Application of Derivatives 6.1 introduces the concept of rate of change of quantities. If a quantity y depends on another quantity x, then the derivative \frac{dy}{dx} represents the rate at which y changes with respect to x.
\displaystyle \text{Rate of change of }y\text{ with respect to }x=\frac{dy}{dx}
The derivative measures how a small change in one quantity affects another related quantity. This idea is widely used in mathematics, physics, economics, engineering, and many real-life situations involving changing quantities.
In this exercise, we will learn how to use derivatives to find rates of change of various quantities involving length, area, volume, and other related variables, and apply these concepts to solve NCERT questions.
Key Concepts
1. Rate of Change of Quantities
If a quantity y depends on another quantity x, then the derivative \frac{dy}{dx} represents the rate of change of y with respect to x.
\displaystyle \text{Rate of change of }y\text{ with respect to }x=\frac{dy}{dx}
It tells us how rapidly one quantity changes when the other quantity changes.
Example: If the radius r of a circle changes with time, then \frac{dr}{dt} represents the rate of change of the radius with respect to time.
2. Positive Rate of Change
If \frac{dy}{dx}>0, then y increases as x increases.
A positive derivative indicates that the two quantities are increasing together.
Example: If the side of a square increases, then its area also increases. Hence the rate of change of area with respect to side length is positive.
3. Negative Rate of Change
If \frac{dy}{dx}\lt 0, then y decreases as x increases.
A negative derivative indicates that one quantity decreases when the other quantity increases.
Example: As air is released from a spherical balloon, its volume decreases with time. Therefore, the rate of change of volume with respect to time is negative.
4. Rate of Change with Respect to Time
In many practical situations, quantities change with time. In such cases, derivatives are taken with respect to the variable t representing time.
Common examples include \frac{ds}{dt}, \frac{dv}{dt} and \frac{dV}{dt}, which represent rates of change of distance, velocity and volume respectively.
5. Finding Rates of Change
- Express one quantity in terms of another.
- Differentiate the resulting relation with respect to the independent variable.
- Substitute the given values, if required.
This method is used extensively to solve problems involving lengths, areas, volumes and other related quantities.
Tip: The interpretation of derivatives, where \frac{dy}{dx}>0 indicates an increasing quantity and \frac{dy}{dx}<0 indicates a decreasing quantity.
Important Mensuration Formulas
The following area, surface area and volume formulas are frequently used in Application of Derivatives. These formulas help in expressing one quantity in terms of another before applying differentiation.
| S.No. | Area / Surface Area / Volume Formula |
|---|---|
| 1. | Area of a square: A=a^2 |
| 2. | Area of a rectangle: A=lb |
| 3. | Area of a triangle: A=\frac{1}{2}bh |
| 4. | Area of a circle: A=\pi r^2 |
| 5. | Circumference of a circle: C=2\pi r |
| 6. | Volume of a cube: V=a^3 |
| 7. | Total surface area of a cube: TSA=6a^2 |
| 8. | Lateral surface area of a cube: LSA=4a^2 |
| 9. | Volume of a cuboid: V=lbh |
| 10. | Total surface area of a cuboid: TSA=2(lb+bh+hl) |
| 11. | Lateral surface area of a cuboid: LSA=2h(l+b) |
| 12. | Volume of a sphere: V=\frac{4}{3}\pi r^3 |
| 13. | Surface area of a sphere: SA=4\pi r^2 |
| 14. | Volume of a hemisphere: V=\frac{2}{3}\pi r^3 |
| 15. | Curved surface area of a hemisphere: CSA=2\pi r^2 |
| 16. | Total surface area of a hemisphere: TSA=3\pi r^2 |
| 17. | Volume of a cylinder: V=\pi r^2h |
| 18. | Curved surface area of a cylinder: CSA=2\pi rh |
| 19. | Total surface area of a cylinder: TSA=2\pi r(r+h) |
| 20. | Volume of a cone: V=\frac{1}{3}\pi r^2h |
| 21. | Curved surface area of a cone: CSA=\pi rl |
| 22. | Total surface area of a cone: TSA=\pi r(l+r) |
Let us now solve all the NCERT questions step by step in this exercise 6.1 of Application of Derivatives.
Question 1: Application of Derivatives 6.1
1. Find the rate of change of the area of a circle with respect to its radius r when (a) r=3\text{ cm} (b) r=4\text{ cm}.
