NCERT Solutions class 12 Maths Exercise 6.1 (Ex 6.1) Chapter 6 Application of Derivatives

NCERT Solutions for Class 12 Maths Exercise 6.1 Chapter 6 Application of Derivatives – FREE PDF Download

Free PDF download of NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.1 (Ex 6.1) and all chapter exercises at one place prepared by expert teacher as per NCERT (CBSE) books guidelines. Class 12 Maths Chapter 6 Application of Derivatives Exercise 6.1 Questions with Solutions to help you to revise complete Syllabus and Score More marks.

NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives (Ex 6.1) Exercise 6.1

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

cm²/sec

(b) When cm, then

cm²/sec

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 and y be the surface area of the cube at any time t

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

is positive = 8

$=> 3 x^{2} \frac{d x}{d t} = 8$

……….(i)

Rate of change of surface area of the cube =

=

= (from (i) ) = 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 r cm be the radius and A be the area of the circle at any time

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

$\frac{d r}{d t}$ is positive and = 3 cm/sec

$\frac{d r}{d t} = 3 . . . . . (1)$ ……….(i)

$N o w, A = π r^{2}$

Rate of change of area of circle = $\frac{d A}{d t} = π 2 r \frac{d r}{d t}$

$=> π 2 r (3) = 6 π r c m^{2} / s e c$ (from (i) )

putting r = 10 cm (given),

$\frac{d A}{d t}$ = cm²/sec

Since $\frac{d A}{d t}$ 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 and be the volume of the variable cube at any time

Rate of increase of edge = 3 cm/sec

is positive and = 3 cm/sec

= 3 cm / sec ……….(i)

Rate of change of volume of cube = =

= (from (i) )

= 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 and be the enclosed area of the circular wave at any time

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

is positive and = 5 cm/sec

= 5 cm / sec ……….(i)

Rate of change of area = =

= (from (i) )

= cm²/sec

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 increase of its circumference?

Ans. Let cm be the radius and be the circumference of the circle at any time

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

is positive and = 0.7 cm/sec

= 0.7 cm / sec ……….(i)

Rate of change of circumference of circle =

= = (from (i) )

= cm/sec

7. The length of a rectangle is decreasing at the rate of 5 cm/minute and width y is increasing at the rate of 4 cm/minute. When = 8 cm and = 6 cm, find the rate 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

=–5 cm/minute ……….(i)

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

is positive= 4 cm/minute

= 4 cm/minute ……….(ii)

let denotes the perimeter of rectangle at any time t

(from (i) and (ii) )

= which is negative.

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

(b) Let denotes the area of rectangle at any time t

= which is positive. (from (i) and (ii) )

Area of the rectangle is increasing at the rate of 2 cm²/min

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,

cm sec
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 =

= =

as x = 10 cm

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

Let CA = meters, CB = meters

It is given that the bottom of the ladder is pulled along the ground away from the wall at the rate of 2 cm/sec

So increases and decreases

and = 2 cm/s

In AC² + BC² = AB² [Using Pythagoras theorem]

……….(i)

( as = 2 cm/s )

……….(ii)

When [From eq. (i)]

From eq. (ii), cm/s

-ve sign indicates that it is decreasing.

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

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

Let be the required point on curve (i)

since y-coordinate is changing 8 times as fast as the x-coordinate

^{$\frac{d y}{d t} = 3 x^{2} \frac{d x}{d t}$} …….(ii)

From eq. (i),

$6 \frac{d y}{d t} = 3 x^{2} \frac{d x}{d t}$

now from (ii) we subsitute the value of $\frac{d y}{d t}$

$6 (8 \frac{d x}{d t}) = 3 x^{2} \frac{d x}{d t}$

( we cancel from both side )

Taking

Required point is (4, 11).

Taking

Required point is .

so the required points are and (4, 11).

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 =

Radius 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) =

=

=

$=> \frac{d V}{d t} = 36 π y^{2} \frac{d y}{d t}$

Now, since

cm/sec

15. The total cost C(x) in rupees associated with the production of units of an item given by C(x) = 0.007x³ – 0.003x² + 15x+4000. 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 Rs 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 Rs 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.

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.