 NCERT solutions for class 12 maths exercise 6.1 of chapter 6-Application of Derivatives - Future Study Point # NCERT solutions for class 12 maths exercise 6.1 of chapter 6-Application of Derivatives # NCERT solutions for class 12 maths exercise 6.1-Application of Derivatives

NCERT Solutions for class 12 maths exercise 6.1 of chapter 6-Application of derivatives are created by an expert CBSE teacher by step by step method. NCERT Solutions for class 12 maths exercise 6.1-Application of derivatives is very easy to understand here because all NCERT solutions are explained here beautifully.Class 12 maths NCERT solutions of exercise 6.1 of chapter 6 are the solutions of unsolved questions of NCERT maths text book prescribed by India’s most popular board CBSE.In chapter 6 you will study about the application of derivatives based on the different kinds of geometrical figures.

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Exercise 6.2 – Application of Derivatives

Exercise 6.3 -Application of Derivatives

Exercise 6.4- Application of Derivatives

Exercise 6.5 – Application of Derivatives

## NCERT solutions for class 12 maths exercise 6.1-Application of Derivatives

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Q1. 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

Ans. The area of circle is given as following

A = π r²

The rate of change of area of the circle is with respect to radius = Differentiation of area with respect to radius

Differentiating area(A) with respect to radius(r)

(a) When r = 3 cm, the rate of change of area is

(b) When r= 4, the rate of change of area is

Q2.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 ?

Ans. Let the side of the cube is = x and its volume is = V

The volume of cube is given as follows

V = x³

Since we are given the rate of change of volume with respect to time

Now, we have to evaluate the increasing rate of surface area

Surface area of the cube  is given by following formula

S = 6x²

Diffrentiating both sides with respect to t

Putting the value of dx/dt = 8/3x²

When the age of cube is = 12 cm, then rate of increasing surface area

Hence,the increasing rate of surface area is= 8/3 cm²/s

Q3.The radius of the circle is increasing uniformly at the rate of 3 cm/s .Find the rate at which the area of circle is increasing when the radius is 10 cm.

Ans. We are given the increasing rate of radius = 3 cm/s

Let the radius of circle is = r

The area of the circle is given by

A = π r²

Putting the value of dr/dt = 3

When the radius of circle is = 10 cm, the rate of increasing area of the circle is

Therefore the increasing rate of area is 60π when radius of circle is 10 cm

Q4.An edge of the variable cube is increasing at the  rate of 3 cm/s .How fast is the volume of cube increasing when the edge if 10 cm long.

Ans.Let the edge of the variable cube at time t is = x

We are given the increasing rate of the edge is

Volume of the cube is given by

V = x³

The rate of the increasing  volume of the cube

Putting the value of dx/dt = 3 cm/s

The rate of increasing volume of cube when side, x= 10 cm

Hence, the rate of increasing volume of the cube is 900 cm³/s when side =10 cm

Q5. A stone is dropped into a quite lake and waves move in circles at the rate of 5 cm/s. At the instant when the radius of circular wave is 8 cm, how fast is the enclosed area increasing?

Ans.Let the radius of circular wave is = r at time t

We are given the increasing rate of change of radius of circular wave = 5 cm/s

The area of circle is given as following

A = π r²

The rate of change of  enclosed area of the circle is with respect to radius = Differentiation of area with respect to time

Differentiating area(A) with respect to radius(t)

Putting the value of dr/dt = 5 cm/s

At the instant when radius is = 8 cm, the rate of increasing enclosed area of circle is

Hence, when the radius is 8 cm the rate of change of the enclosed area of the circular wave is = 80π cm²

### NCERT solutions for class 12 maths exercise 6.1-Application of Derivatives

Q6.The radius of a circle is increasing at the rate of 0.7cm/s. what is the rate of its circumference?

Ans. Consider r cm be the radius of the circle at time t.

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

Rate of the circumference of the circle means, how fast circumference is increasing if the radius is increasing at the rate of 0.7 cm/s

The circumference of a circle is given by

l = 2πr

Putting the value of dr/dt = 0.7 cm/s

Therefore the rate of the circumference is = 1.4 cm/s, if the radius is increasing at the rate of 0.7 cm/s.

Q7.The length x of a rectangle is decreasing at the rate of 5cm/minute and the width,y is increasing at the rate of 4 cm/minute. When x = 8cm and y =6cm, find the rates of change of (a) the perimeter and (b)the area of the rectangle

Ans. The length of the rectangle is =x

We are given the decreasing rate of the length = -5 cm/minute

We are given increasing rate of width of the rectangle

The parameter of the rectangle, z is given by

z = 2x + 2y

Putting the value of dx/dt and dy/dt

Therefore the rate of change of parameter is -2 cm/minute or parameter of rectangle is decreased at the rate of 2 cm/minute.

