Welcome to Class 12 Applied Maths Chapter 6, where we embark on an exciting journey into the world of advanced mathematical concepts tailored for Class 11 students.” Unlock the power of applied mathematics with expert solutions crafted by professionals at AppliedMath.com. Designed to propel students towards academic success, our meticulously curated ML Aggarwal Solutions for Applied Mathematics cater to Class 11 and class 12 students seeking mastery in their examinations. Every query from the CBSE ML Aggarwal Books finds a comprehensive answer on our platform, complete with detailed explanations and step-by-step solutions presented in an easily understandable language.
Dive into the world of applied mathematics and discover how our resources can elevate your understanding and performance. Keep reading to explore the wealth of ML Aggarwal Solutions for Class 11 and Class 12 Applied Mathematics.
Here we provide you with Class 12 Applied Maths Chapter 6, to help you again a comprehensive understanding of the chapter and its concepts.
https://appliedmathsolution.com/wp-admin/post.php?post=6&action=edit
Class 12 Applied Maths Chapter 6 Solutions
Application of Derivatives
EXERCISE- 6.1
Q.1 Find the rate of change of the area of a circle with respect to its radius when the radius is 6 cm.
Ans. Area of circle (A) = πr^2
Differentiate w.r.t . ‘r’,
dA/dr = πd/dr (r^2)
dA/r = 2πr
dA/dr (r = 6) = 2 x π x 6
= 12π cm^2/cm
Q.2 If the radius of a circle is increasing at the rate of 0.7 cm/sec, at what rate is its circumference increasing?
Ans. dr/dt = 0.7 cm/sec, dc/dt = ?
c = 2πr
Differentiate w.r.t.
dc/dt = 2π (dr/dt)
dc/dt = 2π x 0.7
= 1.4π cm/sec
Q.3 If the radius of a circle is increasing at the rate of 3 cm/sec, at what rate is its area increasing when its radius is 10 cm?
Ans. dr/dt = 3 cm/sec
Area of circle = πr^2
Differentiate w.r.t
dA/dt = π (2r)(dr/dt)
dA/dt (r = 10) = π x 2 x 10 x 3
= 60π cm^2/sec
Q.4 If the sides of a square are decreasing at the rate of 1.5 cm/sec, at what rate is its perimeter decreasing?
Ans. Let, Sides of square be ‘s’
dS/dt = – 1.5 cm/sec
Perimeter of square = 4 x s
Differentiate w.r.t ‘t’
dP/dt = 4 (dS/dt)
dP/dt = 4 x (-1.5)
DP/dt = -6 cm/sec
Q.5 If the sides of a square are decreasing at the rate of 1.5 cm/sec, at what rate is its area decreasing when its side is 8 cm?
Ans. ds/dt = -1.5 cm/sec
Area of square = (S)^2
Differentiate w.r.t ‘t’
dA/dt = 2s (ds/dt)
dA/dt = 2 x 8 x (-1.5)
dA/dt = 16 x (-1.5)
= 24 cm^2/sec
Q.6 Find the rate of change of the volume of a cube with respect to its edge when the edge is 5 cm.
Ans. Let, the edge of a cube be ‘a’
Volume of a cube be ‘v’
Volume of cube = (a)^3
Differentiate w.r.t ‘a’
dv/da (a = 5) = 3a^2
dv/da = 3 x 5 x 5
= 75 cm^3/cm
Q.7 If an edge of a variable cube is increasing at the rate of 3 cm/sec, at what rate is its volume increasing when its edge is 10 cm?
Ans. Let, the edge of a cube be ‘a’
Volume of a cube be ‘v’
da/dt = 3 cm/sec
Volume of cube = a^3
Differentiate w.r.t ‘t’
dv/dt = 3a^2 (da/dt)
dv/dt = 3 x (10)^2 x 3
dv/dt = 900 cm^3/sec
Q.8 A balloon which always remains spherical has a variable radius. Find the rate at which its volume is increasing with respect to radius when the radius is 10 cm.
Ans. dv/dr = ?
Volume of sphere = 4/3πr^3
Differentiate w.r.t ‘r’
dv/dr = 4/3π [d/dr(r^3)]
dv/dr = 4/3π x 3r^2
dv/dr (r = 10) = 4π x (10)^2
dv/dr = 400π cm^/cm
Q.9 If the radius of a balloon which always remains spherical is increasing at the rate 1.5 cm/sec, at what rate is its surface area increasing when its radius is 12 cm?
