Application of Derivatives — Class 12 Maths Notes

Application of Derivatives · Class 12 Maths · 21 topics.

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Topics covered in Application of Derivatives

  1. 1.Introduction of Application of Derivative

    Introduction to the Application of Derivatives

    Let's dive into the world of derivatives, a fascinating and powerful tool in mathematics! Imagine you're a photographer aiming to capture a rocket launch. You want to track how fast the rocket ascends at different moments. This is where derivatives come in—they help us understand how things change. In mathematics, a derivative tells us the rate at which one quantity changes with respect to another.

    In everyday life, derivatives help us in many ways, from calculating speed to optimizing resources. They are essential in various fields like engineering, economics, physics, and even in our day-to-day decision making.

    Real-World Examples

    1. Speed and Acceleration: Just like tracking the speed of a rocket, derivatives determine the speed of a car at any moment or how quickly an object is accelerating or decelerating.
    2. Economics: Businesses use derivatives to find the maximum profit or minimum cost by calculating the rate at which costs are changing with respect to production levels.
    3. Medicine: In pharmacology, derivatives determine how quickly a drug's concentration increases or decreases in the bloodstream, which helps in deciding dosages.

    Careers and Industries Using Derivatives

    • Engineering: Engineers use derivatives to design components, understanding how changes in dimensions can affect stress or heat in a structure.
    • Finance: Financial analysts use derivatives to model the changing prices of stocks and commodities to predict future trends and hedge risks.
    • Meteorology: Meteorologists use derivatives to predict changes in weather patterns, including temperature and pressure rates.

    Understanding derivatives and their applications not only enhances your mathematical skills but also opens up numerous career opportunities. Whether you're interested in technology, science, finance, or even sports, derivatives play a key role in making informed decisions and optimizing outcomes.

  2. 2.Rate of Change of Quantities

    Rate of Change of Quantities

    Understanding the rate of change of quantities is like keeping track of how quickly things are happening around us. Let's use a simple example to make this concept clear. Imagine you are filling a balloon with air. The rate at which the balloon expands (its volume increases) as you keep pumping in air is a practical example of the rate of change.

    In mathematics, the rate of change is a way to describe how one quantity changes in relation to another quantity. The derivative is the tool we use to measure this rate of change precisely.

    How it Works

    Think of a car traveling on the highway. The speedometer shows the car's speed, which is the rate at which the car's position changes over time. In mathematical terms, if 𝑠(𝑡)s(t) represents the car's position at time 𝑡t, then the derivative 𝑠′(𝑡)s′(t), or the rate of change of the position, tells us the car's speed at any moment.

    Calculating the Rate of Change

    To calculate the rate of change, you can use the formula:

    Rate of change=Change in quantityChange in timeRate of change=Change in timeChange in quantity​

    For example, if you travel 100 kilometers in 2 hours, your average speed (rate of change of distance with respect to time) is:

    Speed=100 km2 hours=50 km/hourSpeed=2 hours100 km​=50 km/hour

    Real-World Applications

    1. Economics: In economics, the rate of change can indicate how quickly a company's profits are increasing or decreasing over time.
    2. Biology: Biologists might study the rate at which a bacterial population expands to understand and control growth.
    3. Physics: Physicists calculate rates such as acceleration, which is the rate of change of velocity.

    Careers and Industries

    • Data Science: Data scientists analyze changes in data trends over time to make predictions or understand patterns.
    • Environmental Science: Environmental scientists study changes in climate variables like temperature and precipitation rates to predict weather patterns and assess impacts on ecosystems.
    • Sports Science: In sports, analyzing the rate at which a player improves can help in designing personalized training programs.

    Understanding how to calculate and interpret rates of change equips you with a powerful tool for analyzing many aspects of the world around you, enhancing your ability to make informed decisions in daily life and in professional settings.

  3. 3.Example: Calculating Speed as Rate of Change

    Let's explore the concept of rate of change through a simple numerical example involving a car's journey. This example will help clarify how you can calculate the rate of change using actual numbers.

    Example: Calculating Speed as Rate of Change

    Imagine a car travels a certain distance and you're tracking how far it goes over specific time intervals. Here’s the scenario:

    • Distance traveled: The car travels 300 kilometers.
    • Time taken: The journey takes 6 hours.

    Objective

    Calculate the average speed of the car, which is the rate of change of distance with respect to time.

    Formula

    The rate of change is calculated as:

    Rate of change=Change in distanceChange in timeRate of change=Change in timeChange in distance​

    Calculation

    Using the values from our scenario:

    • Change in distance = 300 kilometers
    • Change in time = 6 hours
    Average Speed=300 kilometers6 hours=50 kilometers per hourAverage Speed=6 hours300 kilometers​=50 kilometers per hour

    So, the average speed of the car is 50 kilometers per hour. This means, on average, the car's position changes by 50 kilometers each hour. This is a practical example of how you can use the concept of rate of change in everyday situations, like calculating how fast a vehicle is traveling.

    Real-World Implication

    Understanding this rate of change can help in planning travel times, fuel needs, and can even assist in navigation planning.

  4. 4.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

    To find the rate of change of the area of a circle with respect to its radius, 𝑟r, we start by recalling the formula for the area of a circle, which is:

    𝐴=𝜋𝑟2A=πr2

    Finding the Rate of Change

    To determine how the area changes as the radius changes, we need the derivative of the area 𝐴A with respect to the radius 𝑟r. This derivative will tell us the rate at which the area changes for a small change in radius.

    Derivative Calculation

    The derivative of 𝐴A with respect to 𝑟r is given by:

    𝑑𝐴𝑑𝑟=𝑑𝑑𝑟(𝜋𝑟2)=2𝜋𝑟drdA​=drd​(πr2)=2πr

    This result, 2𝜋𝑟2πr, represents the rate of change of the area with respect to the radius.

    Solving the Problem

    Now, let's use this derivative to find the rate of change at the specific radii given in the problem:

    (a) When 𝑟=3r=3 cm:

    𝑑𝐴𝑑𝑟∣𝑟=3=2𝜋×3=6𝜋 cm2/cmdrdA​∣∣​r=3​=2π×3=6π cm2/cm

    This means that when the radius is 3 cm, the area of the circle is increasing at a rate of 6𝜋6π square centimeters per centimeter of radius increase.

    (b) When 𝑟=4r=4 cm:

    𝑑𝐴𝑑𝑟∣𝑟=4=2𝜋×4=8𝜋 cm2/cmdrdA​∣∣​r=4​=2π×4=8π cm2/cm

    This indicates that when the radius is 4 cm, the area is increasing at a rate of 8𝜋8π square centimeters per centimeter of radius increase.

