RD Sharma Class 12 Ex 13.2 Solutions Chapter 13 Derivative as a Rate Measurer

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TextbookNCERT
ClassClass 12th
SubjectMaths
Chapter13
Exercise13.2
CategoryRD Sharma Solutions

Table of Contents

RD Sharma Class 12 Ex 13.2 Solutions Chapter 13 Derivative as a Rate Measurer

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?

Solution:

Let the side of the square sheet be denoted by ‘a’, then area (A) of the sheet will be a2 cm2.

It is given that side is increasing at the rate of 4cm/min, i.e.,\frac{\mathrm{d}a }{\mathrm{d} t}  = 4cm/min

Since, A = a2

\frac{\mathrm{d}A }{\mathrm{d} t}  = 2a\frac{\mathrm{d}a }{\mathrm{d} t}

\frac{\mathrm{d}A }{\mathrm{d} t}  = 2 x 8 x 4 [ Since, a=8 and\frac{\mathrm{d}a }{\mathrm{d} t}  = 4cm/min]

\frac{\mathrm{d}A }{\mathrm{d} t}  = 64 cm2/min

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?

Solution:

Let the edge of the cube be denoted by symbol ‘a’ and volume of the cube be denoted by ‘V’.

Now, as given the edge of variable cube is increasing, i.e.,\frac{\mathrm{d}a }{\mathrm{d} t}  = 3 cm/sec

Since, V = a3

\frac{\mathrm{d}V }{\mathrm{d} t}  = 3a2\frac{\mathrm{d}a }{\mathrm{d} t}

\frac{\mathrm{d}V }{\mathrm{d} t}  = 3 x 10 x 10 x 3 [Since, a=10 and\frac{\mathrm{d}a }{\mathrm{d} t}  = 3 cm/sec]

\frac{\mathrm{d}V }{\mathrm{d} t}  = 900 cm3/sec

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.

Solution:

Let the side of the square be denoted by a cm and its perimeter (P) = 4a cm

As given, the side is increasing, i.e.,\frac{\mathrm{d}a }{\mathrm{d} t}  = 0.2 cm/sec

Now since, P = 4a

\frac{\mathrm{d}P }{\mathrm{d} t}  = 4\frac{\mathrm{d}a }{\mathrm{d} t}

\frac{\mathrm{d}P }{\mathrm{d} t}  = 4 x 0.2

\frac{\mathrm{d}P }{\mathrm{d} t}  = 0. 8 cm/sec

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?

Solution:

Let the radius of the circle be denoted by ‘r’ cm and its circumference is given by C = 2*\pi  *r

Also given, the radius is increasing i.e.,\frac{\mathrm{d}r }{\mathrm{d} t}  = 0.7 cm/sec at any time t.

Rate of increase of its circumference =\frac{\mathrm{d}C }{\mathrm{d} t}

\frac{\mathrm{d}C }{\mathrm{d} t}  = 2*\pi*\frac{\mathrm{d}r }{\mathrm{d} t}

\frac{\mathrm{d}C }{\mathrm{d} t}  = 2 * 22/7 * 0.7

\frac{\mathrm{d}C }{\mathrm{d} t}  = 4.4 cm/sec

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.

Solution:

Let the radius of the spherical soap be denoted by ‘r’ and its surface area (S) = 4\pi  r2

Also, given the radius is increasing i.e.,\frac{\mathrm{d}r }{\mathrm{d} t}  = 0.2 cm/sec

Therefore, the increase of surface area at any time t is given by\frac{\mathrm{d}S}{\mathrm{d} t}

\frac{\mathrm{d}S }{\mathrm{d} t}  = 4*\pi  *2r*\frac{\mathrm{d}r }{\mathrm{d} t}

\frac{\mathrm{d}S }{\mathrm{d} t}  = 8 * 22/7 * 7 * 0.2

\frac{\mathrm{d}S }{\mathrm{d} t}  = 35.2 cm2/sec

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.