Solution
The area of a circle of radius r is given by
A=\pi r^2
Differentiating both sides with respect to r, we get
\displaystyle \frac{dA}{dr}=\frac{d}{dr}(\pi r^2)
\displaystyle =2\pi r \qquad ...(1)
(a) When r=3\text{ cm}
Substituting r=3 in (1), we get
\displaystyle \frac{dA}{dr}=2\pi(3)=6\pi
Hence, the rate of change of area with respect to radius when r=3\text{ cm} is
\boxed{\displaystyle \frac{dA}{dr}=6\pi\text{ cm}^2/\text{cm}}
(b) When r=4\text{ cm}
Substituting r=4 in (1), we get
\displaystyle \frac{dA}{dr}=2\pi(4)=8\pi
Hence, the rate of change of area with respect to radius when r=4\text{ cm} is
\boxed{\displaystyle \frac{dA}{dr}=8\pi\text{ cm}^2/\text{cm}}
Question 2: Application of Derivatives 6.1
2. The volume of a cube is increasing at the rate of 8\text{ cm}^3/\text{s}. How fast is the surface area increasing when the length of an edge is 12\text{ cm}?
Solution
Let a be the edge length of the cube.
Given,
\displaystyle \frac{dV}{dt}=8\text{ cm}^3/\text{s} \qquad ...(1)
The volume of a cube is
V=a^3
Differentiating both sides with respect to t, we get
\displaystyle \frac{dV}{dt}=3a^2\frac{da}{dt}
Substituting \frac{dV}{dt}=8 and a=12, we get
\displaystyle 8=3(12)^2\frac{da}{dt}
\displaystyle \frac{da}{dt}=\frac{8}{432}=\frac{1}{54}\text{ cm/s} \qquad ...(2)
The surface area of the cube is
S=6a^2
Differentiating both sides with respect to t, we get
\displaystyle \frac{dS}{dt}=12a\frac{da}{dt} \qquad ...(3)
Substituting a=12 and using (2) in (3), we get
\displaystyle \frac{dS}{dt}=12(12)\left(\frac{1}{54}\right)
\displaystyle =\frac{144}{54}=\frac{8}{3}\text{ cm}^2/\text{s}
Hence, the surface area is increasing at the rate
\boxed{\displaystyle \frac{dS}{dt}=\frac{8}{3}\text{ cm}^2/\text{s}}
You may already be following Maths Better for NCERT Solutions for the topics like
- Matrices
- Determinants
- Relations and Functions
- Inverse Trigonometric Functions
- Continuity and Differentiability
Likewise this Exercise 6.1 of Application of Derivatives for Class 12 Maths is designed to strengthen your concepts and improve step-by-step problem-solving skills. Now, let’s proceed to the next question.
Question 3: Rate of Change of Quantities
3. The radius of a circle is increasing uniformly at the rate of 3\text{ cm/s}. Find the rate at which the area of the circle is increasing when the radius is 10\text{ cm}.
Solution
Let A be the area and r be the radius of the circle.
Given,
\displaystyle \frac{dr}{dt}=3\text{ cm/s} \qquad ...(1)
and
r=10\text{ cm} \qquad ...(2)
The area of a circle is given by
A=\pi r^2
Differentiating both sides with respect to t, we get
\displaystyle \frac{dA}{dt}=\frac{d}{dt}(\pi r^2)
\displaystyle =2\pi r\frac{dr}{dt} \qquad ...(3)
Substituting the values from (1) and (2) in (3), we get
\displaystyle \frac{dA}{dt}=2\pi(10)(3)
\displaystyle =60\pi\text{ cm}^2/\text{s}
Hence, the area of the circle is increasing at the rate
\boxed{\displaystyle \frac{dA}{dt}=60\pi\text{ cm}^2/\text{s}}
Question 4: Derivative As a Rate Measure
4. An edge of a variable cube is increasing at the rate of 3\text{ cm/s}. How fast is the volume of the cube increasing when the edge is 10\text{ cm} long?
Solution
Let a be the edge length and V be the volume of the cube.
