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Punjab Board Class 11 Mathematics

Exercise 13.3 Solutions — Differentiation

Complete worked solutions for Exercise 13.3 from Chapter 13 (Differentiation), Punjab Board Class 11 Mathematics. Each question shows every step and names the formula used.

Mathematics 11 (Punjab Board)

Exercise 13.3 Solutions

Worked solutions for Punjab Board Class 11 Mathematics, Chapter 13: Differentiation.

Question 1: A car's position at time \(t\) is given by \(s(t) = 5t^3 - 3t^2 + t\). Find the velocity by differentiating the position function with respect to time.

Velocity position ka derivative hoti hai: \(v(t) = s'(t)\). Power rule lagayein.

\[ v(t) = \frac{d}{dt}(5t^3 - 3t^2 + t) = 15t^2 - 6t + 1 \]
Answer: \(v(t) = 15t^2 - 6t + 1\)

Question 2: Structural stress on a bridge is modeled by the function \(S(x) = 100 - 5x^2\), where \(x\) is the distance from the center of the bridge. Calculate the rate of change of stress at that point.

"At that point" se murad bridge ka center hai, yani \(x = 0\). Rate of change \(= S'(x)\) at \(x = 0\).

\[ S'(x) = -10x, \qquad S'(0) = 0 \]

The stress is momentarily constant at the centre of the bridge (it is the maximum point of the stress curve).

Answer: \(0\) (per unit distance)

Question 3: A company's revenue function is given by \(R(x) = 1000x - 10x^2\), where \(x\) is the number of units produced. The cost function is \(C(x) = 300x + 2000\). (i) Find the profit function \(P(x)\). (ii) Determine the marginal profit when \(x = 15\).

(i) Profit \(= \text{Revenue} - \text{Cost}\). (ii) Marginal profit \(= P'(x)\) at \(x = 15\).

(i)

\[ P(x) = R(x) - C(x) = (1000x - 10x^2) - (300x + 2000) = 700x - 10x^2 - 2000 \]

(ii)

\[ P'(x) = 700 - 20x, \qquad P'(15) = 700 - 300 = 400 \]
Answer: (i) \(P(x) = 700x - 10x^2 - 2000\); (ii) \(400\) per unit

Question 4: An investment grows according to the function \(A(t) = 10000(1 + 0.05t)^3\), where \(A(t)\) is the value of the investment and \(t\) is the time in years. (i) Find the rate of change of the investment after 8 years. (ii) What is the investment value after 8 years?

(i) Rate of change \(= A'(t)\) — chain rule lagegi, phir \(t = 8\) rakhein. (ii) Value ke liye seedha \(t = 8\) original function mein rakhein.

(i) By the chain rule:

\[ A'(t) = 10000 \cdot 3(1 + 0.05t)^2 \cdot 0.05 = 1500(1 + 0.05t)^2 \]
\[ A'(8) = 1500(1 + 0.4)^2 = 1500(1.96) = 2940 \]

(ii)

\[ A(8) = 10000(1.4)^3 = 10000(2.744) = 27400 \]
Answer: (i) \(2940\) per year; (ii) \(27400\)

Question 5: The position of a particle moving along a line is given by \(s(t) = 5t^3 - 12t^2 + 8t\), where \(s(t)\) is the position in meters and \(t\) is the time in seconds. (i) Determine the velocity of the particle at \(t = 4\) seconds. (ii) Find the acceleration at \(t = 4\) seconds. (iii) When is the particle at rest?

Velocity \(= s'(t)\), acceleration \(= s''(t)\). Particle rest par hota hai jab velocity zero ho — quadratic equation solve karein.

(i)

\[ v(t) = s'(t) = 15t^2 - 24t + 8, \qquad v(4) = 15(16) - 96 + 8 = 152 \text{ m/s} \]

(ii)

\[ a(t) = v'(t) = 30t - 24, \qquad a(4) = 120 - 24 = 96 \text{ m/s}^2 \]

(iii) Set \(v(t) = 0\):

\[ 15t^2 - 24t + 8 = 0 \implies t = \frac{24 \pm \sqrt{576 - 480}}{30} = \frac{24 \pm \sqrt{96}}{30} = \frac{12 \pm 2\sqrt{6}}{15} \]
\[ t \approx 0.47 \text{ s} \quad \text{or} \quad t \approx 1.13 \text{ s} \]
Answer: (i) \(152\) m/s; (ii) \(96\) m/s\(^2\); (iii) \(t = \frac{12 \pm 2\sqrt{6}}{15}\) s (\(\approx 0.47\) s and \(\approx 1.13\) s)

Question 6: The position of a car traveling along a straight highway is given by \(x(t) = 30t^2 - 4t^3\), where \(x(t)\) is the distance traveled in kilometres and \(t\) is the time in hours. (i) Find the car's velocity at \(t = 3\) hours. (ii) Determine the car's acceleration at \(t = 3\) hours.

Velocity \(= x'(t)\), acceleration \(= x''(t)\); dono nikal kar \(t = 3\) rakhein. Negative acceleration ka matlab car slow ho rahi hai.

(i)

\[ v(t) = 60t - 12t^2, \qquad v(3) = 180 - 108 = 72 \text{ km/h} \]

(ii)

\[ a(t) = 60 - 24t, \qquad a(3) = 60 - 72 = -12 \text{ km/h}^2 \]
Answer: (i) \(72\) km/h; (ii) \(-12\) km/h\(^2\)

Question 7: The stress on a beam under a varying load is given by \(S(x) = 400x - x^3\), where \(S(x)\) is the stress in pascals (Pa) and \(x\) is the distance from the fixed end in metres. Calculate the rate of change of stress at 6 meters.

Rate of change \(= S'(x)\); derivative nikal kar \(x = 6\) rakhein.

\[ S'(x) = 400 - 3x^2, \qquad S'(6) = 400 - 3(36) = 400 - 108 = 292 \text{ Pa/m} \]
Answer: \(292\) Pa/m

Question 8: The cost \(C(r)\) to construct a cylindrical tank depends on the radius of the base, and is given by \(C(r) = 8000\pi r^2 + \frac{150000}{r}\), where the first term represents the cost of the base and the second term represents the cost of the walls. Determine the rate of change of the cost at \(r = 4\) metres.

\(\frac{150000}{r}\) ko \(150000r^{-1}\) likh kar power rule lagayein, phir \(r = 4\) rakhein.

\[ C'(r) = 16000\pi r - \frac{150000}{r^2} \]
\[ C'(4) = 16000\pi(4) - \frac{150000}{16} = 64000\pi - 9375 \approx 191686.93 \]
Answer: \(64000\pi - 9375 \approx 191687\) (cost units per metre)