Solution:

Let the radius of the spherical balloon be denoted by ‘r’ and volume being inflated at \frac{\mathrm{d} V}{\mathrm{d} t} = 900 cm3/sec

Since, V =\frac{4}{3} \times \pi\times r^3

\frac{\mathrm{d}V }{\mathrm{d} t}  =\frac{4}{3} \times \pi\times 3r^2 \times \frac{\mathrm{d} r}{\mathrm{d} t}

\frac{\mathrm{d}V }{\mathrm{d} t}  =4 \times \pi\times r^2 \times \frac{\mathrm{d} r}{\mathrm{d} t}

⇒ 900 = 4 \times \pi\times (15)^2 \times \frac{\mathrm{d} r}{\mathrm{d} t}

⇒ 900 =900 \pi\times \frac{\mathrm{d} r}{\mathrm{d} t}

\frac{\mathrm{d} r}{\mathrm{d} t}  =\frac{1}{\pi}  cm/sec.

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?

Solution:

Let the radius of the bubble be denoted by ‘r’ and its volume be denoted by V where V =\frac{4}{3} \times \pi\times r^3

Now at any time t, radius is increasing i.e.,\frac{\mathrm{d} r}{\mathrm{d} t}  = 0.5 cm.sec

Therefore,\frac{\mathrm{d} V}{\mathrm{d} t}  = 4\pi  *r2*\frac{\mathrm{d} r}{\mathrm{d} t}

\frac{\mathrm{d} V}{\mathrm{d} t}  = 4\pi  * (1)2 * 0.5

\frac{\mathrm{d} V}{\mathrm{d} t}  = 2\pi  cm3/sec

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.

Solution:

Let MN denote the vertical lamp post of height 6 meter and at any instant t, a man XY of height 2 meter be standing in front of the lamp post at a distance ‘m’ and let ‘n’ be length of his shadow. It can be seen in a figure as:

We can notice\triangle MNO \sim \triangle XYO

\frac{MN}{XY}  =\frac{NO}{YO}

\frac{6}{2}  =\frac{m+n}{n}

⇒ 3n = m + n

⇒ m = 2n

\frac{\mathrm{d} m}{\mathrm{d} t}  = 2\frac{\mathrm{d} n}{\mathrm{d} t}

\frac{\mathrm{d} n}{\mathrm{d} t}  =\frac{5}{2}  km/hr

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?

Solution:

Let the radius of circular wave be denoted by ‘r’ and at any instant t, its radius increasing at\frac{\mathrm{d} r}{\mathrm{d} t}  = 4 cm/sec

Now, area of circular wave (A) =\pi  r2

\frac{\mathrm{d} A}{\mathrm{d} t}  = 2\pi r\frac{\mathrm{d} r}{\mathrm{d} t}

\frac{\mathrm{d} A}{\mathrm{d} t}  = 2*\pi  *10* 4

\frac{\mathrm{d} A}{\mathrm{d} t}  = 80\pi  cm2/sec

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?

Solution:

Let the vertical pole of light be denoted by MN and the man be denoted by XY, then his position with respect to lamp can be drawn as shown in figure:

We can notice,\triangle MNO \sim \triangle XYO

\frac{MN}{XY}  =\frac{NY}{YO}

\frac{6}{1.6}  =\frac{x+y}{y}

\frac{60}{16}  =\frac{x}{y}  + 1

\frac{x}{y}  =\frac{60-16}{16}

\frac{x}{y}  =\frac{44}{16}

⇒ y =\frac{16}{44}x

\frac{\mathrm{d} y}{\mathrm{d} t}  =\frac{16}{44}(\frac{\mathrm{d} x}{\mathrm{d} t})

\frac{\mathrm{d} y}{\mathrm{d} t}  =\frac{16}{44}  * 1.1

\frac{\mathrm{d} y}{\mathrm{d} t}  = 0.4 m/sec

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?

