RD Sharma Class 12 Solutions - Differentials, Errors and Approximations | Mathematics (Maths) Class 12 - JEE PDF Download

 Page 1


14. Differentials, Errors and Approximations
Exercise 14.1
1. Question
If y = sin x and x changes from  to , what is the approximate change in y?
Answer
Given y = sin x and x changes from  to .
Let  so that
On differentiating y with respect to x, we get
We know
When , we have .
Recall that if y = f(x) and ?x is a small increment in x, then the corresponding increment in y, ?y = f(x + ?x)
– f(x), is approximately given as
Here,  and
? ?y = 0
Thus, there is approximately no change in y.
2. Question
The radius of a sphere shrinks from 10 to 9.8 cm. Find approximately the decrease in its volume.
Answer
Given the radius of a sphere changes from 10 cm to 9.8 cm.
Let x be the radius of the sphere and ?x be the change in the value of x.
Hence, we have x = 10 and x + ?x = 9.8
? 10 + ?x = 9.8
Page 2


14. Differentials, Errors and Approximations
Exercise 14.1
1. Question
If y = sin x and x changes from  to , what is the approximate change in y?
Answer
Given y = sin x and x changes from  to .
Let  so that
On differentiating y with respect to x, we get
We know
When , we have .
Recall that if y = f(x) and ?x is a small increment in x, then the corresponding increment in y, ?y = f(x + ?x)
– f(x), is approximately given as
Here,  and
? ?y = 0
Thus, there is approximately no change in y.
2. Question
The radius of a sphere shrinks from 10 to 9.8 cm. Find approximately the decrease in its volume.
Answer
Given the radius of a sphere changes from 10 cm to 9.8 cm.
Let x be the radius of the sphere and ?x be the change in the value of x.
Hence, we have x = 10 and x + ?x = 9.8
? 10 + ?x = 9.8
? ?x = 9.8 – 10
? ?x = –0.2
The volume of a sphere of radius x is given by
On differentiating V with respect to x, we get
We know
When x = 10, we have .
Recall that if y = f(x) and ?x is a small increment in x, then the corresponding increment in y, ?y = f(x + ?x)
– f(x), is approximately given as
Here,  and ?x = –0.2
? ?V = (400p)(–0.2)
? ?V = –80p
Thus, the approximate decrease in the volume of the sphere is 80p cm
3
.
3. Question
A circular metal plate expands under heating so that its radius increases by k%. Find the approximate
increase in the area of the plate, if the radius of the plate before heating is 10 cm.
Answer
Given the radius of a circular plate initially is 10 cm and it increases by k%.
Let x be the radius of the circular plate, and ?x is the change in the value of x.
Hence, we have x = 10 and
? ?x = 0.1k
The area of a circular plate of radius x is given by
A = px
2
On differentiating A with respect to x, we get
Page 3


14. Differentials, Errors and Approximations
Exercise 14.1
1. Question
If y = sin x and x changes from  to , what is the approximate change in y?
Answer
Given y = sin x and x changes from  to .
Let  so that
On differentiating y with respect to x, we get
We know
When , we have .
Recall that if y = f(x) and ?x is a small increment in x, then the corresponding increment in y, ?y = f(x + ?x)
– f(x), is approximately given as
Here,  and
? ?y = 0
Thus, there is approximately no change in y.
2. Question
The radius of a sphere shrinks from 10 to 9.8 cm. Find approximately the decrease in its volume.
Answer
Given the radius of a sphere changes from 10 cm to 9.8 cm.
Let x be the radius of the sphere and ?x be the change in the value of x.
Hence, we have x = 10 and x + ?x = 9.8
? 10 + ?x = 9.8
? ?x = 9.8 – 10
? ?x = –0.2
The volume of a sphere of radius x is given by
On differentiating V with respect to x, we get
We know
When x = 10, we have .
Recall that if y = f(x) and ?x is a small increment in x, then the corresponding increment in y, ?y = f(x + ?x)
– f(x), is approximately given as
Here,  and ?x = –0.2
? ?V = (400p)(–0.2)
? ?V = –80p
Thus, the approximate decrease in the volume of the sphere is 80p cm
3
.
3. Question
A circular metal plate expands under heating so that its radius increases by k%. Find the approximate
increase in the area of the plate, if the radius of the plate before heating is 10 cm.
Answer
Given the radius of a circular plate initially is 10 cm and it increases by k%.
Let x be the radius of the circular plate, and ?x is the change in the value of x.
Hence, we have x = 10 and
? ?x = 0.1k
The area of a circular plate of radius x is given by
A = px
2
On differentiating A with respect to x, we get
We know
When x = 10, we have .
Recall that if y = f(x) and ?x is a small increment in x, then the corresponding increment in y, ?y = f(x + ?x)
– f(x), is approximately given as
Here,  and ?x = 0.1k
? ?A = (20p)(0.1k)
? ?A = 2kp
Thus, the approximate increase in the area of the circular plate is 2kp cm
2
.
4. Question
Find the percentage error in calculating the surface area of a cubical box if an error of 1% is made in
measuring the lengths of the edges of the cube.
Answer
Given the error in the measurement of the edge of a cubical box is 1%.
Let x be the edge of the cubical box, and ?x is the error in the value of x.
Hence, we have
? ?x = 0.01x
The surface area of a cubical box of radius x is given by
S = 6x
2
On differentiating A with respect to x, we get
We know
Page 4