    Conclusion

    These calculations show us how quickly the area of a circle grows as its radius increases. Specifically:

    • At 𝑟=3r=3 cm, the area grows at a rate of 6𝜋6π cm²/cm.
    • At 𝑟=4r=4 cm, the area grows at a rate of 8𝜋8π cm²/cm.

  5. 5.The volume of a cube is increasing at the rate of 8 cm³/s. How fast is the surface area increasing when the length of an edge is 12 cm?

    The volume of a cube is increasing at the rate of 8 cm³/s. How fast is the surface area increasing when the length of an edge is 12 cm?

    Solution: Let's solve this problem using calculus concepts, specifically related rates.

    Step 1: Set up the relationships

    • The volume 𝑉V of a cube with edge length 𝑠s is given by:𝑉=𝑠3V=s3
    • The surface area 𝐴A of a cube is given by:𝐴=6𝑠2A=6s2

    Step 2: Differentiate with respect to time 𝑡t

    • Differentiating both 𝑉V and 𝐴A with respect to time 𝑡t gives us:𝑑𝑉𝑑𝑡=3𝑠2𝑑𝑠𝑑𝑡dtdV​=3s2dtds​𝑑𝐴𝑑𝑡=12𝑠𝑑𝑠𝑑𝑡dtdA​=12sdtds​

    Step 3: Find 𝑑𝑠𝑑𝑡dtds​ using the given rate of change of volume

    Given that 𝑑𝑉𝑑𝑡=8dtdV​=8 cm³/s, we substitute and solve for 𝑑𝑠𝑑𝑡dtds​ when 𝑠=12s=12 cm:

    8=3(12)2𝑑𝑠𝑑𝑡8=3(12)2dtds​8=432𝑑𝑠𝑑𝑡8=432dtds​𝑑𝑠𝑑𝑡=8432=154 cm/sdtds​=4328​=541​ cm/s

    Step 4: Calculate 𝑑𝐴𝑑𝑡dtdA​ using 𝑑𝑠𝑑𝑡dtds​

    Now we use 𝑑𝑠𝑑𝑡dtds​ to find 𝑑𝐴𝑑𝑡dtdA​:

    𝑑𝐴𝑑𝑡=12(12)154dtdA​=12(12)541​𝑑𝐴𝑑𝑡=144×154=14454=83 cm2/sdtdA​=144×541​=54144​=38​ cm2/s

    Conclusion

    The surface area of the cube is increasing at a rate of 8338​ cm²/s when the edge length is 12 cm. This is how we apply the concept of related rates to determine how one quantity changes in relation to another in real-time scenarios.

  6. 6.A stone is dropped into a quiet lake and waves move in circles at the speed of 5 cm/s. At the instant when the radius of the circular wave is 8 cm, how fast is the enclosed area increasing?

    Question: A stone is dropped into a quiet lake and waves move in circles at the speed of 5 cm/s. At the instant when the radius of the circular wave is 8 cm, how fast is the enclosed area increasing?

    Solution: To solve this, we'll use the concept of related rates, which involves calculus, specifically derivatives. Here’s how you can solve it:

    1. Write the formula for the area of the circle: 𝐴=𝜋𝑟2A=πr2 where 𝐴A is the area and 𝑟r is the radius.

    2. Differentiate the area formula with respect to time (t): 𝑑𝐴𝑑𝑡=𝑑𝑑𝑡(𝜋𝑟2)dtdA​=dtd​(πr2)
      Using the chain rule: 𝑑𝐴𝑑𝑡=2𝜋𝑟𝑑𝑟𝑑𝑡dtdA​=2πrdtdr​ Here, 𝑑𝑟𝑑𝑡dtdr​ (the rate at which the radius changes) is given as 5 cm/s.

    3. Substitute the values and solve: At the instant when 𝑟=8r=8 cm: 𝑑𝐴𝑑𝑡=2𝜋×8 cm×5 cm/sdtdA​=2π×8 cm×5 cm/s
      𝑑𝐴𝑑𝑡=80𝜋 cm2/sdtdA​=80π cm2/s

    Thus, at the moment when the radius is 8 cm, the area enclosed by the wave is increasing at a rate of 80𝜋80π square centimeters per second.

    Real-World Application

    This problem is a classic example of related rates in physics and engineering, particularly in fields involving wave dynamics, like oceanography or acoustics, where understanding how changes in one quantity affect another is crucial.

  7. 7.Exercise Questions

    Problem - A balloon, which always remains spherical, has a variable diameter 32(2𝑥+1)23​(2x+1). Find the rate of change of its volume with respect to 𝑥x.

    Solution: To solve this problem, we'll apply calculus to find how quickly the volume of the balloon changes as 𝑥x changes.

    Step-by-Step Solution:

    1. Expression for Radius: Given the diameter of the balloon as 32(2𝑥+1)23​(2x+1), the radius 𝑟r is half of the diameter:

      𝑟=12×32(2𝑥+1)=34(2𝑥+1)r=21​×23​(2x+1)=43​(2x+1)

    2. Volume of a Sphere Formula: The volume 𝑉V of a sphere is given by:

      𝑉=43𝜋𝑟3V=34​πr3

      Substituting the expression for 𝑟r:

      𝑉=43𝜋(34(2𝑥+1))3V=34​π(43​(2x+1))3

    3. Simplify the Volume Expression: Simplifying (34(2𝑥+1))3(43​(2x+1))3 gives:

      𝑉=43𝜋(2764(2𝑥+1)3)V=34​π(6427​(2x+1)3)

      or

      𝑉=2748𝜋(2𝑥+1)3V=4827​π(2x+1)3

    4. Differentiate the Volume with Respect to 𝑥x: Now, differentiate 𝑉V with respect to 𝑥x to find 𝑑𝑉𝑑𝑥dxdV​:

      𝑑𝑉𝑑𝑥=2748𝜋⋅3(2𝑥+1)2⋅2dxdV​=4827​π⋅3(2x+1)2⋅2𝑑𝑉𝑑𝑥=8124𝜋(2𝑥+1)2dxdV​=2481​π(2x+1)2

      or simplified,

      𝑑𝑉𝑑𝑥=8124𝜋(2𝑥+1)2dxdV​=2481​π(2x+1)2

    Conclusion

    The rate of change of the volume 𝑉V of the balloon with respect to 𝑥x is 8124𝜋(2𝑥+1)22481​π(2x+1)2 cubic units per unit change in 𝑥x.

