Ncert Solutions For Class 12 Maths Ex 6.1

Ncert Solutions For Class 12 Maths Chapter 6 Ex 6.1

Q-1) For a circle, with radius given below, find the rate of change of the area w.r.t raduis’a’.

i) a = 2mm,

ii) a = 5mm

Ans.) We know that the area of circle with radius ‘a’ is \(\pi a^{2}\).

i.e. A = \(\pi a^{2}\)

The rate of change of area w.r.t ‘a’ is,

\(\frac{\mathrm{d} A}{\mathrm{d} a} = \frac{\mathrm{d} (\pi a^{2})}{\mathrm{d} a} = 2\pi a\)

a = 2mm

\(\frac{\mathrm{d} A}{\mathrm{d} a} = 2\pi (2) = 4\pi\)

Thus, \(4\pi(mm^{2} /s)\) is the rate of change of area of circle, when a = 2mm.

a = 5mm

\(\frac{\mathrm{d} A}{\mathrm{d} a} = 2\pi (5) = 10\pi\)

Thus, \(10\pi(mm^{2} /s)\) is the rate of change of area of circle, when a = 5mm.

 

 

Q-2) Find the increase in surface area of a cube, given that cube’s volume is increasing at the rate of \(27(mm^{2}/s)\) and the length of the edge is 18mm.

Ans.) Lets assume the length of edge is ymm, surface area as S and volume as V.

So, V = \(y^{3}\), S = \(y^{2}\) ; y is a function of time x.

In the question its is given that \(\frac{\mathrm{d} V}{\mathrm{d} x} = 27(mm^{3}/s)\).

Applying chain rule, we get:

\(27 = \frac{\mathrm{d} V}{\mathrm{d} x} = \frac{\mathrm{d} (y^{3})}{\mathrm{d} x} = \frac{\mathrm{d}( y^{3})}{\mathrm{d} y}*\frac{\mathrm{d} y}{\mathrm{d} x} = 3y^{2}*\frac{\mathrm{d} y}{\mathrm{d} x}\)

\(\frac{\mathrm{d} y}{\mathrm{d} x} = 27/3y^{2}\)……………..(a)

Applying chain rule on surface area, we get

\(\frac{\mathrm{d} S}{\mathrm{d} x} = \frac{\mathrm{d} (6y^{2})}{\mathrm{d} x} = \frac{\mathrm{d} (6y^{2})}{\mathrm{d} y}*\frac{\mathrm{d} y}{\mathrm{d} x} = 12y*\frac{\mathrm{d} y}{\mathrm{d} x}\)

\(=12y( 27/3y^{2}) = 108/y\) (\(because \; from\; (a)\))

Here its is given that y = 18mm.

\( \frac{\mathrm{d} S}{\mathrm{d} x} = 108/18 = 6mm^{2}/s\)

Thus, the surface area of a circle is increasing at the rate of \(6mm^{2}/s\) when the edge length is 18mm.

 

 

Q-3) For a circle, determine the rate of change of area when radius ‘y’ is 20mm. The radius of circle is increase at the rate of 6mm/s.

Ans.) We know that the area of circle with radius ‘y’ is \(\pi y^{2}\).

i.e. A = \(\pi y^{2}\)

The rate of change of area w.r.t time ‘x’ is,

Applying chain rule in above equation, we get

\(\frac{\mathrm{d} A}{\mathrm{d} x} = \frac{\mathrm{d} (\pi y^{2})}{\mathrm{d} x} = \frac{\mathrm{d}(\pi y^{2}) }{\mathrm{d} y}*\frac{\mathrm{d} y}{\mathrm{d} x} = 2\pi y\frac{\mathrm{d} y}{\mathrm{d} x}\)

Here, given that, \(\frac{\mathrm{d} y}{\mathrm{d} x} = 6mm/s\)

\( \frac{\mathrm{d} A}{\mathrm{d} x} = 2\pi y(6) = 12\pi y\)

Also given that radius y = 20mm

\( \frac{\mathrm{d} A}{\mathrm{d} x} = 12\pi (20) = 240mm^{2}/s\)

Thus, \(240mm^{2} /s\) is the rate of change of area of circle, when y = 20mm.

Q-4) For a variable cube, find the increase in volume of a cube when the length of the edge is 20mm, given that cube’s edge is increasing at the rate of 6mm/s.

Ans.) Lets assume the length of edge is ymm, and volume as V.

