Applications of Derivative Ex 6.1: Rate of Change

Understanding Rate of Change with Derivatives

The concept of 'rate of change' is the very foundation of differential calculus. When we talk about the derivative of a function, say y = f(x), with respect to x, denoted as dy/dx or f'(x), we are essentially measuring how sensitive y is to changes in x. More precisely, dy/dx represents the instantaneous rate at which y is changing for a unit change in x. For example, if y is the distance covered and x is time, then dy/dx represents instantaneous speed. If y is the area of a square and x is its side length, then dy/dx tells us how fast the area increases for a small increase in side length.

In Exercise 6.1, we often encounter scenarios where multiple quantities are changing simultaneously, usually with respect to time (t). For instance, if the radius r of a circular oil slick is increasing, and we want to find out how fast its area A is expanding, we need to relate dA/dt to dr/dt. We know the area of a circle is A = πr². Since both A and r are functions of time t (even if t isn't explicitly written in the formula for A), we differentiate both sides of the equation with respect to t using the Chain Rule.

The Chain Rule is a powerful tool for relating different rates. It states that if y is a function of x, and x itself is a function of t, then dy/dt = (dy/dx) * (dx/dt). Applying this to our area example:

1. Start with the formula: A = πr²
2. Differentiate both sides with respect to t: d/dt(A) = d/dt(πr²)
3. Apply the chain rule on the right side: dA/dt = π d/dt(r²) = π (2r) * (dr/dt)

So, dA/dt = 2πr (dr/dt).

This derived expression allows us to calculate the rate of change of area (dA/dt) at any given instant, provided we know the radius r and the rate of change of the radius (dr/dt) at that moment. This systematic approach of identifying variables, establishing relationships, differentiating using the chain rule, and substituting given values is the core strategy for solving all problems in Applications of Derivatives Exercise 6.1.

Key Concepts and Formulas for Exercise 6.1

• Derivative as Rate of Change: dy/dx represents the rate of change of y with respect to x.
• Chain Rule for Time-Based Rates: If y depends on x, and x depends on t, then dy/dt = (dy/dx) * (dx/dt). This is fundamental for problems involving time.
• Implicit Differentiation: When variables are related implicitly (e.g., x² + y² = c), differentiate each term with respect to t (or the relevant variable), remembering to apply the chain rule. For instance, d/dt(x²) = 2x (dx/dt).
• Geometric Formulas: Be familiar with formulas for area, perimeter, surface area, and volume of common shapes (circle, square, cube, sphere, cone, cylinder). You will often need to differentiate these.
• Units: Always pay attention to and include the correct units for rates of change (e.g., cm/s, m³/min, cm²/sec).
• Sign Convention: A positive rate of change indicates an increasing quantity, while a negative rate of change indicates a decreasing quantity.

Step-by-Step Examples: Applying Derivatives to Rate of Change

• Example 1: Expanding Circle The radius of a circle is increasing at the rate of 0.7 cm/s. What is the rate of increase of its circumference? Solution: Step 1: Identify the given information and what needs to be found. Let r be the radius and C be the circumference of the circle. Given: dr/dt = 0.7 cm/s. To find: dC/dt. Step 2: Write down the formula relating the quantities. The circumference of a circle is given by C = 2πr. Step 3: Differentiate the formula with respect to time t. d/dt(C) = d/dt(2πr) dC/dt = 2π (dr/dt) (Using the chain rule) Step 4: Substitute the given values and calculate. dC/dt = 2π (0.7) dC/dt = 1.4π cm/s Final answer: The rate of increase of its circumference is 1.4π cm/s.
• Example 2: Volume of a Cube A cube's side is increasing at the rate of 3 cm/s. How fast is its volume increasing when the side length is 10 cm? Solution: Step 1: Identify given information and what needs to be found. Let s be the side length and V be the volume of the cube. Given: ds/dt = 3 cm/s. To find: dV/dt when s = 10 cm. Step 2: Write down the formula relating the quantities. The volume of a cube is given by V = s³. Step 3: Differentiate the formula with respect to time t. d/dt(V) = d/dt(s³) dV/dt = 3s² (ds/dt) (Using the chain rule) Step 4: Substitute the given values and calculate. We need dV/dt when s = 10 cm. dV/dt = 3 (10)² (3) dV/dt = 3 100 * 3 dV/dt = 900 cm³/s Final answer: The volume of the cube is increasing at a rate of 900 cm³/s when the side length is 10 cm.
• Example 3: Sliding Ladder Problem 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 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: Step 1: Draw a diagram and label variables. Let x be the distance of the foot of the ladder from the wall, and y be the height of the ladder on the wall. The length of the ladder is 5 m (or 500 cm). Given: dx/dt = 2 cm/s (positive because x is increasing). To find: dy/dt when x = 4 m (or 400 cm). Step 2: Write down the relationship between the variables. Using the Pythagorean theorem: x² + y² = 5² (or x² + y² = 500² if using cm). Step 3: Differentiate the equation with respect to time t. d/dt(x² + y²) = d/dt(5²) 2x (dx/dt) + 2y (dy/dt) = 0 (Implicit differentiation and chain rule) x (dx/dt) + y (dy/dt) = 0 Step 4: Find the value of y at the given instant. When x = 4 m, (4)² + y² = 5² 16 + y² = 25 y² = 9 y = 3 m Step 5: Substitute the known values and solve for dy/dt. 4 (0.02 m/s) + 3 (dy/dt) = 0 (Converting dx/dt from cm/s to m/s: 2 cm/s = 0.02 m/s) 0.08 + 3 (dy/dt) = 0 3 (dy/dt) = -0.08 dy/dt = -0.08 / 3 m/s ≈ -0.0267 m/s Final answer: The height on the wall is decreasing at a rate of approximately 0.0267 m/s (or 2.67 cm/s). The negative sign indicates that the height is decreasing.