Given,
\displaystyle \frac{da}{dt}=3\text{ cm/s} \qquad ...(1)
and
a=10\text{ cm} \qquad ...(2)
The volume of a cube is given by
V=a^3
Differentiating both sides with respect to t, we get
\displaystyle \frac{dV}{dt}=\frac{d}{dt}(a^3)
\displaystyle =3a^2\frac{da}{dt} \qquad ...(3)
Substituting the values from (1) and (2) in (3), we get
\displaystyle \frac{dV}{dt}=3(10)^2(3)
\displaystyle =900\text{ cm}^3/\text{s}
Hence, the volume of the cube is increasing at the rate
\boxed{\displaystyle \frac{dV}{dt}=900\text{ cm}^3/\text{s}}
Question 5: Application of Derivatives 6.1
5. A stone is dropped into a quiet lake and waves move in circles at the speed of 5\text{ cm/s}. At the instant when the radius of the circular wave is 8\text{ cm}, how fast is the enclosed area increasing?
Solution
Let A be the area enclosed by the circular wave and r be its radius.
Given,
\displaystyle \frac{dr}{dt}=5\text{ cm/s} \qquad ...(1)
and
r=8\text{ cm} \qquad ...(2)
The area of a circle is given by
A=\pi r^2
Differentiating both sides with respect to t, we get
\displaystyle \frac{dA}{dt}=\frac{d}{dt}(\pi r^2)
\displaystyle =2\pi r\frac{dr}{dt} \qquad ...(3)
Substituting the values from (1) and (2) in (3), we get
\displaystyle \frac{dA}{dt}=2\pi(8)(5)
\displaystyle =80\pi\text{ cm}^2/\text{s}
Hence, the enclosed area is increasing at the rate
\boxed{\displaystyle \frac{dA}{dt}=80\pi\text{ cm}^2/\text{s}}
Important: Derivatives provide a powerful way to measure how one quantity changes with respect to another. This idea forms the foundation of many real-life applications of calculus.
Question 6: Measuring Rate of Change
6. The radius of a circle is increasing at the rate of 0.7\text{ cm/s}. What is the rate of increase of its circumference?
Solution
Let C be the circumference and r be the radius of the circle.
Given,
\displaystyle \frac{dr}{dt}=0.7\text{ cm/s} \qquad ...(1)
The circumference of a circle is given by
C=2\pi r
Differentiating both sides with respect to t, we get
\displaystyle \frac{dC}{dt}=\frac{d}{dt}(2\pi r)
\displaystyle =2\pi\frac{dr}{dt} \qquad ...(2)
Substituting the value from (1) in (2), we get
\displaystyle \frac{dC}{dt}=2\pi(0.7)
\boxed{\displaystyle =1.4\pi\text{ cm/s}}
Further, by using \pi=\frac{22}{7}, we get
\displaystyle \frac{dC}{dt}=1.4\times\frac{22}{7}=4.4\text{ cm/s}
Hence, the circumference is increasing at the rate
\displaystyle \frac{dC}{dt}=4.4\text{ cm/s}
Question 7: Exercise 6.1
7. The length x of a rectangle is decreasing at the rate of 5\text{ cm/min} and the width y is increasing at the rate of 4\text{ cm/min}. When x=8\text{ cm} and y=6\text{ cm}, find the rates of change of (a) the perimeter, and (b) the area of the rectangle.
Solution
Since length is decreasing, so we have
\displaystyle \frac{dx}{dt}=-5\text{ cm/min} \qquad ...(1)
And, since width is increasing, so we have
\displaystyle \frac{dy}{dt}=4\text{ cm/min} \qquad ...(2)
and, we have to find the rates of change when,
x=8\text{ cm},\quad y=6\text{ cm} \qquad ...(3)
(a) Rate of change of perimeter
We know, the perimeter of a rectangle is given by
P=2(x+y)
Differentiating both sides with respect to t, we get
\displaystyle \frac{dP}{dt}=2\left(\frac{dx}{dt}+\frac{dy}{dt}\right) \qquad ...(4)
Substituting the values from (1) and (2) in (4), we get
\displaystyle \frac{dP}{dt}=2(-5+4)
\displaystyle =-2\text{ cm/min}
Hence, because it’s negative, the perimeter is decreasing at the rate
\boxed{\displaystyle \frac{dP}{dt}=-2\text{ cm/min}}
(b) Rate of change of area
Also, the area of the rectangle is given by
A=xy
Differentiating both sides with respect to t, we get
\displaystyle \frac{dA}{dt}=x\frac{dy}{dt}+y\frac{dx}{dt} \qquad ...(5)
Substituting the values from (1), (2) and (3) in (5),
We get, \displaystyle \frac{dA}{dt}=8(4)+6(-5)
\displaystyle =32-30
\displaystyle =2\text{ cm}^2/\text{min}
Hence, the area is increasing at the rate
\boxed{\displaystyle \frac{dA}{dt}=2\text{ cm}^2/\text{min}}
Question 8: Application of Derivatives 6.1
8. A balloon, which always remains spherical on inflation, is being inflated by pumping in 900\text{ cm}^3/\text{s} of gas per second. Find the rate at which the radius of the balloon increases when the radius is 15\text{ cm}.
