NCERT Solutions for Class 12 Maths Chapter 6: Application of Derivatives

Question 1
Find the rate of change of the area of a circle with respect to its radius r when
i) a = 3cm,
ii) a = 4cm
Solution
Area of circle with radius ‘r’ is given by
A = $\pi r^{2}$
The rate of change of area w.r.t ‘r’ is,
$\frac{\mathrm{d} A}{\mathrm{d} r} = \frac{\mathrm{d} (\pi r^{2})}{\mathrm{d} r} = 2\pi r$
At r = 3cm
$\frac{\mathrm{d} A}{\mathrm{d} r} = 2\pi (3) = 6\pi$
Thus, $6\pi(cm^{2} / cm)$ is the rate of change of area of circle, when a = 3cm.
At r = 4cm
$\frac{\mathrm{d} A}{\mathrm{d} r} = 2\pi (4) = 8\pi$
Thus, $8\pi(cm^{2} /cm )$ is the rate of change of area of circle, when a = 4cm.

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’s assume the length of edge is y cm, surface area as S and volume as V.
V = $y^{3}$
S = $y^{2}$
y is a function of time x.
Given that
$\frac{\mathrm{d} V}{\mathrm{d} x} = 8(cm^{3}/s)$.
Applying chain rule, we get:
$8 = \frac{\mathrm{d} V}{\mathrm{d} x} = \frac{\mathrm{d} (y^{3})}{\mathrm{d} x} = \frac{\mathrm{d}( y^{3})}{\mathrm{d} y}*\frac{\mathrm{d} y}{\mathrm{d} x} = 3y^{2}*\frac{\mathrm{d} y}{\mathrm{d} x}$
$\frac{\mathrm{d} y}{\mathrm{d} x} = 8/3y^{2}$……………..(a)
Applying chain rule on surface area, we get
$\frac{\mathrm{d} S}{\mathrm{d} x} = \frac{\mathrm{d} (6y^{2})}{\mathrm{d} x} = \frac{\mathrm{d} (6y^{2})}{\mathrm{d} y}*\frac{\mathrm{d} y}{\mathrm{d} x} = 12y*\frac{\mathrm{d} y}{\mathrm{d} x}$
$=12y( 8/3y^{2}) = 32/y$
Here its is given that y = 18cm.
$\frac{\mathrm{d} S}{\mathrm{d} x} = 32/12 = 8/3 cm^{2}/s$
Thus, the surface area of a circle is increasing at the rate of $8/3cm^{2}/s$ when the edge length is 12cm.

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
 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, $\frac{\mathrm{d} y}{\mathrm{d} x} = 3cm/s$
$\frac{\mathrm{d} A}{\mathrm{d} x} = 2\pi y(3) = 6\pi y$
Also given that radius y = 10cm
$\frac{\mathrm{d} A}{\mathrm{d} x} = 6\pi (10) = 60cm^{2}/s$
Thus, $60cm^{2} /s$ is the rate of change of area of circle, when y = 10cm.

Question 4 F
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’s assume the length of edge is ycm, and volume as V.
So, V = $y^{3}$ ; y is function of time ‘x’.
Then, By applying chain rule we get,
$\frac{\mathrm{d} V}{\mathrm{d} x} = 3y^{2}*\frac{\mathrm{d} y}{\mathrm{d} x}$
It is given in question
$\frac{\mathrm{d} y}{\mathrm{d} x} = 3cm/s$
So we  can say that
$\frac{\mathrm{d} V}{\mathrm{d} x} = 3y^{2}(3) = 9y^{2}$
Also given that edge length is y = 10cm
$\frac{\mathrm{d} V}{\mathrm{d} x} = 9(10^{2}) = 900cm^{3}/s$
Thus, the Volume of a variable cube is increasing at the rate of $900cm^{3}/s$ when the edge length is 10cm.