So, V = \(y^{3}\) ; y is function of time ‘x’.

Then, By applying chain rule we get,

\( \frac{\mathrm{d} V}{\mathrm{d} x} = 3y^{2}*\frac{\mathrm{d} y}{\mathrm{d} x}\)

Here it is given that,

\(\frac{\mathrm{d} y}{\mathrm{d} x} = 6mm/s\) \( \frac{\mathrm{d} V}{\mathrm{d} x} = 3y^{2}(6) = 18y^{2}\)

Also given that edge length is y = 20mm

\( \frac{\mathrm{d} V}{\mathrm{d} x} = 18(20^{2}) = 7200mm^{3}/s\)

Thus, the Volume of a variable cube is increasing at the rate of \(7200mm^{3}/s\) when the edge length is 20mm.

 

 

Q-5) When a stone is thrown into a steady pond the circular waves are created and they started travelling at the rate of 10mm/s. Find how fast the area enclosed by circular wave is increasing at an instant when the radius of the circle is 16mm.

Ans.)We know that the area of circle with radius ‘y’ is \(\pi y^{2}\).

i.e. A = \(\pi y^{2}\)

The rate of change of area w.r.t time ‘x’ is,

Applying chain rule in above equation, we get

\(\frac{\mathrm{d} A}{\mathrm{d} x} = \frac{\mathrm{d} (\pi y^{2})}{\mathrm{d} x} = \frac{\mathrm{d}(\pi y^{2}) }{\mathrm{d} y}*\frac{\mathrm{d} y}{\mathrm{d} x} = 2\pi y\frac{\mathrm{d} y}{\mathrm{d} x}\)

Here, given that, \(\frac{\mathrm{d} y}{\mathrm{d} x} = 10mm/s\)

\( \frac{\mathrm{d} A}{\mathrm{d} x} = 2\pi y(10) = 20\pi y\)

Also given that radius y = 16mm

\( \frac{\mathrm{d} A}{\mathrm{d} x} = 20\pi (16) = 320mm^{2}/s\)

Thus, \(320mm^{2} /s\) is the rate of change of area enclosed by circular wave, when y = 16mm.

 

 

Q-6) For a circle, find the rate at which the circumference is increasing given that its radius is increasing at the rate of 1.4mm/s.

Ans.) We know that the area of circle with radius ‘y’ is

C = \(2\pi y\)

The rate of change of circumference w.r.t time ‘x’ is obtained by applying chain rule,

\(\frac{\mathrm{d} C}{\mathrm{d} x} = \frac{\mathrm{d} (2\pi y)}{\mathrm{d} x} = \frac{\mathrm{d} (2\pi y)}{\mathrm{d} y}*\frac{\mathrm{d} y}{\mathrm{d} x} = 2\pi \frac{\mathrm{d} y}{\mathrm{d} x}\)

Here its is given that, \(\frac{\mathrm{d} y}{\mathrm{d} x} = 1.4mm/s\)

\( \frac{\mathrm{d} C}{\mathrm{d} x} = 2\pi (1.4) = 2.8mm/s\)

Thus, 2.8mm/s is the rate of increase in circumference.

 

 

Q-7) Find the rate at which the perimeter and area of rectangle is changing, when a = 16mm, b = 12mm. Given that the length ‘a’ of a rectangle is increasing at 10mm/min and width of a rectangle is decreasing at 8mm/min.

Ans.)Here given that rectangle is increasing at 10mm/min and width of a rectangle is decreasing at 8mm/min,

\(\frac{\mathrm{d} a}{\mathrm{d} x} = 10mm/min\; and \; \frac{\mathrm{d} b}{\mathrm{d} x} = -8mm/min\)

Perimeter of rectangle, P = 2(a + b)

The rate of change of perimeter w.r.t time ‘x’ is,

\( \frac{\mathrm{d} P}{\mathrm{d} x} = 2(\frac{\mathrm{d} a}{\mathrm{d} x} + \frac{\mathrm{d} b}{\mathrm{d} x}) = 2(10 – 8) = 4mm/min\)

Thus, the perimeter of a rectangle is increasing at 4mm/min.

Area of rectangle, A = a*b

The rate of change of area w.r.t time ‘x’ is,

\( \frac{\mathrm{d} A}{\mathrm{d} x} = \frac{\mathrm{d} a}{\mathrm{d} x}*b + \frac{\mathrm{d} a}{\mathrm{d} x}*a = 10b – 8a\)

Here, when a = 16mm and b = 12mm,

\( \frac{\mathrm{d} A}{\mathrm{d} x} = 10(12) – 8(16) = 120 – 128 = -8mm^{2}/min\)

Thus, the area of rectangle is decreasing at \(8mm^{2}/min\).