Solution
Let r be the radius and V be the volume of the spherical balloon.
Given,
\displaystyle \frac{dV}{dt}=900\text{ cm}^3/\text{s} \qquad ...(1)
and
r=15\text{ cm} \qquad ...(2)
The volume of a sphere is given by
\displaystyle V=\frac{4}{3}\pi r^3
Differentiating both sides with respect to t, we get
\displaystyle \frac{dV}{dt}=\frac{d}{dt}\left(\frac{4}{3}\pi r^3\right)
\displaystyle =4\pi r^2\frac{dr}{dt} \qquad ...(3)
Substituting the values from (1) and (2) in (3),
We get, \displaystyle 900=4\pi(15)^2\frac{dr}{dt}
\displaystyle 900=900\pi\frac{dr}{dt}
\displaystyle \frac{dr}{dt}=\frac{1}{\pi}\text{ cm/s}
Hence, the radius of the balloon is increasing at the rate
\boxed{\displaystyle \frac{dr}{dt}=\frac{1}{\pi}\text{ cm/s}}
Using \pi=\frac{22}{7},
\displaystyle \frac{dr}{dt}=\frac{7}{22}\approx 0.318\text{ cm/s}
NCERT Class 12 Maths includes several important exercises across both Part 1 and Part 2, and I’ll cover them one by one with clear explanations and step-by-step solutions. Many of these questions are also available in video format on my YouTube Channel, @MathsBetter, to help you understand the concepts more visually.
Now, let’s proceed to the next question.
Question 9: Application of Derivatives 6.1
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 latter is 10\text{ cm}.
Solution
Let r be the radius and V be the volume of the spherical balloon.
The volume of a sphere is given by
\displaystyle V=\frac{4}{3}\pi r^3
Differentiating both sides with respect to r, we get
\displaystyle \frac{dV}{dr}=\frac{d}{dr}\left(\frac{4}{3}\pi r^3\right)
\displaystyle =4\pi r^2 \qquad ...(1)
Substituting r=10\text{ cm} in (1), we get
\displaystyle \frac{dV}{dr}=4\pi(10)^2
\displaystyle =400\pi\text{ cm}^3/\text{cm}
Hence, the rate at which the volume is increasing with the radius when r=10\text{ cm} is
\boxed{\displaystyle \frac{dV}{dr}=400\pi\text{ cm}^3/\text{cm}}
Question 10: Rate of Change
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 2\text{ cm/s}. How fast is its height on the wall decreasing when the foot of the ladder is 4\text{ m} away from the wall?
Solution
Let x be the distance of the foot of the ladder from the wall and y be the height of the top of the ladder on the wall.
Given,
\displaystyle \frac{dx}{dt}=2\text{ cm/s}=0.02\text{ m/s} \qquad ...(1)
The length of the ladder remains constant and is equal to 5\text{ m}. Therefore, by Pythagoras theorem,
x^2+y^2=5^2
x^2+y^2=25 \qquad ...(2)
Differentiating both sides of (2) with respect to t,
We get, \displaystyle \frac{d}{dt}(x^2+y^2)=\frac{d}{dt}(25)
\displaystyle 2x\frac{dx}{dt}+2y\frac{dy}{dt}=0
\displaystyle x\frac{dx}{dt}+y\frac{dy}{dt}=0 \qquad ...(3)
When x=4\text{ m}, from (2),
We get, \displaystyle 4^2+y^2=25
\displaystyle y^2=9
\displaystyle y=3\text{ m} \qquad ...(4)
Substituting the values from (1) and (4) in (3),
We get, \displaystyle 4(0.02)+3\frac{dy}{dt}=0
\displaystyle 0.08+3\frac{dy}{dt}=0
\displaystyle \frac{dy}{dt}=-\frac{0.08}{3}=-\frac{8}{3}\text{ cm/s}
The negative sign shows that the height is decreasing.
