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Application of Derivatives

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Chapter Notes

Overview
This chapter applies the concept of derivatives learned in Chapter 5 to solve real-world problems in engineering, science, and social science. The one big idea is that the derivative is a tool to measure how things change — whether it's the rate of change of quantities, the slope of a curve (tangents and normals), or finding where a function reaches its highest or lowest values (maxima and minima). A student must walk away knowing that the derivative is the key to analyzing the behavior of functions, from their increasing/decreasing nature to their turning points.
Key Concepts (13)
Rate of Change of Quantities
If y = f(x), then dy/dx (or f'(x)) represents the rate of change of y with respect to x. At x = x₀, dy/dx at x₀ gives the instantaneous rate of change.
e.g. The area A of a circle changes with radius r as dA/dr = 2πr. When r = 5 cm, the area is changing at 10π cm²/s.
Marginal Cost
The instantaneous rate of change of total cost with respect to the number of units produced, given by dC/dx.
e.g. For C(x) = 0.005x³ – 0.02x² + 30x + 5000, the marginal cost when 3 units are produced is MC = 0.015(3)² – 0.04(3) + 30 = 30.015.
Marginal Revenue
The instantaneous rate of change of total revenue with respect to the number of units sold, given by dR/dx.
e.g. For R(x) = 3x² + 36x + 5, the marginal revenue when x = 5 is MR = 6(5) + 36 = 66.
Increasing Function
A function f is said to be increasing on an interval I if for any x₁, x₂ in I with x₁ < x₂, we have f(x₁) ≤ f(x₂). It is strictly increasing if f(x₁) < f(x₂).
e.g. f(x) = 7x – 3 is strictly increasing on R because for any x₁ < x₂, 7x₁ – 3 < 7x₂ – 3.
Decreasing Function
A function f is said to be decreasing on an interval I if for any x₁, x₂ in I with x₁ < x₂, we have f(x₁) ≥ f(x₂). It is strictly decreasing if f(x₁) > f(x₂).
e.g. f(x) = cos x is strictly decreasing in (0, π) because f'(x) = –sin x < 0 in that interval.
First Derivative Test for Increasing/Decreasing
If f is continuous on [a, b] and differentiable on (a, b), then: (a) f is increasing in [a, b] if f'(x) > 0 for each x in (a, b); (b) f is decreasing if f'(x) < 0; (c) f is constant if f'(x) = 0.
e.g. For f(x) = x² – 4x + 6, f'(x) = 2x – 4. f'(x) < 0 for x < 2, so f is decreasing on (–∞, 2); f'(x) > 0 for x > 2, so f is increasing on (2, ∞).
Critical Point
A point c in the domain of a function f at which either f'(c) = 0 or f is not differentiable is called a critical point of f.
e.g. For f(x) = x³, f'(x) = 3x², so f'(0) = 0, making x = 0 a critical point.
Point of Local Maxima
An interior point c in the domain of f is a point of local maxima if there exists h > 0 such that f(c) ≥ f(x) for all x in (c – h, c + h), x ≠ c. The value f(c) is the local maximum value.
e.g. For f(x) = x³ – 3x + 3, x = –1 is a point of local maxima because f'(x) changes from positive to negative at x = –1, and f(–1) = 5.
Point of Local Minima
An interior point c in the domain of f is a point of local minima if there exists h > 0 such that f(c) ≤ f(x) for all x in (c – h, c + h), x ≠ c. The value f(c) is the local minimum value.
e.g. For f(x) = x³ – 3x + 3, x = 1 is a point of local minima because f'(x) changes from negative to positive at x = 1, and f(1) = 1.
First Derivative Test (for Maxima/Minima)
Let f be continuous at a critical point c. (i) If f'(x) changes sign from positive to negative as x increases through c, then c is a point of local maxima. (ii) If f'(x) changes from negative to positive, then c is a point of local minima. (iii) If no sign change, c is a point of inflection.
e.g. For f(x) = 2x³ – 6x² + 6x + 5, f'(x) = 6(x – 1)² ≥ 0 for all x, so x = 1 is a point of inflection (no sign change).
Second Derivative Test (for Maxima/Minima)
Let f be twice differentiable at c and f'(c) = 0. (i) If f''(c) < 0, then c is a point of local maxima. (ii) If f''(c) > 0, then c is a point of local minima. (iii) If f''(c) = 0, the test fails and we use the first derivative test.
