QBManager

Application of Derivatives

85132 For the curve $y = 5 x - 2 x^{3}$ , if ' $x^{'}$ increases at the rate of 2 units/sec. the rate of change in the slope of the curve at $x = 3$ is /sec

1 72

2 27

3 -72

4 -27

Explanation:

(C): Given,
$\frac{d x}{d t} = 2$
$y = 5 x - 2 x^{3}$
$\frac{d y}{d x} = 5 - 6 x^{2} = g (x)$
$\frac{d}{d t} (g (x)) = \frac{d}{d x} g (x)$
$\frac{d}{d t} g (x) = 0 - 12 x \cdot \frac{d x}{d t}$
$\frac{d}{d t} g (x) = (- 12 x) (2) = - 24 x$
${(\frac{d g (x)}{d t})}_{x = 3} = - 24 \times 3 = - 72$

Shift-II]

Application of Derivatives

85133 The radius of a sphere initially at zero increases at the rate of $5 cm / \sec$ . Then its volume after $1 \sec$ is increasing at the rate 0

1

50 π

2

5 π

3

500 π

4 none of these

Explanation:

(C) : Given,
$\frac{dr}{dt} = 5 cm / \sec$
In 1 second, $r = 5 cm$
$∵$ Volume of sphere $(V) = \frac{4}{3} π r^{3}$
$∴ \frac{dV}{dt} = \frac{4}{3} π \times 3 r^{2} \frac{dr}{dt}$
$\frac{dV}{dt} = 4 π \times 5^{2} \times 5$
$\frac{dV}{dt} = 500 π {cm}^{3} / \sec$

AMU-2001

Application of Derivatives

85134 For a particle moving on a straight line it is observed that the distance ' $S$ ' at a time ' $t$ ' is given by $S = 6 t - \frac{t^{3}}{2}$ . The maximum velocity during the motion is

1 3

2 6

3 9

4 12

Explanation:

(B) : Given,
$S = 6 t - \frac{t^{3}}{2}$
Velocity $(v) = \frac{dS}{dt} = 6 - \frac{3}{2} t^{2} \Rightarrow \frac{dv}{dt} = 0 - 3 t$
For maximum velocity.
$\frac{dv}{dt} = 0 \Rightarrow - 3 t = 0$
Velocity will be maximum at $t = 0$
$v = 6 - \frac{3}{2} t^{2}, v = 6 - 0, v = 6$

AP EAMCET-2022-06.07.2022

Application of Derivatives

85135 The radius of a sphere increases at the rate of $0.04 cm / \sec$ . The rate of increase in the volume of that sphere with respect to its surface area, when its radius is $10 cm$ is

1

16 π

2 25

3 20

4 5

Explanation:

(D) : Given,
$\frac{dr}{dt} = 0.04 cm / s, r = 10 cm$
$∵$ Volume of sphere $(V) = \frac{4}{3} π r^{2}$
$\frac{dV}{dt} = \frac{4}{3} π \times 3 r^{2} \frac{dr}{dt} \Rightarrow \frac{dV}{dt} = 4 π \times 100 \times 0.04$
$\frac{dV}{dt} = 16 π {cm}^{3} / \sec$
We know that,
Surface area of sphere $(A) = 4 π r^{2}$
$\frac{dA}{dt} = 8 π r \frac{dr}{dt} \Rightarrow \frac{dA}{dt} = 8 π \times 10 \times 0.04$
$\frac{dA}{dt} = 3.2 π {cm}^{2} / \sec \Rightarrow \frac{\frac{dV}{dt}}{\frac{dA}{dt}} = \frac{16 π}{3.2 π} = 5$

AP EAMCET-2019-20.04.2019

Application of Derivatives

85132 For the curve $y = 5 x - 2 x^{3}$ , if ' $x^{'}$ increases at the rate of 2 units/sec. the rate of change in the slope of the curve at $x = 3$ is /sec

1 72

2 27

3 -72

4 -27

Explanation:

(C): Given,
$\frac{d x}{d t} = 2$
$y = 5 x - 2 x^{3}$
$\frac{d y}{d x} = 5 - 6 x^{2} = g (x)$
$\frac{d}{d t} (g (x)) = \frac{d}{d x} g (x)$
$\frac{d}{d t} g (x) = 0 - 12 x \cdot \frac{d x}{d t}$
$\frac{d}{d t} g (x) = (- 12 x) (2) = - 24 x$
${(\frac{d g (x)}{d t})}_{x = 3} = - 24 \times 3 = - 72$

Shift-II]

Application of Derivatives

85133 The radius of a sphere initially at zero increases at the rate of $5 cm / \sec$ . Then its volume after $1 \sec$ is increasing at the rate 0

1

50 π

2

5 π

3

500 π

4 none of these

Explanation:

(C) : Given,
$\frac{dr}{dt} = 5 cm / \sec$
In 1 second, $r = 5 cm$
$∵$ Volume of sphere $(V) = \frac{4}{3} π r^{3}$
$∴ \frac{dV}{dt} = \frac{4}{3} π \times 3 r^{2} \frac{dr}{dt}$
$\frac{dV}{dt} = 4 π \times 5^{2} \times 5$
$\frac{dV}{dt} = 500 π {cm}^{3} / \sec$

AMU-2001

Application of Derivatives

85134 For a particle moving on a straight line it is observed that the distance ' $S$ ' at a time ' $t$ ' is given by $S = 6 t - \frac{t^{3}}{2}$ . The maximum velocity during the motion is

1 3

2 6

3 9

4 12

Explanation:

(B) : Given,
$S = 6 t - \frac{t^{3}}{2}$
Velocity $(v) = \frac{dS}{dt} = 6 - \frac{3}{2} t^{2} \Rightarrow \frac{dv}{dt} = 0 - 3 t$
For maximum velocity.
$\frac{dv}{dt} = 0 \Rightarrow - 3 t = 0$
Velocity will be maximum at $t = 0$
$v = 6 - \frac{3}{2} t^{2}, v = 6 - 0, v = 6$

AP EAMCET-2022-06.07.2022

Application of Derivatives

85135 The radius of a sphere increases at the rate of $0.04 cm / \sec$ . The rate of increase in the volume of that sphere with respect to its surface area, when its radius is $10 cm$ is

1

16 π

2 25

3 20

4 5

Explanation:

(D) : Given,
$\frac{dr}{dt} = 0.04 cm / s, r = 10 cm$
$∵$ Volume of sphere $(V) = \frac{4}{3} π r^{2}$
$\frac{dV}{dt} = \frac{4}{3} π \times 3 r^{2} \frac{dr}{dt} \Rightarrow \frac{dV}{dt} = 4 π \times 100 \times 0.04$
$\frac{dV}{dt} = 16 π {cm}^{3} / \sec$
We know that,
Surface area of sphere $(A) = 4 π r^{2}$
$\frac{dA}{dt} = 8 π r \frac{dr}{dt} \Rightarrow \frac{dA}{dt} = 8 π \times 10 \times 0.04$
$\frac{dA}{dt} = 3.2 π {cm}^{2} / \sec \Rightarrow \frac{\frac{dV}{dt}}{\frac{dA}{dt}} = \frac{16 π}{3.2 π} = 5$

AP EAMCET-2019-20.04.2019

Application of Derivatives

85132 For the curve $y = 5 x - 2 x^{3}$ , if ' $x^{'}$ increases at the rate of 2 units/sec. the rate of change in the slope of the curve at $x = 3$ is /sec

1 72

2 27

3 -72

4 -27

Explanation:

(C): Given,
$\frac{d x}{d t} = 2$
$y = 5 x - 2 x^{3}$
$\frac{d y}{d x} = 5 - 6 x^{2} = g (x)$
$\frac{d}{d t} (g (x)) = \frac{d}{d x} g (x)$
$\frac{d}{d t} g (x) = 0 - 12 x \cdot \frac{d x}{d t}$
$\frac{d}{d t} g (x) = (- 12 x) (2) = - 24 x$
${(\frac{d g (x)}{d t})}_{x = 3} = - 24 \times 3 = - 72$

Shift-II]

Application of Derivatives

85133 The radius of a sphere initially at zero increases at the rate of $5 cm / \sec$ . Then its volume after $1 \sec$ is increasing at the rate 0

1

50 π

2

5 π

3

500 π

4 none of these

Explanation:

(C) : Given,
$\frac{dr}{dt} = 5 cm / \sec$
In 1 second, $r = 5 cm$
$∵$ Volume of sphere $(V) = \frac{4}{3} π r^{3}$
$∴ \frac{dV}{dt} = \frac{4}{3} π \times 3 r^{2} \frac{dr}{dt}$
$\frac{dV}{dt} = 4 π \times 5^{2} \times 5$
$\frac{dV}{dt} = 500 π {cm}^{3} / \sec$

AMU-2001

Application of Derivatives

85134 For a particle moving on a straight line it is observed that the distance ' $S$ ' at a time ' $t$ ' is given by $S = 6 t - \frac{t^{3}}{2}$ . The maximum velocity during the motion is

1 3

2 6

3 9

4 12

Explanation:

(B) : Given,
$S = 6 t - \frac{t^{3}}{2}$
Velocity $(v) = \frac{dS}{dt} = 6 - \frac{3}{2} t^{2} \Rightarrow \frac{dv}{dt} = 0 - 3 t$
For maximum velocity.
$\frac{dv}{dt} = 0 \Rightarrow - 3 t = 0$
Velocity will be maximum at $t = 0$
$v = 6 - \frac{3}{2} t^{2}, v = 6 - 0, v = 6$

