This MCQ module is based on: Application of Derivatives
TOPIC 15 OF 16
Application of Derivatives
🎓 Class 12
Mathematics
CBSE
Theory
Ch 6 — Application of Derivatives
⏱ ~15 min
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Targeting Class 12 level in Calculus, with Advanced difficulty.
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Application of Derivatives
Class 12 Mathematics • Chapter 6 • NCERT Part I
6.1 Introduction
The derivative \(dy/dx\) measures the instantaneous rate of change of \(y\) with respect to \(x\). In this chapter we put that geometric/physical idea to work: rates of change in real situations, whether a function is increasing or decreasing, maxima and minima, and solving optimisation problems.
6.2 Rate of Change of Quantities
Interpretation
If a quantity \(y\) depends on another quantity \(x\), then \(dy/dx\) is the rate of change of \(y\) w.r.t. \(x\). If both depend on time \(t\), chain rule gives \(\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}\).
Example 1 — Area of an expanding circle
Radius \(r\) increases at \(dr/dt=2\) cm/s. Area \(A=\pi r^2\). \(\dfrac{dA}{dt}=2\pi r\cdot\dfrac{dr}{dt}=4\pi r.\) At \(r=5\) cm: \(dA/dt=20\pi\) cm²/s.
Fig 6.1 — Expanding circle: area grows at 2πr·dr/dt.
Example 2 — Edge and volume of a cube
Edge \(x\) increasing at 3 cm/s. Volume \(V=x^3\). \(dV/dt=3x^2\cdot dx/dt=9x^2.\) At \(x=10\): \(dV/dt=900\) cm³/s.
Example 3 — Surface area of a sphere
\(S=4\pi r^2\). If \(dr/dt=0.05\) cm/s then \(dS/dt=8\pi r\cdot(0.05)=0.4\pi r.\) At \(r=10\): \(4\pi\) cm²/s.
Example 4 — Spherical balloon
Volume \(V=\tfrac{4}{3}\pi r^3\). If air is pumped in at 900 cm³/s, find \(dr/dt\) when \(r=15\) cm. \(dV/dt=4\pi r^2\,dr/dt\Rightarrow 900=4\pi(225)dr/dt\Rightarrow dr/dt=\dfrac{1}{\pi}\) cm/s.
Example 5 — Ladder sliding
A 5 m ladder leans against a wall. Foot slides out at 0.5 m/s. How fast is the top sliding down when the foot is 3 m from the wall?
Solution. Let \(x\) = distance of foot from wall, \(y\) = height of top. Then \(x^2+y^2=25\). Differentiate: \(2x\dot x+2y\dot y=0\Rightarrow \dot y=-\dfrac{x}{y}\dot x.\) At \(x=3\): \(y=4\). So \(\dot y=-(3/4)(0.5)=-0.375\) m/s. The top slides down at \(0.375\) m/s.
Fig 6.2 — Ladder sliding problem.
Example 6 — Marginal cost / revenue
Economics: if total cost is \(C(x)\) for producing \(x\) units, marginal cost = \(dC/dx\). If revenue is \(R(x)=px\), marginal revenue = \(dR/dx\). These tell the approximate cost/revenue of producing one more unit.
Numerical. \(C(x)=0.007x^3-0.003x^2+15x+4000\). Find marginal cost at \(x=17\). \(C'(x)=0.021x^2-0.006x+15.\) \(C'(17)=0.021(289)-0.006(17)+15=6.069-0.102+15=20.967.\) So ₹20.97 (approx) per additional unit.
Activity 6.1 — Water pouring into a cone
Water is poured into a cone (apex down, radius \(R=6\), height \(H=12\)) at 2 cm³/s.
- Use similar triangles to relate radius \(r\) at water-level to height \(h\): \(r=h/2\).
- Express volume of water \(V\) as a function of \(h\).
- Find \(dh/dt\) when \(h=4\) cm.
\(V=\tfrac13\pi r^2 h=\tfrac13\pi(h/2)^2 h=\dfrac{\pi h^3}{12}.\) \(dV/dt=\dfrac{\pi h^2}{4}\cdot dh/dt.\) At \(h=4\): \(2=\dfrac{\pi(16)}{4}dh/dt=4\pi\,dh/dt\Rightarrow dh/dt=\dfrac{1}{2\pi}\) cm/s.
Worked Examples
Q1. Perimeter of a square \(P=4x\) increases at 0.1 cm/s. Find the rate of change of area when side \(x=10\).
\(dP/dt=4\,dx/dt=0.1\Rightarrow dx/dt=0.025.\) \(A=x^2\Rightarrow dA/dt=2x\,dx/dt=2(10)(0.025)=0.5\) cm²/s.
