RD Sharma Class 12 Solutions - Derivative as a Rate Measurer

# RD Sharma Class 12 Solutions - Derivative as a Rate Measurer | Mathematics (Maths) Class 12 - JEE PDF Download

``` Page 1

13. Derivative as a Rate Measurer
Exercise 13.1
1. Question
Find the rate of change of the total surface area of a cylinder of radius r and height h, when the radius varies.
Total Surface Area of Cylinder = 2 p r
2
+ 2 p r h
Given: Radius of the Cylinder varies.
Therefore, We need to find  where S = Surface Area of Cylinder and r = radius of Cylinder.
Hence, Rate of change of total surface area of the cylinder when the radius is varying is given by (4 p r + 2 p
h).
2. Question
Find the rate of change of the volume of a sphere with respect to its diameter.
The volume of a Sphere =
Where D = diameter of the Sphere
We need to find,  where V = Volume of the sphere and D = Diameter of the Sphere.
Hence, Rate of change of Volume of Sphere with respect to the diameter of the Sphere is given by  .
3. Question
Find the rate of change of the volume of a sphere with respect to its surface area when the radius is 2 cm.
Volume of Sphere =
Surface Area of Sphere = 4 p r
2
We need to find, where V = Volume of the Sphere and S = Surface Area of the Sphere.
Page 2

13. Derivative as a Rate Measurer
Exercise 13.1
1. Question
Find the rate of change of the total surface area of a cylinder of radius r and height h, when the radius varies.
Total Surface Area of Cylinder = 2 p r
2
+ 2 p r h
Given: Radius of the Cylinder varies.
Therefore, We need to find  where S = Surface Area of Cylinder and r = radius of Cylinder.
Hence, Rate of change of total surface area of the cylinder when the radius is varying is given by (4 p r + 2 p
h).
2. Question
Find the rate of change of the volume of a sphere with respect to its diameter.
The volume of a Sphere =
Where D = diameter of the Sphere
We need to find,  where V = Volume of the sphere and D = Diameter of the Sphere.
Hence, Rate of change of Volume of Sphere with respect to the diameter of the Sphere is given by  .
3. Question
Find the rate of change of the volume of a sphere with respect to its surface area when the radius is 2 cm.
Volume of Sphere =
Surface Area of Sphere = 4 p r
2
We need to find, where V = Volume of the Sphere and S = Surface Area of the Sphere.
4. Question
Find the rate of change of the area of a circular disc with respect to its circumference when the radius is 3
cm.
Area of a Circular disc = p r
2
Circumference of a Circular disc = 2 p r
Where r = radius of Circular Disc.
Now we need to find  where A = Area of Circular disc and C = Circumference of the Circular disk.
5. Question
Find the rate of change of the volume of a cone with respect to the radius of its base.
The volume of Cone =
Where r = radius of the cone
h = height of the cone
We need to find, where V = Volume of cone and r = radius of the cone.
6. Question
Find the rate of change of the area of a circle with respect to its radius r when r = 5 cm.
Area of a Circle = p r
2
Where, r = radius of a circle.
We need to find,  , where A = Area of the Circle and r = radius of Circle.
Page 3

13. Derivative as a Rate Measurer
Exercise 13.1
1. Question
Find the rate of change of the total surface area of a cylinder of radius r and height h, when the radius varies.
Total Surface Area of Cylinder = 2 p r
2
+ 2 p r h
Given: Radius of the Cylinder varies.
Therefore, We need to find  where S = Surface Area of Cylinder and r = radius of Cylinder.
Hence, Rate of change of total surface area of the cylinder when the radius is varying is given by (4 p r + 2 p
h).
2. Question
Find the rate of change of the volume of a sphere with respect to its diameter.
The volume of a Sphere =
Where D = diameter of the Sphere
We need to find,  where V = Volume of the sphere and D = Diameter of the Sphere.
Hence, Rate of change of Volume of Sphere with respect to the diameter of the Sphere is given by  .
3. Question
Find the rate of change of the volume of a sphere with respect to its surface area when the radius is 2 cm.
Volume of Sphere =
Surface Area of Sphere = 4 p r
2
We need to find, where V = Volume of the Sphere and S = Surface Area of the Sphere.