  8. 8.Exercise Questions

    Question: 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 4 cm?

    Solution: Step-by-Step Solution

    1. Understand the Relationship and Variables:

      • Let the height of the cone be ℎh, and the radius of the base be 𝑟r.
      • Given: ℎ=16𝑟h=61​r or 𝑟=6ℎr=6h.

    2. Volume of a Cone Formula: 𝑉=13𝜋𝑟2ℎV=31​πr2h Substituting 𝑟r from above: 𝑉=13𝜋(6ℎ)2ℎ=12𝜋ℎ3V=31​π(6h)2h=12πh3

    3. Differentiate the Volume with Respect to Time (t): 𝑑𝑉𝑑𝑡=36𝜋ℎ2𝑑ℎ𝑑𝑡dtdV​=36πh2dtdh​

    4. Solve for 𝑑ℎ𝑑𝑡dtdh​ When ℎ=4h=4 cm and 𝑑𝑉𝑑𝑡=12dtdV​=12 cm³/s: 12=36𝜋(4)2𝑑ℎ𝑑𝑡12=36π(4)2dtdh​ 12=576𝜋𝑑ℎ𝑑𝑡12=576πdtdh​ 𝑑ℎ𝑑𝑡=12576𝜋=148𝜋 cm/sdtdh​=576π12​=48π1​ cm/s

    Conclusion

    When the height of the sand cone is 4 cm, the height is increasing at a rate of 148𝜋48π1​ cm/s.

    Real-World Application

    This type of problem can be applied in fields where material deposition rates and geometric changes are crucial, such as in manufacturing processes or geological studies.

  9. 9.Exercise Questions

    The total cost 𝐶(𝑥)C(x) in Rupees associated with the production of 𝑥x units of an item is given by 𝐶(𝑥)=0.007𝑥3−0.003𝑥2+15𝑥+4000C(x)=0.007x3−0.003x2+15x+4000. Find the marginal cost when 17 units are produced.

    Solution:

    To find the marginal cost, we need to differentiate the total cost function with respect to 𝑥x and then evaluate it at 𝑥=17x=17.

    1. Differentiate the Total Cost Function:

      𝐶(𝑥)=0.007𝑥3−0.003𝑥2+15𝑥+4000C(x)=0.007x3−0.003x2+15x+4000

      The derivative 𝐶′(𝑥)C′(x), which represents the marginal cost function, is:

      𝐶′(𝑥)=0.021𝑥2−0.006𝑥+15C′(x)=0.021x2−0.006x+15

    2. Evaluate at 𝑥=17x=17:

      𝐶′(17)=0.021(17)2−0.006(17)+15C′(17)=0.021(17)2−0.006(17)+15𝐶′(17)=0.021(289)−0.102+15C′(17)=0.021(289)−0.102+15𝐶′(17)=6.069−0.102+15C′(17)=6.069−0.102+15𝐶′(17)=20.967C′(17)=20.967

    Therefore, the marginal cost when 17 units are produced is approximately 20.9720.97 Rupees. This means that to produce one additional unit after the 17th unit, it would cost about 20.97 Rupees more.

  10. 10.Exercise Questions

    The length 𝑥x of a rectangle is decreasing at the rate of 5 cm/minute and the width 𝑦y is increasing at the rate of 4 cm/minute. When 𝑥=8x=8 cm and 𝑦=6y=6 cm, find the rates of change of (a) the perimeter, and (b) the area of the rectangle.

    Solution:

    To solve this, we'll use the concept of related rates to find the rates of change of both the perimeter and the area.

    Part (a): Rate of Change of the Perimeter

    1. Formula for the Perimeter: The perimeter 𝑃P of a rectangle is given by:

      𝑃=2𝑥+2𝑦P=2x+2y

    2. Differentiate with respect to Time: Differentiating the formula with respect to time 𝑡t, we get:

      𝑑𝑃𝑑𝑡=2𝑑𝑥𝑑𝑡+2𝑑𝑦𝑑𝑡dtdP​=2dtdx​+2dtdy​

      Given 𝑑𝑥𝑑𝑡=−5dtdx​=−5 cm/min (since 𝑥x is decreasing) and 𝑑𝑦𝑑𝑡=4dtdy​=4 cm/min (since 𝑦y is increasing):

      𝑑𝑃𝑑𝑡=2(−5)+2(4)=−10+8=−2 cm/mindtdP​=2(−5)+2(4)=−10+8=−2 cm/min

      Thus, the perimeter of the rectangle is decreasing at a rate of 2 cm per minute.

    Part (b): Rate of Change of the Area

    1. Formula for the Area: The area 𝐴A of a rectangle is given by:

      𝐴=𝑥𝑦A=xy

    2. Differentiate with respect to Time: Differentiating the formula with respect to time 𝑡t, using the product rule, we get:

      𝑑𝐴𝑑𝑡=𝑥𝑑𝑦𝑑𝑡+𝑦𝑑𝑥𝑑𝑡dtdA​=xdtdy​+ydtdx​

      Substituting 𝑥=8x=8 cm, 𝑦=6y=6 cm, 𝑑𝑥𝑑𝑡=−5dtdx​=−5 cm/min, and 𝑑𝑦𝑑𝑡=4dtdy​=4 cm/min:

      𝑑𝐴𝑑𝑡=8(4)+6(−5)=32−30=2 cm2/mindtdA​=8(4)+6(−5)=32−30=2 cm2/min

      Thus, the area of the rectangle is increasing at a rate of 2 cm² per minute.

    Conclusion:

    (a) The perimeter of the rectangle is decreasing at a rate of 2 cm/min. (b) The area of the rectangle is increasing at a rate of 2 cm²/min.

    These calculations help in understanding how the dimensions of shapes change over time, which can be important in fields such as manufacturing, where material dimensions may vary during processing.

  11. 11.Increasing and Decreasing Functions

    What are Increasing and Decreasing Functions?

    In simple terms:

    • A function is increasing on an interval if, as you move along the x-axis from left to right within that interval, the function’s value (y-value) keeps going up.
    • A function is decreasing on an interval if, as you move from left to right, the function’s value keeps going down.

    How do we determine if a function is increasing or decreasing?

    To determine if a function is increasing or decreasing over an interval, we use the first derivative of the function. Here’s a step-by-step process:

    1. Find the derivative: First, differentiate the function. The derivative, 𝑓′(𝑥)f′(x), tells us the slope of the tangent to the function at any point 𝑥x.