Solution:

Let MN denote the vertical lamp post of height 6 meter and at any instant t, a man XY of height 2 meter be standing in front of the lamp post at a distance ‘m’ and let ‘n’ be length of his shadow. It can be seen in a figure as:

We can notice

\triangle MNO  ∼ \triangle XYO

\frac{MN}{XY}  =\frac{NO}{YO}

\frac{9}{1.8}  \frac{m+n}{n}

⇒ m = 4n

\frac{\mathrm{d} m}{\mathrm{d} t}  = 4\frac{\mathrm{d} n}{\mathrm{d} t}

\frac{\mathrm{d} n}{\mathrm{d} t}  =\frac{1}{4}  x 2 [Since,\frac{\mathrm{d} m}{\mathrm{d} t}  = 2]

\frac{\mathrm{d} n}{\mathrm{d} t}  = 0.5 m/sec

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.

Solution:

Let the height of the wall that is leaning against a wall be denoted by y meters and the distance of the foot of ladder from base of the wall be x meters.

we can derive tan θ = y/x ⇒ y = xtan θ

Also, using Pythagoras theorem, x2 + y2 = (13)2

⇒ x2 + (xtan θ )2 = 169

⇒ x2 (1+tan2 θ ) = 169

⇒ sec2 θ = 169/x2

⇒ 2 sec θ . tan θ sec θ\frac{\mathrm{d} \theta}{\mathrm{d} t}  = 169.(\frac{-2}{x^3}) \frac{\mathrm{d} x}{\mathrm{d} t}

\frac{\mathrm{d} \theta}{\mathrm{d} t}  =\frac{-338\times1.5}{12^3. 2 sec^2\theta \tan\theta}  ………………(1)

Using Pythagoras Theorem, when x=12, then y=5

Therefore, sec θ = 13/12 and tan θ = 12/5

Then, equation 1 can be written as

\frac{\mathrm{d} \theta}{\mathrm{d} t}  =\frac{-338\times 1.5}{12^3 \times 2\times (\frac{13}{12})^2\times \frac{5}{12}}

\frac{\mathrm{d} \theta}{\mathrm{d} t}  = -0.3 rad/sec

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?

Solution:

We are given y = x2 + 2x

\frac{\mathrm{d} y}{\mathrm{d} t}  = 2x\frac{\mathrm{d} x}{\mathrm{d} t}  + 2\frac{\mathrm{d} x}{\mathrm{d} t}

⇒ \frac{\mathrm{d} y}{\mathrm{d} t}  = (2x+2)\frac{\mathrm{d} x}{\mathrm{d} t}

⇒ 2x + 2 = 1 [Since,\frac{\mathrm{d} y}{\mathrm{d} t}  =\frac{\mathrm{d} x}{\mathrm{d} t}  ]

⇒ x = -1/2

Putting the value of x in our original equation, we get y= -3/4

Hence, the coordinates of the point are\left ( \frac{-1}{2},\frac{-3}{4} \right )

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?

Solution:

We are given y = 7x – x3

\frac{\mathrm{d} y}{\mathrm{d} x}  = 7 – 3x2

Let the slope of the curve be denoted by m, then

⇒ m = 7 – 3x2

\frac{\mathrm{d} m}{\mathrm{d} t}  = -6x\frac{\mathrm{d} x}{\mathrm{d} t}

\frac{\mathrm{d} m}{\mathrm{d} t}  = -6 x 4 x 2

\frac{\mathrm{d} m}{\mathrm{d} t}  = -48

Therefore, the slope of the curve is decreasing at the rate of 48 units/sec.

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.

Solution:

We are given y = x3

\frac{\mathrm{d} y}{\mathrm{d} t}  = 3x2\frac{\mathrm{d} x}{\mathrm{d} t}

Also, the point on y-coordinate changes 3 times more rapidly than x-coordinate, therefore

\frac{\mathrm{d} y}{\mathrm{d} t}  = 3\frac{\mathrm{d} x}{\mathrm{d} t}

⇒ 3\frac{\mathrm{d} x}{\mathrm{d} t}  = 3x2\frac{\mathrm{d} x}{\mathrm{d} t}

⇒ x2 = 1

⇒ x = ±1

Substituting the value of x in y = x3, we get y = ±1

So, the points are (1,1) and (-1,-1).