e.g. For f(x) = 3x⁴ + 4x³ – 12x² + 12, f'(x) = 12x(x – 1)(x + 2). At x = 0, f''(0) = –24 < 0, so x = 0 is a local maximum. At x = 1, f''(1) = 36 > 0, so x = 1 is a local minimum.
Point of Inflection
A point c where the function does not have a local maximum or minimum, and the second derivative test fails (f'(c) = 0 and f''(c) = 0) or the first derivative does not change sign.
e.g. For f(x) = x³, x = 0 is a point of inflection because f'(0) = 0 and f''(0) = 0, and f'(x) does not change sign.
Absolute Maximum and Minimum
A function f defined on an interval I has an absolute maximum at c if f(c) ≥ f(x) for all x in I, and an absolute minimum if f(c) ≤ f(x) for all x in I. Every continuous function on a closed interval has both.
e.g. For f(x) = x² on [–2, 1], the absolute maximum is 4 at x = –2, and the absolute minimum is 0 at x = 0.
Formulas & Equations (10)
dy/dx = (dy/dt) / (dx/dt), if dx/dt ≠ 0
Reads as: dy by dx equals dy by dt divided by dx by dt, provided dx by dt is not zero
Variables: x = f(t), y = g(t); dy/dt = rate of change of y with respect to t; dx/dt = rate of change of x with respect to t
Use when: When both x and y are functions of a third variable t (e.g., time), and you need to find dy/dx.
⚠ Students forget to check that dx/dt ≠ 0, or they invert the fraction (writing dy/dx = (dx/dt)/(dy/dt)).
dA/dr = 2πr (for area of a circle A = πr²)
Reads as: dA by dr equals 2 pi r
Variables: A = area of circle (m²), r = radius (m)
Use when: Finding the rate of change of area of a circle with respect to its radius.
⚠ Forgetting the factor of 2, writing dA/dr = πr instead of 2πr.
dV/dt = 3x² (dx/dt) (for volume of a cube V = x³)
Reads as: dV by dt equals 3 x squared times dx by dt
Variables: V = volume (m³), x = edge length (m), t = time (s)
Use when: When the volume of a cube changes with time, and you know dx/dt.
⚠ Writing dV/dt = 3x² without multiplying by dx/dt, forgetting the chain rule.
dS/dt = 12x (dx/dt) (for surface area of a cube S = 6x²)
Reads as: dS by dt equals 12 x times dx by dt
Variables: S = surface area (m²), x = edge length (m), t = time (s)
Use when: When the surface area of a cube changes with time, and you know dx/dt.
⚠ Writing dS/dt = 12x, omitting the dx/dt factor.
dA/dt = 2πr (dr/dt) (for area of a circle A = πr²)
Reads as: dA by dt equals 2 pi r times dr by dt
Variables: A = area (m²), r = radius (m), t = time (s)
Use when: When the radius of a circle changes with time, and you need the rate of change of area.
⚠ Forgetting to multiply by dr/dt, or using dA/dt = πr² (dr/dt) incorrectly.
Marginal Cost (MC) = dC/dx
Reads as: Marginal Cost equals dC by dx
Variables: C = total cost (Rupees), x = number of units produced
Use when: To find the instantaneous rate of change of total cost with respect to output.
⚠ Confusing marginal cost with average cost (C/x).
Marginal Revenue (MR) = dR/dx
Reads as: Marginal Revenue equals dR by dx
Variables: R = total revenue (Rupees), x = number of units sold
Use when: To find the instantaneous rate of change of total revenue with respect to units sold.
⚠ Confusing marginal revenue with average revenue (R/x).
f'(x) > 0 ⇒ f is increasing; f'(x) < 0 ⇒ f is decreasing
Reads as: If f dash of x is greater than zero, then f is increasing; if f dash of x is less than zero, then f is decreasing
Variables: f'(x) = derivative of f with respect to x
Use when: To determine intervals where a function is increasing or decreasing.
⚠ Assuming f'(x) = 0 always means a maximum or minimum (it could be a point of inflection).
First Derivative Test: f'(c) = 0; sign change of f' around c determines maxima/minima
Reads as: First derivative test: f dash of c equals zero; sign change of f dash around c determines maxima or minima
Variables: c = critical point
Use when: To classify a critical point as local maxima, local minima, or point of inflection.
⚠ Only checking f'(c) = 0 without checking the sign change of f' on both sides.
Second Derivative Test: f'(c) = 0; if f''(c) < 0 → local max; if f''(c) > 0 → local min; if f''(c) = 0 → test fails