 

 

Q-8) A football is being inflated by pumping 1600cc of air per second. Find the rate at which the radius of football will increase at an instant when radius y = 10cm.

Ans.) As we know the shape of a football is spherical.

The volume(V) of sphere is given by,

\(V = \frac{4}{3}\pi y^{3}\)

By applying chain rule in above equation by differentiating w.r.t time ‘x’, we get

\(\frac{\mathrm{d} V}{\mathrm{d} x} = \frac{\mathrm{d} (\frac{4}{3}\pi y^{3})}{\mathrm{d} x} = \frac{\mathrm{d} (\frac{4}{3}\pi y^{3})}{\mathrm{d} y}*\frac{\mathrm{d} y}{\mathrm{d} x} = 4\pi y^{2}\frac{\mathrm{d} y}{\mathrm{d} x}\)

Here, \(\frac{\mathrm{d} V}{\mathrm{d} x} = 1600cc/s\) (given)

\( 1600 = 4\pi y^{2}\frac{\mathrm{d} y}{\mathrm{d} x}\) \( \frac{\mathrm{d} y}{\mathrm{d} x} = \frac{1600}{4\pi y^{2}} = \frac{400}{\pi y^{2}}\)

Here radius y = 10cm (given)

\( \frac{\mathrm{d} y}{\mathrm{d} x} = \frac{400}{\pi 10^{2}} = \frac{4}{\pi }cm/s\)

Thus, radius of football is increasing at\(\frac{4}{\pi }cm/s\) when radius y = 10cm .

 

 

Q-9) Find out the rate of change of volume of a football w.r.t its radius at an instant when its radius is 200mm. The radius of the spherically shaped football is variable.

Ans.)As we know the shape of a football is spherical.

The volume(V) of sphere is given by,

\(V = \frac{4}{3}\pi x^{3}\)

By applying chain rule in above equation by differentiating w.r.t radius ‘x’, we get

\(\frac{\mathrm{d} V}{\mathrm{d} x} = \frac{\mathrm{d} (\frac{4}{3}\pi x^{3})}{\mathrm{d} x} = 4\pi x^{2}\)

Here, radius is given 200mm,

\( \frac{\mathrm{d} V}{\mathrm{d} x} = 4\pi (200)^{2} = 160000\pi \;mm^{3}/s\)

Thus the volume of a football is increasing at the rate of \(160000\pi \;mm^{3}/s\).

 

 

Q-10) An object travels along the curve 12a = \(12b = 2a^{3}+ 4\).Determine the points on the curve where the ‘a’ co-ordinate is (1/8)th of the ‘b’ co-ordinate.

Ans.) Here it is given that the equation of the curve is,

\(12b = 2a^{3}+ 4\)

By differentiating above equation w.r.t. time ‘x, we get

\(12\frac{\mathrm{d} b}{\mathrm{d} x} = 6a^{2}\frac{\mathrm{d} a}{\mathrm{d} x} + 0\)

\(2\frac{\mathrm{d} b}{\mathrm{d} x} = a^{2}\frac{\mathrm{d} a}{\mathrm{d} x}\)………………………….(1)

Here it is given that,‘a’ co-ordinate is (1/8)th of the ‘b’ co-ordinate

i.e. \(\frac{1}{8}\frac{\mathrm{d} b}{\mathrm{d} x} = \frac{\mathrm{d} a}{\mathrm{d} x}\)

\( \frac{\mathrm{d} b}{\mathrm{d} x} = 8\frac{\mathrm{d} a}{\mathrm{d} x}\)

Putting this value in equation (1)

\( 2(8\frac{\mathrm{d} a}{\mathrm{d} x}) = a^{2} \frac{\mathrm{d} a}{\mathrm{d} x}\) \( 16\frac{\mathrm{d} a}{\mathrm{d} x} = a^{2} \frac{\mathrm{d} a}{\mathrm{d} x}\) \( (a^{2} – 16 ) \frac{\mathrm{d} a}{\mathrm{d} x} = 0\) \( a^{2} = 16\) \( a = \pm 4\)

Now, for a = – 4

\( b = \frac{2(-4)^{3} + 4}{12} = \frac{132}{12} = 11\)

For, a = 4

\( b = \frac{2(4)^{3} + 4}{12} = \frac{124}{12} = \frac{-31}{3}\)

Thus, the required co-ordinate or points are \(\left ( -4,\frac{-31}{3} \right )\; and \left ( 4,11 \right )\).

