Rd Sharma XII Vol 1 2020 for Class 12 Science Maths Chapter 12 - Derivative As A Rate Measurer

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Rd Sharma XII Vol 1 2020 Solutions for Class 12 Science Maths Chapter 12 Derivative As A Rate Measurer are provided here with simple step-by-step explanations. These solutions for Derivative As A Rate Measurer are extremely popular among Class 12 Science students for Maths Derivative As A Rate Measurer Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma XII Vol 1 2020 Book of Class 12 Science Maths Chapter 12 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma XII Vol 1 2020 Solutions. All Rd Sharma XII Vol 1 2020 Solutions for class Class 12 Science Maths are prepared by experts and are 100% accurate.

Page No 12.19:

Question 1:

The side of a square sheet is increasing at the rate of 4 cm per minute. At what rate is the area increasing when the side is 8 cm long?

Answer:

$Given : A = x^{2} and \frac{d x}{d t} = 4 cm/min Let x be the side of the square and A be its area at any time t. Then, A = x^{2} \Rightarrow \frac{d A}{d t} = 2 x \frac{d x}{d t} \Rightarrow \frac{d A}{d t} = 2 \times 8 \times 4 [∵ x = 8 cm and \frac{d x}{d t} = 4 cm / \min] \Rightarrow \frac{d A}{d t} = 64 {cm}^{2} / \min$

Page No 12.19:

Question 2:

An edge of a variable cube is increasing at the rate of 3 cm per second. How fast is the volume of the cube increasing when the edge is 10 cm long?

Answer:

$Let x be the side and V be the volume of the cube at any time t . Then, V = x^{3} \Rightarrow \frac{d V}{d t} = 3 x^{2} \frac{d x}{d t} \Rightarrow \frac{d V}{d t} = 3 \times {(10)}^{2} \times 3 [∵ x =10 cm and \frac{d x}{d t} =3 cm/sec] \Rightarrow \frac{d V}{d t} = 900 {cm}^{3} / \sec$

Page No 12.19:

Question 3:

The side of a square is increasing at the rate of 0.2 cm/sec. Find the rate of increase of the perimeter of the square.

Answer:

$Let x be the side and P be the perimeter of the square at any time t . Then, P = 4 x \Rightarrow \frac{d P}{d t} = 4 \frac{d x}{d t} \Rightarrow \frac{d P}{d t} = 4 \times 0.2 [∵ \frac{d x}{d t} = 0.2 cm/sec] \Rightarrow \frac{d P}{d t} = 0.8 cm / \sec$

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Question 4:

The radius of a circle is increasing at the rate of 0.7 cm/sec. What is the rate of increase of its circumference?

Answer:

$Let r be the radius and C be the circumference of the circle at any time t. Then, C = 2 π r \Rightarrow \frac{d C}{d t} = 2 π \frac{d r}{d t} \Rightarrow \frac{d C}{d t} = 2 π \times 0.7 [∵ \frac{d r}{d t} =0.7 cm/sec] \Rightarrow \frac{d C}{d t} = 1.4 π cm / \sec$

Page No 12.19:

Question 5:

The radius of a spherical soap bubble is increasing at the rate of 0.2 cm/sec. Find the rate of increase of its surface area, when the radius is 7 cm.

Answer:

$Let r be the radius and S be the surface area of the spherical ball at any time t. Then, S = 4 π r^{2} \Rightarrow \frac{d S}{d t} = 8 π r \frac{d r}{d t} \Rightarrow \frac{d S}{d t} = 8 π \times 7 \times 0.2 [∵ r = 7 cm and \frac{d r}{d t} = 0.2 cm / \sec] \Rightarrow \frac{d S}{d t} = 11.2 π {cm}^{2} / \sec$

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Question 6:

A balloon which always remains spherical, is being inflated by pumping in 900 cubic centimetres of gas per second. Find the rate at which the radius of the balloon is increasing when the radius is 15 cm.

Answer:

$Let r be the radius and V be the volume of the spherical balloon at any time t . Then, V= \frac{4}{3} π r^{3} ⇒ \frac{d V}{d t} {=4πr}^{2} \frac{d r}{d t} ⇒ \frac{d r}{d t} = (\frac{1}{4 π r^{2}}) \frac{d V}{d t} ⇒ \frac{d r}{d t} = \frac{900}{4 π {(15)}^{2}} [∵ r = 15 cm and \frac{d V}{d t} = 900 {cm}^{3} / \sec] ⇒ \frac{d r}{d t} = \frac{900}{900 π} ⇒ \frac{d r}{d t} = \frac{1}{π} cm/sec$

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Question 7:

The radius of an air bubble is increasing at the rate of 0.5 cm/sec. At what rate is the volume of the bubble increasing when the radius is 1 cm?

Answer:

$Let r be the radius and V be the volume of the air bubble at any time t . Then, V = \frac{4}{3} π r^{3} ⇒ \frac{d V}{d t} = 4 π r^{2} \frac{d r}{d t} ⇒ \frac{d V}{d t} = 4 π {(1)}^{2} × 0.5 (∵ r = 1 cm and \frac{d r}{d t} = 0.5 cm / \sec) ⇒ \frac{d V}{d t} {=2π cm}^{3} /sec$

Page No 12.19:

Question 8:

A man 2 metres high walks at a uniform speed of 5 km/hr away from a lamp-post 6 metres high. Find the rate at which the length of his shadow increases.

Answer:

Let AB be the lamp post. Suppose at any time t, the man CD be at a distance of x km from the lamp post and y m be the length of his shadow CE.

$Since triangles A B E and C D E are similar, \frac{A B}{C D} = \frac{A E}{C E} \Rightarrow \frac{6}{2} = \frac{x + y}{y} \Rightarrow 3 y = x + y \Rightarrow x = 2 y \Rightarrow \frac{d x}{d t} = 2 \frac{d y}{d t} \Rightarrow \frac{d y}{d t} = \frac{1}{2} \frac{d x}{d t} \Rightarrow \frac{d y}{d t} = \frac{1}{2} (5) (\frac{d x}{d t} = 5) \Rightarrow \frac{d y}{d t} = \frac{5}{2} km / hr$

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Question 9:

A stone is dropped into a quiet lake and waves move in circles at a speed of 4 cm/sec. At the instant when the radius of the circular wave is 10 cm, how fast is the enclosed area increasing?

Answer:

$Let r be the radius and A be the area of the circle at any time t . Then, A =π r^{2} ⇒ \frac{d A}{d t} =2π r \frac{d r}{d t} ⇒ \frac{d A}{d t} =2π×4×10 [∵ r = 4 cm and \frac{d r}{d t} = 10 cm / \sec] ⇒ \frac{d A}{d t} {=80π cm}^{2} /sec$

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Question 10:

A man 160 cm tall, walks away from a source of light situated at the top of a pole 6 m high, at the rate of 1.1 m/sec. How fast is the length of his shadow increasing when he is 1 m away from the pole?

Answer:

Let AB be the lamp post. Suppose at any time t, the man CD is at a distance of x km from the lamp post and y m is the length of his shadow CE.

$Since triangles A B E and C D E are similar, \frac{A B}{C D} = \frac{A E}{C E} \Rightarrow \frac{6}{1.6} = \frac{x + y}{y} \Rightarrow \frac{x}{y} = \frac{6}{1.6} - 1 \Rightarrow \frac{x}{y} = \frac{4.4}{1.6} \Rightarrow y = \frac{16}{44} x \Rightarrow \frac{d y}{d t} = \frac{16}{44} (\frac{d x}{d t}) \Rightarrow \frac{d y}{d t} = \frac{16}{44} \times 1.1 (∵ \frac{d x}{d t} = 1.1) \Rightarrow \frac{d y}{d t} = 0.4 m / \sec$

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Question 11:

A man 180 cm tall walks at a rate of 2 m/sec. away, from a source of light that is 9 m above the ground. How fast is the length of his shadow increasing when he is 3 m away from the base of light?

Answer:

Let AB be the lamp post. Suppose at any time t, the man CD is at a distance x km from the lamp post and y m is the length of his shadow CE.

$Since triangles A B E and C D E are similar, \frac{A B}{C D} = \frac{A E}{C E}$

$\Rightarrow \frac{9}{1.8} = \frac{x + y}{y} \Rightarrow \frac{x}{y} = \frac{9}{1.8} - 1 \Rightarrow \frac{x}{y} = \frac{7.2}{1.8} \Rightarrow x = 4 y \Rightarrow \frac{d y}{d t} = \frac{1}{4} (\frac{d x}{d t}) \Rightarrow \frac{d y}{d t} = \frac{1}{4} \times 2 (∵ \frac{d x}{d t} = 2) \Rightarrow \frac{d y}{d t} = 0.5 m / \sec$

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Question 12:

A ladder 13 m long leans against a wall. The foot of the ladder is pulled along the ground away from the wall, at the rate of 1.5 m/sec. How fast is the angle θ between the ladder and the ground is changing when the foot of the ladder is 12 m away from the wall.

Answer:

Let the bottom of the ladder be at a distance of x m from the wall and its top be at a height of y m from the ground.

Then,

$\tan θ = \frac{y}{x} and x^{2} + y^{2} = {(13)}^{2} \Rightarrow x^{2} (1 + \tan^{2} θ) = 169 \Rightarrow \sec^{2} θ = \frac{169}{x^{2}} \Rightarrow 2 \sec^{2} θ \tan θ \frac{d θ}{d t} = 169 (\frac{- 2}{x^{3}}) \frac{d x}{d t} \Rightarrow \frac{d θ}{d t} = \frac{- 338 \times 1.5}{{(12)}^{3} 2 \sec^{2} θ \tan θ} . . . (1) When x = 12, y = \sqrt{169 - 144} = 5 m So, \sec θ = \frac{13}{12} and \tan θ = \frac{12}{5} From eq . (1), we get \frac{d θ}{d t} = \frac{- 338 \times 1.5}{{(12)}^{3} \times 2 \times {(\frac{13}{12})}^{2} \times \frac{5}{12}} = \frac{- 338 \times 1.5}{10 \times 169} = - 0.3 rad / \sec$

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Question 13:

A particle moves along the curve y = x² + 2x. At what point(s) on the curve are the x and y coordinates of the particle changing at the same rate?