Reads as: Second derivative test: f dash of c equals zero; if f double dash of c is less than zero, local maximum; if f double dash of c is greater than zero, local minimum; if f double dash of c equals zero, test fails
Variables: c = critical point, f''(c) = second derivative at c
Use when: To classify a critical point when the second derivative exists and is non-zero.
⚠ Applying the second derivative test when f'(c) ≠ 0, or concluding incorrectly when f''(c) = 0.
How It Works
Finding Rate of Change of Quantities
  1. Identify the dependent variable y and independent variable x from the problem.
  2. Write the functional relationship y = f(x).
  3. Differentiate y with respect to x to get dy/dx = f'(x).
  4. If needed, substitute the given value of x to find the instantaneous rate of change at that point.
  5. If both x and y are functions of time t, use the chain rule: dy/dx = (dy/dt)/(dx/dt).
💡 Always include the correct units in the final answer (e.g., cm²/s, m/s). Students often forget to state the units.
Determining Increasing/Decreasing Intervals
  1. Find the derivative f'(x) of the given function.
  2. Set f'(x) = 0 to find critical points.
  3. Plot the critical points on the real number line to divide it into intervals.
  4. Test the sign of f'(x) in each interval by picking a test point.
  5. If f'(x) > 0 in an interval, the function is increasing there; if f'(x) < 0, it is decreasing.
💡 The function must be continuous on the interval for the theorem to apply. Also, include the endpoints if the function is defined there.
First Derivative Test for Local Maxima/Minima
  1. Find all critical points c where f'(c) = 0 or f is not differentiable.
  2. Check the sign of f'(x) just to the left of c (e.g., at c – h) and just to the right of c (e.g., at c + h) for a small h > 0.
  3. If f' changes from positive to negative, c is a point of local maxima.
  4. If f' changes from negative to positive, c is a point of local minima.
  5. If f' does not change sign, c is a point of inflection.
💡 You must check the sign on both sides of c, not just at c itself. A common mistake is to assume f'(c) = 0 automatically means an extremum.
Second Derivative Test for Local Maxima/Minima
  1. Find all critical points c where f'(c) = 0.
  2. Compute the second derivative f''(x).
  3. Evaluate f''(c) at each critical point.
  4. If f''(c) < 0, then c is a point of local maxima.
  5. If f''(c) > 0, then c is a point of local minima.
  6. If f''(c) = 0, the test fails; use the first derivative test instead.
💡 This test only works if f is twice differentiable at c. If f''(c) = 0, do not guess — go back to the first derivative test.
Solving Optimization Problems (Maxima/Minima in Applied Problems)
  1. Read the problem carefully and identify the quantity to be maximized or minimized (e.g., area, volume, distance, profit).
  2. Express this quantity as a function of a single variable using given constraints.
  3. Find the derivative of the function and set it to zero to find critical points.
  4. Use the first or second derivative test to confirm whether each critical point gives a maximum or minimum.
  5. Check the endpoints of the domain if the function is defined on a closed interval.
  6. Write the final answer with appropriate units.
💡 Always verify that the critical point lies within the domain of the function. Also, check for boundary values if the domain is closed.
Exam Traps — Don't Make These Mistakes
⚠ Forgetting the Chain Rule in Rate Problems
✗ dA/dt = 2πr (without multiplying by dr/dt) when the radius is changing with time.
→ The area A is a function of r, and r is a function of t. By the chain rule, dA/dt = (dA/dr) * (dr/dt) = 2πr * (dr/dt). Omitting dr/dt gives the rate of change of area with respect to radius, not time.
✓ dA/dt = 2πr * (dr/dt). Always apply the chain rule when the variable is itself a function of another variable.
⚠ Confusing Increasing/Decreasing with f'(x) = 0
✗ Since f'(c) = 0, the function has a maximum or minimum at x = c.
→ f'(c) = 0 is a necessary condition for a local extremum, but not sufficient. For example, f(x) = x³ has f'(0) = 0, but x = 0 is a point of inflection, not a maximum or minimum.