    2. Analyze the sign of the derivative:

      • If 𝑓′(𝑥)>0f′(x)>0 (the derivative is positive) for all 𝑥x in an interval, the function is increasing on that interval.
      • If 𝑓′(𝑥)<0f′(x)<0 (the derivative is negative) for all 𝑥x in an interval, the function is decreasing on that interval.

    3. Identify critical points: These are points where 𝑓′(𝑥)=0f′(x)=0 or where the derivative does not exist. These points can be potential places where the function changes from increasing to decreasing or vice versa.

    Real-life Example

    Imagine you're tracking the speed of a car on a road trip. The speed graph over time can tell you periods when the car is accelerating (increasing function) and when it is decelerating (decreasing function). Analyzing this can help in optimizing fuel consumption and travel time.

    Applications in Careers and Industries

    Understanding increasing and decreasing functions is crucial in many fields:

    • Economics: Economists use these concepts to analyze and predict points of maximum profit and minimum cost in business.
    • Engineering: Engineers use derivatives to determine the most efficient structures and processes.
    • Environmental Science: Analyzing changes in population growths or declines in ecosystems involves understanding the behavior of increasing and decreasing functions.

    Activity to Try

    Try finding out whether the function 𝑓(𝑥)=𝑥3−3𝑥+2f(x)=x3−3x+2 is increasing or decreasing in different intervals. Here's how you can start:

    • First, find the derivative of 𝑓(𝑥)f(x).
    • Then, determine where the derivative is positive or negative.
    • Finally, conclude in which intervals 𝑓(𝑥)f(x) is increasing or decreasing.

  12. 12.Exercise Questions

    Question 1

    Show that the function given by 𝑓(𝑥)=3𝑥+17f(x)=3x+17 is increasing on 𝑅R.

    Solution:

    To determine if the function is increasing, we calculate its derivative: 𝑓′(𝑥)=𝑑𝑑𝑥(3𝑥+17)f′(x)=dxd​(3x+17)

    Differentiating 3𝑥+173x+17 with respect to 𝑥x: 𝑓′(𝑥)=3f′(x)=3

    Since 𝑓′(𝑥)=3f′(x)=3 is positive for all 𝑥x in 𝑅R, the function 𝑓(𝑥)=3𝑥+17f(x)=3x+17 is increasing on 𝑅R.

    Question 2

    Show that the function given by 𝑓(𝑥)=𝑒2𝑥f(x)=e2x is increasing on 𝑅R.

    Solution:

    To determine if the function is increasing, we calculate its derivative: 𝑓′(𝑥)=𝑑𝑑𝑥(𝑒2𝑥)f′(x)=dxd​(e2x)

    Using the chain rule to differentiate 𝑒2𝑥e2x: 𝑓′(𝑥)=𝑒2𝑥⋅2=2𝑒2𝑥f′(x)=e2x⋅2=2e2x

    Since 𝑒2𝑥e2x is always positive (as the exponential function is always positive), and multiplying by 2 does not change the sign, 𝑓′(𝑥)=2𝑒2𝑥f′(x)=2e2x is also always positive for all 𝑥x in 𝑅R. Therefore, the function 𝑓(𝑥)=𝑒2𝑥f(x)=e2x is increasing on 𝑅R.

    These solutions confirm that both functions are indeed increasing across all real numbers, 𝑅R.

  13. 13.Exercise Questions

    Question 1

    Find the intervals in which the function given by 𝑓(𝑥)=2𝑥2−3𝑥f(x)=2x2−3x is:

    • (a) increasing
    • (b) decreasing

    Solution for Question 1

    First, find the derivative 𝑓′(𝑥)f′(x): 𝑓′(𝑥)=𝑑𝑑𝑥(2𝑥2−3𝑥)=4𝑥−3f′(x)=dxd​(2x2−3x)=4x−3

    To find where the function is increasing or decreasing, we set 𝑓′(𝑥)=0f′(x)=0 to find critical points: 4𝑥−3=04x−3=0 𝑥=34x=43​

    Now, analyze the sign of 𝑓′(𝑥)f′(x) around this critical point:

    • If 𝑥<34x<43​, pick 𝑥=0x=0 ⇒𝑓′(0)=−3⇒f′(0)=−3 (negative, so decreasing)
    • If 𝑥>34x>43​, pick 𝑥=1x=1 ⇒𝑓′(1)=1⇒f′(1)=1 (positive, so increasing)

    Interval where 𝑓(𝑥)f(x) is:

    • Increasing: (34,∞)(43​,∞)
    • Decreasing: (−∞,34)(−∞,43​)

    Question 2

    Find the intervals in which the function given by 𝑓(𝑥)=2𝑥3−3𝑥2−36𝑥+7f(x)=2x3−3x2−36x+7 is:

    • (a) increasing
    • (b) decreasing

    Solution for Question 2

    First, find the derivative 𝑓′(𝑥)f′(x): 𝑓′(𝑥)=𝑑𝑑𝑥(2𝑥3−3𝑥2−36𝑥+7)=6𝑥2−6𝑥−36f′(x)=dxd​(2x3−3x2−36x+7)=6x2−6x−36

    To find where the function is increasing or decreasing, we set 𝑓′(𝑥)=0f′(x)=0: 6𝑥2−6𝑥−36=06x2−6x−36=0 Divide through by 6: 𝑥2−𝑥−6=0x2−x−6=0 Factoring: (𝑥−3)(𝑥+2)=0(x−3)(x+2)=0 𝑥=3 or 𝑥=−2x=3 or x=−2

    Now, analyze the sign of 𝑓′(𝑥)f′(x) around these critical points:

    • If 𝑥<−2x<−2, pick 𝑥=−3x=−3 ⇒𝑓′(−3)=18+18−36=0⇒f′(−3)=18+18−36=0 (actually zero, move further)
    • If 𝑥=−3x=−3, 𝑓′(−4)=24+24−36=12f′(−4)=24+24−36=12 (positive, so increasing)
    • Between −2−2 and 33, pick 𝑥=0x=0 ⇒𝑓′(0)=−36⇒f′(0)=−36 (negative, so decreasing)
    • If 𝑥>3x>3, pick 𝑥=4x=4 ⇒𝑓′(4)=96−24−36=36⇒f′(4)=96−24−36=36 (positive, so increasing)

    Interval where 𝑓(𝑥)f(x) is:

    • Increasing: (−∞,−2)∪(3,∞)(−∞,−2)∪(3,∞)
    • Decreasing: (−2,3)(−2,3)

  14. 14.Find the intervals in which the following functions are strictly increasing or decreasing:

    Function Analysis

    (a) 𝑓(𝑥)=𝑥2+2𝑥−5f(x)=x2+2x−5

    Find the derivative: 𝑓′(𝑥)=2𝑥+2f′(x)=2x+2

    Set the derivative equal to zero to find critical points: 2𝑥+2=02x+2=0 𝑥=−1x=−1

    Test the sign of 𝑓′(𝑥)f′(x) in intervals divided by the critical point 𝑥=−1x=−1:

    • For 𝑥<−1x<−1 (e.g., 𝑥=−2x=−2), 𝑓′(−2)=−2f′(−2)=−2 (negative, decreasing)
    • For 𝑥>−1x>−1 (e.g., 𝑥=0x=0), 𝑓′(0)=2f′(0)=2 (positive, increasing)

    Intervals:

    • Increasing: (−1,∞)(−1,∞)
    • Decreasing: (−∞,−1)(−∞,−1)

    (b) 𝑓(𝑥)=10−6𝑥−2𝑥2f(x)=10−6x−2x2

    Find the derivative: 𝑓′(𝑥)=−6−4𝑥f′(x)=−6−4x

    Set the derivative equal to zero to find critical points: −6−4𝑥=0−6−4x=0 𝑥=−32x=−23​

    Test the sign of 𝑓′(𝑥)f′(x) in intervals:

    • For 𝑥<−32x<−23​ (e.g., 𝑥=−2x=−2), 𝑓′(−2)=2f′(−2)=2 (positive, increasing)
    • For 𝑥>−32x>−23​ (e.g., 𝑥=0x=0), 𝑓′(0)=−6f′(0)=−6 (negative, decreasing)

    Intervals:

    • Increasing: (−∞,−32)(−∞,−23​)
    • Decreasing: (−32,∞)(−23​,∞)

    (c) 𝑓(𝑥)=−2𝑥3−9𝑥2−12𝑥+1f(x)=−2x3−9x2−12x+1

    Find the derivative: 𝑓′(𝑥)=−6𝑥2−18𝑥−12f′(x)=−6x2−18x−12

    Factor out the greatest common factor: 𝑓′(𝑥)=−6(𝑥2+3𝑥+2)f′(x)=−6(x2+3x+2)

    Factor the quadratic: 𝑓′(𝑥)=−6(𝑥+1)(𝑥+2)f′(x)=−6(x+1)(x+2)

    Critical points are 𝑥=−1x=−1 and 𝑥=−2x=−2:

    • For 𝑥<−2x<−2, 𝑓′(−3)=−6⋅4=−24f′(−3)=−6⋅4=−24 (negative, decreasing)
    • Between −2−2 and −1−1, 𝑓′(−1.5)=−6⋅0.25=−1.5f′(−1.5)=−6⋅0.25=−1.5 (negative, decreasing)
    • For 𝑥>−1x>−1, 𝑓′(0)=−6⋅2=−12f′(0)=−6⋅2=−12 (negative, decreasing)

    Intervals:

    • Decreasing: (−∞,∞)(−∞,∞) (The function is always decreasing)

    (d) 𝑓(𝑥)=6−9𝑥−𝑥2f(x)=6−9x−x2

    Find the derivative: 𝑓′(𝑥)=−9−2𝑥f′(x)=−9−2x

    Set the derivative equal to zero to find critical points: −9−2𝑥=0−9−2x=0 𝑥=−92x=−29​

    Test the sign of 𝑓′(𝑥)f′(x) in intervals:

    • For 𝑥<−92x<−29​ (e.g., 𝑥=−5x=−5), 𝑓′(−5)=1f′(−5)=1 (positive, increasing)
    • For 𝑥>−92x>−29​ (e.g., 𝑥=0x=0), 𝑓′(0)=−9f′(0)=−9 (negative, decreasing)

    Intervals:

    • Increasing: (−∞,−92)(−∞,−29​)
    • Decreasing: (−92,∞)(−29​,∞)

    These solutions indicate the behavior (increasing or decreasing) of each function based on their derivatives.

  15. 15.Exercise Questions

    Question 1: Proving the Logarithmic Function is Increasing on (0,∞)(0,∞)

    The logarithmic function in question is generally assumed to be the natural logarithm, ln⁡(𝑥)ln(x).

    Proof:

    1. Function Definition: 𝑓(𝑥)=ln⁡(𝑥)f(x)=ln(x)
    2. Derivative Calculation: 𝑓′(𝑥)=𝑑𝑑𝑥[ln⁡(𝑥)]=1𝑥f′(x)=dxd​[ln(x)]=x1​
    3. Sign of the Derivative: Since 𝑥>0x>0 in the domain (0,∞)(0,∞), 𝑓′(𝑥)=1𝑥>0f′(x)=x1​>0 for all 𝑥x in this interval.
    4. Conclusion: A positive derivative means that the function is strictly increasing on (0,∞)(0,∞).

    Question 2: Proving the Function 𝑓(𝑥)=𝑥2−𝑥+1f(x)=x2−x+1 is Neither Strictly Increasing Nor Decreasing on (−1,1)(−1,1)

    Proof:

    1. Function Definition: 𝑓(𝑥)=𝑥2−𝑥+1f(x)=x2−x+1
    2. Derivative Calculation: 𝑓′(𝑥)=𝑑𝑑𝑥[𝑥2−𝑥+1]=2𝑥−1f′(x)=dxd​[x2−x+1]=2x−1
    3. Critical Point: Set the derivative equal to zero to find the critical points: 2𝑥−1=02x−1=0 𝑥=12x=21​
    4. Test Intervals:
      • Sign of the Derivative Before 1221​: For 𝑥=0x=0 (within (−1,1)(−1,1)), 𝑓′(0)=−1f′(0)=−1 (negative, decreasing).
      • Sign of the Derivative After 1221​: For 𝑥=1x=1 (also within (−1,1)(−1,1)), 𝑓′(1)=1f′(1)=1 (positive, increasing).
    5. Conclusion: Since 𝑓′(𝑥)f′(x) changes sign around 𝑥=12x=21​, 𝑓(𝑥)f(x) is decreasing before 𝑥=12x=21​ and increasing after 𝑥=12x=21​ within the interval (−1,1)(−1,1). Therefore, the function is neither strictly increasing nor strictly decreasing over the entire interval (−1,1)(−1,1).

  16. 16.Maxima and Minima

    Understanding Maxima and Minima is a key part of calculus, especially when analyzing the behavior of functions. These points represent the peaks and troughs of a function's graph, helping in identifying and solving optimization problems. Let’s delve into the definitions, important theorems, and examples to get a clear picture.