Hence, the height of the ladder on the wall is decreasing at the rate
\boxed{\displaystyle \frac{8}{3}\text{ cm/s}}
Before differentiating, it is often helpful to express all variables in terms of a single quantity. This simplifies the calculation and reduces the chances of error.
Question 11: Application of Derivatives
11. A particle moves along the curve 6y=x^3+2. Find the points on the curve at which the y-coordinate is changing 8 times as fast as the x-coordinate.
Solution
Given,
6y=x^3+2 \qquad ...(1)
Differentiating both sides of (1) with respect to time t, we get
\displaystyle 6\frac{dy}{dt}=3x^2\frac{dx}{dt} \qquad ...(2)
According to the question, the y-coordinate is changing 8 times as fast as the x-coordinate.
Therefore,
\displaystyle \frac{dy}{dt}=8\frac{dx}{dt} \qquad ...(3)
Substituting (3) in (2), we get
\displaystyle 6\left(8\frac{dx}{dt}\right)=3x^2\frac{dx}{dt}
\displaystyle 48\frac{dx}{dt}=3x^2\frac{dx}{dt}
Since \frac{dx}{dt}\neq 0, because then \frac{dy}{dt} would also be 0, which is not possible, thus we can divide by \frac{dx}{dt},
We get, \displaystyle 48=3x^2
\displaystyle x^2=16
\displaystyle x=\pm 4 \qquad ...(4)
Substituting x=4 in (1), we get
\displaystyle 6y=4^3+2=66
\displaystyle y=11
Thus, one point is (4,11).
Substituting x=-4 in (1), we get
\displaystyle 6y=(-4)^3+2=-62
\displaystyle y=-\frac{31}{3}
Thus, the other point is \left(-4,-\frac{31}{3}\right).
Hence, the required points are
\boxed{\displaystyle (4,11)\text{ and }\left(-4,-\frac{31}{3}\right)}
Question 12: Application of Derivatives 6.1
12. The radius of an air bubble is increasing at the rate of \frac{1}{2}\text{ cm/s}. At what rate is the volume of the bubble increasing when the radius is 1\text{ cm}?
Solution
Let r be the radius and V be the volume of the spherical air bubble.
Given,
\displaystyle \frac{dr}{dt}=\frac{1}{2}\text{ cm/s} \qquad ...(1)
and
r=1\text{ cm} \qquad ...(2)
The volume of a sphere is given by
\displaystyle V=\frac{4}{3}\pi r^3
Differentiating both sides with respect to t, we get
\displaystyle \frac{dV}{dt}=\frac{d}{dt}\left(\frac{4}{3}\pi r^3\right)
\displaystyle =4\pi r^2\frac{dr}{dt} \qquad ...(3)
Substituting the values from (1) and (2) in (3), we get
\displaystyle \frac{dV}{dt}=4\pi(1)^2\left(\frac{1}{2}\right)
\displaystyle =2\pi\text{ cm}^3/\text{s}
Hence, the volume of the air bubble is increasing at the rate
\boxed{\displaystyle \frac{dV}{dt}=2\pi\text{ cm}^3/\text{s}}
Question 13: Application of Derivatives Exercise 6.1
13. A balloon, which always remains spherical, has a variable diameter \displaystyle \frac{3}{2}(2x+1). Find the rate of change of its volume with respect to x.
Solution
Let r be the radius and V be the volume of the spherical balloon.
Given, diameter
\displaystyle d=\frac{3}{2}(2x+1) \qquad ...(1)
Since d=2r, from (1)
\displaystyle 2r=\frac{3}{2}(2x+1)
\displaystyle r=\frac{3}{4}(2x+1) \qquad ...(2)
The volume of a sphere is given by
\displaystyle V=\frac{4}{3}\pi r^3
Substituting the value of r from (2), we get
\displaystyle V=\frac{4}{3}\pi\left[\frac{3}{4}(2x+1)\right]^3
\displaystyle V=\frac{9\pi}{16}(2x+1)^3 \qquad ...(3)
Differentiating (3) with respect to x, we get
\displaystyle \frac{dV}{dx}=\frac{9\pi}{16}\cdot 3(2x+1)^2\cdot 2
\displaystyle =\frac{27\pi}{8}(2x+1)^2
Hence, the rate of change of volume with respect to x is
\boxed{\displaystyle \frac{dV}{dx}=\frac{27\pi}{8}(2x+1)^2}
Question 14: Application of Derivatives 6.1
14. Sand is pouring from a pipe at the rate of 12\text{ cm}^3/\text{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\text{ cm}?