Answer:

$Here, y = x^{2} + 2 x \Rightarrow \frac{d y}{d t} = (2 x + 2) \frac{d x}{d t} \Rightarrow 2 x + 2 = 1 [∵ \frac{d y}{d t} = \frac{d x}{d t}] \Rightarrow 2 x = - 1 \Rightarrow x = \frac{- 1}{2} Substituting x = \frac{- 1}{2} in y = x^{2} +2 x, we get y = \frac{- 3}{4} Hence, the coordinates of the point are (\frac{- 1}{2}, \frac{- 3}{4}) .$

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Question 14:

If y = 7x − x³ and x increases at the rate of 4 units per second, how fast is the slope of the curve changing when x = 2?

Answer:

$Here, y = 7 x - x^{3} \Rightarrow \frac{d y}{d x} = 7 x - x^{3} Let s be the slope . Then, s = 7 - 3 x^{2} \Rightarrow \frac{d s}{d t} = - 6 x \frac{d x}{d t} \Rightarrow \frac{d s}{d t} = - 6 (4) (2) [∵ x = 2 and \frac{d x}{d t} = 4 units / \sec] \Rightarrow \frac{d s}{d t} = - 48$

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Question 15:

A particle moves along the curve y = x³. Find the points on the curve at which the y-coordinate changes three times more rapidly than the x-coordinate.

Answer:

$According to the question, \frac{d y}{d t} = 3 \frac{d x}{d t} Now, y = x^{3} \Rightarrow \frac{d y}{d t} = 3 x^{2} \frac{d x}{d t} \Rightarrow 3 \frac{d x}{d t} = 3 x^{2} \frac{d x}{d t} \Rightarrow x^{2} = 1 \Rightarrow x = \pm 1 Substituting x =±1 in y = x^{3}, we get y = \pm 1 So the points are (1, 1) and (- 1, - 1) .$

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Question 16:

Find an angle θ
(i) which increases twice as fast as its cosine.
(ii) whose rate of increase twice is twice the rate of decrease of its cosine.

Answer:

$(i) Let x = \cos θ Differentiating both sides with respect to t, we get \frac{d x}{d t} = \frac{d (\cos θ)}{d t} = - \sin θ \frac{d θ}{d t} But it is given that \frac{d θ}{d t} = 2 \frac{d x}{d t} \Rightarrow \frac{d x}{d t} = - \sin θ (2 \frac{d x}{d t}) \Rightarrow \sin θ = - \frac{1}{2} \Rightarrow θ = π + \frac{π}{6} = \frac{7 π}{6} Hence, θ = \frac{7 π}{6} .$

$(ii) Let x = \cos θ Differentiating both sides with respect to t, we get \frac{d x}{d t} = \frac{d (\cos θ)}{d t} = - \sin θ \frac{d θ}{d t} But it is given that \frac{d θ}{d t} = - 2 \frac{d x}{d t} \Rightarrow \frac{d x}{d t} = - \sin θ (- 2 \frac{d x}{d t}) \Rightarrow \sin θ = \frac{1}{2} \Rightarrow θ = \frac{π}{6} Hence, θ = \frac{π}{6} .$

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Question 17:

The top of a ladder 6 metres long is resting against a vertical wall on a level pavement, when the ladder begins to slide outwards. At the moment when the foot of the ladder is 4 metres from the wall, it is sliding away from the wall at the rate of 0.5 m/sec. How fast is the top-sliding downwards at this instance?
How far is the foot from the wall when it and the top are moving at the same rate?

Answer:

Let the bottom of the ladder be at a distance of x m from the wall and its top be at a height of y m from the ground.

Here,

$x^{2} + y^{2} = 36 \Rightarrow 2 x \frac{d x}{d t} = - 2 y \frac{d y}{d t} . . . (1) When x = 4, y = \sqrt{36 - 16} = 2 \sqrt{5} \Rightarrow 2 \times 4 \times 0.5 = - 2 \times 2 \sqrt{5} \frac{d y}{d t} [∵ \frac{d x}{d t} = 0.5 m / \sec] \Rightarrow \frac{d y}{d t} = \frac{- 1}{\sqrt{5}} m / \sec From eq. (1), we get 2 x \frac{d x}{d t} = - 2 y \frac{d y}{d t} [∵ \frac{d x}{d t} = \frac{d y}{d t}] \Rightarrow x = - y Substituting x=- y in x^{2} + y^{2} =36, we get x^{2} + x^{2} = 36 \Rightarrow x^{2} = 18 \Rightarrow x = 3 \sqrt{2} m$

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Question 18:

A balloon in the form of a right circular cone surmounted by a hemisphere, having a diametre equal to the height of the cone, is being inflated. How fast is its volume changing with respect to its total height h, when h = 9 cm.

Answer:

$Let r be the radius of the hemisphere, h be the height and V be the volume of the cone .$

$Then, H = h + r \Rightarrow H = 3 r [∵ h = 2 r] \Rightarrow \frac{d H}{d t} = 3 \frac{d r}{d t} When H = 9 cm, r = 3 cm Volume = \frac{1}{3} {πr}^{2} h + \frac{2}{3} π r^{3} Substituting h =2 r ⇒ V = \frac{2}{3} π r^{3} + \frac{2}{3} π r^{3} ⇒ V = \frac{4}{3} π r^{3} ⇒ \frac{d V}{d t} =4π r^{2} \frac{d r}{d t} ⇒ \frac{d V}{d t} = \frac{4π r^{2}}{3} \frac{d H}{d t} ⇒ \frac{d V}{d H} = \frac{4π {(3)}^{2}}{3} ⇒ \frac{d V}{d H} {=12π cm}^{3} /sec$

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Question 19:

Water is running into an inverted cone at the rate of π cubic metres per minute. The height of the cone is 10 metres, and the radius of its base is 5 m. How fast the water level is rising when the water stands 7.5 m below the base.

Answer:

$Let r be the radius, h be the height and V be the volume of the cone at any time t .$

$Then, V = \frac{1}{3} π r^{2} h \Rightarrow \frac{d V}{d t} = \frac{1}{3} π r^{2} \frac{d h}{d t} + \frac{2}{3} π rh \frac{d r}{d t} Now, \frac{h}{r} = \frac{10}{5} or r = \frac{h}{2} and \frac{d h}{d t} =2 \frac{d r}{d t} ⇒ \frac{d V}{d t} = \frac{1}{3} π {(\frac{h}{2})}^{2} \frac{d h}{d t} + \frac{2}{3} π (\frac{h}{2}) h \frac{1}{2} \frac{d h}{d t} ⇒ \frac{d V}{d t} = \frac{π}{3} [\frac{h^{2}}{4} \frac{d h}{d t} + \frac{h^{2}}{2} \frac{d h}{d t}] ⇒ \frac{d V}{d t} = \frac{π}{3} × \frac{3 h^{2} d h}{4 d t} ⇒ \frac{d V}{d t} = \frac{π h^{2}}{4} \frac{d h}{d t} \Rightarrow \frac{π h^{2}}{4} \frac{d h}{d t} = π \Rightarrow \frac{d h}{d t} = \frac{4}{h^{2}} \Rightarrow \frac{d h}{d t} = \frac{4}{{(2.5)}^{2}} \Rightarrow \frac{d h}{d t} = 0.64 m / \min$

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Question 20:

A man 2 metres high walks at a uniform speed of 6 km/h away from a lamp-post 6 metres high. Find the rate at which the length of his shadow increases.

Answer:

Let AB be the lamp post. Let at any time t, the man CD be at a distance of x km from the lamp post and y m be the length of his shadow CE.

$Since triangles A B E and C D E are similar, \frac{A B}{C D} = \frac{A E}{C E}$

$\Rightarrow \frac{6}{2} = \frac{x + y}{y} \Rightarrow \frac{x}{y} = \frac{6}{2} - 1 = 2 \Rightarrow \frac{d y}{d t} = \frac{1}{2} \frac{d x}{d t} \Rightarrow \frac{d y}{d t} = \frac{1}{2} (6) \Rightarrow \frac{d y}{d t} = 3 km / hr$

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Question 21:

The surface area of a spherical bubble is increasing at the rate of 2 cm²/s. When the radius of the bubble is 6 cm, at what rate is the volume of the bubble increasing?

Answer:

$Let r be the radius, S be the surface area and V be the volume of the sphere at any time t . Then, S = 4 π r^{2} \Rightarrow \frac{d S}{d t} = 8 π r \frac{d r}{d t} \Rightarrow \frac{d r}{d t} = \frac{1}{8 π r} \frac{d S}{d t} \Rightarrow \frac{d r}{d t} = \frac{2}{8 π \times 6} \Rightarrow \frac{d r}{d t} = \frac{1}{24 π} cm / \sec Now, Volume of sphere = \frac{4}{3} π r^{3} \Rightarrow \frac{d V}{d t} = 4 π r^{2} \frac{d r}{d t} \Rightarrow \frac{d V}{d t} = \frac{4 π {(6)}^{2}}{24 π} \Rightarrow \frac{d V}{d t} = 6 {cm}^{3} / \sec$

Page No 12.20:

Question 22:

The radius of a cylinder is increasing at the rate 2 cm/sec. and its altitude is decreasing at the rate of 3 cm/sec. Find the rate of change of volume when radius is 3 cm and altitude 5 cm.

Answer:

$Let r be the radius, h be the height and V be the volume of the cylinder at any time t . Then, V = π r^{2} h \Rightarrow \frac{d V}{d t} = 2 π r h \frac{d r}{d t} + π r^{2} \frac{d h}{d t} \Rightarrow \frac{d V}{d t} = π r (2 h \frac{d r}{d t} + r \frac{d h}{d t}) \Rightarrow \frac{d V}{d t} = π \times 3 (2 \times 5 \times 2 + 3 \times - 3) \Rightarrow \frac{d V}{d t} = 3 π (20 - 9) \Rightarrow \frac{d V}{d t} = 33 π {cm}^{3} / \sec$

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Question 23:

The volume of metal in a hollow sphere is constant. If the inner radius is increasing at the rate of 1 cm/sec, find the rate of increase of the outer radius when the radii are 4 cm and 8 cm respectively.