Question 16 (i) Find an angle θ which increases twice as fast as its cosine?

(ii) Find an angle θ whose rate of increase twice is twice the rate of decrease of its cosine?

Solution:

(i) Let x = cos∅

Differentiating both sides with respect to t, we get

\frac{\mathrm{d} x}{\mathrm{d} t}  = -sin∅\frac{\mathrm{d} \varnothing }{\mathrm{d} t}

According to the condition given in question:

\frac{\mathrm{d} \varnothing }{\mathrm{d} t}  = 2\frac{\mathrm{d} x}{\mathrm{d} t}

\frac{\mathrm{d} x }{\mathrm{d} t}  = -sin∅(2\frac{\mathrm{d} x }{\mathrm{d} t})

⇒ sin∅ = -1/2

⇒ ∅ = π + π/6 = 7π/6

(ii) Let x = cos∅

Differentiating both sides with respect to t, we get

\frac{\mathrm{d} x}{\mathrm{d} t}  = -sin∅\frac{\mathrm{d} \varnothing }{\mathrm{d} t}

According to the condition given in question:

\frac{\mathrm{d} \varnothing }{\mathrm{d} t}  = -2\frac{\mathrm{d} x}{\mathrm{d} t}

\frac{\mathrm{d} x }{\mathrm{d} t}  = -sin∅(-2\frac{\mathrm{d} x }{\mathrm{d} t})

⇒ sin∅ = 1/2

Question 17. The top of a ladder 6 meters 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 meters 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?

Solution:

Let the foot of the ladder be at a distance of x meter from the base of the wall and its top be at a distance of y meter above the ground.

Using Pythagoras Theorem we can get, x2 + y2 = 36

⇒ 2x\frac{\mathrm{d} x }{\mathrm{d} t}  = -2y\frac{\mathrm{d} y }{\mathrm{d} t}  ……………………..(eqn 1)

when x = 4, y = 2√5

⇒ 2 x 4 x 0.5 = -2 x 2√5\frac{\mathrm{d} y}{\mathrm{d} t}

\frac{\mathrm{d} y}{\mathrm{d} t}  = -1/√5 m/sec

Now using eqn 1, we can write

⇒ 2x\frac{\mathrm{d} x}{\mathrm{d} t}  = -2y\frac{\mathrm{d} y}{\mathrm{d} t}

⇒ x = -y

Putting x = -y in x2 + y2 = 36, we get

⇒ 2x2 = 36 ⇒ x = 3 √2 meter

Question 18. A balloon in the form of a right circular cone surmounted by a hemisphere, having a diameter 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.

Solution:

Let r be the radius of the hemisphere and the cone has height h and volume of the compound arrangement be V, then according to the figure,

⇒ H = h+ r

⇒ H = 3r [Since, h = 2r]

\frac{\mathrm{d} H}{\mathrm{d} t}  = 3\frac{\mathrm{d} r}{\mathrm{d} t}  ———————(eqn. 1)

Now, volume of the compound arrangement is:

V = \frac{1}{3} \pi r^2 h+ \frac{2}{3} \pi r^3

⇒ V = \frac{2}{3} \pi r^3+ \frac{2}{3} \pi r^3  [ h = 2r]

⇒ V = \frac{4}{3} \pi r^3

\frac{\mathrm{d} V}{\mathrm{d} t}  4 \pi r^2\frac{\mathrm{d} r}{\mathrm{d} t}

⇒\frac{\mathrm{d} V}{\mathrm{d} t} = \frac{4}{3} \pi r^2 \frac{\mathrm{d} H}{\mathrm{d} t} [using equation 1]

\frac{dV}{dH}  \frac{4}{3}\pi (3)^2

\frac{dV}{dH}  = 12\pi  cm3/sec

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.

