Put, $x = 2$ in $y^{2} = 8 (x + 4)$ , we get -
$y^{2} = 8 (2 + 4)$
$y^{2} = 8 \times 6$
$y^{2} = 48$
$y^{2} = 16 \times 3$
$y = \sqrt{16 \times 3}$
$y = \pm 4 \sqrt{3}$
Then, intersection points $(2, \pm 4 \sqrt{3})$
$y^{2} = 8 (x + 4)$
$2 \frac{d y}{d x} \cdot y = 8$
$\frac{d y}{d x} = m_{1} = \frac{8}{2 y}$
$m_{1} = \frac{8}{2 \times (\pm 4 \sqrt{3})} = \frac{1}{\pm \sqrt{3}} = \frac{1}{\sqrt{3}}$
$y^{2} = 24 (4 - x)$
And,
$2 y \frac{d y}{d x} = - 24$
$\frac{d y}{d x} = m_{2} = \frac{- 24}{2 y} \Rightarrow \frac{d y}{d x} = \frac{- 24}{2 \times (\pm 4 \sqrt{3})}$
$\frac{d y}{d x} = \pm \sqrt{3} = - \sqrt{3} = m_{2}$
Therefore, $m_{1} m_{2}$
$\frac{1}{\sqrt{3}} \times - \sqrt{3} = - 1$
$θ = 90^{\circ}$
So, angle between curves is $\frac{π}{2}$ .

AP EAMCET-2019-22.04.2019

Application of Derivatives

85793 A ladder $5 m$ long is leaning against a wall. If the top of the ladder slides downwards at a rate of $10 cm . \sec^{- 1}$ , then the rate at which the angle between the floor and the ladder decreases, when the lower end of ladder is $2 m$ from the wall, is radian. sec

1

\frac{1}{10}

2

\frac{1}{20}

3 20

4 10

Explanation:

(B) : Given, length of the ladder $= 500 cm$ consider the horizontal length covered between the wall and the ladder be $x$ and vertical length covered between the wall and the ladder be $y$ .

And, let the angle between the floor and ladder be $θ$ .
Then, $\sin θ = \frac{y}{500}$
On differentiating w.r.t. t, we get-
$\cos θ \frac{d θ}{dt} = \frac{1}{500} \frac{dy}{dt}$
$∵$ Given that,
$\frac{dy}{dt} = - 10 cm / \sec .$
And also, $\cos θ = \frac{x}{500}$
When, $x = 200 cm, \cos θ = \frac{200}{500} = \frac{2}{5}$
Substituting equation (ii) and (iii) in equation (i), we get-
$\frac{2}{5} \frac{d θ}{dt} = \frac{1}{500} (- 10)$
$\frac{d θ}{dt} = \frac{- 1}{20} radian / second$
So, the angle between the floor and the ladder is decreasing at the rate of $\frac{1}{20}$ radian / second.

Shift-II

Application of Derivatives

85794 The angle between the curves $y^{2} = 4 ax$ and ay $=$

1

\tan^{- 1} \frac{3}{4}

2

\tan^{- 1} \frac{3}{5}

3

\tan^{- 1} \frac{4}{3}

4

\tan^{- 1} \frac{5}{3}

Explanation:

(B) : Given that, $y^{2} = 4 ax$
and, ay $= 2 x^{2}$
Solving equation (i) and equation (ii), we get -
$x = a and y = 2 a$
Differentiating equation (i) w.r.t.'x', we get -
$2 y \frac{d y}{d x} = 4 a$
Now, ${[\frac{dy}{dx}]}_{(a .2 a)} = \frac{4 a}{2 (2 a)} = 1$
$m_{1} = 1$
Now, differentiating equation (ii) w.r.t. 'x', we get -
$a \frac{dy}{dx} = 4 x$
${[\frac{dy}{dx}]}_{(a, 2 a)} = \frac{4 a}{a} = 4$
$m_{2} = 4$
Angle between curves is equal to angle between their tangents.
$\tan θ = | \frac{m_{2} - m_{1}}{1 + m_{1} m_{2}} |$
$\tan θ = | \frac{4 - 1}{1 + 4 \times (1)} |$
$\tan θ = \frac{3}{5}$
Or $θ = \tan^{- 1} (\frac{3}{5})$

COMEDK-2015

Application of Derivatives

85795 If the angle between the curves $y^{2} = 4 x$ and $y =$ $e^{- x / 2}$ is $θ$ , then ${cosec}^{2} (θ / 2) =$

1 2

2 3

3

\sqrt{3}

4

\sqrt{2}

Explanation:

(A) : $y^{2} = 4 x$
$\frac{dy}{dx} = \frac{2 y}{y}$
$m_{1} = {(\frac{dy}{dx})}_{(x_{1} y_{1})} = \frac{2}{y_{1}}$
Also, $y = e^{- \frac{x}{2}}$
$\frac{d y}{d x} = e^{- \frac{x}{2}} \times - \frac{1}{2}$
$m_{2} = {(\frac{d y}{d x})}_{(x_{1} y_{1})} = - \frac{1}{2} \times e^{- \frac{x_{1}}{2}} = \frac{- 1}{2} y_{1}$
$\tan θ = | \frac{m_{2} - m_{1}}{1 + m_{1} m_{2}} |$
$= | \frac{- \frac{y_{1}}{2} - \frac{2}{y_{1}}}{1 + (\frac{- y_{1}}{2} \times \frac{2}{y_{1}})} | = \infty$
$θ = \frac{π}{2} = 90^{\circ}$
Then ${cosec}^{2} (\frac{θ}{2}) = {cosc}^{2} (\frac{90^{\circ}}{2}) = {cosec}^{2} 45^{\circ} = 2$

AP EAMCET-2022-08.07.2022

Application of Derivatives

85791 If $θ$ is an angle between the curves $x^{2} + 4 y = 0$ and $x y = 2$ , then $\tan θ =$

1 -1

2

\frac{1}{3}

3

\frac{1}{2}

4 3

Explanation:

(D) : Given that,
The curves $x^{2} + 4 y = 0$
And $x y = 2$
From $x^{2} + 4 y = 0$
$4 y = - x^{2}$
$y = \frac{- x^{2}}{4}$
Putting the value of $y$ in equation (ii), we get-
$x (\frac{- x^{2}}{4}) = 2$
$x^{3} = - 8$
$x = - 2$
Substitute value of $x$ in $x y = 2$ , we get -
$(- 2) y = 2$
$y = - 1$
So, the point of intersection is -
$(x, y) = (- 2, - 1)$
Again, differentiate $x y = 2$ w. r. t. $x$ both sides-
$\frac{x d y}{d x} + y = 0 \Rightarrow \frac{d y}{d x} = \frac{- y}{x}$
Then, $m_{1} = \frac{dy}{dx} = \frac{- y}{x} \Rightarrow m_{1} = \frac{- 1}{2}$
And, differentiate $x^{2} + 4 y = 0$ w.r.t $x$ both sides-
$2 x + 4 \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = \frac{- 2 x}{4}$
$\frac{d y}{d x} = \frac{- x}{2}$
Then, $m_{2} = \frac{dy}{dx} = \frac{- x}{2} \Rightarrow m_{2} = + \frac{2}{2}$
$m_{2} = 1 ∵ \tan θ = | \frac{(m_{1} - m_{2})}{1 + m_{1} m_{2}} |$
$∴ \tan θ = | \frac{\frac{- 1}{2} - 1}{1 - \frac{1}{2}} | = | \frac{\frac{- 3}{2}}{\frac{1}{2}} |$
$\tan θ = | - 3 |$
$\tan θ = 3$

Shift-I

Application of Derivatives

85792 The angle between the curves $y^{2} = 8 (x + 4)$ and $y^{2} = 24 (4 - x)$ is

1

\tan^{- 1} (\frac{1}{6})

2

\tan^{- 1} (3)

3

\frac{π}{2}

4

\frac{π}{4}

Explanation:

(C) : Given,
The curves $y^{2} = 8 (x + 4)$ and $y^{2} = 24 (4 - x)$
Then, $8 (x + 4) = 24 (4 - x)$
$(x + 4) = 3 (4 - x)$
$x + 4 = 12 - 3 x$
$4 x = 12 - 4$
$4 x = 8$
$x = 2$
Put, $x = 2$ in $y^{2} = 8 (x + 4)$ , we get -
$y^{2} = 8 (2 + 4)$
$y^{2} = 8 \times 6$
$y^{2} = 48$
$y^{2} = 16 \times 3$
$y = \sqrt{16 \times 3}$
$y = \pm 4 \sqrt{3}$
Then, intersection points $(2, \pm 4 \sqrt{3})$
$y^{2} = 8 (x + 4)$
$2 \frac{d y}{d x} \cdot y = 8$
$\frac{d y}{d x} = m_{1} = \frac{8}{2 y}$
$m_{1} = \frac{8}{2 \times (\pm 4 \sqrt{3})} = \frac{1}{\pm \sqrt{3}} = \frac{1}{\sqrt{3}}$
$y^{2} = 24 (4 - x)$
And,
$2 y \frac{d y}{d x} = - 24$
$\frac{d y}{d x} = m_{2} = \frac{- 24}{2 y} \Rightarrow \frac{d y}{d x} = \frac{- 24}{2 \times (\pm 4 \sqrt{3})}$
$\frac{d y}{d x} = \pm \sqrt{3} = - \sqrt{3} = m_{2}$
Therefore, $m_{1} m_{2}$
$\frac{1}{\sqrt{3}} \times - \sqrt{3} = - 1$
$θ = 90^{\circ}$
So, angle between curves is $\frac{π}{2}$ .

