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Class 12 Science (PCM)

Mathematics Part 1

Application of Derivatives

Application of Derivatives

Action PlanNotesQuestionsMock-Test
Study Action Plan (Sorted by Difficulty)
Optimization Modelling Steps
Accuracy: 0%
Total Attempted2
Total Correct0
Total Incorrect2
Difficulty Level:
High
First Derivative Test for Local Extrema
Accuracy: 20%
Total Attempted5
Total Correct1
Total Incorrect4
Difficulty Level:
High
Monotonic Functions
Accuracy: 38%
Total Attempted8
Total Correct3
Total Incorrect5
Difficulty Level:
High
Applications of Derivatives in Geometry
Accuracy: 40%
Total Attempted5
Total Correct2
Total Incorrect3
Difficulty Level:
High
Marginal Cost and Marginal Revenue
Accuracy: 42%
Total Attempted12
Total Correct5
Total Incorrect7
Difficulty Level:
High
Absolute Maxima and Minima on Closed Intervals
Accuracy: 45%
Total Attempted11
Total Correct5
Total Incorrect6
Difficulty Level:
High
Closed Interval Method for Absolute Extrema
Accuracy: 50%
Total Attempted2
Total Correct1
Total Incorrect1
Difficulty Level:
Medium
Finding Intervals of Increase and Decrease
Accuracy: 50%
Total Attempted6
Total Correct3
Total Incorrect3
Difficulty Level:
Medium
Local Maxima and Minima
Accuracy: 50%
Total Attempted8
Total Correct4
Total Incorrect4
Difficulty Level:
Medium
Related Rates Using Chain Rule
Accuracy: 50%
Total Attempted8
Total Correct4
Total Incorrect4
Difficulty Level:
Medium
Rate of Change and Derivatives
Accuracy: 54%
Total Attempted13
Total Correct7
Total Incorrect6
Difficulty Level:
Medium
Maximum, Minimum and Extreme Values
Accuracy: 58%
Total Attempted12
Total Correct7
Total Incorrect5
Difficulty Level:
Medium
Second Derivative Test for Local Extrema
Accuracy: 67%
Total Attempted3
Total Correct2
Total Incorrect1
Difficulty Level:
Medium
Tangents and Normals to Curves
Accuracy: 67%
Total Attempted6
Total Correct4
Total Incorrect2
Difficulty Level:
Medium
Slope of Tangent and Geometrical Meaning of Derivative
Accuracy: 75%
Total Attempted4
Total Correct3
Total Incorrect1
Difficulty Level:
Low
First Derivative Test for Monotonicity
Accuracy: 80%
Total Attempted5
Total Correct4
Total Incorrect1
Difficulty Level:
Low
Increasing, Decreasing and Constant Functions
Accuracy: 80%
Total Attempted5
Total Correct4
Total Incorrect1
Difficulty Level:
Low
Approximation Using Differentials
Accuracy: 100%
Total Attempted3
Total Correct3
Total Incorrect0
Difficulty Level:
Low
Geometrical Optimization Problems
Accuracy: 100%
Total Attempted2
Total Correct2
Total Incorrect0
Difficulty Level:
Low

Learning Objectives

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Learning Objectives

  • Understand the applications of derivatives in various fields such as engineering, science, and social science.
  • Determine the rate of change of quantities using derivatives.
  • Find the equations of tangent and normal to a curve at a point.
  • Identify turning points on the graph of a function to locate local maxima and minima.
  • Analyze intervals on which a function is increasing or decreasing.
  • Use derivatives to approximate values of certain quantities.

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Revision Notes & Summary

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Chapter 6: Application of Derivatives

6.1 Introduction

  • Study applications of derivatives in various fields such as engineering, science, and social science.
  • Key applications include:
    • Determining the rate of change of quantities.
    • Finding equations of tangent and normal to curves.
    • Locating turning points on graphs to identify local maxima and minima.
    • Identifying intervals of increase or decrease for functions.
    • Approximating values of certain quantities.

6.2 Rate of Change of Quantities

  • The derivative $ rac{ds}{dt}$ represents the rate of change of distance S with respect to time t.
  • For a function y=f(x)y = f(x), the derivative $ rac{dy}{dx}$ represents the rate of change of y with respect to x.
  • If two variables x and y vary with respect to another variable t, then by the Chain Rule:
    • $ rac{dy}{dx} = rac{dy}{dt} imes rac{dt}{dx}$

6.4 Maxima and Minima

  • Use derivatives to find maximum or minimum values of functions.
  • Definitions:
    • A function f has a maximum value in an interval I if there exists a point c in I such that f(c)≥f(x)f(c) \geq f(x) for all x∈Ix \in I.
    • A function f has a minimum value in an interval I if there exists a point c in I such that f(c)≤f(x)f(c) \leq f(x) for all x∈Ix \in I.
    • A critical point is where f′(c)=0f'(c) = 0 or f is not differentiable.