 

 

Q-11) A wall has a stick of 13m leaning on it. The base of the stick is pulled away from wall along the ground at 4cm/s. Find out at which rate the height of the stick will decrease when the bottom point of the stick is 5m far from the wall.

Ans.) Let a m be the distance from bottom of stick to wall and b m be the height of the wall where the stick is in contact with the wall.

Thus, we can apply Pythagoras theorem in this case, then we get

\(a^{2}+ b^{2} = 169\)

( stick is of 13m length)

\( b = \sqrt{169 – a^{2}}\)

By differentiating above equation w.r.t time ‘x’ we will get rate of change of height,

\(\frac{\mathrm{d} b}{\mathrm{d} x} = \frac{-a}{\sqrt{169 – a^{2}}}*\frac{\mathrm{d} a}{\mathrm{d} x}\)

Here we have \(\frac{\mathrm{d} a}{\mathrm{d} x} = 4cm/s\)

\( \frac{\mathrm{d} b}{\mathrm{d} x} = \frac{-4a}{\sqrt{169 – a^{2}}}\)

Here, the bottom point of the stick is 5m far from the wall, a = 5m

\( \frac{\mathrm{d} b}{\mathrm{d} x} = \frac{-4(5)}{\sqrt{169 – (5)^{2}}} = \frac{-20}{12} = \frac{-5}{3}\)

Thus, the height of the stick is decreasing at the rate of (-5/3)cm/s.

 

 

Q-12) Find the rate at which the volume of a balloon is increasing at an instant when its radius is 2cm. The radius of balloon is increasing at the rate of 1cm/s.

Ans.)As we know the shape of a balloon is spherical.

The volume(V) of sphere with radius ‘y’ is given by,

\(V = \frac{4}{3}\pi y^{3}\)

By applying chain rule in above equation by differentiating w.r.t time ‘x’, we get

\(\frac{\mathrm{d} V}{\mathrm{d} x} = \frac{\mathrm{d} (\frac{4}{3}\pi y^{3})}{\mathrm{d} x} = \frac{\mathrm{d} (\frac{4}{3}\pi y^{3})}{\mathrm{d} y}*\frac{\mathrm{d} y}{\mathrm{d} x} = 4\pi y^{2}\frac{\mathrm{d} y}{\mathrm{d} x}\)

Here, given that,

\(\frac{\mathrm{d} y}{\mathrm{d} x} = 1cm/s\) \( \frac{\mathrm{d} V}{\mathrm{d} x} = 4\pi y^{2}(1) = 4\pi y^{2}cm^{3}/s\)

Also given that the radius y = 2cm,

\( \frac{\mathrm{d} V}{\mathrm{d} x} = 4\pi (2)^{2} = 8\pi cm^{3}/s\)

Thus, the volume of balloon is increasing at the rate of \(8\pi cm^{3}/s\).

 

 

Q-13) A football is having radius \(\frac{5}{2}(4a + 1)\) which is variable. Find the rate of change in volume of football w.r.t. ‘a’.

Ans.) As we know the shape of a football is spherical.

The volume(V) of sphere with radius ‘y’ is given by,

\(V = \frac{4}{3}\pi y^{3}\)

Here it is given that y = \(\frac{5}{2}(4a + 1)\)

\( V = \frac{4}{3}\pi (\frac{5}{2}(4a + 1))^{3} = \frac{125}{6}\pi (4a + 1)^{3}\)

Differentiating above equation w.r.t ‘a’, we get,

\( \frac{\mathrm{d} V}{\mathrm{d} a} = \frac{\mathrm{d} \frac{125}{6}\pi(4a + 1)^{3}}{\mathrm{d} a} = \frac{125}{6}\pi *3(4a + 1)^{2}*4 = 250(4a + 1)^{2}\)

 

 

Q-14) A wheat flour is being pouring from a sprue at \(24cm^{2}/s\). The fallen wheat flour forms a conical shape in the ground in such a way that the radius of the cone at base is 6 times the height of the cone. Then find out the rate of increase in the height of cone at an instant when the height is 8cm.

Ans.) Let V be the volume of cone , x be the radius of base of cone and y be the height of the cone which is increasing.