AP EAMCET-2019-22.04.2019

Application of Derivatives

85793 A ladder $5 m$ long is leaning against a wall. If the top of the ladder slides downwards at a rate of $10 cm . \sec^{- 1}$ , then the rate at which the angle between the floor and the ladder decreases, when the lower end of ladder is $2 m$ from the wall, is radian. sec

1

\frac{1}{10}

2

\frac{1}{20}

3 20

4 10

Explanation:

(B) : Given, length of the ladder $= 500 cm$ consider the horizontal length covered between the wall and the ladder be $x$ and vertical length covered between the wall and the ladder be $y$ .

And, let the angle between the floor and ladder be $θ$ .
Then, $\sin θ = \frac{y}{500}$
On differentiating w.r.t. t, we get-
$\cos θ \frac{d θ}{dt} = \frac{1}{500} \frac{dy}{dt}$
$∵$ Given that,
$\frac{dy}{dt} = - 10 cm / \sec .$
And also, $\cos θ = \frac{x}{500}$
When, $x = 200 cm, \cos θ = \frac{200}{500} = \frac{2}{5}$
Substituting equation (ii) and (iii) in equation (i), we get-
$\frac{2}{5} \frac{d θ}{dt} = \frac{1}{500} (- 10)$
$\frac{d θ}{dt} = \frac{- 1}{20} radian / second$
So, the angle between the floor and the ladder is decreasing at the rate of $\frac{1}{20}$ radian / second.

Shift-II

Application of Derivatives

85794 The angle between the curves $y^{2} = 4 ax$ and ay $=$

1

\tan^{- 1} \frac{3}{4}

2

\tan^{- 1} \frac{3}{5}

3

\tan^{- 1} \frac{4}{3}

4

\tan^{- 1} \frac{5}{3}

Explanation:

(B) : Given that, $y^{2} = 4 ax$
and, ay $= 2 x^{2}$
Solving equation (i) and equation (ii), we get -
$x = a and y = 2 a$
Differentiating equation (i) w.r.t.'x', we get -
$2 y \frac{d y}{d x} = 4 a$
Now, ${[\frac{dy}{dx}]}_{(a .2 a)} = \frac{4 a}{2 (2 a)} = 1$
$m_{1} = 1$
Now, differentiating equation (ii) w.r.t. 'x', we get -
$a \frac{dy}{dx} = 4 x$
${[\frac{dy}{dx}]}_{(a, 2 a)} = \frac{4 a}{a} = 4$
$m_{2} = 4$
Angle between curves is equal to angle between their tangents.
$\tan θ = | \frac{m_{2} - m_{1}}{1 + m_{1} m_{2}} |$
$\tan θ = | \frac{4 - 1}{1 + 4 \times (1)} |$
$\tan θ = \frac{3}{5}$
Or $θ = \tan^{- 1} (\frac{3}{5})$

COMEDK-2015

Application of Derivatives

85795 If the angle between the curves $y^{2} = 4 x$ and $y =$ $e^{- x / 2}$ is $θ$ , then ${cosec}^{2} (θ / 2) =$

1 2

2 3

3

\sqrt{3}

4

\sqrt{2}

Explanation:

(A) : $y^{2} = 4 x$
$\frac{dy}{dx} = \frac{2 y}{y}$
$m_{1} = {(\frac{dy}{dx})}_{(x_{1} y_{1})} = \frac{2}{y_{1}}$
Also, $y = e^{- \frac{x}{2}}$
$\frac{d y}{d x} = e^{- \frac{x}{2}} \times - \frac{1}{2}$
$m_{2} = {(\frac{d y}{d x})}_{(x_{1} y_{1})} = - \frac{1}{2} \times e^{- \frac{x_{1}}{2}} = \frac{- 1}{2} y_{1}$
$\tan θ = | \frac{m_{2} - m_{1}}{1 + m_{1} m_{2}} |$
$= | \frac{- \frac{y_{1}}{2} - \frac{2}{y_{1}}}{1 + (\frac{- y_{1}}{2} \times \frac{2}{y_{1}})} | = \infty$
$θ = \frac{π}{2} = 90^{\circ}$
Then ${cosec}^{2} (\frac{θ}{2}) = {cosc}^{2} (\frac{90^{\circ}}{2}) = {cosec}^{2} 45^{\circ} = 2$

AP EAMCET-2022-08.07.2022

Application of Derivatives

85791 If $θ$ is an angle between the curves $x^{2} + 4 y = 0$ and $x y = 2$ , then $\tan θ =$

1 -1

2

\frac{1}{3}

3

\frac{1}{2}

4 3

Explanation:

(D) : Given that,
The curves $x^{2} + 4 y = 0$
And $x y = 2$
From $x^{2} + 4 y = 0$
$4 y = - x^{2}$
$y = \frac{- x^{2}}{4}$
Putting the value of $y$ in equation (ii), we get-
$x (\frac{- x^{2}}{4}) = 2$
$x^{3} = - 8$
$x = - 2$
Substitute value of $x$ in $x y = 2$ , we get -
$(- 2) y = 2$
$y = - 1$
So, the point of intersection is -
$(x, y) = (- 2, - 1)$
Again, differentiate $x y = 2$ w. r. t. $x$ both sides-
$\frac{x d y}{d x} + y = 0 \Rightarrow \frac{d y}{d x} = \frac{- y}{x}$
Then, $m_{1} = \frac{dy}{dx} = \frac{- y}{x} \Rightarrow m_{1} = \frac{- 1}{2}$
And, differentiate $x^{2} + 4 y = 0$ w.r.t $x$ both sides-
$2 x + 4 \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = \frac{- 2 x}{4}$
$\frac{d y}{d x} = \frac{- x}{2}$
Then, $m_{2} = \frac{dy}{dx} = \frac{- x}{2} \Rightarrow m_{2} = + \frac{2}{2}$
$m_{2} = 1 ∵ \tan θ = | \frac{(m_{1} - m_{2})}{1 + m_{1} m_{2}} |$
$∴ \tan θ = | \frac{\frac{- 1}{2} - 1}{1 - \frac{1}{2}} | = | \frac{\frac{- 3}{2}}{\frac{1}{2}} |$
$\tan θ = | - 3 |$
$\tan θ = 3$

Shift-I

Application of Derivatives

85792 The angle between the curves $y^{2} = 8 (x + 4)$ and $y^{2} = 24 (4 - x)$ is

1

\tan^{- 1} (\frac{1}{6})

2

\tan^{- 1} (3)

3

\frac{π}{2}

4

\frac{π}{4}

Explanation:

(C) : Given,
The curves $y^{2} = 8 (x + 4)$ and $y^{2} = 24 (4 - x)$
Then, $8 (x + 4) = 24 (4 - x)$
$(x + 4) = 3 (4 - x)$
$x + 4 = 12 - 3 x$
$4 x = 12 - 4$
$4 x = 8$
$x = 2$
Put, $x = 2$ in $y^{2} = 8 (x + 4)$ , we get -
$y^{2} = 8 (2 + 4)$
$y^{2} = 8 \times 6$
$y^{2} = 48$
$y^{2} = 16 \times 3$
$y = \sqrt{16 \times 3}$
$y = \pm 4 \sqrt{3}$
Then, intersection points $(2, \pm 4 \sqrt{3})$
$y^{2} = 8 (x + 4)$
$2 \frac{d y}{d x} \cdot y = 8$
$\frac{d y}{d x} = m_{1} = \frac{8}{2 y}$
$m_{1} = \frac{8}{2 \times (\pm 4 \sqrt{3})} = \frac{1}{\pm \sqrt{3}} = \frac{1}{\sqrt{3}}$
$y^{2} = 24 (4 - x)$
And,
$2 y \frac{d y}{d x} = - 24$
$\frac{d y}{d x} = m_{2} = \frac{- 24}{2 y} \Rightarrow \frac{d y}{d x} = \frac{- 24}{2 \times (\pm 4 \sqrt{3})}$
$\frac{d y}{d x} = \pm \sqrt{3} = - \sqrt{3} = m_{2}$
Therefore, $m_{1} m_{2}$
$\frac{1}{\sqrt{3}} \times - \sqrt{3} = - 1$
$θ = 90^{\circ}$
So, angle between curves is $\frac{π}{2}$ .