Ans. dr/dt = ?
Surface area of sphere = 4πr^2
Differentiate w.r.t ‘t’
ds/dt = 4π(2r)(dr/dt), (r = 12)
ds/dt = 4π x 2 x 12 x (1.5)
ds/dt = 6 x 2 x 12 x π
ds/dt = 144π cm^2/sec
10. If the radius of a soap bubble is increasing at the rate of 1/2 cm/sec, at what rate is its volume increasing when the radius is 1 cm?
Ans. dr/dt = 1/2 cm/s
Volume of sphere (Soap bubble) = 4/3πr^3
Differentiate w.r.t. ‘t’
dv/dt = 4/3π (3r^2) (dr/dt)
dv/dt = 4π x (1)^2 x (1/2)
dv/dt = 2π cm^3/sec
Q.11 A cylindrical tank of radius 10 m is being filled with wheat at the rate of 314 cubic metre per hour. Find the rate at which the depth of the wheat is increasing, take л = 3.14.
Ans. dv/dt = 314 m^3/hr, dh/dt = ?, dept = h
Volume of cylinder = πr^2h
Differentiate w.r.t ‘t’
dv/dt = πr^2h (dh/dt)
314 = 3.14 x (10)^2 x dh/dt
314 = 3.14 x 100 x dh/dt
dh/dt = 1 m/hr
Q.12 Find the rate of change of the whole surface of a closed circular cylinder of radius r and height h with respect to change in radius.
Ans. Total surface area of cylinder = 2πrh + 2πr^2
Differentiate w.r.t ‘r’
ds/dr = 2πh + 2π(2r)
ds/dr = 2πh + 4πr
ds/dr = 2π(h + 2r) cm^2/cm
Q.13 A stone is dropped into a quiet lake and waves move in circles at the speed of 5 cm/sec. At the instant when the radius of the circular wave is 8 cm, how fast is the enclosed area increasing?
Ans. dr/dt = 5 cm/sec, r = 8 cm
Area of circle = πr^2
Differentiate w.r.t ‘t’
dA/dt = π (2r) (dr/dt)
dA/dt = π x 2 x 8 x 5
dA/dt = 80π cm^2/sec
Q.14 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 rate of change of
(i) the perimeter
(ii) the area of the rectangle.
Ans.(i) dx/dt = -5 cm/min, dy/dt = 4 cm/min
Perimeter of rectangle = 2(x + y)
Differentiate w.r.t ‘t’
dP/dt = 2[(dx/dt) + dy/dt)]
dP/dt = 2(-5 + 4)
dP/dt = -2 cm/min
Perimeter is ‘decreasing’ at the rate of 2 cm/min when x = 8 cm and y = 6 cm.
(ii) Area of rectangle = x (y)
Differentiate w.r.t ‘t’
dA/dt = (dx/dt )(y) + (x)(dy/dt)
dA/dt = (-5) x 6 + 8 x 4
dA/dt = -30 + 32
dA/dt = 2 cm^2/min
Q.15 The volume of a cube is increasing at the rate of 9 cubic centimetres per second. How fast is the surface area increasing when the length of the edge is 10 centimetres?
Ans. dv/dt = 9 cm^3/sec, Edge (a) = 10 cm
Volume of cube = (a)^3
Differentiate w.r.t ‘t’
dv/dt = 3a^2 (da/dt)
9 = 3 x (10)^2 (da/dt)
3/100 = da/dt
da/dt = 0.03 cm/sec
Now, Surface area of cube = 6 x (a)^2
Differentiate w.r.t ‘t’
ds/dt = 6 x 2 x 10 x 0.03
ds/dt = 12 x 0.3
ds/dt = 3.6 cm^2/sec
Q.16 A particle moves along the curve y = 2/3 x^3 + 1. Find the points on the curve at which the y-coordinate is changing twice as fast as the x-coordinate.
Ans. dy/dt = 2 (dx/dt) [Given]
y = 2/3x^3 + 1
Differentiate w.r.t ‘t’
dy/dt = 2/3 (3x^2) (dx/dt)
2dx/dt = 2x^2 (dx/dt)
x^2 = 1
So, x = ± 1
When x = 1,
y = 2/3x^3 + 1
y = 2/3 + 1
y = 5/3
Point (1, 5/3)
When x = -1,
y = 2/3 (-1)^3 + 1
y = -2/3 + 1
y = 1/3
Points (-1, 1/3)
Q.17 A man of height 2 metres walks at a uniform speed of 5 km/h away from a lamp post which is 6 metres high. Find the rate at which the length of shadow increases.