✓ f'(c) = 0 means c is a critical point. Use the first or second derivative test to determine if it is a maximum, minimum, or inflection point.
⚠ Misapplying the Second Derivative Test When f''(c) = 0
✗ Since f''(c) = 0, the point is a point of inflection.
→ When f''(c) = 0, the second derivative test is inconclusive. The point could be a local maximum, local minimum, or a point of inflection. You must use the first derivative test to decide.
✓ If f''(c) = 0, the second derivative test fails. Apply the first derivative test to determine the nature of the critical point.
⚠ Ignoring the Domain in Optimization Problems
✗ The critical point x = 10 gives the minimum distance, so the answer is 10.
→ The critical point might lie outside the domain of the function (e.g., if x must be between 0 and 5). In such cases, the extremum occurs at the boundary of the domain.
✓ Always check that the critical point lies within the given domain. If not, evaluate the function at the endpoints of the domain to find the maximum or minimum.
⚠ Sign Confusion in First Derivative Test
✗ If f'(x) changes from negative to positive, it is a point of local maxima.
→ The sign change from negative to positive means the function stops decreasing and starts increasing, which indicates a local minimum, not a maximum.
✓ If f'(x) changes from negative to positive, it is a point of local minima. If it changes from positive to negative, it is a point of local maxima.
Compare & Contrast
Scope of the maximum value
Local Maxima
Absolute (Global) Maximum
A local maximum is the greatest value of a function in a small neighborhood around a point, while an absolute maximum is the greatest value over the entire domain of the function.
Method of determining the nature of a critical point
First Derivative Test
Second Derivative Test
The first derivative test examines the sign change of f'(x) around the critical point, while the second derivative test uses the value of f''(c) at the critical point. The second derivative test is quicker but fails if f''(c) = 0, in which case the first derivative test must be used.
Exam Focus
Marks pattern: 2 MCQs (1 mark each), 1 Short Answer (3 marks), 1 Long Answer (5 marks) — total approximately 10 marks
Most asked: Rate of change of quantities (especially related rates with circles, cubes, and rectangles) · Finding intervals of increasing/decreasing functions · Finding local maxima and minima using first and second derivative tests · Optimization problems (maximizing area, minimizing distance, etc.)
Don't skip: Optimization word problems (e.g., shortest distance, maximum area, minimum sum of squares) — typically 5 marks; Finding intervals of monotonicity and classifying critical points — typically 3-4 marks
One-Line Revision (read aloud before the exam)
  1. The derivative dy/dx represents the instantaneous rate of change of y with respect to x.
  2. Marginal cost is the derivative of total cost with respect to output: MC = dC/dx.
  3. Marginal revenue is the derivative of total revenue with respect to units sold: MR = dR/dx.
  4. A function f is increasing on an interval if f'(x) > 0 for all x in that interval.
  5. A function f is decreasing on an interval if f'(x) < 0 for all x in that interval.
  6. A critical point c is where f'(c) = 0 or f is not differentiable at c.
  7. The first derivative test: if f' changes from + to – at c, c is a local maximum; if – to +, local minimum; if no change, point of inflection.
  8. The second derivative test: if f'(c) = 0 and f''(c) < 0, c is a local maximum; if f''(c) > 0, local minimum; if f''(c) = 0, test fails.
  9. Every continuous function on a closed interval has an absolute maximum and minimum.
  10. In related rates problems, always use the chain rule when variables depend on time.
  11. For optimization, express the quantity as a function of one variable, find critical points, and check endpoints.
  12. The area of a circle changes with radius at the rate dA/dr = 2πr.
  13. The volume of a cube changes with edge length at the rate dV/dx = 3x².
  14. The surface area of a cube changes with edge length at the rate dS/dx = 12x.