Solution:

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

V = πr2h/3

\frac{\mathrm{d} V}{\mathrm{d} t}  \frac{1}{3} \pi r^2\frac{\mathrm{d} h}{\mathrm{d} t} + \frac{2}{3}\pi rh\frac{\mathrm{d} r}{\mathrm{d} t}

From the image we can conclude,

h = 2r and\frac{\mathrm{d} h}{\mathrm{d} t} = 2 \frac{\mathrm{d} r}{\mathrm{d} t}

\frac{\mathrm{d} V}{\mathrm{d} t}  \frac{1}{3}\pi(\frac{h}{2})^2h\frac{\mathrm{d} r}{\mathrm{d} t}  +\frac{2}{3}\pi(\frac{h}{2})h\frac{1}{2}\frac{\mathrm{d} h}{\mathrm{d} t}

\frac{\mathrm{d} V}{\mathrm{d} t}  \frac{\pi h^2}{6}\left [ \frac{1}{2}\frac{\mathrm{d} h}{\mathrm{d} t} + \frac{\mathrm{d} h}{\mathrm{d} t} \right ]

\frac{\mathrm{d} V}{\mathrm{d} t}  \frac{\pi h^2}{6} \times \left [ \frac{3}{2}\frac{\mathrm{d} h}{\mathrm{d} t} \right ]

\frac{\mathrm{d} V}{\mathrm{d} t}  \frac{\pi h^2}{4}.\frac{\mathrm{d} h}{\mathrm{d} t}

\frac{\pi h^2}{4}.\frac{\mathrm{d} h}{\mathrm{d} t}  \pi

\frac{\mathrm{d} h}{\mathrm{d} t} = \frac{4}{h^2}

\frac{\mathrm{d} h}{\mathrm{d} t} = \frac{4}{(2.5)^2}

\frac{\mathrm{d} h}{\mathrm{d} t}  = 0.64 m/min

Question 20. A man 2 meters high walks at a uniform speed of 6 km/h away from a lamp-post 6 meters high. Find the rate at which the length of his shadow increases ?

Solution:

Since,\triangle  MNO ∼\triangle  XYO,

\frac{MN}{XY}  \frac{NO}{YO}

\frac{6}{2}  \frac{m+n}{m}

⇒ m/n = 2

⇒ m = 2n

\frac{\mathrm{d} m}{\mathrm{d} t} = 2 \frac{\mathrm{d} n}{\mathrm{d} t}

⇒6 = 2\frac{\mathrm{d} n}{\mathrm{d} t}

\frac{\mathrm{d} n}{\mathrm{d} t}  = 3 km/hr

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?

Solution:

Let r, S, V be the radius, surface area, volume of the spherical bubble respectively.

The surface area is increasing, therefore\frac{\mathrm{d} S}{\mathrm{d} t}  = 2 cm2/s

Since, surface area of a spherical bubble is given by S = 4πr2

\frac{\mathrm{d} S}{\mathrm{d} t}  = 8πr\frac{\mathrm{d} r}{\mathrm{d} t}

⇒ 2 = 8πx6\frac{\mathrm{d} r}{\mathrm{d} t}

\frac{\mathrm{d} r}{\mathrm{d} t}  \frac{1}{24\pi}  cm/sec

We know, V = \frac{4}{3}\pi r^3

\frac{\mathrm{d} V}{\mathrm{d} t}  = 4πr2\frac{\mathrm{d} r}{\mathrm{d} t}

\frac{\mathrm{d} V}{\mathrm{d} t}  = 4π x 36 x\frac{1}{24 \pi}

\frac{\mathrm{d} V}{\mathrm{d} t}  = 6 cm3/sec

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.