Examples of Applications

  1. Maximizing Profit: Given profit function P(x)=ax+bx2P(x) = ax + bx^2, find the number of trees per acre that maximizes profit.
  2. Projectile Motion: For a ball thrown from a height, find the maximum height it reaches.
  3. Distance Minimization: Find the nearest distance from a point to a curve.

Important Problems

  • Find maximum and minimum values of various functions, such as:
    • f(x)=x4−62x2+ax+9f(x) = x^4 - 62x^2 + ax + 9 at x=1x = 1.
    • f(x)=x+extsin(2x)f(x) = x + ext{sin}(2x) on [0,2extπ][0, 2 ext{π}].
    • Two numbers whose sum is 24 and product is maximized.
    • Dimensions of a box formed from a rectangular sheet to maximize volume.

Conclusion

  • Understanding derivatives is crucial for solving real-world problems involving optimization and rates of change.
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Practice Test – MCQs, True/False

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Multiple Choice Questions

A.

30.00530.005

B.

29.9829.98

C.

30.1530.15

D.

30.0230.02
Correct Answer: A

Solution:

The marginal cost is the derivative of the cost function with respect to xx. Thus, MC=dCdx=0.015x2−0.04x+30MC = \frac{dC}{dx} = 0.015x^2 - 0.04x + 30. Substituting x=10x = 10, we get MC=0.015(10)2−0.04(10)+30=30.005MC = 0.015(10)^2 - 0.04(10) + 30 = 30.005.

Chapter Concept:

Marginal Cost and Marginal Revenue

A.

10π10\pi cm²/s

B.

5π5\pi cm²/s

C.

20π20\pi cm²/s

D.

15π15\pi cm²/s
Correct Answer: A

Solution:

The rate of change of the area AA of a circle with respect to its radius rr is given by dAdr=2πr\frac{dA}{dr} = 2\pi r. When r=5r = 5 cm, dAdr=10π\frac{dA}{dr} = 10\pi cm²/s.

Chapter Concept:

Rate of Change and Derivatives

A.

f(x)f(x) has a local minimum at x=cx = c.

B.

f(x)f(x) has a local maximum at x=cx = c.

C.

f(x)f(x) is constant at x=cx = c.

D.

f(x)f(x) has a point of inflection at x=cx = c.
Correct Answer: B

Solution:

If f′(x)f'(x) changes from positive to negative as xx passes through cc, then f(x)f(x) has a local maximum at x=cx = c.

Chapter Concept:

First Derivative Test for Monotonicity

A.

14 m each

B.

16 m for the square, 12 m for the circle

C.

12 m for the square, 16 m for the circle

D.

10 m for the square, 18 m for the circle
Correct Answer: C

Solution:

Let the length of the wire used for the square be xx m, then the length for the circle is 28−x28-x m. The area of the square is (x4)2\left(\frac{x}{4}\right)^2 and the area of the circle is π(28−x2π)2\pi \left(\frac{28-x}{2\pi}\right)^2. To minimize the total area, differentiate the sum of these areas with respect to xx and set the derivative to zero to solve for xx. The optimal lengths are 12 m for the square and 16 m for the circle.

Chapter Concept:

Geometrical Optimization Problems

A.

The function has a local maximum at x=2x = 2.

B.

The function has a local minimum at x=2x = 2.

C.

The function is neither increasing nor decreasing at x=2x = 2.

D.

The function is constant at x=2x = 2.
Correct Answer: B

Solution:

The derivative f′(x)=2x−4f'(x) = 2x - 4. At x=2x = 2, f′(x)=0f'(x) = 0. The second derivative f′′(x)=2>0f''(x) = 2 > 0, indicating a local minimum at x=2x = 2.

Chapter Concept:

Finding Intervals of Increase and Decrease

A.

1.6 cm/s

B.

2.4 cm/s

C.

3.2 cm/s

D.

4.8 cm/s
Correct Answer: A

Solution:

Using the Pythagorean theorem, x2+y2=25x^2 + y^2 = 25. Differentiating with respect to time, 2xdxdt+2ydydt=02x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0. Substituting x=4x = 4 m, dxdt=2\frac{dx}{dt} = 2 cm/s, and y=3y = 3 m, we solve for dydt\frac{dy}{dt} to get 1.61.6 cm/s.

Chapter Concept:

Related Rates Using Chain Rule

A.