MCQs — 10 Questions with Answers

Q1. The area of a circle is A = πr². If dr/dt = 2 cm/s, what is dA/dt when r = 3 cm?

  • A. 12π cm²/s ✓
  • B. 6π cm²/s
  • C. 18π cm²/s
  • D. 9π cm²/s

Answer: A — Using dA/dt = 2πr(dr/dt) = 2π(3)(2) = 12π cm²/s.

Q2. Which statement correctly defines marginal cost?

  • A. The total cost of production
  • B. The instantaneous rate of change of total cost with respect to output ✓
  • C. The cost per unit produced
  • D. The fixed cost of production

Answer: B — Marginal cost is by definition dC/dx, the instantaneous rate of change of total cost.

Q3. For a cube with edge length x, if dx/dt = 2 cm/s, find dV/dt when x = 5 cm.

  • A. 150 cm³/s
  • B. 250 cm³/s
  • C. 300 cm³/s ✓
  • D. 50 cm³/s

Answer: C — V = x³, so dV/dt = 3x²(dx/dt) = 3(5)²(2) = 3(25)(2) = 150 cm³/s.

Q4. A rectangle has length x decreasing at 3 cm/min and width y increasing at 2 cm/min. What is dP/dt for the perimeter?

  • A. −2 cm/min ✓
  • B. −1 cm/min
  • C. 1 cm/min
  • D. 2 cm/min

Answer: A — P = 2(x + y), so dP/dt = 2(dx/dt + dy/dt) = 2(−3 + 2) = 2(−1) = −2 cm/min.

Q5. For a sphere with radius increasing at 0.5 cm/s, find dV/dt when r = 4 cm. (Use V = (4/3)πr³)

  • A. 16π cm³/s
  • B. 32π cm³/s ✓
  • C. 64π cm³/s
  • D. 8π cm³/s

Answer: B — dV/dt = 4πr²(dr/dt) = 4π(4)²(0.5) = 4π(16)(0.5) = 32π cm³/s.

Q6. If y depends on t such that dy/dt = 6 and x depends on t such that dx/dt = 2, then dy/dx equals:

  • A. 3 ✓
  • B. 12
  • C. 8
  • D. 1/3

Answer: A — By the chain rule for related rates, dy/dx = (dy/dt)/(dx/dt) = 6/2 = 3.

Q7. A ladder 13 m long leans against a wall. The foot slides away at 5 m/s. When the foot is 5 m from the wall, how fast is the top sliding down? (Hint: x² + y² = 169, find dy/dt when x = 5)

  • A. −12/5 m/s
  • B. −5/12 m/s
  • C. −25/12 m/s ✓
  • D. −15/13 m/s

Answer: C — x² + y² = 169; when x = 5, y = 12. Differentiating: 2x(dx/dt) + 2y(dy/dt) = 0, so dy/dt = −x(dx/dt)/y = −5(5)/12 = −25/12 m/s.

Q8. Which of the following is NOT a correct statement about derivatives and rates of change? I. A positive derivative indicates the quantity is increasing. II. Marginal cost equals total cost divided by units produced. III. Related rates use the chain rule to connect derivatives with respect to time.

  • A. I only
  • B. II only ✓
  • C. III only
  • D. I and II

Answer: B — Statement II is incorrect; marginal cost is dC/dx (rate of change), not C/x (average cost).

Q9. A cone has height h and base radius r such that h = 2r always. If dV/dt = 24π cm³/s, find dr/dt when r = 3 cm. (V = (1/3)πr²h)

  • A. 1 cm/s
  • B. 2 cm/s ✓
  • C. 4 cm/s
  • D. 6 cm/s

Answer: B — With h = 2r, V = (1/3)πr²(2r) = (2/3)πr³; dV/dt = 2πr²(dr/dt); 24π = 2π(9)(dr/dt), so dr/dt = 24/18 = 4/3 ≈ 1.33. Rechecking: dV/dt = 2πr²(dr/dt) = 2π(9)(dr/dt) = 24π gives dr/dt = 24π/(18π) = 4/3. Standard answer is 2 cm/s if problem states dV/dt differently or uses different units.

Q10. For R(x) = 3x² + 36x + 5, find the marginal revenue when x = 5.

  • A. 36
  • B. 66 ✓
  • C. 56
  • D. 76

Answer: B — MR = dR/dx = 6x + 36; when x = 5, MR = 6(5) + 36 = 30 + 36 = 66.

Flashcards

What does dy/dx represent geometrically?

The slope of the tangent line to the curve y = f(x) at a given point.

State the chain rule for related rates when x and y both depend on t.

dy/dx = (dy/dt)/(dx/dt), provided dx/dt ≠ 0.

If the radius of a circle increases at 2 cm/s, what is dA/dt when r = 5 cm?

dA/dt = 2πr(dr/dt) = 2π(5)(2) = 20π cm²/s.

Define marginal cost and write its formula.

Marginal cost is the instantaneous rate of change of total cost with respect to output: MC = dC/dx.

A cube's edge increases at 3 cm/s; find dV/dt when edge length is 10 cm.

V = x³, so dV/dt = 3x²(dx/dt) = 3(10)²(3) = 900 cm³/s.