The area of rectangle, A is given by

A = xy

Putting the value of dx/dt , dy/dt, x= 8 and y= 6

Since rate of change of area is positive,so the area of rectangle is increased at the rate of 2 cm²/minute.

Q8.A balloon, which always remains spherical on inflammation, 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 15cm

Ans. Since the balloon always remains spherical on inflammation,so the volume of balloon, V is given as follows

The rate of change of volume of balloon is given 900 cubic centimeter/second

Differentiating the volume equation of baloon

Putting the value of dV/dt

The rate of change of area = 225/πr²

Hence the rate of change of radius (dr/dt), when radius is 15 cm

Q9.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 10cm.

Ans.

The volume of balloon(V)  with variable radius is given as follows

The rate of increasing its volume with respect to radius is

The rate of increasing the volume of the balloon when the radius is 10 cm

Hence, the rate of the increasing volume of the radius is 400 cm³/ cm.

Q10. A ladder 5m long is leaning against a wall. The bottom of the ladder is pulled among 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 the height of the wall at which the ladder touches = y m and the distance of foot of the ladder on the ground = x m

We are given that the distance of foot of the ladder is increasing at the rate of 2 cm/s.

Now,let’s find out the rate of the increasing height of the ladder i.e dy/dt

Applying the Pythagoras theorem

x² + y² = 5²

x² + y² =25

The rate of increasing height when x = 4 m

Q11.A particle moves along the curve 6y = x³ + 2. Find the points on the curve at which the y-coordinate is changing 8 times as fast as the x-coordinate.

Ans.We are given the equation of the curve

6y = x³ + 2

Let’s find the rate of change of the position of the particle on x and y-axis

We are given the condition that the y-coordinate is changing 8 times as fast as the x-coordinate.

Substituting this value in the differential equation we got

x² = 16 ⇒ x = ±4

Putting value of x = 4 and x = -4 in the equation of the curve

y =(4³ +2)/6 = 66/6 = 11 and y=(-4³ +2)/6 = -62/6=-31/3

Hence the points on the curve at which the y-coordinate is changing 8 times as fast as the x-coordinate are (4,11) and (-4, -31/3)

### NCERT solutions for class 12 maths exercise 6.1-Application of Derivatives

Q12.The radius of an air bubble is increasing at the rate of the  1/2 cm/s. At what rate is the volume of the bubble increasing when the radius is 1cm?

Ans.We are given the radius of an air bubble is increasing at the rate of 1/2 cm/s.

Volume of the bubble (V) is given as follows

Putting the value of dr/dt = 1/2

Rate of change of volume is =2πr²

At r = 1 cm the rate of change of volume is = 2π(1)²=2π cm³/s

Q13.A balloon, which always remains spherical , has a variable diameter 3/2(2x+1) .Find the rate of change of its volume with respect to x.

Ans. We are given variable diameter(2r)

Volume (V) of the balloon is given as follows

The rate of change of volume is

Putting the value of dr/dx =3/2 and r= 3/4(2x+1)

Q14.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 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.

Ans.Let the height of cone is = h and radius of base is = r at time t

The given relation between height and radius of base of the cone is

h = r/6

r = 6h

We are given the rate of increasing the volume of cone =12 cm³/s

The volume of cone is given as follows

Putting r = 6h

V = 12πh³

Differentiating Volume with respect time

Putting the value of dV/dt and h = 4 cm

Hence the rate of increasing height of the cone is (1/48π) cm/s when height is 4 cm.

Q15.The total cost C(x) in rupees associated with the production of x units of an item given by C(x) = 0.007x³ – 0.003x² + 15x + 4000. Find the marginal cost when 17 units are produced.

Ans. We are given the relationship between total cost C(x) and the number of units (x) produced

C = 0.007x³ – 0.003x² + 15x + 4000

We know the marginal cost is the rate of change in cost with respect to items produced,so differentiating C with respect to items produced(x)

Putting x = 17, we have

Hence the marginal cost of the production is Rs 20.987 when 17 units of the items are produced.

Q16.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.

Ans. We are given total revenue (R) from the sale of (x) units.

R = 13x² + 26x  + 15

Marginal revenue is the rate of change in revenue with respect to sold units of the product, so differentiating R with respect to x.

Marginal revenue ,when x = 7

Hence,the marginal revenue is Rs 208.

### NCERT solutions for class 12 maths exercise 6.1-Application of Derivatives

Q17.The rate of change the area of a circle with respect to its radius r at r = 6 cm is:

(A) 10 π

(B) 12π

(C)8π

(D) 11π

Ans. The area (A) of circle is given as follows

A = π r²

Rate of change of area with respect to radius,when r =6 cm is

Hence,the correct answer is (B) 12π cm²

Q18.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

Ans. We are given total revenue R(x)

R= 3x² + 36x + 5.

Marginal revenue when x= 15 is

Hence, the correct answer is (D) Rs 126.

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