NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.1 (Applications of Derivatives)
Class 12 Maths Chapter 6 Exercise 6.1 (Applications of Derivatives):- Chapter 6 of Class 12 Maths, titled "Applications of Derivatives," focuses on the various real-world applications of derivatives, a fundamental concept in calculus. Derivatives help in understanding the rate of change of quantities and are crucial in fields such as physics, engineering, and economics.
This chapter explores several practical applications, including finding the slope of a curve at a point, determining the maxima and minima of functions, and solving problems related to increasing and decreasing functions. Exercise 6.1 introduces these concepts with problems designed to enhance your understanding and ability to apply derivatives in different scenarios. Get NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.1 (Applications of Derivatives) in the below article.
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NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.1 (Applications of Derivatives)
Here are the NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.1 (Applications of Derivatives) below. Practice CBSE Class 12th Maths with our faculty solutions:-
Question 1. Find the rate of change of the area of a circle with respect to its radius r when
(a)r = 3 cm
(b)r = 4 cm
Solution : The area of a circle (A) with radius (r) is given by,
A = πr2
Now, the rate of change of the area with respect to its radius is given by,
=
(a) When r = 3 cm, then
(b) When r = 4 cm, then
Question 2. The volume of a cube is increasing at the rate of 8 cm3/sec. How fast is the surface area increasing when the length of an edge is 12 cm?
Solution : Let x be the length of a side, V be the volume, and s be the surface area of the cube.
Given: Rate of increase of volume of cube = 8 cm3/sec
Then, V = x3 and S = 6x2 where x is a function of time t
It is given tha
Hence, if the length of the edge of the cube is 12 cm, then the surface area is increasing at the rate of 8/3 cm2/s.
Question 3. The radius of the circle is increasing uniformly at the rate of 3 cm per second. Find the rate at which the area of the circle is increasing when the radius is 10 cm.
Solution : The area of a circle (A) with radius (r) is given by,
A = πr2
Now, the rate of change of the area with respect to its radius is given by,
=
It is given that,
Hence, the rate at which the area of the circle is increasing when the radius is 10 cm is 60π cm2/s.
Question 4. 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 if 10 cm long?
Solution : Let x be the length of a side and V be the volume of the cube. Then,
V = x3.
Hence, the volume of the cube is increasing at the rate of 900 cm3/s when the edge is 10 cm long.
Question 5. A stone is dropped into a quite lake and waves move in circles at the rate of 5 cm/sec. At the instant when radius of the circular wave is 8 cm, how fast is the enclosed area increasing?
Solution: The area of a circle (A) with radius (r) is given by
A = πr2
Therefore, the rate of change of area (A) with respect to time (t) is given by,
Hence, when the radius of the circular wave is 8 cm, the enclosed area is increasing at the rate of 80π cm2/s.
Question 6. The radius of a circle is increasing at the rate of 0.7 cm/s. What is the rate of its circumference?
Solution : The circumference of a circle (C) with radius (r) is given by
C = 2πr.
Therefore, the rate of change of circumference (C) with respect to time (t) is given by,
Hence, the rate of increase of the circumference is 2π(0.7) = 1.4π cm/s
Question 7. The length x of a rectangle is decreasing at the rate of 5 cm/minute. When x = 8 cm and y = 6 cm, find the rates of change of (a) the perimeter and (b) the area of the rectangle.
Solution : Since the length (x) is decreasing at the rate of 5 cm/minute and the width (y) is increasing at the rate of 4 cm/minute, we have:
Hence, the area of the rectangle is increasing at the rate of 2 cm2/min.
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Question 8. A balloon, which always remains spherical on inflation, is being inflated by pumping in 900 cubic centimeters of gas per second. Find the rate at which the radius of the balloon increases when the radius is 15 cm.
Solution : The volume of a sphere (V) with radius (r) is given by,
V = 4/3πr3
∴Rate of change of volume (V) with respect to time (t) is given by,
Hence, the rate at which the radius of the balloon increases when the radius is 15 cm is 1/π cm/s.
Question 9. A balloon, which always remains spherical has a variables radius. Find the rate at which its volume is increasing with the radius when the later is 10 cm.
Solution : Since, V = 4/3πr3
Rate of change of volume (V) with respect to its radius (r) is given by,
Hence, the volume of the balloon is increasing at the rate of 400π cm2.
Question 10. A ladder 5 cm 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 cm/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 y m be the height of the wall at which the ladder touches. Also, let the foot of the ladder be x maway from the wall.
Then, by Pythagoras theorem, we have:
x2 + y2 = 25 [Length of the ladder = 5 m]
⇒ y = √25 – x2
Then, the rate of change of height (y) with respect to time (t) is given by,
Hence, the height of the ladder on the wall is decreasing at the rate of 8/3 cm/s.
Question 11. 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 : Given: Equation of the curve 6y = x3 + 2……….(i)
The rate of change of the position of the particle with respect to time (t) is given by,
Question 12. The radius of an air bubble is increasing at the rate of 1/2 cm/s. At what rate is the volume of the bubble increasing when the radius is 1 cm?
Solution : The air bubble is in the shape of a sphere.