Navigating Common Mistakes and Scoring High

To excel in problems from Exercise 6.1, be mindful of these common pitfalls:

1. Forgetting the Chain Rule: This is the most frequent error. Remember, if A is a function of r, and r is changing with t, you must use dA/dt = (dA/dr) * (dr/dt). Don't just write dA/dt = d(formula)/dr.
2. Substituting Values Prematurely: Always differentiate the formula first with respect to time (t), and then substitute the given numerical values for variables and rates. Substituting values before differentiation will lead to incorrect results (as the derivative of a constant is zero).
3. Incorrect Units: A rate of change must always have units that reflect the change in the numerator quantity per unit of the denominator quantity (e.g., area in cm², time in s, so rate is cm²/s). Make sure all units are consistent (e.g., don't mix cm and m in the same calculation without conversion).
4. Misinterpreting Signs: A negative rate means the quantity is decreasing, and a positive rate means it is increasing. Your final answer should clearly state 'decreasing' or 'increasing' if the question asks for the 'rate of decrease' or 'rate of increase'.
5. Algebraic Errors: After setting up the derivative equation, ensure you perform the algebraic manipulations carefully to solve for the required rate.

By systematically applying the Chain Rule, maintaining unit consistency, and performing careful calculations, you can confidently solve these real-world derivative problems.

Practice Questions with Solutions

• Q: The volume of a sphere is increasing at the rate of 8 cm³/s. Find the rate at which its surface area is increasing when the radius is 12 cm. A: Step 1: Identify given information and what to find. Let V be the volume, S be the surface area, and r be the radius of the sphere. Given: dV/dt = 8 cm³/s. To find: dS/dt when r = 12 cm. Step 2: Write down relevant formulas and differentiate. Volume: V = (4/3)πr³ => dV/dt = (4/3)π 3r² (dr/dt) = 4πr² (dr/dt). Surface Area: S = 4πr² => dS/dt = 4π 2r (dr/dt) = 8πr (dr/dt). Step 3: Use the dV/dt equation to find dr/dt. 8 = 4π(12)² (dr/dt) 8 = 4π(144) (dr/dt) 8 = 576π (dr/dt) dr/dt = 8 / (576π) = 1 / (72π) cm/s. Step 4: Substitute dr/dt into the dS/dt equation. dS/dt = 8π(12) * (1 / (72π)) dS/dt = 96π / (72π) dS/dt = 96 / 72 = 4/3 cm²/s. Final answer: The surface area is increasing at a rate of 4/3 cm²/s.
• Q: A man 2 m tall walks at a speed of 5 km/h away from a lamppost which is 6 m high. At what rate is the tip of his shadow moving? A: Step 1: Draw a diagram and define variables. Let the man be M and the lamppost be L. Let x be the distance of the man from the lamppost, and s be the length of his shadow. Let y be the total distance of the tip of the shadow from the lamppost. y = x + s. Given: dx/dt = 5 km/h (speed of the man). To find: dy/dt (rate at which the tip of the shadow is moving). Step 2: Use similar triangles to relate x and s. From the diagram, (6 / (x + s)) = (2 / s) 6s = 2x + 2s 4s = 2x s = (1/2)x Step 3: Find the relation for y and differentiate. y = x + s = x + (1/2)x = (3/2)x. Differentiate with respect to t: dy/dt = (3/2) (dx/dt). Step 4: Substitute the given rate. dy/dt = (3/2) 5 km/h dy/dt = 7.5 km/h. Final answer: The tip of his shadow is moving at a rate of 7.5 km/h.
• Q: A particle moves along the curve y = (2/3)x³ + 1. Find the points on the curve at which the y-coordinate is changing twice as fast as the x-coordinate. A: Step 1: Identify the given condition. Given: dy/dt = 2 (dx/dt). Step 2: Differentiate the curve equation with respect to t. d/dt(y) = d/dt((2/3)x³ + 1) dy/dt = (2/3) 3x² (dx/dt) + 0 dy/dt = 2x² (dx/dt). Step 3: Substitute the condition from Step 1 into the differentiated equation. 2 (dx/dt) = 2x² (dx/dt). Step 4: Solve for x. Assuming dx/dt ≠ 0 (particle is moving), 2 = 2x² x² = 1 x = ±1. Step 5: Find the corresponding y coordinates. If x = 1, y = (2/3)(1)³ + 1 = 2/3 + 1 = 5/3. If x = -1, y = (2/3)(-1)³ + 1 = -2/3 + 1 = 1/3. Final answer: The points on the curve are (1, 5/3) and `(-1, 1/3).
• Q: A balloon, which always remains spherical, has a variable radius. Find the rate at which its volume is increasing with respect to the radius when the radius is 10 cm. A: Step 1: Identify the quantities and what is asked. Let V be the volume and r be the radius of the spherical balloon. To find: dV/dr when r = 10 cm. Step 2: Write down the formula for the volume of a sphere. V = (4/3)πr³. Step 3: Differentiate V with respect to r. dV/dr = d/dr((4/3)πr³) dV/dr = (4/3)π * 3r² dV/dr = 4πr². Step 4: Substitute the given radius. When r = 10 cm, dV/dr = 4π(10)² dV/dr = 4π(100) dV/dr = 400π cm³/cm. Final answer: The rate at which its volume is increasing with respect to the radius is 400π cm³/cm.

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