**NCERT Solutions for Maths Class 12 Exercise 6.1**.

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**NCERT Solutions for Maths Class 12 Exercise 6.2**

**NCERT Solutions for Maths Class 12 Exercise 6.3**

**NCERT Solutions for Maths Class 12 Exercise 6.1**

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**Maths Class
12 Ex 6.1 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. Then, A = Ï€*r*^{2}

Therefore, the rate of change of the
area with respect to its radius is given by

a. When r = 3 cm, then

dA/dr = 2Ï€(3) = 6Ï€

Hence, the
rate of change of the area of the circle is 6Ï€.

b. When r = 4
cm, then

dA/dr = 2Ï€(4) = 8Ï€

Hence, the
rate of change of the area of the circle is 8Ï€.

**Maths Class
12 Ex 6.1 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 side
length of the cube, then volume V = x³, and surface area S = 6x^{2}

Hence, the
surface area is increasing at the rate of 8/3 cm^{3}/s.

**Maths Class
12 Ex 6.1 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 the circle, A = Ï€*r*², and also
dr/dt = 3 cm/s.

Hence, the area
of the circle is increasing at the rate of 60Ï€ cm^{2}/s.

**Maths Class
12 Ex 6.1 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 be x cm. Then, dx/dt = 3 cm/s

Now, volume of
the cube, V = x^{3} and dV/dt = 3x^{2}

Hence, the volume
of the cube is increasing at the rate of 900 cm^{3}/s.

**Maths Class
12 Ex 6.1 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. Given,
dr/dt = 5 cm/s.

We have, the area
of the circle, A = Ï€*r*² and dA/dr = 2Ï€*r*

Hence, the
enclosed area is increasing at the rate of 80Ï€ cm^{2}/s.

**Maths Class
12 Ex 6.1 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 the radius of a circle is given by 0.7 cm/s, i.e., if the radius is
r, then dr/dt = 0.7 cm/s.

Now, circumference of the circle, C = 2Ï€r

Hence, the rate of increase of its circumference is 1.4Ï€ cm/s.

**Maths Class
12 Ex 6.1 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
the rate of 5 cm/min.

Therefore,
dx/dt = –5 cm/min … (i)

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

**Maths Class
12 Ex 6.1 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 = (4/3)Ï€*r*^{3}

**Maths Class
12 Ex 6.1 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.

Let the volume of
the balloon be V. Then,

Hence, the volume
is increasing at the rate of 400Ï€ cm^{3}/s.

**Maths Class
12 Ex 6.1 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 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?

**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,

**Maths Class
12 Ex 6.1 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)

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

The radius of an
air bubble is increasing at the rate of 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 of the air bubble, then
the volume, V = (4/3)Ï€*r*^{3}

^{3}/s.

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

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.

**Solution:**

Diameter of the
sphere = (3/2)[2x + 1]

∴ Radius = (3/4)[2x + 1]

**Maths Class
12 Ex 6.1 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 = *r*/6 … (i)

**Maths Class
12 Ex 6.1 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 output = dC/dx

So, when 17 units are produced, the
marginal cost is Rs 20.967.

**Maths Class
12 Ex 6.1 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 = dR/dx

We know that, R(x) = 13x² + 26x + 15,

MR = dR/dx = 26x + 26 = 26(x + 1)

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

Hence, the marginal revenue is Rs 208.

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

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

(A) 10Ï€ (B)
12Ï€ (C) 8Ï€ (D) 11Ï€

**Solution:**

(B) ∵ A = Ï€r²

Therefore, dA/dr = 2Ï€r = 2Ï€ × 6 = 12Ï€ cm²/s

Hence, option (B) is the correct answer.

**Maths Class
12 Ex 6.1 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) We have, R(x)
= 3x² + 36x + 5

MR = dR/dx = 6x + 36

At x = 15; dR/dx = 6 × 15 + 36 = 90 + 36 = 126

Hence, option (D) is the correct answer.

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