Q2. A stone dropped into a pond makes ripples whose radius grows at 4 cm/s. Rate of area of disturbed water when \(r=8\)?
\(dA/dt=2\pi r\,dr/dt=2\pi(8)(4)=64\pi\) cm²/s.
Q3. Volume of a cube grows at 8 cm³/s. How fast is edge changing when edge = 12 cm?
\(dV/dt=3x^2\,dx/dt\Rightarrow 8=3(144)dx/dt\Rightarrow dx/dt=\dfrac{1}{54}\) cm/s.
Q4. For the total revenue \(R(x)=3x^2+36x+5\), marginal revenue at \(x=5\)?
\(R'(x)=6x+36\). \(R'(5)=30+36=66.\)
Competency-Based Questions
An inverted conical tank (radius 3 m at top, height 6 m) is being filled with water at 5 m³/min.
Q1. Express water volume \(V\) in terms of water depth \(h\).
L3By similar triangles \(r/h=3/6=1/2\). \(V=\tfrac13\pi r^2 h=\tfrac{\pi h^3}{12}.\)
Q2. How fast is \(h\) rising when \(h=2\) m?
L3\(dV/dt=\tfrac{\pi h^2}{4}\,dh/dt\Rightarrow 5=\tfrac{\pi(4)}{4}\,dh/dt=\pi\,dh/dt\Rightarrow dh/dt=5/\pi\) m/min.
Q3. Argue why \(dh/dt\) is larger when the tank is nearly empty than when nearly full.
L4\(dh/dt=\dfrac{4}{\pi h^2}\cdot dV/dt\). Since \(dV/dt\) is constant and denominator \(\propto h^2\) grows, \(dh/dt\) decreases as \(h\) increases. Geometrically: more surface area higher up, so each mm of depth requires more volume.
Q4. Design an experiment/model for a non-conical container (e.g. cylindrical + conical bottom) and analyse \(dh/dt\).
L6Piecewise \(V(h)\): conical \(\tfrac{\pi h^3}{12}\) for \(h\le 6\), and \(\tfrac{\pi(3)^2}{1}(h-6)+V(6)\) above. \(dh/dt\) is continuous but its formula switches at \(h=6\); compute both branches.
Assertion–Reason
A: If \(V=\tfrac43\pi r^3\) and \(dV/dt\) is constant, then \(dr/dt\) decreases as \(r\) grows. R: \(dV/dt=4\pi r^2\,dr/dt\).
A — R explains A: \(dr/dt=(dV/dt)/(4\pi r^2)\) shrinks as \(r^2\) grows.
A: Marginal cost equals total cost divided by number of units. R: Marginal = derivative of cost.
D — A false (that's average cost), R true.
A: Related-rates problems always require the chain rule. R: Quantities evolve in time jointly.
A — both correct, R explains A.
Frequently Asked Questions — Application of Derivatives
What is Application of Derivatives in NCERT Class 12 Mathematics?
Application of Derivatives is a key concept covered in NCERT Class 12 Mathematics, Chapter 6: Application of Derivatives. This lesson builds the student's foundation in the chapter by explaining the core ideas with worked examples, definitions, and step-by-step methods aligned to the CBSE curriculum.
How do I solve problems on Application of Derivatives step by step?
To solve problems on Application of Derivatives, follow the NCERT method: identify the given quantities, choose the relevant formula or theorem, substitute values carefully, and simplify. Class 12 exercises gradually increase in difficulty — start with solved NCERT examples before attempting exercise questions, and always verify your answer by substitution or diagram.
What are the most important formulas for Chapter 6: Application of Derivatives?
The essential formulas of Chapter 6 (Application of Derivatives) are listed in the chapter summary and highlighted throughout the lesson in formula boxes. Memorise them and practise at least 2–3 problems per formula. CBSE board exams frequently test direct application as well as combined use of multiple formulas from this chapter.
Is Application of Derivatives important for the Class 12 board exam?
Application of Derivatives is part of the NCERT Class 12 Mathematics syllabus and appears in CBSE board exams. Questions typically include short-answer, long-answer, and competency-based items. Review the NCERT examples, exercise questions, and previous-year board problems on this topic to prepare confidently.
What mistakes should students avoid in Application of Derivatives?
Common mistakes in Application of Derivatives include skipping steps, misapplying formulas, sign errors, and losing track of units. Write each step clearly, double-check algebraic manipulations, and re-read the question after solving to verify that your answer matches what was asked.
Where can I find more NCERT practice questions on Application of Derivatives?
End-of-chapter NCERT exercises for Application of Derivatives cover all difficulty levels tested in CBSE exams. After completing them, try the examples again without looking at the solutions, attempt the NCERT Exemplar questions for Chapter 6, and solve at least one previous-year board paper to consolidate your understanding.
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