4. Question
Find the rate of change of the area of a circular disc with respect to its circumference when the radius is 3
cm.
Area of a Circular disc = p r
2
Circumference of a Circular disc = 2 p r
Where r = radius of Circular Disc.
Now we need to find  where A = Area of Circular disc and C = Circumference of the Circular disk.
5. Question
Find the rate of change of the volume of a cone with respect to the radius of its base.
The volume of Cone =
Where r = radius of the cone
h = height of the cone
We need to find, where V = Volume of cone and r = radius of the cone.
6. Question
Find the rate of change of the area of a circle with respect to its radius r when r = 5 cm.
Area of a Circle = p r
2
Where, r = radius of a circle.
We need to find,  , where A = Area of the Circle and r = radius of Circle.
At r = 5 cm we have,
7. Question
Find the rate of change of the volume of a ball with respect to its radius r. How fast is the volume changing
Volume of a Ball = Volume of Sphere =
Where r = radius of the ball.
We need to find  , where V = Volume of Ball and r = radius of the ball.
At r = 2 cm,
8. Question
The total cost C (x) associated with the production of x units of an item is given by C (x) = 0.007 x
3
– 0.003x
2
+ 15x + 4000. Find the marginal cost when 17 units are produced.
Given: Total Cost, C (x) = 0.007 x
3
– 0.003x
2
+ 15x + 4000
Where x = Number of units
Marginal Cost is given by,
Marginal Cost M(x) =
Therefore, M(x) = 0.007 × 3 x
2
– 0.003 × 2 x + 15
At Number of units = 17, x = 17.
So, M(17) = 0.021 (17)
2
– 0.006 (17) + 15
M(17) = 20.97
Hence, the Marginal Cost of 17 units is 20.97.
9. Question
The total revenue received from the sale of x units of a product is given by R(x) = 13 x
2
+ 26 x + 15. Find
the marginal revenue when x = 7?
Given: Total Revenue, R(x) = 13 x
2
+ 26 x + 15
Page 4

13. Derivative as a Rate Measurer
Exercise 13.1
1. Question
Find the rate of change of the total surface area of a cylinder of radius r and height h, when the radius varies.
Total Surface Area of Cylinder = 2 p r
2
+ 2 p r h
Given: Radius of the Cylinder varies.
Therefore, We need to find  where S = Surface Area of Cylinder and r = radius of Cylinder.
Hence, Rate of change of total surface area of the cylinder when the radius is varying is given by (4 p r + 2 p
h).
2. Question
Find the rate of change of the volume of a sphere with respect to its diameter.
The volume of a Sphere =
Where D = diameter of the Sphere
We need to find,  where V = Volume of the sphere and D = Diameter of the Sphere.
Hence, Rate of change of Volume of Sphere with respect to the diameter of the Sphere is given by  .
3. Question
Find the rate of change of the volume of a sphere with respect to its surface area when the radius is 2 cm.
Volume of Sphere =
Surface Area of Sphere = 4 p r
2
We need to find, where V = Volume of the Sphere and S = Surface Area of the Sphere.
4. Question
Find the rate of change of the area of a circular disc with respect to its circumference when the radius is 3
cm.
Area of a Circular disc = p r
2
Circumference of a Circular disc = 2 p r
Where r = radius of Circular Disc.
Now we need to find  where A = Area of Circular disc and C = Circumference of the Circular disk.
5. Question
Find the rate of change of the volume of a cone with respect to the radius of its base.
The volume of Cone =
Where r = radius of the cone
h = height of the cone
We need to find, where V = Volume of cone and r = radius of the cone.
6. Question
Find the rate of change of the area of a circle with respect to its radius r when r = 5 cm.
Area of a Circle = p r
2
Where, r = radius of a circle.
We need to find,  , where A = Area of the Circle and r = radius of Circle.
At r = 5 cm we have,
7. Question
Find the rate of change of the volume of a ball with respect to its radius r. How fast is the volume changing
Volume of a Ball = Volume of Sphere =
Where r = radius of the ball.
We need to find  , where V = Volume of Ball and r = radius of the ball.