    Definitions

    1. Local Maximum: A function 𝑓(𝑥)f(x) has a local maximum at 𝑥=𝑐x=c if 𝑓(𝑐)f(c) is greater than all other values of 𝑓(𝑥)f(x) near 𝑐c. This means there's a small interval around 𝑐c where 𝑓(𝑐)f(c) is the highest point.

    2. Local Minimum: A function 𝑓(𝑥)f(x) has a local minimum at 𝑥=𝑐x=c if 𝑓(𝑐)f(c) is less than all other values of 𝑓(𝑥)f(x) near 𝑐c. Here, 𝑓(𝑐)f(c) is the lowest point in a small interval around 𝑐c.

    3. Global (or Absolute) Maximum: 𝑓(𝑥)f(x) has a global maximum at 𝑥=𝑐x=c if 𝑓(𝑐)f(c) is the highest value of 𝑓(𝑥)f(x) over the entire domain of the function.

    4. Global (or Absolute) Minimum: 𝑓(𝑥)f(x) has a global minimum at 𝑥=𝑐x=c if 𝑓(𝑐)f(c) is the lowest value of 𝑓(𝑥)f(x) over the entire domain of the function.

    Important Theorems

    1. Fermat’s Theorem on Stationary Points: If 𝑓(𝑥)f(x) has a local maximum or minimum at 𝑥=𝑐x=c, and if 𝑓f is differentiable at 𝑐c, then 𝑓′(𝑐)=0f′(c)=0. This means the derivative (slope) at 𝑐c is zero, indicating a horizontal tangent.

    2. The First Derivative Test: This test helps determine whether a point 𝑐c is a maximum or a minimum:

      • If 𝑓′(𝑥)f′(x) changes from positive to negative at 𝑐c, 𝑓(𝑐)f(c) is a local maximum.
      • If 𝑓′(𝑥)f′(x) changes from negative to positive at 𝑐c, 𝑓(𝑐)f(c) is a local minimum.

    3. The Second Derivative Test: If 𝑓′(𝑐)=0f′(c)=0 and:

      • 𝑓′′(𝑐)<0f′′(c)<0, then 𝑓(𝑐)f(c) is a local maximum.
      • 𝑓′′(𝑐)>0f′′(c)>0, then 𝑓(𝑐)f(c) is a local minimum.
      • 𝑓′′(𝑐)=0f′′(c)=0, the test is inconclusive, and one may need to use the first derivative test or higher derivatives.

    Examples

    1. Quadratic Function: Consider 𝑓(𝑥)=−𝑥2+4𝑥−3f(x)=−x2+4x−3.

      • Derivative: 𝑓′(𝑥)=−2𝑥+4f′(x)=−2x+4.
      • Setting 𝑓′(𝑥)=0f′(x)=0 gives 𝑥=2x=2.
      • Second derivative: 𝑓′′(𝑥)=−2f′′(x)=−2 (always negative).
      • Thus, 𝑓(𝑥)f(x) has a local (and global) maximum at 𝑥=2x=2.

    2. Cubic Function: Consider 𝑔(𝑥)=𝑥3−3𝑥2+3𝑥−1g(x)=x3−3x2+3x−1.

      • Derivative: 𝑔′(𝑥)=3𝑥2−6𝑥+3g′(x)=3x2−6x+3.
      • Setting 𝑔′(𝑥)=0g′(x)=0 gives 𝑥=1x=1 (repeated root).
      • Second derivative: 𝑔′′(𝑥)=6𝑥−6g′′(x)=6x−6, 𝑔′′(1)=0g′′(1)=0.
      • Neither a maximum nor a minimum; use higher derivative tests or graphing.

    These concepts are crucial in fields like engineering, economics, and science, where optimization problems are common. For instance, finding the maximum profit or minimum cost, or determining critical points in mechanical structures can all be analyzed using these principles.

  17. 17.Exercise Questions

    Question

    Find the local minimum value of the function 𝑓(𝑥)=3+∣𝑥∣f(x)=3+∣x∣, where 𝑥∈𝑅x∈R.

    Solution

    The function given is 𝑓(𝑥)=3+∣𝑥∣f(x)=3+∣x∣. To find the local minimum value of this function, we analyze the behavior of the absolute value component.

    1. Understanding the Function: The absolute value function ∣𝑥∣∣x∣ measures the distance of 𝑥x from zero on the real number line. Thus, ∣𝑥∣∣x∣ is always non-negative, and it reaches its minimum value of 0 when 𝑥=0x=0.

    2. Evaluating 𝑓(𝑥)f(x) at 𝑥=0x=0: 𝑓(0)=3+∣0∣=3+0=3f(0)=3+∣0∣=3+0=3

    3. Behavior of 𝑓(𝑥)f(x) for 𝑥≠0x=0:

      • For 𝑥>0x>0, ∣𝑥∣=𝑥∣x∣=x. Hence, 𝑓(𝑥)=3+𝑥f(x)=3+x, which increases as 𝑥x increases.
      • For 𝑥<0x<0, ∣𝑥∣=−𝑥∣x∣=−x. Hence, 𝑓(𝑥)=3−𝑥f(x)=3−x, which increases as 𝑥x becomes more negative.

      In both cases, as 𝑥x moves away from 0, whether in the positive or negative direction, the function value 𝑓(𝑥)f(x) increases.

    4. Conclusion: Since 𝑓(𝑥)=3+∣𝑥∣f(x)=3+∣x∣ attains its lowest value at 𝑥=0x=0 and increases as 𝑥x moves away from 0, the point 𝑥=0x=0 is where 𝑓(𝑥)f(x) has its local minimum.

    5. Local Minimum Value: The local minimum value of 𝑓(𝑥)f(x) is 3, occurring at 𝑥=0x=0.

    This example illustrates the graphical behavior of the absolute value function, where the vertex of the V-shaped graph represents the local minimum. The graph and the algebraic analysis confirm that the local minimum value of 𝑓(𝑥)=3+∣𝑥∣f(x)=3+∣x∣ is indeed 3.

  18. 18.Maximum and Minimum Values of a Function in a Closed Interval

    Understanding the maximum and minimum values of a function within a closed interval is a fundamental concept in calculus, particularly useful in practical scenarios where constraints are present, such as in engineering, economics, and physical sciences. Here’s a breakdown of the concepts and methods used to determine these values:

    Definitions

    1. Closed Interval: A closed interval [𝑎,𝑏][a,b] includes both endpoints 𝑎a and 𝑏b. A function 𝑓(𝑥)f(x) defined on this interval has values for every 𝑥x between and including 𝑎a and 𝑏b.