Ans. △ABD and △ECD (A.A similarity), dx/dt = ?, dy/dt = 5 km/h
AB/EC = BD/CD
6/2 = (x + y)/x
3x = x + y
3x – x = y
2x = y
2 dx/dt = dy/dt
dx/dt = 5/2 km/h
Q.18 A kite is 120 m high and 130 m of string is out. If the kite is moving horizontally at the rate of 5.2 m/sec, find the rate at which the string is being paid out at that instant.
Ans. dx/dt = 5.2 m/sec, dy/dt = ?, y = 130 m
y^2 = x^2 + (120)^2
(130)^2 = x^2 + 14400
16900 – 14400 = x^2
x^2 = 2500
x = 50 m
Now, y^2 = x^2 + (120)^2
Differentiate w.r.t ‘t’
2y (dy/dt) = 2x(dx/dt) + 0
2 x 130 (dy/dt = 2 x 50 x 5.2
dy/dt = (100 x 5.2)/260
dy/dt = 52/26
dy/dt = 2 m/sec
Q.19 Sand is pouring from a pipe at the rate of 12 cm^3/sec. 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?
Ans. h = 1/6 r, r = 6h, dh/dt = ?, h = 4 cm, dv/dt = 12 cm^3/sec
Volume of cone = 1/3πr^2h
Volume = 1/3
π(6h)^2 x h
Volume = 1/3π x 36h
Volume = 12πh^3
Differentiate w.r.t ‘t’
dv/dt = 12π (3h^2)(dh/dt)
12 = 12π x 3 x (4)^2 x dh/dt
1 = 48π x dh/dt
dh/dt = 1/48 cm/sec
Q.20 A conical vessel whose height is 4 metres and of base radius 2 metres is being filled with water at the rate of 0.75 cubic metres per minute. Find the rate at which the level of the water is rising when the depth of water is 1.5 metres.
Ans. dv/dt = 0.75 m^3/min, dh/dt = ?, h = 1.5 m
△BAC and △DEC are similar (A.A similarity)
BA/DE = AC/EC
2/r = AC/EC
2/r = 4/h
2r = h
r = h/2
Volume of cone = 1/3πr^2h
Volume = 1/3π(h/2)^2 x h
Volume = 1/3π x h^2/4 x h
Volume = 1/2π x h^3
Differentiate w.r.t ‘t’
dv/dt = 1/12π(3h^2) x dh/dt
0.75 = π/4 x (1.5)^2 x dh/dt
(0.75 x 4)/(π x 1.5)^2 = dh/dt
3/2.25π = dh/dt
300/225 = dh/dt
dh/dt = 4/3π m/min
FAQ’s related to Class 12 Applied Maths Chapter 6 on Application of Derivatives :
Q.1 What are the main applications of derivatives?
Ans. Derivatives have the following applications:
(i) Finding approximations using differentials.
(ii) Finding rate of change of a quantity.
(iii) Determining tangents and normals to a curve.
(iv) Solving optimization problems (maximum and minimum values).
(v) Identifying the increasing or decreasing nature of functions.
Q.2 How do you find the rate of change of a quantity?
Ans. The rate of change of a quantity y with respect to another quantity xxx is given by dy/dx. This represents how y changes as x varies.
Q.3 What is the geometrical significance of a derivative?
Ans. The derivative dy/dx represents the slope of the tangent to the curve at a given point. It indicates how steep the curve is at that point.
Q.4 What are tangents and normals, and how are they calculated?
Ans. A tangent is a line that touches a curve at one point without crossing it.
A normal is a line perpendicular to the tangent at the point of contact.
Slope of normal: m(n) = −1/m(t)
Slope of tangent: m(t) = dy/dx at the given point.
These are a few Frequently Asked Questions relating to Class 12 Applied Maths Chapter 13
In Class 12 Applied Maths chapter 6, you will explore fascinating topics that form the backbone of practical problem-solving techniques. Through clear explanations, illustrative examples, and step-by-step solutions, you’ll grasp complex concepts effortlessly. Whether you’re preparing for exams or simply eager to deepen your mathematical understanding, Class 12 Applied Maths Chapter 6 promises an enriching learning experience that will set you on the path to success. Class 12 Applied Maths Chapter 6, we delve deep into advanced mathematical concepts that are crucial for understanding.