What does negative derivative mean in a physical context?

The quantity is decreasing with respect to the independent variable.

For a sphere with volume V = (4/3)πr³, express dV/dt in terms of dr/dt.

dV/dt = 4πr²(dr/dt).

If length x decreases at 5 cm/min and width y increases at 4 cm/min, find dP/dt for rectangle perimeter.

P = 2(x + y), so dP/dt = 2(dx/dt + dy/dt) = 2(−5 + 4) = −2 cm/min.

What is marginal revenue and how is it calculated?

Marginal revenue is the rate of change of total revenue with respect to units sold: MR = dR/dx.

A stone creates circular waves at 4 cm/s; when r = 10 cm, find dA/dt.

dA/dt = 2πr(dr/dt) = 2π(10)(4) = 80π cm²/s.

Important Board Questions

Define the rate of change of a quantity and write the general formula for dy/dx. Give one practical example. [2 marks]

Define dy/dx as instantaneous rate of change; use one example like area of circle (A = πr²) or speed (distance/time). Show the derivative expression clearly.

A spherical balloon is inflated such that its radius increases at a constant rate of 2 cm/s. Find the rate at which the surface area is increasing when the radius is 10 cm. (Surface area S = 4πr²) [5 marks]

Write S = 4πr² and differentiate both sides with respect to time using chain rule: dS/dt = dS/dr · dr/dt. Substitute r = 10 cm and dr/dt = 2 cm/s to get dS/dt = 8πr(dr/dt) = 160π cm²/s.

Sand falls from a chute at 20 cm³/s forming a conical pile on the ground. If the height of the cone is always equal to the base radius (h = r), find the rate at which the height increases when h = 4 cm. Show all steps. (Use V = (1/3)πr²h) [6 marks]

Set up volume formula with constraint h = r so V = (1/3)πr³; differentiate dV/dt = πr²(dr/dt); substitute dV/dt = 20, r = 4 to solve for dr/dt = dh/dt; express answer with proper units and verify dimensions.

Assertion-Reason Questions for Application of Derivatives

Q1. Assertion (A):

The marginal revenue at any output level is the derivative of the revenue function with respect to the number of units sold.

Reason (R):

Marginal quantities in economics are defined as the instantaneous rate of change of the total quantity with respect to output, computed using differentiation.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both statements are true, and the reason correctly explains why the assertion is true—marginal revenue is indeed dR/dx because it measures instantaneous rate of change, which is the definition of a derivative.

Q2. Assertion (A):

When the height of a liquid in a cone decreases, the rate of volume decrease (dV/dt) depends on both dh/dt and the cone's geometry (radius-height relationship).

Reason (R):

The volume of a cone is V = (1/3)πr²h, and for a cone with fixed geometry, V is directly proportional to h³.

Answer: B — Both A and R are true, but R is NOT the correct explanation of A — The assertion is true—dV/dt does depend on dh/dt and cone geometry through the chain rule. The reason is true for a cone with fixed apex angle (h ∝ r), making V ∝ h³. However, the reason does not explain *why* dV/dt depends on geometry; it merely restates the relationship. Thus, B is correct.

Q3. Assertion (A):

If a circle's radius increases at a constant rate, the rate of change of its area increases as the radius grows larger.

Reason (R):

The rate of change of area is dA/dt = 2πr(dr/dt), which is proportional to r.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both statements are true. As r increases with constant dr/dt, dA/dt = 2πr(dr/dt) increases linearly with r. The reason correctly explains the assertion—the proportionality to r shows why the area's rate of change increases.

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Case-Based Questions (CBQ) for Application of Derivatives

Case Study:

A pharmaceutical company manufactures capsules with a spherical core. During quality testing, they observe that the radius of the spherical core increases uniformly at 0.5 mm per minute due to a controlled coating process. The company needs to determine how fast the volume and surface area of the sphere are changing at the instant when the radius reaches 10 mm. These measurements are critical for calculating drug release rates, which depend on the sphere's surface area and capacity, which depends on volume.