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
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, $\frac{\mathrm{d} y}{\mathrm{d} x} = 5cm/s$
$\frac{\mathrm{d} A}{\mathrm{d} x} = 2\pi y(5) =10\pi y$
Now we need to find the increase at  radius y = 8cm
$\frac{\mathrm{d} A}{\mathrm{d} x} = 10\pi (8) = 80cm^{2}/s$


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
 We know that the circumference 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,
$\frac{\mathrm{d} C}{\mathrm{d} x} = \frac{\mathrm{d} (2\pi y)}{\mathrm{d} x} = \frac{\mathrm{d} (2\pi y)}{\mathrm{d} y}*\frac{\mathrm{d} y}{\mathrm{d} x} = 2\pi \frac{\mathrm{d} y}{\mathrm{d} x}$
Given $\frac{\mathrm{d} y}{\mathrm{d} x} = .7 cm/s$
Therefore
$\frac{\mathrm{d} C}{\mathrm{d} x} = 2\pi (0.7) = 1.4cm/s$

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 = 8cm and y = 6cm, find
the rates of change of (a) the perimeter, and (b) the area of the rectangle
Solution
Let a and b are the length and width of the rectangle
Then given   in question
$\frac{\mathrm{d} a}{\mathrm{d} x} = -5cm/min\; and \; \frac{\mathrm{d} b}{\mathrm{d} x} = 4cm/min$
Perimeter of rectangle, P = 2(a + b)
The rate of change of perimeter w.r.t time ‘x’ is,
$\frac{\mathrm{d} P}{\mathrm{d} x} = 2(\frac{\mathrm{d} a}{\mathrm{d} x} + \frac{\mathrm{d} b}{\mathrm{d} x}) = 2(-5 +4) = -2cm/min$
Thus, the perimeter of a rectangle is decreasing  at 4cm/min.
Area of rectangle, A = a*b
The rate of change of area w.r.t time ‘x’ is,
$\frac{\mathrm{d} A}{\mathrm{d} x} = \frac{\mathrm{d} a}{\mathrm{d} x}*b + \frac{\mathrm{d} a}{\mathrm{d} x}*a = -5b + 4a$
Here, when a = 8cm and b = 6cm,
$\frac{\mathrm{d} A}{\mathrm{d} x} = -5(6) +4(8) =  -2cm^{2}/min$
Thus, the area of rectangle is decreasing at $12cm^{2}/min$.

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
As we know the shape of a balloon is spherical.
The volume(V) of Balloon 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
$\frac{\mathrm{d} V}{\mathrm{d} x} = \frac{\mathrm{d} (\frac{4}{3}\pi y^{3})}{\mathrm{d} x} = \frac{\mathrm{d} (\frac{4}{3}\pi y^{3})}{\mathrm{d} y}*\frac{\mathrm{d} y}{\mathrm{d} x} = 4\pi y^{2}\frac{\mathrm{d} y}{\mathrm{d} x}$
Given in question , $\frac{\mathrm{d} V}{\mathrm{d} x} = 900cc/s$ (given)
$900 = 4\pi y^{2}\frac{\mathrm{d} y}{\mathrm{d} x}$
$\frac{\mathrm{d} y}{\mathrm{d} x} = \frac{900}{4\pi y^{2}} = \frac{225}{\pi y^{2}}$
Here radius y = 15cm (given)
$\frac{\mathrm{d} y}{\mathrm{d} x} = \frac{225}{\pi 15^{2}} = \frac{1}{\pi }cm/s$
Thus, radius of football is increasing at$\frac{1}{\pi }cm/s$ when radius y = 15cm .