    2. Absolute Maximum and Minimum: The absolute maximum value of 𝑓(𝑥)f(x) on the interval [𝑎,𝑏][a,b] is the greatest value that 𝑓(𝑥)f(x) attains within the interval. Similarly, the absolute minimum value is the smallest value attained. These values can occur at critical points inside the interval or at the endpoints.

    Theorem: Extreme Value Theorem

    The Extreme Value Theorem states that if a function 𝑓(𝑥)f(x) is continuous on a closed interval [𝑎,𝑏][a,b], then 𝑓(𝑥)f(x) must attain both an absolute maximum and an absolute minimum at least once each within the interval. This theorem ensures that the search for extreme values is not in vain in such conditions.

    Procedure to Find Maximum and Minimum Values

    1. Identify the Interval: Ensure that the function is defined and continuous on the closed interval [𝑎,𝑏][a,b].

    2. Find the Derivative: Compute 𝑓′(𝑥)f′(x), the first derivative of 𝑓(𝑥)f(x). This derivative helps identify the critical points where potential maxima or minima might occur (where 𝑓′(𝑥)=0f′(x)=0 or where 𝑓′(𝑥)f′(x) does not exist).

    3. Evaluate Critical Points: Solve 𝑓′(𝑥)=0f′(x)=0 to find critical points. These are points within the interval where the function’s rate of change is zero and thus possible locations of maxima or minima.


    4. Evaluate Endpoints: Compute 𝑓(𝑎)f(a) and 𝑓(𝑏)f(b). Since the interval is closed, the function’s values at these points must also be considered for extreme values.

    5. Compare Values: Compare the function values at all critical points and endpoints. The highest of these values will be the absolute maximum, and the lowest will be the absolute minimum on the interval.

    Example

    Consider the function 𝑓(𝑥)=𝑥3−3𝑥2+4f(x)=x3−3x2+4 defined on the closed interval [−1,3][−1,3].

    1. Derivative: 𝑓′(𝑥)=3𝑥2−6𝑥f′(x)=3x2−6x.

    2. Critical Points: Solve 𝑓′(𝑥)=0f′(x)=0. 3𝑥2−6𝑥=0⇒𝑥(𝑥−2)=0⇒𝑥=0 or 𝑥=2.3x2−6x=0⇒x(x−2)=0⇒x=0 or x=2.

    3. Evaluate at Critical Points and Endpoints:

      • 𝑓(−1)=(−1)3−3(−1)2+4=−1−3+4=0f(−1)=(−1)3−3(−1)2+4=−1−3+4=0
      • 𝑓(0)=03−3×02+4=4f(0)=03−3×02+4=4
      • 𝑓(2)=23−3×22+4=8−12+4=0f(2)=23−3×22+4=8−12+4=0
      • 𝑓(3)=33−3×32+4=27−27+4=4f(3)=33−3×32+4=27−27+4=4

    4. Determine Maximum and Minimum:

      • The minimum values, 𝑓(−1)=0f(−1)=0 and 𝑓(2)=0f(2)=0, occur at −1−1 and 22.
      • The maximum values, 𝑓(0)=4f(0)=4 and 𝑓(3)=4f(3)=4, occur at 00 and 33.

    This procedure and example highlight the importance of checking all critical points and endpoints to accurately determine the extreme values of a function over a closed interval.

  19. 19.Exercise Questions

    Find the maximum and minimum values, if any, of the following functions given by:

    1. 𝑓(𝑥)=(2𝑥−1)2+3f(x)=(2x−1)2+3
    2. 𝑓(𝑥)=9𝑥2+12𝑥+2f(x)=9x2+12x+2
    3. 𝑓(𝑥)=−(𝑥−1)2+10f(x)=−(x−1)2+10
    4. 𝑔(𝑥)=𝑥3+1g(x)=x3+1

    Solutions

    1. 𝑓(𝑥)=(2𝑥−1)2+3f(x)=(2x−1)2+3

    • Function Form: Quadratic function in standard form where 𝑎>0a>0. It opens upwards.
    • Vertex Form: The given function is already in vertex form, 𝑓(𝑥)=𝑎(𝑥−ℎ)2+𝑘f(x)=a(x−h)2+k where 𝑎=1a=1, ℎ=12h=21​, and 𝑘=3k=3.
    • Minimum Value: The minimum occurs at the vertex. 𝑥=12x=21​ gives 𝑓(12)=0+3=3f(21​)=0+3=3.
    • Conclusion: The minimum value is 3, no maximum value since it opens upwards and goes to infinity.

    2. 𝑓(𝑥)=9𝑥2+12𝑥+2f(x)=9x2+12x+2

    • Completing the Square: To find the vertex, rewrite the function:𝑓(𝑥)=9(𝑥2+43𝑥)+2=9(𝑥2+43𝑥+49)−9⋅49+2=9(𝑥+23)2−2f(x)=9(x2+34​x)+2=9(x2+34​x+94​)−9⋅94​+2=9(x+32​)2−2
    • Minimum Value: Occurs at 𝑥=−23x=−32​, where 𝑓(−23)=−2f(−32​)=−2.
    • Conclusion: The minimum value is -2, no maximum value as the function opens upwards.

    3. 𝑓(𝑥)=−(𝑥−1)2+10f(x)=−(x−1)2+10

    • Function Form: This is an inverted parabola.
    • Vertex Form: The vertex form is 𝑓(𝑥)=−1(𝑥−1)2+10f(x)=−1(x−1)2+10 where ℎ=1h=1, 𝑘=10k=10.
    • Maximum Value: The maximum occurs at the vertex. 𝑥=1x=1 gives 𝑓(1)=10f(1)=10.
    • Conclusion: The maximum value is 10, no minimum value as the function opens downwards.

    4. 𝑔(𝑥)=𝑥3+1g(x)=x3+1

    • Analysis of Critical Points:
      • First Derivative: 𝑔′(𝑥)=3𝑥2g′(x)=3x2.
      • Critical Points: 𝑔′(𝑥)=0g′(x)=0 gives 𝑥=0x=0.
      • Second Derivative: 𝑔′′(𝑥)=6𝑥g′′(x)=6x.
        • At 𝑥=0x=0, 𝑔′′(0)=0g′′(0)=0, which is inconclusive (second derivative test fails here).
      • Behavior Analysis: Since 𝑔′(𝑥)=3𝑥2≥0g′(x)=3x2≥0 for all 𝑥x and 𝑔′′(𝑥)g′′(x) changes sign at 𝑥=0x=0, 𝑔(𝑥)g(x) has an inflection point at 𝑥=0x=0, not a max or min.
    • Conclusion: 𝑔(𝑥)g(x) does not have a maximum or minimum value. It decreases for 𝑥<0x<0 and increases for 𝑥>0x>0.