Now, the volume of an air bubble (V) with radius (r) is given by,
V = 4/3πr3
The rate of change of volume (V) with respect to time (t) is given by,
Hence, the rate at which the volume of the bubble increases is 2π cm3/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 : The volume of a sphere (V) with radius (r) is given by,
V = 4/3πr3
It is given that:
Diameter 3/2 (2x + 1)
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 4cm?
Solution : The volume of a cone (V) with radius (r) and height (h) is given by,
V = 4/3πr3
It is given that,
h = 1/6r ⇒r = 6h
Hence, when the height of the sand cone is 4 cm, its height is increasing at the rate of 1/48π cm/s.
Question 15. The total cost C(x) in rupees associated with the production of x units of an item given by 
Solution : Marginal cost is the rate of change of total cost with respect to output.
Hence, when 17 units are produced, the marginal cost is Rs. 20.967.
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Question 16. The total revenue in rupees received from the sale of x units of a product is given by
Solution : Marginal Revenue (MR) =
When x = 7,
MR = 26(7) + 26 = 182 + 26 = 208
Hence, the required marginal revenue is Rs 208.
Choose the correct answer in Exercises 17 and 18.
Question 17. The rate of change of the area of a circle with respect to its radius r at r = 6 cm is
(A) 10π
(B) 12π
(C) 8π
(D) 11π
Solution :
The area of a circle (A) with radius (r) is given by,
A = πr2
Therefore, the rate of change of the area with respect to its radius r is
Hence, the required rate of change of the area of a circle is 12π cm2/s.
The correct answer is B.
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 is the rate of change of total revenue with respect to the number of units sold.
∴Marginal Revenue (MR)= dR/dx= 3(2x) + 36 = 6x + 36
∴When x = 15,
MR = 6(15) + 36 = 90 + 36 = 126
Hence, the required marginal revenue is Rs 126.
The correct answer is D.
Class 12 Maths Chapter 6 Exercise 6.1 (Applications of Derivatives) Summary
Rate of Change:
-
Derivatives represent the rate of change of a quantity. For example, if s(t)s(t)s(t) is the position of an object at time ttt, then s′(t)s'(t)s′(t) is the velocity, representing the rate of change of position.
Tangents and Normals:
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The derivative of a function at a point gives the slope of the tangent to the function's graph at that point. The equation of the tangent line can be found using y−y1=m(x−x1)y - y_1 = m(x - x_1)y−y1=m(x−x1), where mmm is the slope.
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The normal line is perpendicular to the tangent. Its slope is the negative reciprocal of the tangent's slope.
Increasing and Decreasing Functions:
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A function is increasing on an interval if its derivative is positive over that interval, and decreasing if its derivative is negative.
-
To determine where a function is increasing or decreasing, find the intervals where the derivative is positive or negative.
Maxima and Minima:
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Critical points occur where the derivative is zero or undefined. These points are potential candidates for local maxima or minima.
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Use the First Derivative Test or the Second Derivative Test to classify critical points as local maxima, minima, or points of inflection.
Concavity and Points of Inflection:
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A function is concave up where its second derivative is positive, and concave down where its second derivative is negative.
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Points of inflection occur where the concavity changes, which is where the second derivative changes sign.
Optimization Problems:
-
Derivatives are used to solve optimization problems, where you need to find the maximum or minimum values of a function subject to certain constraints.
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Class 12 Maths Chapter 6 Exercise 6.1 (Applications of Derivatives) FAQs
Q1. What is the rate of change, and how is it determined using derivatives?
Ans. The rate of change of a quantity is represented by its derivative. For example, if y=f(x)y = f(x)y=f(x), the rate of change of yyy with respect to xxx is f′(x)f'(x)f′(x). It indicates how yyy changes as xxx changes.
Q2. How do you find the equation of a tangent line to a curve at a given point?
Ans. To find the equation of the tangent line to the curve y=f(x)y = f(x)y=f(x) at the point (x1,y1)(x_1, y_1)(x1,y1), use the formula y−y1=m(x−x1)y - y_1 = m(x - x_1)y−y1=m(x−x1), where m=f′(x1)m = f'(x_1)m=f′(x1) is the slope of the tangent.
Q3. What is the normal line, and how do you determine its equation?
Ans. The normal line is perpendicular to the tangent at a given point on the curve. If the slope of the tangent is mmm, the slope of the normal line is −1m-\frac{1}{m}−m1. The equation is y−y1=−1m(x−x1)y - y_1 = -\frac{1}{m}(x - x_1)y−y1=−m1(x−x1).
Q4. How can you determine if a function is increasing or decreasing?
Ans. A function f(x)f(x)f(x) is increasing on an interval if f′(x)>0f'(x) > 0f′(x)>0 for all xxx in that interval, and decreasing if f′(x)<0f'(x) < 0f′(x)<0. Analyze the sign of the derivative over the interval.
Q5. What are critical points, and how are they found?
Ans. Critical points are points where the derivative f′(x)f'(x)f′(x) is zero or undefined. To find critical points, solve f′(x)=0f'(x) = 0f′(x)=0 or determine where f′(x)f'(x)f′(x) does not exist.
