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## NCERT Solutions for Class 12 Maths Application of Derivatives

1. Find the rate of change of the area of a circle with respect to its radius  when

(a)  = 3 cm

(b)  = 4 cm

Ans. Let  denote the area of the circle of variable radius

Area of circle

Rate of change of area  w.r.t.

=

(a) When  cm, then

sq. cm

(b) When  cm, then

sq. cm

### 2. The volume of a cube is increasing at the rate of 8 cm3/sec. How fast is the surface area increasing when the length of an edge is 12 cm?

Ans. Let  cm be the edge of the cube.

Given: Rate of increase of volume of cube = 8 cm3/sec

is positive = 8

……….(i)

Let  be the surface area of the cube, i.e.,

Rate of change of surface area of the cube =

=  =

=  cm2/sec

Putting  = 12 cm (given),

cm2/sec

Since  is positive, therefore surface area is increasing at the rate of  cm2/sec.

### 3. The radius of the circle is increasing uniformly at the rate of 3 cm per second. Find the rate at which the area of the circle is increasing when the radius is 10 cm.

Ans. Let  cm be the radius of the circle at time

Rate of increase of radius of circle = 3 cm/sec

is positive and = 3 cm/sec

Let  be the area of the circle.

Rate of change of area of circle =  =

=

=

Putting  = 10 cm (given),

=  cm2/sec

Since  is positive, therefore surface area is increasing at the rate of  cm2/sec.

### 4. 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 if 10 cm long?

Ans. Let  cm be the edge of variable cube at time

Rate of increase of edge = 3 cm/sec

is positive and = 3 cm/sec

Let  be the volume of the cube.

Rate of change of volume of cube =  =

=

=  cm3/sec

Putting  = 10 cm (given),

=  cm3/sec

Since  is positive, therefore volume of cube is increasing at the rate of 900 cm3/sec.

### 5. A stone is dropped into a quite lake and waves move in circles at the rate of 5 cm/sec. At the instant when radius of the circular wave is 8 cm, how fast is the enclosed area increasing?

Ans. Let  cm be the radius of the circular wave at time

Rate of increase of radius of circular wave = 5 cm/sec

is positive and = 5 cm/sec

Let  be the enclosed area of the circular wave.

Rate of change of area =  =

=  =

Putting  = 8 cm (given),

=  cm2/sec

Since  is positive, therefore area of circular wave is increasing at the rate of  cm2/sec.

### 6. The radius of a circle is increasing at the rate of 0.7 cm/s. What is the rate of its circumference?

Ans. Let  cm be the radius of the circle at time

Rate of increase of radius of circle = 0.7 cm/sec

is positive and = 0.7 cm/sec

Let  be the circumference of the circle.

Rate of change of circumference of circle =

=  =

=  cm/sec

### 7. The length  of a rectangle is decreasing at the rate of 5 cm/minute. When  = 8 cm and  = 6 cm, find the rates of change of (a) the perimeter and (b) the area of the rectangle.

Ans. Given: Rate of decrease of length  of rectangle is 5 cm/minute.

is negative = –5 cm/minute

Also, Rate of increase of width  of rectangle is 4 cm/minute

is positive

= 4 cm/minute

(a) Let  denotes the perimeter of rectangle.

=  is negative.

Perimeter of the rectangle is decreasing at the rate of 2 cm/sec.

(b) Let  denotes the area of rectangle.

=  is positive.

Area of the rectangle is increasing at the rate of 2 cm2/sec.

### 8. A balloon, which always remains spherical on inflation, is being inflated by pumping in 900 cubic centimeters of gas per second. Find the rate at which the radius of the balloon increases when the radius is 15 cm.

Ans. Let  cm be the radius of the spherical balloon at time

According to the question,

Radius of balloon is increasing at the rate of  cm sec.

### 9. A balloon, which always remains spherical has a variables radius. Find the rate at which its volume is increasing with the radius when the later is 10 cm.

Ans. Since, V =

=  =

Therefore, the volume is increasing at the rate of  cm3/sec.

### 10. A ladder 5 cm long is leaning against a wall. The bottom of the ladder is pulled along the ground, away from the wall, at the rate of 2 cm/s. How fast is its height on the wall decreasing when the foot of the ladder is 4 m away from the wall?

Ans. Let AB be the ladder and C is the junction of wall and ground, AB = 5 m  B

Let CA =  meters, CB =  meters

According to the equation,

increases,  decreases

and  = 2 cm/s

In AC2 + BC2 = AB2  [Using Pythagoras theorem]

……….(i)

……….(ii)

When

[From eq. (i)]

From eq. (ii),   cm/s

### 11. A particle moves along the curve  Find the points on the curve at which the coordinate is changing 8 times as fast as the coordinate.

Ans. Given: Equation of the curve  ……….(i)

Let  be the required point on curve (i)

According to the question,  ……….(ii)

From eq. (i),

[From eq. (ii)]

Taking

Required point is (4, 11).

Taking

Required point is .

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

Ans. Let  cm be the radius of the air bubble at time

According to question,   is positive =  cm/sec  ……….(i)

Volume of air bubble

=

=

=

Therefore, required rate of increase of volume of air bubble is  cm3/sec.

### 13. A balloon which always remains spherical, has a variable diameter  Find the rate of change of its volume with respect to

Ans. Given: Diameter of the balloon =

Volume of the balloon =

=  cu. units

Rate of change of volume w.r.t.  =

=

=

=

### 14. Sand is pouring from a pipe at the rate of 12 cm3/s. The falling sand forms a cone on the ground in such a way that the height of the cone is always one-sixth of the radius of the base. How fast is the height of the sand cone increasing when the height is 4cm?

Ans. Let the height and radius of the sand-cone formed at time  second be  cm and  cm respectively.

According to question,

Volume of cone (V) =

=

=

Now, since

cm/sec

### 15. The total cost C in rupees associated with the production of  units of an item given by  C Find the marginal cost when 17 units are produced.

Ans. Marginal cost =

=

=

Now, when MC

=

= 6.069 – 0.102 + 15 = 20.967

Therefore, required Marginal cost is  20.97.

### 16. The total revenue in rupees received from the sale of  units of a product is given by

R  Find the marginal revenue when

Ans. Marginal Revenue (MR) =

=

=

Now, when MR = 26 x 7 + 26 = 208

Therefore, the required marginal revenue is  208.

Choose the correct answer in Exercises 17 and 18.

### 17. The rate of change of the area of a circle with respect to its radius  at  =6 cm is:

(A)

(B)

(C)

(D)

Ans. Area of circle (A) =

=

Therefore, option (B) is correct.

### NCERT Solutions class 12 Maths Exercise 6.1

18. The total revenue in Rupees received from the sale of  units of a product is given by

R  The marginal revenue, when  = 15 is:

(A) 116

(B) 96

(C) 90

(D) 126

Ans. Total revenue

Marginal revenue =

= 6 x 15 + 36 = 126

Therefore, option (D) is correct.

## NCERT Solutions class 12 Maths Exercise 6.1

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