Answer:

$Let r_{1} be the inner radius and r_{2} be the outer radius and V be the volume of the hollow sphere at any time t . Then, V = \frac{4}{3} π ({r_{1}}^{3} - {r_{2}}^{3}) \Rightarrow \frac{d V}{d t} =4π ({r_{1}}^{2} \frac{d r_{1}}{d t} - {r_{2}}^{2} \frac{d r_{2}}{d t}) ⇒ {r_{1}}^{2} \frac{d r_{1}}{d t} = {r_{2}}^{2} \frac{d r_{2}}{d t} [∵ \frac{d V}{d t} = 0] ⇒ {(4)}^{2} ×1= {(8)}^{2} \frac{d r_{2}}{d t} ⇒ \frac{d r_{2}}{d t} = \frac{1}{4} cm/sec$

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Question 24:

Sand is being poured onto a conical pile at the constant rate of 50 cm³/ minute such that the height of the cone is always one half of the radius of its base. How fast is the height of the pile increasing when the sand is 5 cm deep.

Answer:

$Let r be the radius, h be the height and V be the volume of the conical pile at any time t . Then, V = \frac{1}{3} π r^{2} h \Rightarrow V = \frac{1}{3} π {(2 h)}^{2} h [∵ h = \frac{r}{2}] \Rightarrow V = \frac{4}{3} {πh}^{3} \Rightarrow \frac{d V}{d t} = 4 π h^{2} \frac{d h}{d t} \Rightarrow 50 = 4 π h^{2} \frac{d h}{d t} \Rightarrow \frac{d h}{d t} = \frac{50}{4 π {(5)}^{2}} \Rightarrow \frac{d h}{d t} = \frac{1}{2 π} cm / \min$

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Question 25:

A kite is 120 m high and 130 m of string is out. If the kite is moving away horizontally at the rate of 52 m/sec, find the rate at which the string is being paid out.

Answer:

$In the right triangle A B C,$

$Here, A B^{2} + B C^{2} = A C^{2} \Rightarrow x^{2} + {(120)}^{2} = y^{2} \Rightarrow 2 x \frac{d x}{d t} = 2 y \frac{d y}{d t} \Rightarrow \frac{d y}{d t} = \frac{x}{y} \frac{d x}{d t} \Rightarrow \frac{d y}{d t} = \frac{50}{130} \times 52 [∵ x = \sqrt{{(130)}^{2} - {(120)}^{2}} = 50] \Rightarrow \frac{d y}{d t} = 20 m / \sec$

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Question 26:

A particle moves along the curve y = (2/3)x³ + 1. Find the points on the curve at which the y-coordinate is changing twice as fast as the x-coordinate.

Answer:

$Here, y = \frac{2}{3} x^{3} + 1 \Rightarrow \frac{d y}{d t} = 2 x^{2} \frac{d x}{d t} \Rightarrow 2 \frac{d x}{d t} = 2 x^{2} \frac{d x}{d t} [∵ \frac{d y}{d t} = 2 \frac{d x}{d t}] \Rightarrow x = \pm 1 Substituting the value of x =1 and x =-1 in y = \frac{2}{3} x^{3} + 1, we get \Rightarrow y = \frac{5}{3} and y = \frac{1}{3} So, the points are (1, \frac{5}{3}) and (- 1, \frac{1}{3}) .$

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Question 27:

Find the point on the curve y² = 8x for which the abscissa and ordinate change at the same rate.

Answer:

$Here, y^{2} = 8 x . . . (1) ⇒2 y \frac{d y}{d t} =8 \frac{d x}{d t} ⇒2 y =8 [∵ \frac{d y}{d t} = \frac{d x}{d t}] ⇒ y =4 ⇒ x = \frac{y^{2}}{8} [From eq. (1)] ⇒ x = \frac{16}{8} =2 So, the point is (2, 4) .$

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Question 28:

The volume of a cube is increasing at the rate of 9 cm³/sec. How fast is the surface area increasing when the length of an edge is 10 cm?

Answer:

$Let x be the side and V be the volume of the cube at any time t . Then, V = x^{3} ⇒ \frac{d V}{d t} =3 x^{2} \frac{d x}{d t} ⇒9=3 {(10)}^{2} \frac{d x}{d t} [∵ x = 10 cm and \frac{d V}{d t} = {cm}^{3} / \sec] ⇒ \frac{d x}{d t} =0.03 cm/sec Let S be the surface area of the cube at any time t . Then, S = 6 x^{2} ⇒ \frac{d S}{d t} =12 x \frac{d x}{d t} ⇒ \frac{d S}{d t} =12×10×0.03 [∵ x = 10 cm and \frac{d x}{d t} = 0.03 cm/sec] ⇒ \frac{d S}{d t} {=3.6 cm}^{2} /sec$

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Question 29:

The volume of a spherical balloon is increasing at the rate of 25 cm³/sec. Find the rate of change of its surface area at the instant when radius is 5 cm.

Answer:

$Let r be the radius and V be the volume of the sphere at any time t . Then, V = \frac{4}{3} π r^{3} \Rightarrow \frac{d V}{d t} = 4 π r^{2} \frac{d r}{d t} \Rightarrow \frac{d r}{d t} = \frac{1}{4 π r^{2}} \frac{d V}{d t} \Rightarrow \frac{d r}{d t} = \frac{25}{4 π {(5)}^{2}} [∵ r = 5 cm and \frac{d V}{d t} = 25 {cm}^{3} / \sec] \Rightarrow \frac{d r}{d t} = \frac{1}{4 π} cm / \sec Now, let S be the surface area of the sphere at any time t. Then, S = 4π r^{2} ⇒ \frac{d S}{d t} =8πr \frac{d r}{d t} ⇒ \frac{d S}{d t} =8π (5) × \frac{1}{4 π} [∵ r = 5 cm and \frac{d r}{d t} = \frac{1}{4 π} cm / \sec] ⇒ \frac{d S}{d t} {=10 cm}^{2} /sec$

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Question 30:

The length x of a rectangle is decreasing at the rate of 5 cm/minute and the width y is increasing at the rate of 4 cm/minute. When x = 8 cm and y = 6 cm, find the rates of change of (i) the perimeter (ii) the area of the rectangle.

Answer:

$(i) Let P be the perimeter of the rectangle at any time t . Then, P =2 (x + y) ⇒ \frac{d P}{d t} =2 (\frac{d x}{d t} + \frac{d y}{d t}) ⇒ \frac{d P}{d t} =2 (- 5 + 4) [∵ \frac{d x}{d t} = - 5 cm / \min and \frac{d y}{d t} = 4 cm / \min] ⇒ \frac{d P}{d t} =-2 cm/min (i i) Let A be the a rea of the rectangle at any time t. Then, A = x y \Rightarrow \frac{d A}{d t} = x \frac{d y}{d t} + y \frac{d x}{d t} \Rightarrow \frac{d A}{d t} = 8 (4) + 6 (- 5) [∵ x = 8 cm, y = 6 cm \frac{d x}{d t} = - 5 cm / \min and \frac{d y}{d t} = 4 cm / \min] \Rightarrow \frac{d A}{d t} = 32 - 30 \Rightarrow \frac{d A}{d t} = 2 {cm}^{2} / \min$

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Question 31:

A circular disc of radius 3 cm is being heated. Due to expansion, its radius increases at the rate of 0.05 cm/sec. Find the rate at which its area is increasing when radius is 3.2 cm.

Answer:

$Let r be the radius and A be the area of the circular disc at any time t. Then, A =π r^{2} ⇒ \frac{d A}{d t} =2π r \frac{d r}{d t} ⇒ \frac{d A}{d t} =2π×3.2×0.05 [∵ r = 3.2 cm and \frac{d r}{d t} = 0.05 cm / \sec] ⇒ \frac{d A}{d t} {=0.32π cm}^{2} /sec$

Page No 12.24:

Question 1:

If $V = \frac{4}{3} π r^{3}$ , at what rate in cubic units is V increasing when r = 10 and $\frac{d r}{d t} = 0.01$ ?
(a) π
(b) 4π
(c) 40π
(d) 4π/3

Answer:

(b) 4π
$Given: V = \frac{4}{3} π r^{3}, r = 10 and \frac{d r}{d t} = 0.01 \Rightarrow \frac{d V}{d t} = 4 π r^{2} \frac{d r}{d t} \Rightarrow \frac{d V}{d t} = 4 π {(10)}^{2} \times 0.01 \Rightarrow \frac{d V}{d t} = 4 π$

Page No 12.24:

Question 2:

Side of an equilateral triangle expands at the rate of 2 cm/sec. The rate of increase of its area when each side is 10 cm is
(a) $10 \sqrt{2} {cm}^{2} / \sec$
(b) $10 \sqrt{3} {cm}^{2} / \sec$
(c) 10 cm²/sec
(d) 5 cm²/sec

Answer:

(b) $10 \sqrt{3} {cm}^{2} / \sec$

$Let x be the side and A be the area of the equilateral triangle at any time t. Then, A = \frac{\sqrt{3}}{4} x^{2} \Rightarrow \frac{d A}{d t} = 2 \times \frac{\sqrt{3}}{4} x^{} \frac{d x}{d t} ⇒ \frac{d A}{d t} = \frac{\sqrt{3}}{2} ×2×10 ⇒ \frac{d A}{d t} =10 \sqrt{3} {cm}^{2} /sec$

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Question 3:

The radius of a sphere is changing at the rate of 0.1 cm/sec. The rate of change of its surface area when the radius is 200 cm is
(a) 8π cm²/sec
(b) 12π cm²/sec
(c) 160π cm²/sec
(d) 200 cm²/sec

Answer:

(c) 160π cm²/sec

$Let r be the radius and S be the surface area of the sphere at any time t. Then, S = 4π r^{2} ⇒ \frac{d S}{d t} =8π r \frac{d r}{d t} ⇒ \frac{d S}{d t} =8π (200) (0.1) ⇒ \frac{d S}{d t} {=160π cm}^{2} /sec$

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Question 4:

A cone whose height is always equal to its diameter is increasing in volume at the rate of 40 cm³/sec. At what rate is the radius increasing when its circular base area is 1 m²?
(a) 1 mm/sec
(b) 0.001 cm/sec
(c) 2 mm/sec
(d) 0.002 cm/sec

Answer:

(d) 0.002 cm/sec

$Let r be the radius, h be the height and V be the volume of the cone at any time t . Then, V = \frac{1}{3} π r^{2} h \Rightarrow V = \frac{2}{3} π r^{3} [∵ h = 2 r] \Rightarrow \frac{d V}{d t} = 2 \times 10^{4} \frac{d r}{d t} [∵ π r^{2} = 1 m^{2} or 10^{4} {cm}^{2}] \Rightarrow \frac{d r}{d t} = \frac{1}{2 \times 10^{4}} \frac{d V}{d t} \Rightarrow \frac{d r}{d t} = \frac{40}{2 \times 10^{4}} \Rightarrow \frac{d r}{d t} = 0.002 cm / \sec$

Page No 12.24:

Question 5:

A cylindrical vessel of radius 0.5 m is filled with oil at the rate of 0.25 π m³/minute. The rate at which the surface of the oil is rising, is
(a) 1 m/minute
(b) 2 m/minute
(c) 5 m/minute
(d) 1.25 m/minute

Answer:

(a) 1 m/minute

$Let r be the radius, h be the height and V be the volume of the cylindrical vessel at any time t. Then, V = π r^{2} h \Rightarrow \frac{d V}{d t} = π r^{2} \frac{d h}{d t} \Rightarrow \frac{d h}{d t} = \frac{1}{π r^{2}} \frac{d V}{d t} \Rightarrow \frac{d h}{d t} = \frac{0.25 π}{π {(0.5)}^{2}} \Rightarrow \frac{d h}{d t} = \frac{0.25}{0.25} \Rightarrow \frac{d h}{d t} = 1 m / \min$

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Question 6:

The distance moved by the particle in time t is given by x = t³− 12t² + 6t + 8. At the instant when its acceleration is zero, the velocity is
(a) 42
(b) −42
(c) 48
(d) −48

Answer:

(b) −42

$x = t^{3} - 12 t^{2} + 6 t + 8 \Rightarrow \frac{d x}{d t} = 3 t^{2} - 24 t + 6 \Rightarrow \frac{d^{2} x}{d t^{2}} = 6 t - 24 \Rightarrow 6 t - 24 = 0 [∵ acceleration is zero] \Rightarrow t = 4 So, V elocity at t =4 \Rightarrow \frac{d x}{d t} = 3 {(4)}^{2} - 24 \times 4 + 6 \Rightarrow \frac{d x}{d t} = 48 - 96 + 6 \Rightarrow \frac{d x}{d t} = - 42$

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Question 7:

The altitude of a cone is 20 cm and its semi-vertical angle is 30°. If the semi-vertical angle is increasing at the rate of 2° per second, then the radius of the base is increasing at the rate of
(a) 30 cm/sec
(b) $\frac{160}{3} cm / \sec$
(c) 10 cm/sec
(d) 160 cm/sec

Answer:

(b) $\frac{160}{3} cm / \sec$

$Let r be the radius, h be the height and α be the semi - vertical angle of the cone .$

$Then, \tan α = \frac{r}{h} \Rightarrow s e c^{2} α (\frac{d α}{d t}) = \frac{d r}{h d t} \Rightarrow \frac{d r}{d t} = h \times s e c^{2} α (\frac{d α}{d t}) \Rightarrow \frac{d r}{d t} = 20 \times s e c^{2} 30 \times 2 [∵ h = 20 cm, α = 30 ° and \frac{d α}{d t} = 2 ° per second] \Rightarrow \frac{d r}{d t} = 40 \times {(\frac{2}{\sqrt{3}})}^{2} \Rightarrow \frac{d r}{d t} = \frac{160}{3} cm / \sec$

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Question 8:

For what values of x is the rate of increase of x³− 5x² + 5x + 8 is twice the rate of increase of x?
(a) $- 3, - \frac{1}{3}$

(b) $- 3, \frac{1}{3}$

(c) $3, - \frac{1}{3}$

(d) $3, \frac{1}{3}$

Answer:

(d) $3, \frac{1}{3}$

$Let y = x^{3} - 5 x^{2} + 5 x + 8 \Rightarrow \frac{d y}{d t} = (3 x^{2} - 10 x + 5) \frac{d x}{d t} According to the question, \Rightarrow 2 \frac{d x}{d t} = (3 x^{2} - 10 x + 5) \frac{d x}{d t} \Rightarrow 3 x^{2} - 10 x + 5 = 2 \Rightarrow 3 x^{2} - 10 x + 3 = 0 \Rightarrow 3 x^{2} - 9 x - x + 3 = 0 \Rightarrow 3 x (x - 3) - 1 (x - 3) = 0 \Rightarrow (x - 3) = 0 or (3 x - 1) = 0 \Rightarrow x = 3 or x = \frac{1}{3}$

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Question 9:

The coordinates of the point on the ellipse 16x² + 9y² = 400 where the ordinate decreases at the same rate at which the abscissa increases, are
(a) (3, 16/3)
(b) (−3, 16/3)
(c) (3, −16/3)
(d) (3, −3)

Answer:

(a) (3, 16/3)

$According to the question, \frac{d y}{d t} = \frac{- d x}{d t} 16 x^{2} + 9 y^{2} = 400 \Rightarrow 32 x \frac{d x}{d t} + 18 y \frac{d y}{d t} = 0 \Rightarrow 32 x \frac{d x}{d t} = - 18 y \frac{d y}{d t} \Rightarrow 32 x = 18 y \Rightarrow x = \frac{9 y}{16} . . . (1) Now, 16 {(\frac{9 y}{16})}^{2} + 9 y^{2} = 400 \Rightarrow \frac{81 y^{2}}{16} + 9 y^{2} = 400 \Rightarrow 81 y^{2} + 144 y^{2} = 6400 \Rightarrow 225 y^{2} = 6400 \Rightarrow y^{2} = \frac{6400}{225} \Rightarrow y = \sqrt{\frac{6400}{225}} \Rightarrow y = \frac{16}{3} or - \frac{16}{3} So, x = \frac{9}{16} \times \frac{16}{3} [Using (1)] or x = - \frac{9}{16} \times \frac{16}{3} \Rightarrow x = 3 or - 3 So, the required point is (3, \frac{16}{3}) .$

Page No 12.24:

Question 10:

The radius of the base of a cone is increasing at the rate of 3 cm/minute and the altitude is decreasing at the rate of 4 cm/minute. The rate of change of lateral surface when the radius = 7 cm and altitude 24 cm is
(a) 54π cm²/min
(b) 7π cm²/min
(c) 27 cm²/min
(d) none of these

Answer:

$Let r be the radius, h be the height and S be the lateral surface area of the cone at any time t .$

$Given : \frac{d r}{d t} = 3 cm / \min and \frac{d h}{d t} = - 4 cm / \min Here, l^{2} = h^{2} + r^{2} \Rightarrow l = \sqrt{{(24)}^{2} + {(7)}^{2}} \Rightarrow l = \sqrt{625} \Rightarrow l = 25 S =πrl \Rightarrow S^{2} = {(πrl)}^{2} \Rightarrow S^{2} = π^{2} r^{2} (h^{2} + r^{2}) \Rightarrow S^{2} = π^{2} r^{4} + π^{2} h^{2} r^{2} \Rightarrow 2 S \frac{d S}{d t} = 4 π^{2} r^{3} \frac{d r}{d t} + 2 π^{2} r^{2} h \frac{d h}{d t} + 2 π^{2} h^{2} r \frac{d r}{d t} \Rightarrow 2 π r l \frac{d S}{d t} = 2 π^{2} r h [\frac{2 r^{2}}{h} \frac{d r}{d t} + r \frac{d h}{d t} + h \frac{d r}{d t}] \Rightarrow 25 \frac{d S}{d t} = 24 π [\frac{2 {(7)}^{2}}{24} \times 3 - 7 \times 4 + 24 \times 3] [Given : r = 7, h = 24] \Rightarrow 25 \frac{d S}{d t} = 24 π [\frac{49}{4} - 28 + 72] \Rightarrow 25 \frac{d S}{d t} = 24 π [\frac{49 + 288 - 112}{4}] \Rightarrow \frac{d S}{d t} = 24 π [\frac{225}{100}] \Rightarrow \frac{d S}{d t} = 24 π (2.25) \Rightarrow \frac{d S}{d t} = 54 π {cm}^{2} / \sec$

Page No 12.24:

Question 11:

The radius of a sphere is increasing at the rate of 0.2 cm/sec. The rate at which the volume of the sphere increase when radius is 15 cm, is
(a) 12π cm³/sec
(b) 180π cm³/sec
(c) 225π cm³/sec
(d) 3π cm³/sec

Answer:

(b) 180π cm³/sec

$Let r be the radius and V be the volume of the sphere at any time t. Then, V= \frac{4}{3} π r^{3} ⇒ \frac{d V}{d t} =4π r^{2} \frac{d r}{d t} ⇒ \frac{d V}{d t} = 4π {(15)}^{2} \times 0.2 ⇒ \frac{d V}{d t} {=180π cm}^{3} /sec$

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Question 12:

The volume of a sphere is increasing at 3 cm³/sec. The rate at which the radius increases when radius is 2 cm, is
(a) $\frac{3}{32 π} cm / \sec$

(b) $\frac{3}{16 π} cm / \sec$

(c) $\frac{3}{48 π} cm / \sec$

(d) $\frac{1}{24 π} cm / \sec$

Answer:

(b) $\frac{3}{16 π} cm / \sec$

$Let r be the radius and V be the volume of the sphere at any time t. Then, V= \frac{4}{3} {πr}^{3} ⇒ \frac{d V}{d t} =4π r^{2} \frac{d r}{d t} ⇒ \frac{d r}{d t} = \frac{1}{4π r^{2}} \frac{d V}{d t} ⇒ \frac{d r}{d t} = \frac{3}{4π {(2)}^{2}} ⇒ \frac{d r}{d t} = \frac{3}{16 π} cm/sec$

Page No 12.24:

Question 13:

The distance moved by a particle travelling in straight line in t seconds is given by s = 45t + 11t²− t³. The time taken by the particle to come to rest is
(a) 9 sec
(b) 5/3 sec
(c) 3/5 sec
(d) 2 sec

Answer:

(a) 9 sec

$s = 45 t + 11 t^{2} - t^{3} \Rightarrow \frac{d s}{d t} = 45 + 22 t - 3 t^{2} According to the question, 3 t^{2} - 22 t - 45 = 0 \Rightarrow 3 t^{2} - 27 t + 5 t - 45 = 0 \Rightarrow 3 t (t - 9) + 5 (t - 9) = 0 \Rightarrow (t - 9) (3 t + 5) = 0 \Rightarrow (t - 9) = 0 or (3 t + 5) = 0 As time can't be negative, t = 9 \sec$

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Question 14:

The volume of a sphere is increasing at the rate of 4π cm³/sec. The rate of increase of the radius when the volume is 288 π cm³, is
(a) 1/4
(b) 1/12
(c) 1/36
(d) 1/9

Answer:

(c) 1/36

$Let r be the radius and V be the volume of the sphere at any time t . Then, V = \frac{4}{3} π r^{3} \Rightarrow \frac{4}{3} π r^{3} =288π \Rightarrow r^{3} = \frac{288 \times 3}{4} \Rightarrow r^{3} = 216 \Rightarrow r = 6 \Rightarrow \frac{d V}{d t} = 4 π r^{2} \frac{d r}{d t} \Rightarrow \frac{d V}{d t} = 4 π {(6)}^{2} \frac{d r}{d t} \Rightarrow 4 π = 144 π \frac{d r}{d t} \Rightarrow \frac{d r}{d t} = \frac{1}{36}$

Page No 12.25:

Question 15:

If the rate of change of volume of a sphere is equal to the rate of change of its radius, then its radius is equal to

(a) 1 unit

(b) $\sqrt{2 π} units$

(c) $\frac{1}{\sqrt{2 π}} unit$

(d) $\frac{1}{2 \sqrt{π}} unit$

Answer:

(d) $\frac{1}{2 \sqrt{π}} unit$

$Let r be the radius and V be the volume of the sphere at any time t . Then, V = \frac{4}{3} π r^{3} ⇒ \frac{d V}{d t} = \frac{4}{3} (3 π r^{2}) \frac{d r}{d t} ⇒ \frac{d V}{d t} =4 π r^{2} \frac{d r}{d t} \Rightarrow 4 π r^{2} = 1 [∵ \frac{d V}{d t} = \frac{d r}{d t}] \Rightarrow r^{2} = \frac{1}{4 π} \Rightarrow r = \sqrt{\frac{1}{4 π}} \Rightarrow r = \frac{1}{2 \sqrt{π}} unit$

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Question 16:

If the rate of change of area of a circle is equal to the rate of change of its diameter, then its radius is equal to
(a) $\frac{2}{π} unit$

(b) $\frac{1}{π} unit$

(c) $\frac{π}{2} units$

(d) π units

Answer:

(b) $\frac{1}{π} unit$

$Let r be the radius and A be the area of the circle at any time t . Then, A = π r^{2} \Rightarrow A = \frac{π D^{2}}{4} [∵ r = \frac{D}{2}] \Rightarrow \frac{d A}{d t} = \frac{π D}{2} \frac{d D}{d t} \Rightarrow \frac{d D}{d t} = \frac{π D}{2} \frac{d D}{d t} [∵ \frac{d A}{d t} = \frac{d D}{d t}] \Rightarrow \frac{D}{2} = \frac{1}{π} \Rightarrow r = \frac{1}{π} units$

Page No 12.25:

Question 17:

Each side of an equilateral triangle is increasing at the rate of 8 cm/hr. The rate of increase of its area when side is 2 cm, is

(a) $8 \sqrt{3} {cm}^{2} / hr$

(b) $4 \sqrt{3} {cm}^{2} / hr$

(c) $\frac{\sqrt{3}}{8} {cm}^{2} / hr$

(d) none of these

Answer:

(a) $8 \sqrt{3} {cm}^{2} / hr$

$Let x be the side and A be the area of the equilateral triangle at any time t . Then, A = \frac{\sqrt{3}}{4} x^{2} \Rightarrow \frac{d A}{d t} = \frac{\sqrt{3}}{2} x (\frac{d x}{d t}) \Rightarrow \frac{d A}{d t} = \frac{\sqrt{3}}{2} (2) (8) \Rightarrow \frac{d A}{d t} = 8 \sqrt{3} {cm}^{2} / hr$

Page No 12.25:

Question 18:

If s = t³ − 4t² + 5 describes the motion of a particle, then its velocity when the acceleration vanishes, is

(a) $\frac{16}{9} unit / \sec$

(b) $- \frac{32}{3} unit / \sec$

(c) $\frac{4}{3} unit / \sec$

(d) $- \frac{16}{3} unit / \sec$

Answer:

(d) $- \frac{16}{3} unit / \sec$

$According to the question, s = t^{3} - 4 t^{2} + 5 \Rightarrow \frac{d s}{d t} = 3 t^{2} - 8 t \Rightarrow \frac{d^{2} s}{d t^{2}} = 6 t - 8 \Rightarrow 6 t - 8 = 0 [As velocity deminishes, then \frac{d^{2} s}{d t^{2}} =0] \Rightarrow t = \frac{4}{3} Now, {(\frac{d s}{d t})}_{t = \frac{4}{3}} = 3 {(\frac{4}{3})}^{2} - 8 (\frac{4}{3}) \Rightarrow \frac{d s}{d t} = \frac{16}{3} - \frac{32}{3} \Rightarrow \frac{d s}{d t} = - \frac{16}{3} unit / \sec$

Page No 12.25:

Question 19:

The equation of motion of a particle is s = 2t² + sin 2t, where s is in metres and t is in seconds. The velocity of the particle when its acceleration is 2 m/sec², is
(a) $π + \sqrt{3} m / \sec$

(b) $\frac{π}{3} + \sqrt{3} m / \sec$

(c) $\frac{2 π}{3} + \sqrt{3} m / \sec$

(d) $\frac{π}{3} + \frac{1}{\sqrt{3}} m / \sec$

Answer:

(b) $\frac{π}{3} + \sqrt{3} m / \sec$

$According to the question, s = 2 t^{2} + \sin 2 t \Rightarrow \frac{d s}{d t} = 4 t + 2 \cos 2 t \Rightarrow \frac{d^{2} s}{d t^{2}} = 4 - 4 \sin 2 t \Rightarrow 4 - 4 \sin 2 t = 2 \Rightarrow 4 \sin 2 t = 2 \Rightarrow \sin 2 t = \frac{1}{2} \Rightarrow 2 t = \frac{π}{6} Now, \frac{d s}{d t} = 4 (\frac{π}{12}) + 2 \cos (\frac{π}{6}) \Rightarrow \frac{d s}{d t} = \frac{π}{3} + \sqrt{3} m / \sec$

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Question 20:

The radius of a circular plate is increasing at the rate of 0.01 cm/sec. The rate of increase of its area when the radius is 12 cm, is
(a) 144 π cm²/sec
(b) 2.4 π cm²/sec
(c) 0.24 π cm²/sec
(d) 0.024 π cm²/sec

Answer:

(c) 0.24 π cm²/sec

$Let r be the radius and A be the area of the circular plate at any time t. Then, A = π r^{2} ⇒ \frac{d A}{d t} =2π r \frac{d r}{d t} ⇒ \frac{d A}{d t} =2π (12) (0.01) ⇒ \frac{d A}{d t} {=0.24π cm}^{2} /sec$

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Question 21:

The diameter of a circle is increasing at the rate of 1 cm/sec. When its radius is π, the rate of increase of its area is
(a) π cm²/sec
(b) 2π cm²/sec
(c) π² cm²/sec
(d) 2π² cm²/sec²

Answer:

(c) π² cm²/sec

$Let D be the diameter and A be the area of the circle at any time t . Then, A = π r^{2} (where r is the radius of the cicle) \Rightarrow A =π \frac{D^{2}}{4} [∵ r = \frac{D}{2}] \Rightarrow \frac{d A}{d t} = 2 π \frac{D}{4} \frac{d D}{d t} \Rightarrow \frac{d A}{d t} = \frac{π}{2} \times 2 π \times 1 [∵ \frac{d D}{d t} = 1 cm / \sec] \Rightarrow \frac{d A}{d t} = π^{2} {cm}^{2} / \sec$

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Question 22:

A man 2 metres tall walks away from a lamp post 5 metres height at the rate of 4.8 km/hr. The rate of increase of the length of his shadow is
(a) 1.6 km/hr
(b) 6.3 km/hr
(c) 5 km/hr
(d) 3.2 km/hr

Answer:

Let AB be the lamp post. Suppose at any time t, the man CD be at a distance of x km from the lamp post and y m be the length of his shadow CE.

$Since triangles A B E and C D E are similar, \frac{A B}{C D} = \frac{A E}{C E}$

$\Rightarrow \frac{5}{2} = \frac{x + y}{y} \Rightarrow \frac{x}{y} = \frac{5}{2} - 1 \Rightarrow \frac{x}{y} = \frac{3}{2} \Rightarrow y = \frac{2}{3} x \Rightarrow \frac{d y}{d t} = \frac{2}{3} (\frac{d x}{d t}) \Rightarrow \frac{d y}{d t} = \frac{2}{3} \times 4.8 \Rightarrow \frac{d y}{d t} = 3.2 km / hr$

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Question 23:

A man of height 6 ft walks at a uniform speed of 9 ft/sec from a lamp fixed at 15 ft height. The length of his shadow is increasing at the rate of
(a) 15 ft/sec
(b) 9 ft/sec
(c) 6 ft/sec
(d) none of these

Answer:

(c) 6 ft/sec

Let AB be the lamp post. Suppose at any time t, the man CD be at a distance of x km from the lamp post and y ft be the length of his shadow CE.