Solution
Let r be the radius, h be the height and V be the volume of the sand cone.
Given,
\displaystyle \frac{dV}{dt}=12\text{ cm}^3/\text{s} \qquad ...(1)
The height of the cone is always one-sixth of the radius of the base.
\displaystyle h=\frac{r}{6}
or
\displaystyle r=6h \qquad ...(2)
The volume of a cone is given by
\displaystyle V=\frac{1}{3}\pi r^2h
Substituting r=6h from (2), we get
\displaystyle V=\frac{1}{3}\pi(6h)^2h
\displaystyle =12\pi h^3 \qquad ...(3)
Differentiating both sides of (3) with respect to t, we get
\displaystyle \frac{dV}{dt}=36\pi h^2\frac{dh}{dt} \qquad ...(4)
Substituting \frac{dV}{dt}=12 and h=4 in (4),
We get, \displaystyle 12=36\pi(4)^2\frac{dh}{dt}
\displaystyle 12=576\pi\frac{dh}{dt}
\displaystyle \frac{dh}{dt}=\frac{12}{576\pi}=\frac{1}{48\pi}\text{ cm/s}
Hence, the height of the sand cone is increasing at the rate
\boxed{\displaystyle \frac{dh}{dt}=\frac{1}{48\pi}\text{ cm/s}}
Derivatives are frequently used in economics to study marginal cost, marginal revenue and other quantities that measure the effect of small changes in production or sales.
Question 15: Application of Derivatives
15. The total cost C(x) in Rupees associated with the production of x units of an item is given by
C(x)=0.007x^3-0.003x^2+15x+4000
Find the marginal cost when 17 units are produced.
Solution
The marginal cost is the rate of change of the total cost with respect to the number of units produced.
Therefore,
\displaystyle \text{Marginal Cost}=\frac{dC}{dx}
Given,
\displaystyle C(x)=0.007x^3-0.003x^2+15x+4000
Differentiating with respect to x, we get
\displaystyle \frac{dC}{dx}=\frac{d}{dx}(0.007x^3-0.003x^2+15x+4000)
\displaystyle \frac{dC}{dx}=0.021x^2-0.006x+15 \qquad ...(1)
Substituting x=17 in (1),
We get, \displaystyle \frac{dC}{dx}=0.021(17)^2-0.006(17)+15
\displaystyle =0.021(289)-0.102+15
\displaystyle =6.069-0.102+15
i.e. \displaystyle \frac{dC}{dx} =20.967
Hence, the marginal cost when 17 units are produced is
\boxed{\displaystyle \text{Marginal Cost}=₹\,20.967}
Marginal Cost is the rate of change of the total cost with respect to the number of units produced. It is obtained by differentiating the cost function C(x) with respect to x.
Marginal Cost = \displaystyle \frac{dC}{dx}
Question 16: Application of Derivatives
16. The total revenue in Rupees received from the sale of x units of a product is given by
R(x)=13x^2+26x+15
Find the marginal revenue when x=7.
Solution
The marginal revenue is the rate of change of the total revenue with respect to the number of units sold.
Therefore,
\displaystyle \text{Marginal Revenue}=\frac{dR}{dx}
Given,
\displaystyle R(x)=13x^2+26x+15
Differentiating with respect to x, we get
\displaystyle \frac{dR}{dx}=\frac{d}{dx}(13x^2+26x+15)
\displaystyle \frac{dR}{dx}=26x+26 \qquad ...(1)
Substituting x=7 in (1),
We get, \displaystyle \frac{dR}{dx}=26(7)+26
\displaystyle =182+26
\displaystyle =208
Hence, the marginal revenue when 7 units are sold is
\boxed{\displaystyle \text{Marginal Revenue}=₹\,208}
Marginal Revenue is the rate of change of the total revenue with respect to the number of units sold.