14. Differentials, Errors and Approximations
Exercise 14.1
1. Question
If y = sin x and x changes from  to , what is the approximate change in y?
Answer
Given y = sin x and x changes from  to .
Let  so that
On differentiating y with respect to x, we get
We know
When , we have .
Recall that if y = f(x) and ?x is a small increment in x, then the corresponding increment in y, ?y = f(x + ?x)
– f(x), is approximately given as
Here,  and
? ?y = 0
Thus, there is approximately no change in y.
2. Question
The radius of a sphere shrinks from 10 to 9.8 cm. Find approximately the decrease in its volume.
Answer
Given the radius of a sphere changes from 10 cm to 9.8 cm.
Let x be the radius of the sphere and ?x be the change in the value of x.
Hence, we have x = 10 and x + ?x = 9.8
? 10 + ?x = 9.8
? ?x = 9.8 – 10
? ?x = –0.2
The volume of a sphere of radius x is given by
On differentiating V with respect to x, we get
We know
When x = 10, we have .
Recall that if y = f(x) and ?x is a small increment in x, then the corresponding increment in y, ?y = f(x + ?x)
– f(x), is approximately given as
Here,  and ?x = –0.2
? ?V = (400p)(–0.2)
? ?V = –80p
Thus, the approximate decrease in the volume of the sphere is 80p cm
3
.
3. Question
A circular metal plate expands under heating so that its radius increases by k%. Find the approximate
increase in the area of the plate, if the radius of the plate before heating is 10 cm.
Answer
Given the radius of a circular plate initially is 10 cm and it increases by k%.
Let x be the radius of the circular plate, and ?x is the change in the value of x.
Hence, we have x = 10 and
? ?x = 0.1k
The area of a circular plate of radius x is given by
A = px
2
On differentiating A with respect to x, we get
We know
When x = 10, we have .
Recall that if y = f(x) and ?x is a small increment in x, then the corresponding increment in y, ?y = f(x + ?x)
– f(x), is approximately given as
Here,  and ?x = 0.1k
? ?A = (20p)(0.1k)
? ?A = 2kp
Thus, the approximate increase in the area of the circular plate is 2kp cm
2
.
4. Question
Find the percentage error in calculating the surface area of a cubical box if an error of 1% is made in
measuring the lengths of the edges of the cube.
Answer
Given the error in the measurement of the edge of a cubical box is 1%.
Let x be the edge of the cubical box, and ?x is the error in the value of x.
Hence, we have
? ?x = 0.01x
The surface area of a cubical box of radius x is given by
S = 6x
2
On differentiating A with respect to x, we get
We know
Recall that if y = f(x) and ?x is a small increment in x, then the corresponding increment in y, ?y = f(x + ?x)
– f(x), is approximately given as
Here,  and ?x = 0.01x
? ?S = (12x)(0.01x)
? ?S = 0.12x
2
The percentage error is,
? Error = 0.02 × 100%
? Error = 2%
Thus, the error in calculating the surface area of the cubical box is 2%.
5. Question
If there is an error of 0.1% in the measurement of the radius of a sphere, find approximately the percentage
error in the calculation of the volume of the sphere.
Answer
Given the error in the measurement of the radius of a sphere is 0.1%.
Let x be the radius of the sphere and ?x be the error in the value of x.
Hence, we have
? ?x = 0.001x
The volume of a sphere of radius x is given by
On differentiating V with respect to x, we get
We know
Recall that if y = f(x) and ?x is a small increment in x, then the corresponding increment in y, ?y = f(x + ?x)
– f(x), is approximately given as
Page 5