$Since the triangles A B E and C D E are similar, \frac{A B}{C D} = \frac{A E}{C E}$

$\Rightarrow \frac{15}{6} = \frac{x + y}{y} \Rightarrow \frac{x}{y} = \frac{15}{6} - 1 \Rightarrow \frac{x}{y} = \frac{3}{2} \Rightarrow y = \frac{2}{3} x \Rightarrow \frac{d y}{d t} = \frac{2}{3} (\frac{d x}{d t}) \Rightarrow \frac{d y}{d t} = \frac{2}{3} \times 9 \Rightarrow \frac{d y}{d t} = 6 ft / \sec$

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Question 24:

In a sphere the rate of change of volume is
(a) π times the rate of change of radius
(b) surface area times the rate of change of diameter
(c) surface area times the rate of change of radius
(d) none of these

Answer:

(c) surface area times the rate of change of radius

$Let r be the radius and V be the volume of sphere at any time t. Then, V = \frac{4}{3} π r^{3} \Rightarrow \frac{d V}{d t} = \frac{4}{3} (3 π r^{2}) (\frac{d r}{d t}) \Rightarrow \frac{d V}{d t} = 4 π r^{2} (\frac{d r}{d t}) Thus, t he rate of change of volume is surface area times the rate of change of the radius.$

Page No 12.25:

Question 25:

In a sphere the rate of change of surface area is
(a) 8π times the rate of change of diameter
(b) 2π times the rate of change of diameter
(c) 2π times the rate of change of radius
(d) 8π times the rate of change of radius

Answer:

(d) 8π times the rate of change of radius

$Let r be the radius and S be the surface area of the sphere at any time t. Then, S = 4 π r^{2} \Rightarrow \frac{d S}{d t} = 8 π r \frac{d r}{d t} ∴ The rate of change of surface area is 8π times the rate of change of the radius.$

Page No 12.25:

Question 26:

A cylindrical tank of radius 10 m is being filled with wheat at the rate of 314 cubic metre per hour. Then the depth of the wheat is increasing at the rate of
(a) 1 m/hr
(b) 0.1 m/hr
(c) 1.1 m/hr
(d) 0.5 m/hr

Answer:

(a) 1 m/hr

$Let r, h and V be the radius, height and volume of the cylinder at any time t . Then, V = π r^{2} h \Rightarrow \frac{d V}{d t} = π r^{2} \frac{d h}{d t} \Rightarrow 314 = 3.14 \times {(10)}^{2} \frac{d h}{d t} \Rightarrow \frac{d h}{d t} = \frac{314}{314} \Rightarrow \frac{d h}{d t} = 1 m / hr$

Page No 12.26:

Question 1:

The rate of change of $\sqrt{x^{2} + 16}$ with respect to $\frac{x}{x - 1}$ at x = 3 is ________________.

Answer:

Let $u (x) = \sqrt{x^{2} + 16}$ and $v (x) = \frac{x}{x - 1}$ .

$u (x) = \sqrt{x^{2} + 16}$

Differentiating both sides with respect to x, we get

$\frac{d u}{d x} = \frac{1}{2 \sqrt{x^{2} + 16}} \times 2 x = \frac{x}{\sqrt{x^{2} + 16}}$

$v (x) = \frac{x}{x - 1}$

Differentiating both sides with respect to x, we get

$\frac{d v}{d x} = \frac{(x - 1) \times 1 - x \times 1}{{(x - 1)}^{2}} = - \frac{1}{{(x - 1)}^{2}}$

Now,

Rate of change of u(x) with respect to v(x)

$= \frac{d u}{d v}$

$= \frac{\frac{d u}{d x}}{\frac{d v}{d x}}$

$= \frac{\frac{x}{\sqrt{x^{2} + 16}}}{- \frac{1}{{(x - 1)}^{2}}}$

$= - \frac{x {(x - 1)}^{2}}{\sqrt{x^{2} + 16}}$

∴ Rate of change of u(x) with respect to v(x) at x = 3

$= {(\frac{d u}{d v})}_{x = 3}$

$= - \frac{3 \times {(3 - 1)}^{2}}{\sqrt{3^{2} + 16}}$

$= - \frac{12}{5}$

Thus, the rate of change of $\sqrt{x^{2} + 16}$ with respect to $\frac{x}{x - 1}$ at x = 3 is $- \frac{12}{5}$ .

The rate of change of $\sqrt{x^{2} + 16}$ with respect to $\frac{x}{x - 1}$ at x = 3 is $- \frac{12}{5}$ .

Page No 12.26:

Question 2:

The rate of change of the surface are of a sphere of radius r when the radius is increasing at the rate of 2 cm/sec is _________________.

Answer:

Let r be the radius of sphere at any time t.

It is given that,

$\frac{d r}{d t}$ = 2 cm/sec

Surface area of the sphere, S = 4 $π$ r²

S = 4 $π$ r²

Differentiating both sides with respect to t, we get

$\frac{d S}{d t} = 4 π \frac{d}{d t} r^{2}$

$\Rightarrow \frac{d S}{d t} = 4 π \times 2 r \frac{d r}{d t}$

$\Rightarrow \frac{d S}{d t} = 8 π r \frac{d r}{d t}$

Putting $\frac{d r}{d t}$ = 2 cm/sec, we get

$\frac{d S}{d t} = 8 π r \times 2$ = 16 $π$ r cm²/sec

Thus, the rate of change of the surface are of a sphere of radius r when the radius is increasing at the rate of 2 cm/sec is 16 $π$ r cm²/sec.

The rate of change of the surface are of a sphere of radius r when the radius is increasing at the rate of 2 cm/sec is ___16 $π$ r cm²/sec___.

Page No 12.26:

Question 3:

The diagonal of a square is changing at the rate of $\frac{1}{2} cm / \sec .$ Then the rate of change of area, when the area is 400 cm², is equal to ____________________.

Answer:

Let the side of the square be x cm at any time t.

∴ Length of the diagonal of square, l = $\sqrt{2}$ × Side of square = $\sqrt{2} x$

$l = \sqrt{2} x$

Differentiating both sides with respect to t, we get

$\frac{d l}{d t} = \sqrt{2} \times \frac{d x}{d t}$

It is given that, $\frac{d l}{d t} = \frac{1}{2}$ cm/sec

$∴ \frac{1}{2} = \sqrt{2} \times \frac{d x}{d t}$

$\Rightarrow \frac{d x}{d t} = \frac{1}{2 \sqrt{2}}$ cm/sec

Now,

Area of the square, A = x²

A = x²

Differentiating both sides with respect to t, we get

$\frac{d A}{d t} = 2 x \frac{d x}{d t}$ .....(1)

When A = 400 cm²,

x² = 400 cm² = (20 cm)²

⇒ x = 20 cm

Putting x = 20 cm and $\frac{d x}{d t} = \frac{1}{2 \sqrt{2}}$ cm/sec in (1), we get

$\frac{d A}{d t} = 2 \times 20 \times \frac{1}{2 \sqrt{2}} = 10 \sqrt{2}$ cm²/sec

Thus, the rate of change of area of the square is $10 \sqrt{2}$ cm²/sec.

The diagonal of a square is changing at the rate of $\frac{1}{2} cm / \sec .$ Then the rate of change of area, when the area is 400 cm², is equal to $10 \sqrt{2} {cm}^{2} / \sec$ .

Page No 12.26:

Question 4:

The rate of change of volume of a sphere with respect to its surface area, when the radius is 2 cm, is _________________.

Answer:

Let r be the radius of the sphere at any time t.

Volume of the sphere, V = $\frac{4}{3} π r^{3}$

$V = \frac{4}{3} π r^{3}$

Differentiating both sides with respect to r, we get

$\frac{d V}{d r} = \frac{4}{3} π \times \frac{d}{d r} r^{3}$

$\Rightarrow \frac{d V}{d r} = \frac{4}{3} π \times 3 r^{2}$

$\Rightarrow \frac{d V}{d r} = 4 π r^{2}$

Surface area of the sphere, S = $4 π r^{2}$

$S = 4 π r^{2}$

Differentiating both sides with respect to r, we get

$\frac{d S}{d r} = 4 π \times \frac{d}{d r} r^{2}$

$\Rightarrow \frac{d S}{d r} = 4 π \times 2 r$

$\Rightarrow \frac{d S}{d r} = 8 π r$

Now, rate of change of volume of a sphere with respect to its surface area

$= \frac{d V}{d S} = \frac{\frac{d V}{d r}}{\frac{d S}{d r}} = \frac{4 π r^{2}}{8 π r} = \frac{r}{2}$

When r = 2 cm, we get

${(\frac{d V}{d S})}_{r = 2 cm} = \frac{2 cm}{2}$ = 1 cm

Thus, the rate of change of volume of a sphere with respect to its surface area when the radius is 2 cm is 1 cm.

The rate of change of volume of a sphere with respect to its surface area, when the radius is 2 cm, is ____1 cm____.

Page No 12.26:

Question 5:

The angle $θ, 0 < θ < \frac{π}{2},$ which increases twice as fast as its sine, is _________________.

Answer:

Let angle θ increase twice as fast as its sine.

It is given that,

$\frac{d θ}{d t} = 2 \frac{d}{d t} (\sin θ)$

$\Rightarrow \frac{d θ}{d t} = 2 \cos θ \frac{d θ}{d t}$

$\Rightarrow 2 \cos θ = 1$

$\Rightarrow \cos θ = \frac{1}{2} = \cos \frac{π}{3}$

$\Rightarrow θ = \frac{π}{3} (0 < θ < \frac{π}{2})$

Thus, the angle θ is $\frac{π}{3}$ .

The angle $θ, 0 < θ < \frac{π}{2},$ which increases twice as fast as its sine, is $\frac{π}{3}$ .

