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આ MCQ મોડ્યુલ આના પર આધારિત છે: Intuitive Idea of Derivatives
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Intuitive Idea of Derivatives
🎓 Class 11
Mathematics
CBSE
Theory
Ch 12 — Limits and Derivatives
⏱ ~30 min
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આ મૂલ્યાંકન આના પર આધારિત હશે: Intuitive Idea of Derivatives
Class 11 સ્તર, Calculus માં, Advanced કઠિનતા સાથે લક્ષ્ય.
મૂલ્યાંકન બનાવવામાં તેમની સામગ્રી સામેલ કરવા ચિત્રો, PDF અથવા Word દસ્તાવેજ અપલોડ કરો.
12.1 Introduction
This chapter is an introduction to Calculus? — the branch of mathematics that deals with change and motion. Two fundamental questions are at its heart:
- If a quantity changes with time, how fast is it changing at a particular instant?
- If a curve is drawn on graph paper, what is the slope of the tangent to the curve at a given point?
Both questions lead to the concept of a derivative?. Before we define it rigorously, we build an intuitive picture using a familiar example: a freely-falling body.
Sir Isaac Newton
1642 – 1727
Sir Isaac Newton, the English mathematician and physicist, co-invented calculus in the late 17th century (independently and simultaneously with Gottfried Leibniz). His notation \(\dot y\) for rate of change, and his use of "fluxions" to describe instantaneous velocity in the Principia Mathematica (1687), laid the foundation for modern calculus, mechanics, and mathematical physics.
12.2 Intuitive Idea of Derivatives
Consider a stone that is dropped from a height of 45 m above the ground. The distance \(y\) (in metres) fallen in time \(t\) (in seconds) is given by: \[y=f(t)=4.9\,t^2.\] This is the well-known law of free fall (with \(g\approx 9.8\,\text{m/s}^2\)).
Average Velocity Over an Interval
The average speed between \(t=t_1\) and \(t=t_2\) is simply: \[\bar v=\frac{f(t_2)-f(t_1)}{t_2-t_1}=\frac{\text{distance covered}}{\text{time taken}}.\] Let us tabulate average velocities for shorter and shorter intervals around \(t=2\) s.
| Interval \([t_1,t_2]\) | \(f(t_1)\) | \(f(t_2)\) | Average velocity (m/s) |
|---|---|---|---|
| [1.00, 2.00] | 4.9 | 19.6 | 14.70 |
| [1.50, 2.00] | 11.025 | 19.6 | 17.15 |
| [1.90, 2.00] | 17.689 | 19.6 | 19.11 |
| [1.99, 2.00] | 19.40 | 19.6 | 19.551 |
| [2.00, 2.01] | 19.6 | 19.800 | 19.649 |
| [2.00, 2.10] | 19.6 | 21.609 | 20.09 |
| [2.00, 2.50] | 19.6 | 30.625 | 22.05 |
| [2.00, 3.00] | 19.6 | 44.1 | 24.50 |
As the interval shrinks towards \(t=2\) from both sides, the average velocity approaches 19.6 m/s. This value is called the instantaneous velocity at \(t=2\) s.
Instantaneous Velocity
The instantaneous velocity at time \(t=c\) is:
\[v(c)=\lim_{h\to 0}\frac{f(c+h)-f(c)}{h}.\]
Geometrically, this is the slope of the tangent line to the graph of \(y=f(t)\) at \(t=c\).
Fig 12.1: As \(Q\to P\), the secant \(PQ\) tends to the tangent at \(P\); its slope is the derivative.
Example 1
Find the average velocity of the falling stone over the intervals (i) \([1, 2]\), (ii) \([2, 3]\); and its instantaneous velocity at \(t=2\) s.
(i) \(\bar v_{[1,2]}=\dfrac{f(2)-f(1)}{2-1}=\dfrac{19.6-4.9}{1}=14.7\) m/s.
(ii) \(\bar v_{[2,3]}=\dfrac{44.1-19.6}{1}=24.5\) m/s.
Instantaneous velocity at \(t=2\): \(v(2)=\lim_{h\to0}\dfrac{4.9(2+h)^2-4.9(4)}{h}=\lim_{h\to0}\dfrac{4.9(4+4h+h^2-4)}{h}=\lim_{h\to0}4.9(4+h)=19.6\) m/s.
(ii) \(\bar v_{[2,3]}=\dfrac{44.1-19.6}{1}=24.5\) m/s.
Instantaneous velocity at \(t=2\): \(v(2)=\lim_{h\to0}\dfrac{4.9(2+h)^2-4.9(4)}{h}=\lim_{h\to0}\dfrac{4.9(4+4h+h^2-4)}{h}=\lim_{h\to0}4.9(4+h)=19.6\) m/s.
Geometric Meaning
For a curve \(y=f(x)\), the slope of the tangent at \(x=c\) is defined as the limiting value of the slope of secants through \((c,f(c))\) and \((c+h,f(c+h))\) as \(h\to 0\):
\[\text{slope of tangent}=\lim_{h\to 0}\frac{f(c+h)-f(c)}{h}.\]
This number, when it exists, is called the derivative of \(f\) at \(c\), denoted \(f'(c)\) or \(\left.\dfrac{df}{dx}\right|_{x=c}\).
Why Two Names: Tangent Slope vs Instantaneous Rate?
The remarkable fact of calculus is that the physical question of instantaneous velocity and the geometric question of tangent slope have the same answer — the derivative. Whether we are asking:
- How fast is temperature rising at 3 p.m.?