    These analyses provide the extreme values for each function over their entire domain, employing both vertex form evaluations for quadratics and derivative tests for higher-degree polynomials.

  20. 20.Exercise Questions

    1. Find the maximum and minimum values of 𝑓(𝑥)=𝑥+sin⁡2𝑥f(x)=x+sin2x on [0,2𝜋][0,2π].

    Solution:

    • Derivative: Compute the derivative to find critical points.𝑓′(𝑥)=1+2cos⁡2𝑥f′(x)=1+2cos2x
    • Find Critical Points: Solve 𝑓′(𝑥)=0f′(x)=0 for 𝑥x.1+2cos⁡2𝑥=0⇒cos⁡2𝑥=−121+2cos2x=0⇒cos2x=−21​This occurs at 2𝑥=2𝜋/32x=2π/3 and 2𝑥=4𝜋/32x=4π/3 within one cycle, so 𝑥=𝜋/3,2𝜋/3x=π/3,2π/3.
    • Evaluate at Critical Points and Endpoints:𝑓(0)=0+sin⁡0=0,𝑓(𝜋/3)=𝜋/3+sin⁡(2𝜋/3)=𝜋/3+3/2f(0)=0+sin0=0,f(π/3)=π/3+sin(2π/3)=π/3+3​/2𝑓(2𝜋/3)=2𝜋/3+sin⁡(4𝜋/3)=2𝜋/3−3/2,𝑓(2𝜋)=2𝜋+sin⁡(4𝜋)=2𝜋f(2π/3)=2π/3+sin(4π/3)=2π/3−3​/2,f(2π)=2π+sin(4π)=2π
    • Determine Max and Min: Comparing these values, 𝑓(2𝜋)=2𝜋f(2π)=2π is the maximum and 𝑓(2𝜋/3)=2𝜋/3−3/2f(2π/3)=2π/3−3​/2 is the minimum.

    2. Find two numbers whose sum is 24 and whose product is as large as possible.

    Solution:

    • Let the numbers be 𝑥x and 𝑦y such that 𝑥+𝑦=24x+y=24.
    • Express 𝑦y in terms of 𝑥x: 𝑦=24−𝑥y=24−x.
    • Product Function: 𝑃(𝑥)=𝑥(24−𝑥)=24𝑥−𝑥2P(x)=x(24−x)=24x−x2.
    • Vertex Formula: The maximum of a quadratic 𝑎𝑥2+𝑏𝑥+𝑐ax2+bx+c occurs at 𝑥=−𝑏/(2𝑎)x=−b/(2a).𝑥=−24/(−2)=12x=−24/(−2)=12
    • Maximum Product: 𝑃(12)=12(24−12)=144P(12)=12(24−12)=144.
    • The two numbers are 12 and 12.

    3. Find two positive numbers 𝑥x and 𝑦y such that 𝑥+𝑦=60x+y=60 and 𝑥𝑦xy is maximum.

    Solution:

    • Express 𝑦y in terms of 𝑥x: 𝑦=60−𝑥y=60−x.
    • Product Function: 𝑃(𝑥)=𝑥(60−𝑥)=60𝑥−𝑥2P(x)=x(60−x)=60x−x2.
    • Vertex Formula: Maximum occurs at 𝑥=30x=30.
    • Maximum Product: 𝑃(30)=30(60−30)=900P(30)=30(60−30)=900.
    • The two numbers are 30 and 30.

  21. 21.Exercise Questions

    Question 01: Show that the right circular cone of least curved surface area and given volume has an altitude equal to

    22​ times the radius of the base.

    Solution 01:

    1. Formulas Used:

      • Volume of a cone, 𝑉=13𝜋𝑟2ℎV=31​πr2h
      • Curved Surface Area (CSA) of a cone, 𝑆=𝜋𝑟ℓS=πrℓ, where ℓℓ is the slant height ℓ=𝑟2+ℎ2ℓ=r2+h2​

    2. Given: Volume 𝑉V is constant.

      • We need to minimize 𝑆S under the constraint 𝑉=13𝜋𝑟2ℎV=31​πr2h.

    3. Using Lagrange Multipliers:

      • Set 𝑉=13𝜋𝑟2ℎV=31​πr2h and express ℎh in terms of 𝑟r: ℎ=3𝑉𝜋𝑟2h=πr23V​
      • Substitute ℎh in ℓℓ: ℓ=𝑟2+(3𝑉𝜋𝑟2)2ℓ=r2+(πr23V​)2​
      • Curved Surface Area in terms of 𝑟r becomes: 𝑆=𝜋𝑟𝑟2+(3𝑉𝜋𝑟2)2S=πrr2+(πr23V​)2​
      • Differentiate 𝑆S with respect to 𝑟r, set to zero, and solve for 𝑟r for minimum 𝑆S.

    4. Simplification: After differentiating and simplifying (a detailed calculus workout is skipped for brevity), it can be shown that the minimum surface area occurs when ℎ=2𝑟h=2​r.


    Question 02: Show that the semi-vertical angle 𝜃θ of the cone of the maximum volume and of given slant height is tan⁡−1(2)tan−1(2​).

    Solution 02:

    1. Geometry Relation:

      • ℎ=ℓcos⁡(𝜃)h=ℓcos(θ), 𝑟=ℓsin⁡(𝜃)r=ℓsin(θ)
      • Volume 𝑉=13𝜋𝑟2ℎ=13𝜋(ℓsin⁡(𝜃))2(ℓcos⁡(𝜃))V=31​πr2h=31​π(ℓsin(θ))2(ℓcos(θ))
      • Simplify volume as function of 𝜃θ: 𝑉=13𝜋ℓ3sin⁡2(𝜃)cos⁡(𝜃)V=31​πℓ3sin2(θ)cos(θ)

    2. Maximize Volume:

      • Differentiate 𝑉V with respect to 𝜃θ, set to zero, and solve for 𝜃θ: 𝑑𝑉𝑑𝜃=13𝜋ℓ3(2sin⁡(𝜃)cos⁡2(𝜃)−sin⁡3(𝜃))=0dθdV​=31​πℓ3(2sin(θ)cos2(θ)−sin3(θ))=0
      • Solve 2cos⁡2(𝜃)−sin⁡2(𝜃)=02cos2(θ)−sin2(θ)=0 gives tan⁡(𝜃)=2tan(θ)=2​.

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