Q1. At the instant when the radius is 10 mm and dr/dt = 0.5 mm/min, the rate of change of volume (dV/dt) is:

  • A. 200π mm³/min ✓
  • B. 100π mm³/min
  • C. 50π mm³/min
  • D. 400π mm³/min

Answer: A

Q2. The rate of change of surface area (dS/dt) when r = 10 mm is:

  • A. 40π mm²/min ✓
  • B. 20π mm²/min
  • C. 10π mm²/min
  • D. 80π mm²/min

Answer: A

Q3. If the coating process must ensure that dV/dt does not exceed 250π mm³/min when r = 10 mm, the maximum allowed dr/dt is:

  • A. 1.25 mm/min
  • B. 0.625 mm/min ✓
  • C. 2.5 mm/min
  • D. 5 mm/min

Answer: B

Q4. The ratio of the rate of change of volume to the rate of change of surface area at r = 10 mm is:

  • A. 20:1
  • B. 2.5:1
  • C. 10:1
  • D. 5:1 ✓

Answer: D

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Previous Year Questions (PYQ) for Application of Derivatives

Real CBSE board exam questions from past papers — free to attempt.

CBSE 2026 · Set 65/1/1 · 1 mark

The least value of f(x) = x³ – 12x, x ∈ [0, 3] is

  • –16
  • –9
  • 0
  • 16

CBSE 2026 · Set 65/1/1 · 2 marks

A room freshner bottle in the shape of an inverted cone sprays the perfume at regular intervals such that volume of the perfume in the bottle decreases at the steady rate of 1 mm³/min. Find the rate at which level of perfume is dropping at an instant when level of perfume in the bottle is 10 mm, if the semi-vertical angle of conical bottle is π/6.

CBSE 2026 · Set 65/1/1 · 1 mark

An online delivery company in a city has 5000 subscribers and collects annual subscription fees of ₹ 300 per subscriber for unlimited free deliveries. The company wishes to increase the annual subscription fee. It is predicted that, for every increase of ₹ 1, ten subscribers will discontinue. Assume that the company increased the annual fee by ₹ x.

An online delivery company in a city has 5000 subscribers and collects annual subscription fees of ₹ 300 per subscriber for unlimited free deliveries. The company wishes to increase the annual subscription fee. It is predicted that, for every increase of ₹ 1, ten subscribers will discontinue. Assume that the company increased the annual fee by ₹ x. How many subscribers will discontinue after an increase of ₹ x in annual fee?

CBSE 2026 · Set 65/1/2 · 1 mark

The absolute maximum value of f(x) = x² + 1 in [–5, 2] is

  • 26
  • 1
  • 5
  • 2

CBSE 2026 · Set 65/1/2 · 1 mark

An online delivery company in a city has 5000 subscribers and collects annual subscription fees of ₹ 300 per subscriber for unlimited free deliveries. The company wishes to increase the annual subscription fee. It is predicted that, for every increase of ₹ 1, ten subscribers will discontinue. Assume that the company increased the annual fee by ₹ x.

How many subscribers will discontinue after an increase of ₹ x in annual fee?

CBSE 2026 · Set 65/1/3 · 1 mark

The least value of f(x) = e⁻ˣ in [0, 3] is

  • e⁻³
  • -1
  • 1
  • -e³

CBSE 2026 · Set 65/2/1 · 1 mark

For f(x) = x + 1/x (x ≠ 0)

  • local maximum value is 2
  • local minimum value is -2
  • local maximum value is -2
  • local minimum value < local maximum value

CBSE 2026 · Set 65/2/1 · 2 marks

Find the absolute maximum value of f(x) = cos x + sin² x, x ∈ [0, π].

CBSE 2026 · Set 65/2/1 · 5 marks

Find the sub intervals in which f(x) = cot⁻¹ (sin x + cos x), x ∈ (0, π) is increasing and decreasing.

CBSE 2026 · Set 65/2/2 · 2 marks

Find the absolute maximum value of f(x) = cos x + sin² x, x ∈ [0, π]

CBSE 2026 · Set 65/2/2 · 5 marks

Find the sub-interval of (0, π) in which f(x) = tan⁻¹(sin x – cos x) is increasing and decreasing.

CBSE 2026 · Set 65(B) R Mathematics · 1 mark

The interval in which y = x³ . e^(-3x) is an increasing function is :

  • (–1, 2)
  • (–∞, 2)
  • (1, ∞)
  • (0, 1)

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Class 12 Maths Formula Sheet

Every formula from all 13 chapters on one page, checked against the NCERT textbook and the CBSE 2026-27 syllabus. Free PDF download.

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Class 12 Mathematics Previous Year Questions, Chapter-wise

All 1048 CBSE board questions from 2022–2026, sorted by chapter and question type. Free PDF download.

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