14. Differentials, Errors and Approximations
Exercise 14.1
1. Question
If y = sin x and x changes from  to , what is the approximate change in y?
Answer
Given y = sin x and x changes from  to .
Let  so that
On differentiating y with respect to x, we get
We know
When , we have .
Recall that if y = f(x) and ?x is a small increment in x, then the corresponding increment in y, ?y = f(x + ?x)
– f(x), is approximately given as
Here,  and
? ?y = 0
Thus, there is approximately no change in y.
2. Question
The radius of a sphere shrinks from 10 to 9.8 cm. Find approximately the decrease in its volume.
Answer
Given the radius of a sphere changes from 10 cm to 9.8 cm.
Let x be the radius of the sphere and ?x be the change in the value of x.
Hence, we have x = 10 and x + ?x = 9.8
? 10 + ?x = 9.8
? ?x = 9.8 – 10
? ?x = –0.2
The volume of a sphere of radius x is given by
On differentiating V with respect to x, we get
We know
When x = 10, we have .
Recall that if y = f(x) and ?x is a small increment in x, then the corresponding increment in y, ?y = f(x + ?x)
– f(x), is approximately given as
Here,  and ?x = –0.2
? ?V = (400p)(–0.2)
? ?V = –80p
Thus, the approximate decrease in the volume of the sphere is 80p cm
3
.
3. Question
A circular metal plate expands under heating so that its radius increases by k%. Find the approximate
increase in the area of the plate, if the radius of the plate before heating is 10 cm.
Answer
Given the radius of a circular plate initially is 10 cm and it increases by k%.
Let x be the radius of the circular plate, and ?x is the change in the value of x.
Hence, we have x = 10 and
? ?x = 0.1k
The area of a circular plate of radius x is given by
A = px
2
On differentiating A with respect to x, we get
We know
When x = 10, we have .
Recall that if y = f(x) and ?x is a small increment in x, then the corresponding increment in y, ?y = f(x + ?x)
– f(x), is approximately given as
Here,  and ?x = 0.1k
? ?A = (20p)(0.1k)
? ?A = 2kp
Thus, the approximate increase in the area of the circular plate is 2kp cm
2
.
4. Question
Find the percentage error in calculating the surface area of a cubical box if an error of 1% is made in
measuring the lengths of the edges of the cube.
Answer
Given the error in the measurement of the edge of a cubical box is 1%.
Let x be the edge of the cubical box, and ?x is the error in the value of x.
Hence, we have
? ?x = 0.01x
The surface area of a cubical box of radius x is given by
S = 6x
2
On differentiating A with respect to x, we get
We know
Recall that if y = f(x) and ?x is a small increment in x, then the corresponding increment in y, ?y = f(x + ?x)
– f(x), is approximately given as
Here,  and ?x = 0.01x
? ?S = (12x)(0.01x)
? ?S = 0.12x
2
The percentage error is,
? Error = 0.02 × 100%
? Error = 2%
Thus, the error in calculating the surface area of the cubical box is 2%.
5. Question
If there is an error of 0.1% in the measurement of the radius of a sphere, find approximately the percentage
error in the calculation of the volume of the sphere.
Answer
Given the error in the measurement of the radius of a sphere is 0.1%.
Let x be the radius of the sphere and ?x be the error in the value of x.
Hence, we have
? ?x = 0.001x
The volume of a sphere of radius x is given by
On differentiating V with respect to x, we get
We know
Recall that if y = f(x) and ?x is a small increment in x, then the corresponding increment in y, ?y = f(x + ?x)
– f(x), is approximately given as
Here,  and ?x = 0.001x
? ?V = (4px
2
)(0.001x)
? ?V = 0.004px
3
The percentage error is,
? Error = 0.003 × 100%
? Error = 0.3%
Thus, the error in calculating the volume of the sphere is 0.3%.
6. Question
The pressure p and the volume v of a gas are connected by the relation pv
1.4
 = const. Find the percentage
error in p corresponding to a decrease of in v.
Answer
Given pv
1.4
 = constant and the decrease in v is .
Hence, we have
? ?v = –0.005v
We have pv
1.4
 = constant
Taking log on both sides, we get
log(pv
1.4
) = log(constant)
? log p + log v
1.4
 = 0 [? log(ab) = log a + log b]
? log p + 1.4 log v = 0 [ ? log(a
m
) = m log a]
On differentiating both sides with respect to v, we get
We know