Solution:

Le r be the radius and h be the altitude and V be the volume of the cylinder, then as per given

\frac{\mathrm{d} r}{\mathrm{d} t}  = 2 cm/sec and\frac{\mathrm{d} h}{\mathrm{d} t}  = -3 cm/sec

Volume of cylinder is given by:

V = πr2h

\frac{\mathrm{d} V}{\mathrm{d} t}  = 2πrh\frac{\mathrm{d} r}{\mathrm{d} t}  + πr2\frac{\mathrm{d} h}{\mathrm{d} t}

\frac{\mathrm{d} V}{\mathrm{d} t}  = πr\left (2h\frac{\mathrm{d} r}{\mathrm{d} t}+r \frac{\mathrm{d} h}{\mathrm{d} t} \right )

\frac{\mathrm{d} V}{\mathrm{d} t}  = π x 3 (2 x 5 x 2 + 2 x (-3) )

\frac{\mathrm{d} V}{\mathrm{d} t}  = 33π cm3/sec

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.

Solution:

Let the outer radius be represented by R and inner radius be represented by r. Now, Volume of the hollow sphere is given by

V = \frac{4}{3}\pi [R^3 - r^3]

\frac{\mathrm{d} V}{\mathrm{d} t}  = 4π\left [ R^2 \frac{\mathrm{d} R}{\mathrm{d} t} - r^2 \frac{\mathrm{d} r}{\mathrm{d} t} \right ]

Since, volume is constant,\frac{\mathrm{d} V}{\mathrm{d} t}  = 0

R^2 \frac{\mathrm{d} R}{\mathrm{d} t} = r^2 \frac{\mathrm{d} r}{\mathrm{d} t}

⇒ 64\frac{\mathrm{d} R}{\mathrm{d} t}  = 16 x 1

\frac{\mathrm{d} R}{\mathrm{d} t}  \frac{1}{4}  cm/sec

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 ?

Solution:

Let r be the radius, h be the height and V be the volume of conical pile. Now, volume of conical pile is given as:

V = \frac{1}{3}\pi r^2h

⇒ V = \frac{1}{3}\pi (2h)^2 h  [Since, h = r/2]

⇒ \frac{\mathrm{d} V}{\mathrm{d} t}  \frac{4}{3}\pi h^3

\frac{\mathrm{d} V}{\mathrm{d} t}  = 4πh2\frac{\mathrm{d} h}{\mathrm{d} t}

⇒ 50 = 4πh2\frac{\mathrm{d} h}{\mathrm{d} t}

\frac{\mathrm{d} h}{\mathrm{d} t}  \frac{50}{100\pi}

\frac{\mathrm{d} h}{\mathrm{d} t}  \frac{1}{2\pi}  cm/min

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.

Solution:

From the above figure, we can infer using Pythagoras Theorem

MN2 + NO2 = MO2

⇒ x2 + (120) = y2

⇒ 2\frac{\mathrm{d} x}{\mathrm{d} t}  = 2y\frac{\mathrm{d} y}{\mathrm{d} t}

\frac{\mathrm{d} y}{\mathrm{d} t}  \frac{x}{y}\frac{\mathrm{d} x}{\mathrm{d} t}

Now, x = \sqrt{(130)^2 - (120)^2}  = 50

\frac{\mathrm{d} y}{\mathrm{d} t}  \frac{50}{130}  x 52

\frac{\mathrm{d} y}{\mathrm{d} t}  = 20 m/sec

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.

Solution:

Given y = \frac{2}{3}  x3 + 1

\frac{\mathrm{d} y}{\mathrm{d} t}  = 2x2\frac{\mathrm{d} x}{\mathrm{d} t}

It is also given that, y-coordinate is changing twice as fast as the x-coordinate, therefore\frac{\mathrm{d} y}{\mathrm{d} t} = 2\frac{\mathrm{d} x}{\mathrm{d} t}

2\frac{\mathrm{d} x}{\mathrm{d} t}  2x^2\frac{\mathrm{d} x}{\mathrm{d} t}

Solving the above equation, we get x = ±1

Substituting the value of x in above given equation y = 5/3 and y = 1/3

So the coordinates of the point are\left ( 1,\frac{5}{3} \right )  and\left ( -1,\frac{1}{3} \right )

Question 27. Find the point on the curve y2 = 8x for which the abscissa and ordinate change at the same rate?