Local maxima at x=1x = 1 and local minima at x=−1x = -1

B.

Local maxima at x=−1x = -1 and local minima at x=1x = 1

C.

Local maxima at x=0x = 0 and no local minima

D.

No local maxima or minima
Correct Answer: B

Solution:

To find the local maxima and minima, we first calculate the derivative f′(x)=3x2−3=3(x−1)(x+1)f'(x) = 3x^2 - 3 = 3(x - 1)(x + 1). Setting f′(x)=0f'(x) = 0 gives critical points at x=1x = 1 and x=−1x = -1. Using the first derivative test, f′(x)f'(x) changes from negative to positive at x=−1x = -1, indicating a local maxima, and from positive to negative at x=1x = 1, indicating a local minima.

Chapter Concept:

Approximation Using Differentials

A.

Local maximum at x=1x = 1, local minimum at x=−1x = -1

B.

Local minimum at x=1x = 1, local maximum at x=−1x = -1

C.

Local maximum at x=−1x = -1, no local minimum

D.

No local extrema
Correct Answer: B

Solution:

The critical points are x=1x = 1 and x=−1x = -1. Using the first derivative test, f′(x)f'(x) changes from negative to positive at x=1x = 1, indicating a local minimum, and from positive to negative at x=−1x = -1, indicating a local maximum.

Chapter Concept:

Maximum, Minimum and Extreme Values

A.

(−∞,0)(-\infty, 0)

B.

(1,∞)(1, \infty)

C.

(0,1)(0, 1)

D.

(−∞,1)(-\infty, 1)
Correct Answer: B

Solution:

The derivative f′(x)=4x3−12x2+12x=4x(x2−3x+3)f'(x) = 4x^3 - 12x^2 + 12x = 4x(x^2 - 3x + 3) is positive for x>1x > 1, indicating the function is increasing in the interval (1,∞)(1, \infty).

Chapter Concept:

Monotonic Functions

A.

1

B.

5

C.

0

D.

3
Correct Answer: A

Solution:

The local minimum value is f(1)=1f(1) = 1.

Chapter Concept:

Absolute Maxima and Minima on Closed Intervals

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Application of Derivatives

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True or False

Correct Answer: False

Solution:

The second derivative test fails if the second derivative is zero at the critical point, in which case the first derivative test must be used to determine the nature of the critical point.

Chapter Concept :

Applications of Derivatives in Geometry

Correct Answer: True

Solution:

According to the first derivative test, if f′(x)f'(x) changes from positive to negative as xx increases through a point, then that point is a local maximum.

Chapter Concept :

Local Maxima and Minima

Correct Answer: True

Solution:

Derivatives are used to find turning points on the graph of a function, which help in locating points where the function attains its maximum or minimum values.

Chapter Concept :

Applications of Derivatives in Geometry

Correct Answer: True

Solution:

The area AA of a circle with radius rr is given by A=πr2A = \pi r^2. Therefore, the rate of change of the area AA with respect to its radius rr is dAdr=2πr\frac{dA}{dr} = 2\pi r.

Chapter Concept :

Related Rates Using Chain Rule

Correct Answer: True

Solution:

According to the definition of an increasing function, for any two points x1<x2x_1 < x_2 in the interval II, the function values satisfy f(x1)≤f(x2)f(x_1) \leq f(x_2).

Chapter Concept :

Increasing, Decreasing and Constant Functions

Correct Answer: True

Solution:

According to the theorem, if a function has a local maxima or minima at a point, then the derivative at that point is either zero or the function is not differentiable there.

Chapter Concept :

Maximum, Minimum and Extreme Values

Correct Answer: True

Solution:

According to the formal definition of increasing functions, a function ff is increasing on an interval II if for any x1<x2x_1 < x_2 in II, f(x1)≤f(x2)f(x_1) \leq f(x_2). This is consistent with the definition provided in the text.

Chapter Concept :

Increasing, Decreasing and Constant Functions

Correct Answer: False

Solution:

The correct expression for the rate of change of the area of a circle with respect to its radius rr is dAdt=2πrdrdt\frac{dA}{dt} = 2\pi r \frac{dr}{dt}, not πr2drdt\pi r^2 \frac{dr}{dt}.

Chapter Concept :

Related Rates Using Chain Rule

Correct Answer: True

Solution:

According to the definition of increasing functions, a function is increasing on an interval if the derivative f′(x)>0f'(x) > 0 for all xx in that interval.

Chapter Concept :

Monotonic Functions

Correct Answer: False

Solution:

According to the First Derivative Test, if f′(x)f'(x) changes from negative to positive at cc, then cc is a point of local minimum.

Chapter Concept :

First Derivative Test for Local Extrema

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