So, volume of cone is given by,

\(V = \frac{1}{3}\pi x^{2}y\)

Here , x = 6y is given.

\( V = \frac{1}{3}\pi (6y)^{2}y = 12\pi y^{3}\)

Now, the rate of increase in the height of cone is obtained by differentiating above equation w.r.t. time ‘t’

\(\frac{\mathrm{d} V}{\mathrm{d} t} = 12\pi \frac{\mathrm{d} y^{3}}{\mathrm{d} y}*\frac{\mathrm{d} y}{\mathrm{d} t} = 36\pi y^{2}\frac{\mathrm{d} y}{\mathrm{d} t}\)

Here, \(\frac{\mathrm{d} V}{\mathrm{d} t} = 24cm^{3}/s\)

\( 24 = 36\pi y^{2}\frac{\mathrm{d} y}{\mathrm{d} t}\)

Also the y = 8cm is given,

\( 24 = 36\pi (8)^{2}\frac{\mathrm{d} y}{\mathrm{d} t}\) \( \frac{\mathrm{d} y}{\mathrm{d} t} = \frac{1}{96\pi }\)

Thus, the height of wheat flour cone is increasing at the rate of \(\frac{1}{96\pi } cm/s\).

 

 

Q-15) S(x) = \(0.014a^{3} – 0.006a^{2} + 30a + 8000\)

The above equation represents the total cost of the ‘a’ units manufactured.

Give the value of the marginal cost when 34 units are manufactured.

Ans.) Marginal cost(MC) is defined as the rate of change of total cost w.r.t. the numbers of units manufactured.

MC = \(\frac{\mathrm{d}C }{\mathrm{d} a} = \frac{\mathrm{d} (0.014a^{3} – 0.006a^{2} + 30a + 8000)}{\mathrm{d} a} = 0.042a^{2} – 0.012a +30\)

Here, a = 34 is given.

\( MC = \frac{\mathrm{d}C }{\mathrm{d} a} = 0.042a^{2} – 0.012a +30 = 0.042(34)^{2} – 0.012(34) + 30 = 78.144\)

Thus, marginal cost for 34 units manufactured is Rs.78.144.

 

 

Q-16) C(x) = \(36y^{2} + 52y + 30\)

The above equation shows the revenue generated from y units sold.

Then give the value of the marginal revenue for 14 units sold.

Ans.)Marginal Revenue(MR) is defined as the rate of change of total revenue generated w.r.t. numbers of units sold.

\(MR = \frac{\mathrm{d} C}{\mathrm{d} y} = \frac{\mathrm{d} (36y^{2} + 52y + 30)}{\mathrm{d} y} = 72y + 52\)

Here, y = 14 units (given)

MR = 72(14) + 52 = 1060

Thus, marginal revenue for 14 units sold is Rs.1060.

 

 

Q-17) For a circle the values for rate of change of area w.r.t. its radius at an instant when its radius is 12cm is given, find the correct option from the options given below.

(i) \(20\pi\)

(ii) \(10\pi\)

(iii) \(40\pi\)

(iv) \(24\pi\)

Ans.)We know that the area of circle with radius ‘a’ is \(\pi a^{2}\).

i.e. A = \(\pi a^{2}\)

The rate of change of area w.r.t ‘a’ is,

\(\frac{\mathrm{d} A}{\mathrm{d} a} = \frac{\mathrm{d} (\pi a^{2})}{\mathrm{d} a} = 2\pi a\)

Here a = 12cm is given,

\( \frac{\mathrm{d} A}{\mathrm{d} a} = 2\pi (12) = 24cm^{2}/s\)

Thus, Option (iv) is correct.

 

 

Q-18) C(x) = \(6y^{2} + 72y + 10\)

The above equation shows the revenue generated from y units sold.

Then give the value of the marginal revenue for 30 units sold from the options given below.

(i) 118

(ii) 126

(iii) 432

(iv) 180

Ans.)Marginal Revenue(MR) is defined as the rate of change of total revenue generated w.r.t. numbers of units sold.

\(MR = \frac{\mathrm{d} C}{\mathrm{d} y} = \frac{\mathrm{d} (6y^{2} + 72y + 10)}{\mathrm{d} y} = 12y + 72\)

Here, y = 30 units (given)

MR = 12(30) + 72 = 432

Thus, marginal revenue for 30 units sold is Rs.432

Thus, option (iii) is correct.