AP EAMCET-2022-06.07.2022

Application of Derivatives

85135 The radius of a sphere increases at the rate of $0.04 cm / \sec$ . The rate of increase in the volume of that sphere with respect to its surface area, when its radius is $10 cm$ is

1

16 π

2 25

3 20

4 5

Explanation:

(D) : Given,
$\frac{dr}{dt} = 0.04 cm / s, r = 10 cm$
$∵$ Volume of sphere $(V) = \frac{4}{3} π r^{2}$
$\frac{dV}{dt} = \frac{4}{3} π \times 3 r^{2} \frac{dr}{dt} \Rightarrow \frac{dV}{dt} = 4 π \times 100 \times 0.04$
$\frac{dV}{dt} = 16 π {cm}^{3} / \sec$
We know that,
Surface area of sphere $(A) = 4 π r^{2}$
$\frac{dA}{dt} = 8 π r \frac{dr}{dt} \Rightarrow \frac{dA}{dt} = 8 π \times 10 \times 0.04$
$\frac{dA}{dt} = 3.2 π {cm}^{2} / \sec \Rightarrow \frac{\frac{dV}{dt}}{\frac{dA}{dt}} = \frac{16 π}{3.2 π} = 5$

AP EAMCET-2019-20.04.2019

Application of Derivatives

85132 For the curve $y = 5 x - 2 x^{3}$ , if ' $x^{'}$ increases at the rate of 2 units/sec. the rate of change in the slope of the curve at $x = 3$ is /sec

1 72

2 27

3 -72

4 -27

Explanation:

(C): Given,
$\frac{d x}{d t} = 2$
$y = 5 x - 2 x^{3}$
$\frac{d y}{d x} = 5 - 6 x^{2} = g (x)$
$\frac{d}{d t} (g (x)) = \frac{d}{d x} g (x)$
$\frac{d}{d t} g (x) = 0 - 12 x \cdot \frac{d x}{d t}$
$\frac{d}{d t} g (x) = (- 12 x) (2) = - 24 x$
${(\frac{d g (x)}{d t})}_{x = 3} = - 24 \times 3 = - 72$

Shift-II]

Application of Derivatives

85133 The radius of a sphere initially at zero increases at the rate of $5 cm / \sec$ . Then its volume after $1 \sec$ is increasing at the rate 0

1

50 π

2

5 π

3

500 π

4 none of these

Explanation:

(C) : Given,
$\frac{dr}{dt} = 5 cm / \sec$
In 1 second, $r = 5 cm$
$∵$ Volume of sphere $(V) = \frac{4}{3} π r^{3}$
$∴ \frac{dV}{dt} = \frac{4}{3} π \times 3 r^{2} \frac{dr}{dt}$
$\frac{dV}{dt} = 4 π \times 5^{2} \times 5$
$\frac{dV}{dt} = 500 π {cm}^{3} / \sec$

AMU-2001

Application of Derivatives

85134 For a particle moving on a straight line it is observed that the distance ' $S$ ' at a time ' $t$ ' is given by $S = 6 t - \frac{t^{3}}{2}$ . The maximum velocity during the motion is

1 3

2 6

3 9

4 12

Explanation:

(B) : Given,
$S = 6 t - \frac{t^{3}}{2}$
Velocity $(v) = \frac{dS}{dt} = 6 - \frac{3}{2} t^{2} \Rightarrow \frac{dv}{dt} = 0 - 3 t$
For maximum velocity.
$\frac{dv}{dt} = 0 \Rightarrow - 3 t = 0$
Velocity will be maximum at $t = 0$
$v = 6 - \frac{3}{2} t^{2}, v = 6 - 0, v = 6$

AP EAMCET-2022-06.07.2022

Application of Derivatives

85135 The radius of a sphere increases at the rate of $0.04 cm / \sec$ . The rate of increase in the volume of that sphere with respect to its surface area, when its radius is $10 cm$ is

1

16 π

2 25

3 20

4 5

Explanation:

(D) : Given,
$\frac{dr}{dt} = 0.04 cm / s, r = 10 cm$
$∵$ Volume of sphere $(V) = \frac{4}{3} π r^{2}$
$\frac{dV}{dt} = \frac{4}{3} π \times 3 r^{2} \frac{dr}{dt} \Rightarrow \frac{dV}{dt} = 4 π \times 100 \times 0.04$
$\frac{dV}{dt} = 16 π {cm}^{3} / \sec$
We know that,
Surface area of sphere $(A) = 4 π r^{2}$
$\frac{dA}{dt} = 8 π r \frac{dr}{dt} \Rightarrow \frac{dA}{dt} = 8 π \times 10 \times 0.04$
$\frac{dA}{dt} = 3.2 π {cm}^{2} / \sec \Rightarrow \frac{\frac{dV}{dt}}{\frac{dA}{dt}} = \frac{16 π}{3.2 π} = 5$

AP EAMCET-2019-20.04.2019

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