Marginal Revenue = \displaystyle \frac{dR}{dx}
Question 17: Application of Derivatives 6.1 – MCQ
17. The rate of change of the area of a circle with respect to its radius r at r=6\text{ cm} is
- (A) 10\pi
- (B) 12\pi
- (C) 8\pi
- (D) 11\pi
Solution
The area of a circle is given by
A=\pi r^2
Differentiating with respect to r, we get
\displaystyle \frac{dA}{dr}=2\pi r
Substituting r=6, we get
\displaystyle \frac{dA}{dr}=2\pi(6)=12\pi
✅️ Hence, the correct answer is
(B) 12\pi
Question 18: Application of Derivatives 6.1 – MCQ
18. The total revenue in Rupees received from the sale of x units of a product is given by
R(x)=3x^2+36x+5
The marginal revenue when x=15 is
- (A) 116
- (B) 96
- (C) 90
- (D) 126
Solution
The marginal revenue is given by
\displaystyle \frac{dR}{dx}
Given,
\displaystyle R(x)=3x^2+36x+5
Differentiating with respect to x, we get
\displaystyle \frac{dR}{dx}=6x+36
Substituting x=15,
We get, \displaystyle \frac{dR}{dx}=6(15)+36
\displaystyle =90+36
\displaystyle =126
✅️ Hence, the correct answer is
(D) 126
Common Mistakes to Avoid
- Using the wrong variable for differentiation: In rate of change problems, identify the independent variable carefully. If quantities change with time, differentiate with respect to t, not with respect to x or r.
- Ignoring units: Always write the correct units with the final answer. For example, \frac{dA}{dr} has units \text{cm}^2/\text{cm}, while \frac{dV}{dt} has units \text{cm}^3/\text{s}.
- Forgetting the Chain Rule: When quantities depend on time, expressions such as A=\pi r^2 and V=\frac{4}{3}\pi r^3 must be differentiated using the Chain Rule. Remember to multiply by \frac{dr}{dt}.
- Not expressing one variable in terms of another: In many questions, relations such as r=6h or h=\frac{r}{6} must be used before differentiation to reduce the number of variables.
- Substituting values too early: Differentiate the general formula first and substitute the given values only after obtaining the required derivative.
- Ignoring negative rates: If a quantity is decreasing, its rate of change should be taken as negative. For example, if a length decreases at 5\text{ cm/min}, then \frac{dx}{dt}=-5\text{ cm/min}.
- Using incorrect mensuration formulas: Errors in area, volume, CSA or TSA formulas lead to incorrect derivatives. Verify the required formula before differentiating.
- Mixing different units: Convert all quantities to consistent units before substitution. For example, metres and centimetres should not be used together in the same calculation.
- Confusing rate with actual value: Quantities such as \frac{dV}{dt}, \frac{dA}{dt} and \frac{dr}{dt} represent rates of change, not the actual volume, area or radius.
- Ignoring the meaning of the final sign: A positive derivative indicates an increasing quantity, while a negative derivative indicates a decreasing quantity. Interpret the answer accordingly.
Continue Learning
After completing Exercise 6.1 of Application of Derivatives, you should now be familiar with using derivatives to find rates of change of related quantities. You have also seen how differentiation can be applied to problems involving areas, volumes, surface areas, economics and real-life situations where one quantity changes with respect to another.
To strengthen your understanding further, make sure that you revise:
- The meaning of rate of change of quantities
- The interpretation of \frac{dy}{dx} as a rate of change
- Positive and negative rates of change
- The Chain Rule and differentiation with respect to time
- Area and circumference formulas for circles
- Volume, CSA and TSA formulas for common solids
- Related rates problems involving rectangles, spheres, cones and cubes
- Applications of the Product Rule in area-related questions
- Economic applications such as marginal cost and marginal revenue
- Correct interpretation of units in rate of change problems
Explore More
Try solving a variety of rate of change problems on your own. Begin by identifying the quantities involved and expressing one quantity in terms of another. Differentiate carefully with respect to the appropriate variable and substitute the given values only after obtaining the required derivative. Practise questions involving areas, volumes, surface areas, related rates and economic applications such as marginal cost and marginal revenue. Regular practice will help you apply differentiation correctly and improve your ability to model real-life situations using derivatives.
All the best and keep learning 👍
Discover more from Maths Better
Subscribe to get the latest posts sent to your email.