Page No 12.26:

Question 6:

The sides of an equilateral triangle are increasing at the rate of 2 cm/sec. The rate at which the area increases, when the side is 10 cm, is _____________.

Answer:

Let x be the side and A be the area of an equilateral triangle at any time t.

It is given that,

$\frac{d x}{d t}$ = 2 cm/sec

Area of the equilateral triangle, A = $\frac{\sqrt{3}}{4} {(Side)}^{2}$ = $\frac{\sqrt{3}}{4} x^{2}$

$A = \frac{\sqrt{3}}{4} x^{2}$

Differentiating both sides with respect to t, we get

$\frac{d A}{d t} = \frac{\sqrt{3}}{4} \times \frac{d}{d t} x^{2}$

$\Rightarrow \frac{d A}{d t} = \frac{\sqrt{3}}{4} \times 2 x \frac{d x}{d t}$

$\Rightarrow \frac{d A}{d t} = \frac{\sqrt{3}}{2} x \frac{d x}{d t}$

When x = 10 cm and $\frac{d x}{d t}$ = 2 cm/sec, we get

$\frac{d A}{d t} = \frac{\sqrt{3}}{2} \times 10 \times 2$

$\Rightarrow \frac{d A}{d t} = 10 \sqrt{3}$ cm²/sec

Thus, the area of the equilateral triangle increases at the rate of $10 \sqrt{3}$ cm²/sec.

The sides of an equilateral triangle are increasing at the rate of 2 cm/sec. The rate at which the area increases, when the side is 10 cm, is $10 \sqrt{3} {cm}^{2} / \sec$ .

Page No 12.26:

Question 7:

Gas is being pumped into a spherical balloon at the rate of 30 cm³/min. The rate at which the radius increases when it reaches the value 15 cm, is ___________.

Answer:

Let r be the radius and V be the volume of the balloon at time t.

It is given that, $\frac{d V}{d t}$ = 30 cm³/min

Now,

Volume of the spherical balloon, V = $\frac{4}{3} π r^{3}$

$V = \frac{4}{3} π r^{3}$

Differentiating both sides with respect to t, we get

$\frac{d V}{d t} = \frac{d}{d t} (\frac{4}{3} π r^{3})$

$\Rightarrow \frac{d V}{d t} = \frac{4}{3} π \times 3 r^{2} \frac{d r}{d t}$

$\Rightarrow \frac{d V}{d t} = 4 π r^{2} \frac{d r}{d t}$

When r = 15 cm and $\frac{d V}{d t}$ = 30 cm³/min, we get

$30 = 4 π \times {(15)}^{2} \frac{d r}{d t}$

$\Rightarrow \frac{d r}{d t} = \frac{30}{4 π \times {(15)}^{2}}$

$\Rightarrow \frac{d r}{d t} = \frac{1}{30 π}$ cm/min

Thus, the radius of the balloon is increasing at the rate of $\frac{1}{30 π}$ cm/min.

Gas is being pumped into a spherical balloon at the rate of 30 cm³/min. The rate at which the radius increases when it reaches the value 15 cm, is $\frac{1}{30 π} cm / \min$ .

Page No 12.26:

Question 8:

The distance s described by a particle in t seconds is given by $s = a e^{t} + \frac{b}{e^{t}}$ . Then the acceleration of the particle at time t is equal to _____________.

Answer:

It is given that, the distance s described by a particle in t seconds is given by $s = a e^{t} + \frac{b}{e^{t}}$ .

The acceleration of the particle at time t is given by $\frac{d^{2} s}{d t^{2}}$ .

$s = a e^{t} + \frac{b}{e^{t}}$

Differentiating both sides with respect to t, we get

$\frac{d s}{d t} = \frac{d}{d t} (a e^{t} + \frac{b}{e^{t}})$

$\Rightarrow \frac{d s}{d t} = \frac{d}{d t} (a e^{t} + b e^{- t})$

$\Rightarrow \frac{d s}{d t} = a e^{t} + b e^{- t} \times (- 1)$

$\Rightarrow \frac{d s}{d t} = a e^{t} - b e^{- t}$

Again differentiating both sides with respect to t, we get

$\frac{d}{d t} (\frac{d s}{d t}) = \frac{d}{d t} (a e^{t} - b e^{- t})$

$\Rightarrow \frac{d^{2} s}{d t^{2}} = a e^{t} - b e^{- t} \times (- 1)$

$\Rightarrow \frac{d^{2} s}{d t^{2}} = a e^{t} + \frac{b}{e^{t}}$ = s

Thus, the acceleration of the particle at time t is $a e^{t} + \frac{b}{e^{t}}$ .

The distance s described by a particle in t seconds is given by $s = a e^{t} + \frac{b}{e^{t}}$ . Then the acceleration of the particle at time t is equal to $a e^{t} + \frac{b}{e^{t}}$ .

Page No 12.26:

Question 9:

The volume V and depth x of water in a vessel are connected by the relation V $= 5 x - \frac{x^{2}}{6}$ and the volume of water is increasing the rate of 5 cm³/sec, when x = 2 cm. The rate of which the depth of water is increasing is equal to _____________________.

Answer:

It is given that, $\frac{d V}{d t}$ = 5 cm³/sec

The volume V and depth x of water in a vessel are connected by the relation V $= 5 x - \frac{x^{2}}{6}$ .

$V = 5 x - \frac{x^{2}}{6}$

Differentiating both sides with respect to t, we get

$\frac{d V}{d t} = 5 \frac{d x}{d t} - \frac{2 x}{6} \frac{d x}{d t}$

$\Rightarrow \frac{d V}{d t} = (5 - \frac{x}{3}) \frac{d x}{d t}$

When x = 2 cm and $\frac{d V}{d t}$ = 5 cm³/sec, we get

$5 = (5 - \frac{2}{3}) \frac{d x}{d t}$

$\Rightarrow \frac{d x}{d t} = \frac{15}{13}$ cm/sec

Thus, depth of water is increasing at the rate of $\frac{15}{13}$ cm/sec.

The volume V and depth x of water in a vessel are connected by the relation V $= 5 x - \frac{x^{2}}{6}$ and the volume of water is increasing the rate of 5 cm³/sec, when x = 2 cm. The rate of which the depth of water is increasing is equal to $\frac{15}{13} cm / \sec$ .

Page No 12.26:

Question 10:

Water is flowing into a vertical cylindrical tank of radius 2 ft at the rate of 8 cubic/minute. The rate at which the water level is rising, is _______________.

Answer:

Let h be the water level in the cylindrical tank at time t minutes.

Radius of the cylinder, r = 2 ft

∴ Volume of the water in the cylindrical tank at time t, V = $π r^{2} h$ = $π \times {(2)}^{2} \times h$

$V = 4 π h$

Differentiating both sides with respect to t, we get

$\frac{d V}{d t} = 4 π \times \frac{d h}{d t}$

Now,

$\frac{d V}{d t}$ = 8 cubic feet/minute (Given)

$∴ 8 = 4 π \times \frac{d h}{d t}$

$\Rightarrow \frac{d h}{d t} = \frac{8}{4 π} = \frac{2}{π}$ ft/min

Thus, the water level in the tank is rising at the rate of $\frac{2}{π}$ feet/minute.

Water is flowing into a vertical cylindrical tank of radius 2 ft at the rate of 8 cubic/minute. The rate at which the water level is rising, is $\frac{2}{π} feet / minute$ .

Page No 12.26:

Question 1:

If a particle moves in a straight line such that the distance travelled in time t is given by s = t³ − 6t² + 9t + 8. Find the initial velocity of the particle.

Answer:

$s = t^{3} - 6 t^{2} + 9 t + 8 \Rightarrow \frac{d s}{d t} = 3 t^{2} - 12 t + 9 Initial velocity=Velocity at t =0 \Rightarrow \frac{d s}{d t} = 3 {(0)}^{2} - 12 (0) + 9 \Rightarrow \frac{d s}{d t} =9 units/unit time$

Page No 12.26:

Question 2:

The volume of a sphere is increasing at 3 cubic centimeter per second. Find the rate of increase of the radius, when the radius is 2 cms.

Answer:

$Let r be the radius and V be the volume of the sphere at any time t. Then, V = \frac{4}{3} π r^{3} \Rightarrow \frac{d V}{d t} = 4 π r^{2} \frac{d r}{d t} \Rightarrow \frac{d r}{d t} = \frac{1}{4 π r^{2}} \frac{d V}{d t} \Rightarrow \frac{d r}{d t} = \frac{3}{4 π {(2)}^{2}} [∵ r = 2 cm and \frac{d V}{d t} = 3 {cm}^{3} / \sec] \Rightarrow \frac{d r}{d t} = \frac{3}{16 π} cm / \sec$

Page No 12.26:

Question 3:

The sides of an equilateral triangle are increasing at the rate of 2 cm/sec. How far is the area increasing when the side is 10 cms?

Answer:

$Let x be the side and A be the area of the equilateral triangle at any time t. Then, A = \frac{\sqrt{3}}{4} x^{2} \Rightarrow \frac{d A}{d t} = 2 \times \frac{\sqrt{3}}{4} x^{} \frac{d x}{d t} ⇒ \frac{d A}{d t} = \frac{\sqrt{3}}{2} ×2×10 ⇒ \frac{d A}{d t} =10 \sqrt{3} {cm}^{2} /sec$

Page No 12.26:

Question 4:

The side of a square is increasing at the rate of 0.1 cm/sec. Find the rate of increase of its perimeter.

Answer:

$Let x be the side and P be the perimeter of the square at any time t . Then, P = 4 x \Rightarrow \frac{d P}{d t} = 4 \frac{d x}{d t} \Rightarrow \frac{d P}{d t} = 4 \times 0.1 [∵ \frac{d x}{d t} =0.1 cm/sec] \Rightarrow \frac{d P}{d t} = 0.4 cm / \sec$

Page No 12.26:

Question 5:

The radius of a circle is increasing at the rate of 0.5 cm/sec. Find the rate of increase of its circumference.