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
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
$\frac{\mathrm{d} V}{\mathrm{d} x} = \frac{\mathrm{d} (\frac{4}{3}\pi x^{3})}{\mathrm{d} x} = 4\pi x^{2}$
We need to find the increase at r=10 cm
$\frac{\mathrm{d} V}{\mathrm{d} x} = 4\pi (10)^{2} = 400\pi \;cm^{3}/s$

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 4 m away from the wall ?
Solution
Let a  be the distance from bottom of ladder to wall and b  be the height of the wall where the ladder is in contact with the wall.
Thus, we can apply Pythagoras theorem in this case, then we get
$a^{2}+ b^{2} = 25$
( As ladder is of 5m length)
$b = \sqrt{25 – a^{2}}$
By differentiating above equation w.r.t time ‘x’ we will get rate of change of height,
$\frac{\mathrm{d} b}{\mathrm{d} x} = \frac{-a}{\sqrt{25 – a^{2}}}*\frac{\mathrm{d} a}{\mathrm{d} x}$
Here we have $\frac{\mathrm{d} a}{\mathrm{d} x} = 2cm/s$
$\frac{\mathrm{d} b}{\mathrm{d} x} = \frac{-2a}{\sqrt{25 – a^{2}}}$
Here, the foot of the ladder is 4m far from the wall, a = 4m
$\frac{\mathrm{d} b}{\mathrm{d} x} = \frac{-2(4)}{\sqrt{25 – (4)^{2}}} = \frac{8}{3} = \frac{-8}{3}$
Thus, the height of the Ladder is decreasing at the rate of (6/3)cm/s.


Question 12
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.
Solution
$6y = x^{3}+ 2$
By differentiating above equation w.r.t. time ‘t, we get
$6\frac{\mathrm{d} y}{\mathrm{d} t} = 3x^{2}\frac{\mathrm{d} x}{\mathrm{d} t} + 0$
$2\frac{\mathrm{d} y}{\mathrm{d} x} = x^{2}\frac{\mathrm{d} x}{\mathrm{d} t}$………………………….(1)
Given y-coordinate is changing 8 times as fast as the x-coordinate
i.e. $\frac{1}{8}\frac{\mathrm{d} y}{\mathrm{d} t} = \frac{\mathrm{d} x}{\mathrm{d} t}$
$\frac{\mathrm{d} y}{\mathrm{d} t} = 8\frac{\mathrm{d} x}{\mathrm{d} t}$
Putting this value in equation (1)
$2(8\frac{\mathrm{d} x}{\mathrm{d} t}) =x^{2} \frac{\mathrm{d} x}{\mathrm{d} t}$ $16\frac{\mathrm{d} x}{\mathrm{d} t} = a^{2} \frac{\mathrm{d} x}{\mathrm{d} t}$
$(x^{2} – 16 ) \frac{\mathrm{d} x}{\mathrm{d} t} = 0$
$x^{2} = 16$
 $x = \pm 4$
Now, for x = 4
$y=11$
For, x = -4
$y=-31/3$
Thus, the required co-ordinate or points are $\left ( -4,\frac{-31}{3} \right )\; and \left ( 4,11 \right )$.

Question 12
The radius of an air bubble is increasing at the rate of ½ cm/s. At what rate is the volume of the bubble increasing when the radius is 1 cm?
Solution
As we know the shape of an air bubble 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,
$\frac{\mathrm{d} y}{\mathrm{d} x} = 1/2cm/s$
$\frac{\mathrm{d} V}{\mathrm{d} x} =2\pi y^{2}cm^{3}/s$
Also given that the radius y = 1cm,
$\frac{\mathrm{d} V}{\mathrm{d} x} = 2\pi (1)^{2} = 2\pi cm^{3}/s$

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
We know the shape of a balloon is spherical.
The volume(V) of sphere with radius ‘y’ is given by,
$V = \frac{4}{3}\pi y^{3}$
Here it is given that y = $\frac{3}{2}(2x + 1)$
$V = \frac{4}{3}\pi (\frac{3}{2}(2x + 1))^{3} = \frac{9}{16}\pi (4a + 1)^{3}$
Differentiating above equation w.r.t ‘x’, we get,
$\frac{\mathrm{d} V}{\mathrm{d} x} = \frac{\mathrm{d} \frac{9}{16}\pi(2x + 1)^{3}}{\mathrm{d} x} = \frac{27}{8}\pi (2x + 1)^{2}$