At r = 2 cm,
8. Question
The total cost C (x) associated with the production of x units of an item is given by C (x) = 0.007 x
3
– 0.003x
2
+ 15x + 4000. Find the marginal cost when 17 units are produced.
Given: Total Cost, C (x) = 0.007 x
3
– 0.003x
2
+ 15x + 4000
Where x = Number of units
Marginal Cost is given by,
Marginal Cost M(x) =
Therefore, M(x) = 0.007 × 3 x
2
– 0.003 × 2 x + 15
At Number of units = 17, x = 17.
So, M(17) = 0.021 (17)
2
– 0.006 (17) + 15
M(17) = 20.97
Hence, the Marginal Cost of 17 units is 20.97.
9. Question
The total revenue received from the sale of x units of a product is given by R(x) = 13 x
2
+ 26 x + 15. Find
the marginal revenue when x = 7?
Given: Total Revenue, R(x) = 13 x
2
+ 26 x + 15
Where x = number of units.
Marginal Revenue is given by,
M(x) =
Therefore,
M(x) = 26 (x + 1)
For x = 7,
M(x) = 26(7 + 1)
M(x) = 26 × 8 = Rs. 207
10. Question
The money to be spent for the welfare of the employees of a firm is proportional to the rate of change of its
total revenue (Marginal Revenue). If the total revenue ( in rupees) received from the sale of x units of a
product is given by
R(x) = 3 x
2
+ 36 x + 5, find the marginal revenue, when x = 5, and write which value does the question
indicate.
Given: Marginal Revenue, M(x) =
M(x) proportional to Money spent on the welfare of employees.
Thus, M(x) = k (Money spent on the welfare of employees)
M(x) = 6 x + 36
At x = 5, M(x) = 66
Now at x = 6, M(x) = 72
Thus Marginal Revenue for x = 5 is 66
And when x increased from x = 5 to x = 6,
Marginal revenue also increases and thus the money spent on the welfare of employees increases.
Exercise 13.2
1. Question
The side of a square sheet is increasing at the rate of 4 cm per minute. At what rate is the area increasing
when the side is 8 cm long?
Given: the side of a square sheet is increasing at the rate of 4 cm per minute.
To find rate of area increasing when the side is 8 cm long
let the side of the given square sheet be x cm at any instant time.
Then according to the given criteria,
Rate of side of the sheet increasing is,
Then the area of the square sheet at any time t will be
A = x
2
cm
2
.
Applying derivative with respect to time on both sides we get,
Page 5

13. Derivative as a Rate Measurer
Exercise 13.1
1. Question
Find the rate of change of the total surface area of a cylinder of radius r and height h, when the radius varies.
Total Surface Area of Cylinder = 2 p r
2
+ 2 p r h
Given: Radius of the Cylinder varies.
Therefore, We need to find  where S = Surface Area of Cylinder and r = radius of Cylinder.
Hence, Rate of change of total surface area of the cylinder when the radius is varying is given by (4 p r + 2 p
h).
2. Question
Find the rate of change of the volume of a sphere with respect to its diameter.
The volume of a Sphere =
Where D = diameter of the Sphere
We need to find,  where V = Volume of the sphere and D = Diameter of the Sphere.
Hence, Rate of change of Volume of Sphere with respect to the diameter of the Sphere is given by  .
3. Question
Find the rate of change of the volume of a sphere with respect to its surface area when the radius is 2 cm.
Volume of Sphere =
Surface Area of Sphere = 4 p r
2
We need to find, where V = Volume of the Sphere and S = Surface Area of the Sphere.
4. Question
Find the rate of change of the area of a circular disc with respect to its circumference when the radius is 3
cm.
Area of a Circular disc = p r
2
Circumference of a Circular disc = 2 p r
Where r = radius of Circular Disc.
Now we need to find  where A = Area of Circular disc and C = Circumference of the Circular disk.
5. Question
Find the rate of change of the volume of a cone with respect to the radius of its base.
The volume of Cone =
Where r = radius of the cone
h = height of the cone
We need to find, where V = Volume of cone and r = radius of the cone.
6. Question
Find the rate of change of the area of a circle with respect to its radius r when r = 5 cm.
Area of a Circle = p r
2
Where, r = radius of a circle.