Answer:

$Let r be the radius and C be the circumference of the circle at any time t. Then, C = 2 π r \Rightarrow \frac{d C}{d t} = 2 π \frac{d r}{d t} \Rightarrow \frac{d C}{d t} = 2 π \times 0.5 [∵ \frac{d r}{d t} =0.5 cm/sec] \Rightarrow \frac{d C}{d t} = π cm / \sec$

Page No 12.26:

Question 6:

The side of an equilateral triangle is increasing at the rate of $\frac{1}{3}$ cm/sec. Find the rate of increase of its perimeter.

Answer:

$Let x be the side and P be the perimeter of the equilateral triangle at any time t. Then, P = 3 x \Rightarrow \frac{d P}{d t} = 3 \frac{d x}{d t} \Rightarrow \frac{d P}{d t} = 3 \times \frac{1}{3} [∵ \frac{d x}{d t} = \frac{1}{3} cm / \sec] \Rightarrow \frac{d P}{d t} = 1 cm / \sec$

Page No 12.26:

Question 7:

Find the surface area of a sphere when its volume is changing at the same rate as its radius.

Answer:

$Let r be the radius and V be the volume of the sphere at any time t . Then, V = \frac{4}{3} π r^{3} \Rightarrow \frac{d V}{d t} = 4 π r^{2} (\frac{d r}{d t}) \Rightarrow \frac{d V}{d t} = 4 π r^{2} (\frac{d V}{d t}) [∵ \frac{d V}{d t} = \frac{d r}{d t}] \Rightarrow 4 π r^{2} = 1 \Rightarrow Surface area of sphere =1 square unit$

Page No 12.27:

Question 8:

If the rate of change of volume of a sphere is equal to the rate of change of its radius, find the radius of the sphere.

Answer:

$Let r be the radius and V be the volume of the sphere at any time t . Then, V = \frac{4}{3} π r^{3} \Rightarrow \frac{d V}{d t} = 4 π r^{2} \frac{d r}{d t} \Rightarrow 4 π r^{2} = 1 [∵ \frac{d V}{d t} = \frac{d r}{d t}] \Rightarrow r^{2} = \frac{1}{4 π} \Rightarrow r = \sqrt{\frac{1}{4 π}} \Rightarrow r = \frac{1}{2 \sqrt{π}} units$

Page No 12.27:

Question 9:

The amount of pollution content added in air in a city due to x diesel vehicles is given by P(x) = 0.005x³ + 0.02x² + 30x. Find the marginal increase in pollution content when 3 diesel vehicles are added and write which value is indicated in the above questions.

Answer:

Since, marginal increase in the pollution content is the rate of change of total pollution with respect to the number of diesel vehicles, we have

Marginal increase in pollution = $\frac{d P}{d x} = 0.015 x^{2} + 0.04 x + 30$

When x = 3, marginal increase in pollution = $0.015 (9) + 0.04 (3) + 30 = 0.135 + 0.12 + 30 = 30.255$

Hence, the required marginal increase in pollution is 30.255 units.

It indicates the pollution level due to x diesel vehicles.

Page No 12.27:

Question 10:

A ladder, 5 metre long, standing on a horizontal floor, leans against a vertical wall. If the top of the ladder slides down wards at the rate of 10 cm/sec, then find the rate at which the angle between the floor and ladder is decreasing when lower end of ladder is 2 metres from the wall.

Answer:

$Length of the ladder = 500 c m Let the horizontal length covered between the wall and the ladder be x and vertical length covered between the wall and the ladder be y . And let the angle between the floor and ladder be θ . Then, \sin θ = \frac{y}{500} On differentiating with respect to t, we get \cos θ \frac{d θ}{d t} = \frac{1}{500} \frac{d y}{d t} . . . (1) It is given that \frac{d y}{d t} = - 10 c m / s e c . . . . (2) Also, \cos θ = \frac{x}{500} When x = 200 c m, \cos θ = \frac{200}{500} = \frac{2}{5} . . . (3) Substituting (2) and (3) in (1), we get \frac{2}{5} \frac{d θ}{d t} = \frac{1}{500} (- 10) \frac{d θ}{d t} = - \frac{1}{20} radian / second Hence, the angle between the floor and the ladder is decreasing at the rate of \frac{1}{20} radian / second .$

Page No 12.4:

Question 1:

Find the rate of change of the total surface area of a cylinder of radius r and height h, when the radius varies.

Answer:

Let T be the total surface area of a cylinder. Then,

T = $2 π r (r + h)$

Since the radius varies, we differentiate the total surface area w.r.t. radius r.

Now,

$\frac{d T}{d r} = \frac{d}{d r} [2 π r (r + h)] \Rightarrow \frac{d T}{d r} = \frac{d}{d r} (2 π r^{2}) + \frac{d}{d r} (2 π r h) \Rightarrow \frac{d T}{d r} = 4 π r + 2 π h \Rightarrow \frac{d T}{d r} = 2 π (r + h)$

Page No 12.4:

Question 2:

Find the rate of change of the volume of a sphere with respect to its diameter.

Answer:

Let V and r be the volume and diameter of the sphere, respectively. Then,
V = $\frac{4}{3} π {(radius)}^{3}$

$\Rightarrow V = \frac{4}{3} π {(\frac{r}{2})}^{3} = \frac{1}{6} π r^{3}$

$\Rightarrow \frac{d V}{d r} = \frac{1}{2} π r^{2}$

Page No 12.4:

Question 3:

Find the rate of change of the volume of a sphere with respect to its surface area when the radius is 2 cm.

Answer:

Let V be the volume of the sphere. Then,

V = $\frac{4}{3} π r^{3}$

$\Rightarrow \frac{d V}{d r} = 4 π r^{2}$

Let S be the total surface area of sphere. Then,

S = $4 π r^{2}$

$\Rightarrow \frac{d S}{d r} = 8 π r$
$∴ \frac{d V}{d S} = \frac{\frac{d V}{d r}}{\frac{d S}{d r}} \Rightarrow \frac{d V}{d S} = \frac{4 π r^{2}}{8 π r} = \frac{r}{2} \Rightarrow {(\frac{d V}{d S})}_{r = 2} = \frac{2}{2} = 1 cm$

Page No 12.4:

Question 4:

Find the rate of change of the area of a circular disc with respect to its circumference when the radius is 3 cm.

Answer:

Let A be the area of the circular disc. Then,

A = $π r^{2}$

$\Rightarrow \frac{d A}{d r} = 2 π r$
Let C be the circumference of the circular disc. Then,

C = $2 π r$

$\Rightarrow \frac{d C}{d r} = 2 π$

$∴ \frac{d A}{d C} = \frac{\frac{d A}{d r}}{\frac{d C}{d r}} \Rightarrow \frac{d A}{d C} = \frac{2 π r}{2 π} = r \Rightarrow {(\frac{d A}{d C})}_{r = 3} = 3 cm$

Page No 12.4:

Question 5:

Find the rate of change of the volume of a cone with respect to the radius of its base.

Answer:

Let V be the volume of the cone. Then,

V = $\frac{1}{3} π r^{2} h$

$\Rightarrow \frac{d V}{d r} = \frac{2}{3} π r h$

Page No 12.4:

Question 6:

Find the rate of change of the area of a circle with respect to its radius r when r = 5 cm.

Answer:

Let A be area of the circle. Then,
A = $π r^{2}$

$\Rightarrow \frac{d A}{d r} = 2 π r$
Hence, the rate of change of the area of the circle is $2 π r$ .
When r = 5 cm,

${(\frac{d A}{d r})}_{r = 5} = 2 π (5) = 10 π {cm}^{2} / cm$

Page No 12.4:

Question 7:

Find the rate of change of the volume of a ball with respect to its radius r. How fast is the volume changing with respect to the radius when the radius is 2 cm?

Answer:

Let V be the volume of the spherical ball. Then,
V = $\frac{4}{3} π r^{3}$

$\Rightarrow \frac{d V}{d r} = 4 π r^{2}$
Thus, the rate of change of the volume of the sphere is $4 π r^{2}$ .

$When r = 2 cm, {(\frac{d V}{d r})}_{r = 2} = 4 π {(2)}^{2}_{} = 16 π {cm}^{3} / cm$

Page No 12.4:

Question 8:

The total cost C (x) associated with the production of x units of an item is given by C (x) = 0.007x³ − 0.003x² + 15x + 4000. Find the marginal cost when 17 units are produced.

Answer:

Since the marginal cost is the rate of change of total cost with respect to its output,

Marginal Cost (MC) = $\frac{d C}{d x} (x) = \frac{d}{d x} (0.007 x^{3} - 0.003 x^{2} + 15 x + 4000) = 0.021 x^{2} - 0.006 x + 15$
When x = 17,
Marginal Cost (MC) = $= 0.021 {(17)}^{2} - 0.006 (17) + 15 = 6.069 - 0.102 + 15 = Rs 20.967$

Page No 12.4:

Question 9:

The total revenue received from the sale of x units of a product is given by R (x) = 13x² + 26x + 15. Find the marginal revenue when x = 7.

Answer:

Since the marginal revenue is the rate of change of total revenue with respect to its output,
Marginal Revenue (MR) = $\frac{d R}{d x} (x) = \frac{d}{d x} (13 x^{2} + 26 x + 15) = 26 x + 26$

When x = 7,
Marginal Revenue (MR) $= 26 (7) + 26 = 182 + 26 = Rs 208$

Page No 12.4:

Question 10:

The money to be spent for the welfare of the employees of a firm is proportional to the rate of change of its total revenue (Marginal revenue). If the total revenue (in rupees) recieved from the sale of x units of a product is given by R(x) = 3x² + 36x + 5, find the marginal revenue, when x = 5, and write which value does the question indicate.

Answer:

Since, marginal revenue is the rate of change of total revenue with respect to the number of units sold, we have

$Marginal revenue (MR) = \frac{d R}{d x} = 6 x + 36 When x = 5, MR = 6 (5) + 36 = 66$

Hence, the required marginal revenue is ₹66.

It indicates the extra money spent when number of employees increased from 5 to 6.

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