AP EAMCET-2019-22.04.2019

Application of Derivatives

85793 A ladder $5 m$ long is leaning against a wall. If the top of the ladder slides downwards at a rate of $10 cm . \sec^{- 1}$ , then the rate at which the angle between the floor and the ladder decreases, when the lower end of ladder is $2 m$ from the wall, is radian. sec

1

\frac{1}{10}

2

\frac{1}{20}

3 20

4 10

Explanation:

(B) : Given, length of the ladder $= 500 cm$ consider the horizontal length covered between the wall and the ladder be $x$ and vertical length covered between the wall and the ladder be $y$ .

And, let the angle between the floor and ladder be $θ$ .
Then, $\sin θ = \frac{y}{500}$
On differentiating w.r.t. t, we get-
$\cos θ \frac{d θ}{dt} = \frac{1}{500} \frac{dy}{dt}$
$∵$ Given that,
$\frac{dy}{dt} = - 10 cm / \sec .$
And also, $\cos θ = \frac{x}{500}$
When, $x = 200 cm, \cos θ = \frac{200}{500} = \frac{2}{5}$
Substituting equation (ii) and (iii) in equation (i), we get-
$\frac{2}{5} \frac{d θ}{dt} = \frac{1}{500} (- 10)$
$\frac{d θ}{dt} = \frac{- 1}{20} radian / second$
So, the angle between the floor and the ladder is decreasing at the rate of $\frac{1}{20}$ radian / second.

Shift-II

Application of Derivatives

85794 The angle between the curves $y^{2} = 4 ax$ and ay $=$

1

\tan^{- 1} \frac{3}{4}

2

\tan^{- 1} \frac{3}{5}

3

\tan^{- 1} \frac{4}{3}

4

\tan^{- 1} \frac{5}{3}

Explanation:

(B) : Given that, $y^{2} = 4 ax$
and, ay $= 2 x^{2}$
Solving equation (i) and equation (ii), we get -
$x = a and y = 2 a$
Differentiating equation (i) w.r.t.'x', we get -
$2 y \frac{d y}{d x} = 4 a$
Now, ${[\frac{dy}{dx}]}_{(a .2 a)} = \frac{4 a}{2 (2 a)} = 1$
$m_{1} = 1$
Now, differentiating equation (ii) w.r.t. 'x', we get -
$a \frac{dy}{dx} = 4 x$
${[\frac{dy}{dx}]}_{(a, 2 a)} = \frac{4 a}{a} = 4$
$m_{2} = 4$
Angle between curves is equal to angle between their tangents.
$\tan θ = | \frac{m_{2} - m_{1}}{1 + m_{1} m_{2}} |$
$\tan θ = | \frac{4 - 1}{1 + 4 \times (1)} |$
$\tan θ = \frac{3}{5}$
Or $θ = \tan^{- 1} (\frac{3}{5})$

COMEDK-2015

Application of Derivatives

85795 If the angle between the curves $y^{2} = 4 x$ and $y =$ $e^{- x / 2}$ is $θ$ , then ${cosec}^{2} (θ / 2) =$

1 2

2 3

3

\sqrt{3}

4

\sqrt{2}

Explanation:

(A) : $y^{2} = 4 x$
$\frac{dy}{dx} = \frac{2 y}{y}$
$m_{1} = {(\frac{dy}{dx})}_{(x_{1} y_{1})} = \frac{2}{y_{1}}$
Also, $y = e^{- \frac{x}{2}}$
$\frac{d y}{d x} = e^{- \frac{x}{2}} \times - \frac{1}{2}$
$m_{2} = {(\frac{d y}{d x})}_{(x_{1} y_{1})} = - \frac{1}{2} \times e^{- \frac{x_{1}}{2}} = \frac{- 1}{2} y_{1}$
$\tan θ = | \frac{m_{2} - m_{1}}{1 + m_{1} m_{2}} |$
$= | \frac{- \frac{y_{1}}{2} - \frac{2}{y_{1}}}{1 + (\frac{- y_{1}}{2} \times \frac{2}{y_{1}})} | = \infty$
$θ = \frac{π}{2} = 90^{\circ}$
Then ${cosec}^{2} (\frac{θ}{2}) = {cosc}^{2} (\frac{90^{\circ}}{2}) = {cosec}^{2} 45^{\circ} = 2$

AP EAMCET-2022-08.07.2022