- What is the steepness of a curve at a point?
- How quickly does the volume of a balloon grow with its radius?
every such "rate at an instant" is computed by the same limiting process.
Activity: From Secant to Tangent
L3 ApplyMaterials: Graph paper, ruler, calculator.
Predict: For \(f(x)=x^2\) at \(x=1\), what will the slope of the tangent turn out to be?
- Plot \(y=x^2\) on graph paper for \(x\in[0,3]\) using a scale of 1 cm = 0.5 units.
- Mark the point \(P(1,1)\). Draw secants from \(P\) to \(Q_1(2,4)\), \(Q_2(1.5, 2.25)\), \(Q_3(1.1, 1.21)\), \(Q_4(1.01, 1.0201)\).
- Compute the slope of each secant: \(m=(y_Q-y_P)/(x_Q-x_P)\).
- Record the slopes in a table. What number are they approaching?
- Using the same method with \(Q\) to the left of \(P\) (e.g. at \(0.9, 0.99\)), confirm the limit from the other side.
The secant slopes are \(3, 2.5, 2.1, 2.01\) — converging to 2. This matches the calculation \(f'(1)=2x|_{x=1}=2\). Both sides give the same limit, confirming the tangent slope at \(x=1\) is 2.
Competency-Based Questions
Scenario: A balloon is inflated so that at time \(t\) seconds, its volume is \(V(t)=\tfrac{4}{3}\pi t^3\) cm³. A student wants to know how fast the balloon grows at \(t=2\) s.
Q1. The average rate of growth over \([1,3]\) is:
L3 ApplyAnswer: (b). \(\dfrac{V(3)-V(1)}{3-1}=\dfrac{\tfrac{4\pi}{3}(27-1)}{2}=\dfrac{52}{3}\pi\) cm³/s.
Q2. Analyse: why does the average rate on \([1,3]\) not equal the instantaneous rate at \(t=2\)? Compute \(V'(2)\) by the limit definition to compare.
L4 AnalyseAnswer: \(V'(2)=\lim_{h\to 0}\dfrac{\tfrac{4\pi}{3}((2+h)^3-8)}{h}=\dfrac{4\pi}{3}\lim_{h\to 0}(12+6h+h^2)=\dfrac{4\pi}{3}\times12=16\pi\). The average over \([1,3]\) is \(\tfrac{52}{3}\pi\approx17.33\pi\), while the instantaneous rate is \(16\pi\). They differ because \(V\) grows faster near \(t=3\) than near \(t=2\); averaging over the interval biases the mean upward.
Q3. Evaluate the claim: "If the instantaneous rate at \(t=2\) is \(16\pi\) cm³/s, then in the next 0.01 s the balloon grows by exactly \(0.16\pi\) cm³."
L5 EvaluateAnswer: The claim is approximately correct, not exact. Actual growth \(=V(2.01)-V(2)=\tfrac{4\pi}{3}(2.01^3-8)\approx\tfrac{4\pi}{3}(0.12060)\approx 0.1608\pi\) cm³. The linear approximation \(V'(2)\cdot 0.01=0.16\pi\) is very close but ignores higher-order terms.
Q4. Create a real-world quantity other than balloons whose "rate at an instant" can be modelled in the same way. Give the function and compute its instantaneous rate at one time.
L6 CreateSample Answer: Population of a bacterial colony: \(P(t)=100\,t^2\) (thousands). Rate at \(t=5\): \(P'(5)=\lim_{h\to 0}\dfrac{100((5+h)^2-25)}{h}=100\cdot 10=1000\) thousand per hour. Other valid examples: fuel consumption vs distance, temperature vs time, revenue vs units sold.
Assertion–Reason Questions
Assertion (A): The instantaneous velocity of a freely-falling stone at \(t=2\) s is 19.6 m/s.
Reason (R): For \(y=4.9t^2\), \(\lim_{h\to 0}\frac{y(t+h)-y(t)}{h}=9.8\,t\).
Reason (R): For \(y=4.9t^2\), \(\lim_{h\to 0}\frac{y(t+h)-y(t)}{h}=9.8\,t\).
Answer: (a). At \(t=2\), \(9.8\times 2=19.6\). R is the formula that yields A, so R explains A.
Assertion (A): The slope of the tangent to \(y=f(x)\) at \(x=c\) equals the derivative \(f'(c)\).
Reason (R): Secant slopes approach the tangent slope as the second point approaches the first.
Reason (R): Secant slopes approach the tangent slope as the second point approaches the first.
Answer: (a). Both statements are true and the second is the precise reason for the first.
Frequently Asked Questions
What does the derivative represent physically?
If s(t) is position, the derivative ds/dt is the instantaneous velocity. If v(t) is velocity, dv/dt is the instantaneous acceleration.
How is the derivative related to a tangent line?
The value of the derivative at x = a equals the slope of the tangent line to the curve y = f(x) at the point (a, f(a)).
What is the first-principle definition of the derivative?
f '(x) = limit as h approaches 0 of [ f(x + h) - f(x) ] / h, provided this limit exists.
What is the difference between average and instantaneous rate?
Average rate uses a finite interval [a, b]; instantaneous rate is the limit of the average rate as b approaches a, giving the derivative at a.
Why do we need limits to define a derivative?
Because the instantaneous rate is an idealised zero-interval ratio - only a limit can handle the 0/0 that appears in a raw difference quotient.
How are derivatives used in daily life?
Derivatives measure growth rates in economics, reaction rates in chemistry, velocities in physics, marginal cost in business and rate of infection in epidemiology.
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