NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.1 are part of NCERT Solutions for Class 12 Maths. Here we have given NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.1.

- Application of Derivatives Class 12 Ex 6.2
- Application of Derivatives Class 12 Ex 6.3
- Application of Derivatives Class 12 Ex 6.4
- Application of Derivatives Class 12 Ex 6.5

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 12 |

Subject |
Maths |

Chapter |
Chapter 6 |

Chapter Name |
Application of Derivatives |

Exercise |
Ex 6.1 |

Number of Questions Solved |
18 |

Category |
NCERT Solutions |

## NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.1

**Ex 6.1 Class 12 Maths Question 1.**

Find the rate of change of the area of a circle with respect to its radius r when (a) r = 3 cm (b) r = 4 cm

**Solution:**

Let A be the area of the circle

**Ex 6.1 Class 12 Maths Question 2.**

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

**Solution:**

Let x be the length of the cube volume V = x³,

**Ex 6.1 Class 12 Maths Question 3.**

The radius of a circle is increasing uniformly at the rate of 3 cm/s. Find the rate at which the area of the circle is increasing when the radius is 10 cm.

**Solution:**

Let r be the radius of the circle.

Area of circle = πr² = A also \(\frac { dr }{ dt }\) = 3 cm/ sec.

**Ex 6.1 Class 12 Maths Question 4.**

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

**Solution:**

Let the edge of the cube = x cm

∴ \(\frac { dx }{ dt }\) = 3

**Ex 6.1 Class 12 Maths Question 5.**

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

**Solution:**

Let r be the radius of a wave circle: \(\frac { dx }{ dt }\) = 5cm/sec.

**Ex 6.1 Class 12 Maths Question 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?

**Solution:**

The rate of change of circle w.r.t time t is given

to be 0.7 cm/sec. i.e. \(\frac { dr }{ dt }\) = 0.7 cm/sec.

Now, circumference of the circle is c = 2πr

\(\therefore \frac { dc }{ dt } =\left[ \frac { d }{ dr } \left( 2\pi r \right) .\frac { dr }{ dt } \right] =2\pi \frac { dr }{ dt } =1.4\pi cm/sec\)

**Ex 6.1 Class 12 Maths Question 7.**

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

(a) the perimeter, and

(b) the area of the rectangle.

**Solution:**

(a) The length x of a rectangle is decreasing at dx the rate of 5cm/min. => \(\frac { dx }{ dt }\) = – 5cm min …(i)

The width y is increasing at the rate of 4cm/min.

**Ex 6.1 Class 12 Maths Question 8.**

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

**Solution:**

Volume of the spherical balloon = V = \(\frac { 4 }{ 3 } \pi { r }^{ 3 }\)

**Ex 6.1 Class 12 Maths Question 9.**

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

**Solution:**

Let r be the variable radius of the balloon which is in the form of sphere Vol. of the sphere

**Ex 6.1 Class 12 Maths Question 10.**

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

**Solution:**

Let AB be the ladder and OB be the wall. At an instant,

let OA = x, OB = y,

x² + y² = 25 …(i)

On differentiating,

**Ex 6.1 Class 12 Maths Question 11.**

A particle moves along the curve 6y = x^{3} + 2. Find the points on the curve at which the y-coordinate is changing 8 times as fast as the x-coordinate.

**Solution:**

We have

6y = x^{3} + 2…(i)

**Ex 6.1 Class 12 Maths Question 12.**

The radius of an air bubble is increasing at the rate of \(\frac { 1 }{ 2 }\) cm/s. At what rate is the volume of the bubble increasing when the radius is 1 cm?

**Solution:**

Let r be the radius then V = \(\frac { 4 }{ 3 } \pi { r }^{ 3 }\)

\(\frac { dr }{ dt } =\frac { 1 }{ 2 } cm/sec\)

\(\frac { dv }{ dt } =\frac { d }{ dt } \left( \frac { 4 }{ 3 } \pi { r }^{ 3 } \right) =\frac { 4 }{ 3 } { \pi .3r }^{ 2 }.\frac { dr }{ dt } ={ 2\pi r }^{ 2 }\)

Hence, the rate of increase of volume when radius is 1 cm = 2π x 1² = 2π cm^{3}/sec.

**Ex 6.1 Class 12 Maths Question 13.**

A balloon, which always remains spherical, has a variable diameter \(\frac { 3 }{ 2 }(2x+1)\) . Find the rate of change of its volume with respect to x.

**Solution:**

Dia of sphere = \(\frac { 3 }{ 2 }(2x+1)\)

∴ Radius = \(\frac { 3 }{ 4 }(2x+1)\)

**Ex 6.1 Class 12 Maths Question 14.**

Sand is pouring from a pipe at the rate of 12 cm^{3}/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 4 cm?

**Solution:**

Let r and h be the radius and height of the sand

– cone at time t respectively, h = \(\frac { r }{ 6 } \) …(i)

**Ex 6.1 Class 12 Maths Question 15.**

The total cost C (x) in Rupees associated with the production of x units of an item is given by C (x) = 0.007 x^{3} – 0.003 x^{2} + 15x + 4000. Find the marginal cost when 17 units are produced.

**Solution:**

Marginal cost MC = Instantaneous rate of change

of total cost at any level of out put = \(\frac { dC }{ dx } \)

**Ex 6.1 Class 12 Maths Question 16.**

The total revenue in Rupees 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.

**Solution:**

Marginal Revenue (MR)

= Rate of change of total revenue w.r.t. the

number of items sold at an instant = \(\frac { dR }{ dx } \)

We know R(x) = 13x² + 26x + 15,

MR = \(\frac { dR }{ dx } \) = 26x + 26 = 26(x+1)

Now x = 7, MR = 26 (x + 1) = 26 (7 + 1) = 208

Hence, Marginal Revenue = Rs 208.

**Choose the correct answer in the Exercises 17 and 18.**

**Ex 6.1 Class 12 Maths Question 17.**

The rate of change of the area of a circle with respect to its radius r at r = 6 cm is

(a) 10π

(b) 12π

(c) 8π

(d) 11π

**Solution:**

(b) ∵ A = πr² => \(\frac { dA }{ dr } \) = 2π x 6 = 12π cm²/radius

**Ex 6.1 Class 12 Maths Question 18.**

The total revenue in Rupees received from the sale of x units of a product is given by R (x) = 3x² + 36x + 5. The marginal revenue, when x = 15 is

(a) 116

(b) 96

(c) 90

(d) 126

**Solution:**

(d) R(x) = 3x² + 36x + 5 ,

MR = \(\frac { dR }{ dx } \) = 6x + 36 ,

At x = 15; \(\frac { dR }{ dx } \) = 6 x 15 + 36 = 90 + 36 = 126

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