Question 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 4 cm?
Solution
 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,
$V = \frac{1}{3}\pi x^{2}y$
Given , y = (1/6)x  or 6y=x
$V = \frac{1}{3}\pi (6y)^{2}y = 12\pi y^{3}$
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, $\frac{\mathrm{d} V}{\mathrm{d} t} = 24cm^{3}/s$
$12 = 36\pi y^{2}\frac{\mathrm{d} y}{\mathrm{d} t}$
Also the y = 4cm is given,
$12 = 36\pi (4)^{2}\frac{\mathrm{d} y}{\mathrm{d} t}$
$\frac{\mathrm{d} y}{\mathrm{d} t} = \frac{1}{48\pi }$
Thus, the height of wheat flour cone is increasing at the rate of $\frac{1}{48\pi } cm/s$.

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 x³– 0.003x² + 15x + 4000.
Find the marginal cost when 17 units are produced.
Solution
We know that Marginal cost(MC) is defined as the rate of change of total cost w.r.t. the numbers of units manufactured.
MC = $\frac{\mathrm{d}C }{\mathrm{d} x} = \frac{\mathrm{d} (0.007x^{3} – 0.003x^{2} + 15x + 4000)}{\mathrm{d} x} = 0.021x^{2} – 0.006x +15$
Here, a = 17 is given.
$MC = \frac{\mathrm{d}C }{\mathrm{d} a} = 0.021x^{2} – 0.006x +15 = 0.021(17)^{2} – 0.003(17) + 15 = 20.967$
Thus, marginal cost for 17 units manufactured is Rs. 20.967.

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) is defined as the rate of change of total revenue generated w.r.t. numbers of units sold.
$MR = \frac{\mathrm{d} R}{\mathrm{d} x} = \frac{\mathrm{d} (13x^{2} + 26x + 15)}{\mathrm{d} x} = 26x + 26$
Here, x = 7 units (given)
MR = 26(7) + 26 = 208

Question 17
The rate of change of the area of a circle with respect to its radius r at r = 6 cm is
(A) 10p
(B) 12p
(C) 8p
(D) 11p
Solution
We know that the area of circle with radius ‘r’ is $\pi r^{2}$.
A = $\pi r^{2}$
The rate of change of area w.r.t ‘r’ is,
$\frac{\mathrm{d} A}{\mathrm{d} r} = \frac{\mathrm{d} (\pi r^{2})}{\mathrm{d} r} = 2\pi r$
Here r = 6cm is given,
$\frac{\mathrm{d} A}{\mathrm{d} a} = 2\pi (6) = 12\pi cm^{2}/s$
Thus, Option (B) is correct.

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
Marginal Revenue(MR) is defined as the rate of change of total revenue generated w.r.t. numbers of units sold.
$MR = \frac{\mathrm{d} R}{\mathrm{d}xy} = \frac{\mathrm{d} (3x^{2} + 36x + 5)}{\mathrm{d} x} = 6x + 36$
Here, x = 15 units (given)
MR = 6(15) + 36 = 126
Thus, marginal revenue for 15 units sold is Rs.126
Thus, option (D) is correct.
 
 
 
 
 

 

Related Topics

Also Read

---

• Notes
  • Rate of Change
  • Increasing and Decreasing Functions
  • Tangents and Normals
  • Approximations of Functions
  • Maxima and Minima of Functions
• NCERT Solutions & Assignments
  • NCERT Solution for Application of Derivatives Exercise 6.1
  • NCERT Solution for Application of Derivatives Exercise 6.2
  • Application of Derivatives Class 12 Important Questions

---

---

Go back to Class 12 Main Page using below links
Class 12 Maths Class 12 Physics Class 12 Chemistry Class 12 Biology

Latest Updates Classification of Elements JEE MCQ

---

Chemical Equilibrium Class 11 MCQ

---

Redox Reactions JEE Main MCQ

---

Chemical Equilibrium Class 11 MCQ

---

Chemical Thermodynamics JEE Advanced MCQ

---