Solving Problems On Application of Differentiation

 

Applications of Differentiation: Core Concepts

Differentiation is not just an abstract algebraic exercise; it is a powerful tool used to analyze behavior, rates, and limits in the real world. Here are the three most common applications found in calculus:

• 1. Rates of Change: Since a derivative measures how fast a dependent variable changes relative to an independent variable, it represents real-world rates. For example, if an object's position is given by a function of time, its velocity is the first derivative, and its acceleration is the second derivative.
• 2. Turning Points (Maxima and Minima): To find where a curve reaches its highest or lowest point, we look for where the slope of the tangent line equals zero (^dy⁄_dx = 0). These are called stationary points.
• 3. Optimization: This is the business and engineering application of finding turning points. It involves constructing a mathematical model to calculate maximum possible profit, minimum material waste, or optimal structural dimensions under specific constraints.

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Guided Examples

Example 1 (Kinematics / Rate of Change):
The distance s (in meters) traveled by a particle in time t (in seconds) is given by the equation:
s = 2t³ - 5t + 3
Find the velocity of the particle at t = 3 seconds.

Solution:
Velocity (v) is defined as the instantaneous rate of change of distance with respect to time: v = ^ds⁄_dt.
1. Differentiate the equation with respect to t using the Power Rule:
v = ^ds⁄_dt = 6t² - 5
2. Substitute t = 3 into the velocity derivative:
v = 6(3)² - 5 = 6(9) - 5 = 54 - 5 = 49 m/s.
The velocity at 3 seconds is 49 m/s.

Example 2 (Stationary Points):
Find the coordinates of the turning point for the curve:
y = x² - 4x + 7

Solution:
1. Turning points occur exclusively where the gradient/slope is zero (^dy⁄_dx = 0). Differentiate the function:
^dy⁄_dx = 2x - 4
2. Set the derivative equal to zero and isolate x:
2x - 4 = 0 → 2x = 4 → x = 2
3. Find the corresponding y value by substituting x = 2 back into the original curve equation:
y = (2)² - 4(2) + 7 = 4 - 8 + 7 = 3.
The stationary turning point coordinates are (2, 3).

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Application of Differentiation Practice Quiz

Question 1: The displacement of an object is given by s = t² - 6t + 8. At what time t does the object momentarily come to rest (velocity = 0)?

• A) t = 2 seconds
• B) t = 3 seconds
• C) t = 6 seconds
• D) t = 0 seconds

View Answer & mSolution

Correct Answer: B

Solution: Velocity is the derivative of displacement: v = ^ds⁄_dt = 2t - 6.
The term "at rest" means velocity equals 0. Set the expression to 0:
2t - 6 = 0 → 2t = 6 → t = 3 seconds.

Question 2: Find the x-coordinate of the maximum turning point for the curve y = -x² + 8x - 12.

• A) x = 4
• B) x = -4
• C) x = 8
• D) x = 2

View Answer & Solution

Correct Answer: A

Solution: Differentiate the function to find the gradient expression:
^dy⁄_dx = -2x + 8
Set the derivative to zero for a stationary point:
-2x + 8 = 0 → 2x = 8 → x = 4.

Question 3: A profit function for selling x items is given by P(x) = 400x - 2x². How many items must be sold to maximize profit?

• A) 400 items
• B) 200 items
• C) 100 items
• D) 50 items

View Answer & Solution

Correct Answer: C

Solution: To find maximum profit, find the marginal profit derivative P'(x) and set it to 0:
P'(x) = 400 - 4x
Set equal to zero:
400 - 4x = 0 → 4x = 400 → x = 100 items.

Question 4: The velocity of a moving body is given by v = 3t² + 2t. Find its acceleration at t = 2 seconds.

• A) 14 m/s²
• B) 16 m/s²
• C) 12 m/s²
• D) 8 m/s²

View Answer & Solution

Correct Answer: A

Solution: Acceleration (a) is the derivative of velocity over time: a = ^dv⁄_dt.
Differentiate the velocity function:
a = 6t + 2
Substitute t = 2:
a = 6(2) + 2 = 12 + 2 = 14 m/s².

Question 5: If a curve is given by y = x³ - 3x, what are the x-values of its two stationary points?

• A) x = 0 and x = 3
• B) x = 1 and x = -1
• C) x = 0 and x = 1
• D) x = 3 and x = -3

View Answer & Solution

Correct Answer: B

Solution: Differentiate the cubic equation to find its slope formula:
^dy⁄_dx = 3x² - 3
Set the expression equal to 0:
3x² - 3 = 0 → 3x² = 3 → x² = 1
Taking the square root gives: x = 1 and x = -1.