Solution:

Given, y2 = 8x

⇒ 2y\frac{\mathrm{d} y}{\mathrm{d} t}  = 8\frac{\mathrm{d} x}{\mathrm{d} t}

Now, since abscissa and ordinate change at the same rate, therefore\frac{\mathrm{d} y}{\mathrm{d} t} = \frac{\mathrm{d} x}{\mathrm{d} t}

⇒ 2y = 8

⇒ y = 4

Therefore, substituting the value of y in original equation. we get

16 = 8x ⇒ x = 2

Hence, the co-ordinate of the point is (2,4)

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?

Solution:

Let the edge of the cube be denoted by a and its volume be denoted by V.

We know, Volume of cube, V = a3

⇒ \frac{\mathrm{d} V}{\mathrm{d} t}  = 3a2\frac{\mathrm{d} a}{\mathrm{d} t}

⇒ 9 = 3 x (10)2\frac{\mathrm{d} a}{\mathrm{d} t}

\frac{\mathrm{d} a}{\mathrm{d} t}  = 0.03 cm/sec

Now, let the surface area of cube be given by A = 6a2

\frac{\mathrm{d} A}{\mathrm{d} t}  = 12a\frac{\mathrm{d} a}{\mathrm{d} t}

\frac{\mathrm{d} A}{\mathrm{d} t}  = 12 x 10 x 0.03

\frac{\mathrm{d} A}{\mathrm{d} t}  = 3.6 cm2/sec

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?

Solution:

Let r be the radius, V be the volume and S be the surface area of the spherical balloon.

We know, V = \frac{4}{3}  πr3

\frac{\mathrm{d} V}{\mathrm{d} t}  = 4πr2\frac{\mathrm{d} r}{\mathrm{d} t}

⇒ 25 = 4π (5)2\frac{\mathrm{d} r}{\mathrm{d} t}

\frac{\mathrm{d} r}{\mathrm{d} t}  \frac{1}{4\pi}  cm/sec

Also, surface area, A = 4πr2

\frac{\mathrm{d} A}{\mathrm{d} t}  = 8πr\frac{\mathrm{d} r}{\mathrm{d} t}

⇒ \frac{\mathrm{d} A}{\mathrm{d} t}  = 8π x 5 x\frac{1}{4\pi}

\frac{\mathrm{d} A}{\mathrm{d} t}  = 10 cm2/sec

Question 30. (i) 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 the perimeter ?

(ii) 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 the area of the rectangle ?

Solution:

We are given,\frac{\mathrm{d} x}{\mathrm{d} t}  = -5cm/min and\frac{\mathrm{d} y}{\mathrm{d} t}  = 4cm/min

(i) We know, perimeter of a rectangle P = 2(x + y)

⇒ \frac{\mathrm{d} P}{\mathrm{d} t}  = 2 (\frac{\mathrm{d} x}{\mathrm{d} t}  +\frac{\mathrm{d} y}{\mathrm{d} t}  )

\frac{\mathrm{d} P}{\mathrm{d} t}  = 2 (-5 + 4)

\frac{\mathrm{d} P}{\mathrm{d} t}  = -2 cm/min

(ii) Also, Area of rectangle A = xy

\frac{\mathrm{d} A}{\mathrm{d} t}  = x\frac{\mathrm{d} y}{\mathrm{d} t}  + y\frac{\mathrm{d} x}{\mathrm{d} t}

⇒ \frac{\mathrm{d} A}{\mathrm{d} t}  = 8 x 4 + 6 x (-5)

\frac{\mathrm{d} A}{\mathrm{d} t}  = 2 cm2/min

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.

Solution:

Let r be the radius and A be the area of circular disc respectively.

Then A = πr2

\frac{\mathrm{d} A}{\mathrm{d} t}  = 2πr\frac{\mathrm{d} r}{\mathrm{d} t}

\frac{\mathrm{d} A}{\mathrm{d} t}  = 2π x 3.2 x 0.05

\frac{\mathrm{d} A}{\mathrm{d} t}  = 0.32π cm2/sec

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