We need to find,  , where A = Area of the Circle and r = radius of Circle.
At r = 5 cm we have,
7. Question
Find the rate of change of the volume of a ball with respect to its radius r. How fast is the volume changing
Volume of a Ball = Volume of Sphere =
Where r = radius of the ball.
We need to find  , where V = Volume of Ball and r = radius of the ball.
At r = 2 cm,
8. Question
The total cost C (x) associated with the production of x units of an item is given by C (x) = 0.007 x
3
– 0.003x
2
+ 15x + 4000. Find the marginal cost when 17 units are produced.
Given: Total Cost, C (x) = 0.007 x
3
– 0.003x
2
+ 15x + 4000
Where x = Number of units
Marginal Cost is given by,
Marginal Cost M(x) =
Therefore, M(x) = 0.007 × 3 x
2
– 0.003 × 2 x + 15
At Number of units = 17, x = 17.
So, M(17) = 0.021 (17)
2
– 0.006 (17) + 15
M(17) = 20.97
Hence, the Marginal Cost of 17 units is 20.97.
9. Question
The total revenue received from the sale of x units of a product is given by R(x) = 13 x
2
+ 26 x + 15. Find
the marginal revenue when x = 7?
Given: Total Revenue, R(x) = 13 x
2
+ 26 x + 15
Where x = number of units.
Marginal Revenue is given by,
M(x) =
Therefore,
M(x) = 26 (x + 1)
For x = 7,
M(x) = 26(7 + 1)
M(x) = 26 × 8 = Rs. 207
10. Question
The money to be spent for the welfare of the employees of a firm is proportional to the rate of change of its
total revenue (Marginal Revenue). If the total revenue ( in rupees) received from the sale of x units of a
product is given by
R(x) = 3 x
2
+ 36 x + 5, find the marginal revenue, when x = 5, and write which value does the question
indicate.
Given: Marginal Revenue, M(x) =
M(x) proportional to Money spent on the welfare of employees.
Thus, M(x) = k (Money spent on the welfare of employees)
M(x) = 6 x + 36
At x = 5, M(x) = 66
Now at x = 6, M(x) = 72
Thus Marginal Revenue for x = 5 is 66
And when x increased from x = 5 to x = 6,
Marginal revenue also increases and thus the money spent on the welfare of employees increases.
Exercise 13.2
1. Question
The side of a square sheet is increasing at the rate of 4 cm per minute. At what rate is the area increasing
when the side is 8 cm long?
Given: the side of a square sheet is increasing at the rate of 4 cm per minute.
To find rate of area increasing when the side is 8 cm long
let the side of the given square sheet be x cm at any instant time.
Then according to the given criteria,
Rate of side of the sheet increasing is,
Then the area of the square sheet at any time t will be
A = x
2
cm
2
.
Applying derivative with respect to time on both sides we get,
[from from equation(i)]
So when the side is 8cm long, the rate of area increasing will become
[from equation(ii)]
Hence the area is increasing at the rate of 64cm
2
/min when the side is 8 cm long
2. Question
An edge of a variable cube is increasing at the rate of 3 cm per second. How fast is the volume of the cube
increasing when the edge is 1 cm long?
Given: the edge of a variable cube is increasing at the rate of 3 cm per second.
To find rate of volume of the cube increasing when the edge is 1 cm long
Let the edge of the given cube be x cm at any instant time.
Then according to the given criteria,
Rate of edge of the cube increasing is,
Then the volume of the cube at any time t will be
V = x
3
cm
3
.
Applying derivative with respect to time on both sides we get,
[from equation(i)]
When the edge of the cube is 1cm long the rate of volume increasing becomes
Hence the volume of the cube increasing at the rate of 9cm
3
/sec when the edge of the cube is 1 cm long
3. Question
The side of a square is increasing at the rate of 0.2 cm/sec. Find the rate of increase of the perimeter of the
square.
Given: the side of a square is increasing at the rate of 0.2 cm/sec.
To find rate of increase of the perimeter of the square
Let the edge of the given cube be x cm at any instant time.
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## Mathematics (Maths) Class 12

204 videos|288 docs|139 tests

## Mathematics (Maths) Class 12

204 videos|288 docs|139 tests

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