Questions: Application of Derivatives

Q1. Define the rate of change of a quantity with respect to time.

Answer (English): The rate of change of a quantity y with respect to time is dy by dt, the derivative of y with respect to t. When y depends on another variable x which itself depends on t, the chain rule gives dy by dt equal to dy by dx times dx by dt. The sign tells whether the quantity increases or decreases at that instant, and the unit is the unit of y per unit time.

Hinglish: Rate of change dy/dt hota hai. Chain rule se dy/dt = (dy/dx)(dx/dt). Sign batata hai quantity badh rahi hai ya ghat rahi hai, aur unit likhna zaroori hai.

Q2. How do you decide whether a function is increasing or decreasing on an interval?

Answer (English): Differentiate the function and study the sign of the derivative on the interval. If the derivative is positive at every interior point, the function is strictly increasing there; if it is negative, the function is strictly decreasing. The critical points, where the derivative is zero or does not exist, split the domain into intervals, and one test value per interval decides the sign.

Hinglish: Derivative ka sign dekho: positive matlab increasing, negative matlab decreasing. Critical points se intervals banao aur har interval me ek test value daal ke sign check karo.

Q3. State the second derivative test for a local maximum.

Answer (English): If the first derivative vanishes at a point c and the second derivative is negative there, the function has a local maximum at c. The negative second derivative means the slope is decreasing, so the curve bends downward around the point. If the second derivative is positive, the point is a local minimum, and if it is zero the test is inconclusive.

Hinglish: f'(c) = 0 aur f''(c) < 0 ho to local maximum. f'' positive ho to local minimum, aur zero ho to test fail, phir first derivative sign test lagao.

Q4. A circle's radius grows by 2 cm each second. Work out how quickly its area is growing at the instant the radius equals 5 cm.

Answer (English): The area is pi r squared. Differentiating with respect to time gives dA by dt equal to 2 pi r dr by dt. Substituting r equal to 5 and dr by dt equal to 2 gives 20 pi square centimetres per second. The unit comes from multiplying the radius by the radius rate, and the direction is an increase because the answer is positive.

Hinglish: Area = pi r², isliye dA/dt = 2 pi r dr/dt = 2 pi (5)(2) = 20 pi cm²/s. Positive answer matlab area badh raha hai.

Q5. Find the local minimum value of f(x) = x squared minus 4x plus 7.

Answer (English): The derivative is 2x minus 4, which is zero at x equal to 2. The second derivative is 2, positive everywhere, so the critical point is a local minimum. The minimum value is f(2) = 4 minus 8 plus 7, which is 3. This is also the absolute minimum on the whole real line because the parabola opens upward.

Hinglish: f'(x) = 2x - 4 = 0 se x = 2. f''(x) = 2 > 0, isliye local minimum. Value f(2) = 3.

Q6. Find the interval in which f(x) = x cubed minus 3x squared plus 4 is decreasing.

Answer (English): The derivative is 3x squared minus 6x, which factors as 3x(x minus 2). The critical points are 0 and 2. Testing values gives a positive derivative before 0, a negative derivative between 0 and 2, and a positive derivative after 2. So the function decreases on the interval (0, 2).

Hinglish: f'(x) = 3x² - 6x = 3x(x - 2). Signs: 0 se pehle plus, 0 aur 2 ke beech minus, 2 ke baad plus. Isliye decreasing interval (0, 2).

Q7. A ball is thrown upward with height h(t) = 20t minus 5t squared. Find the maximum height.

Answer (English): The velocity is the derivative of the height, so v(t) = 20 minus 10t. Setting the velocity to zero gives t equal to 2 seconds. The second derivative of the height is minus 10, negative, so t equal to 2 is a maximum. Substituting into the height gives 40 minus 20, which is 20 metres.

Hinglish: v(t) = 20 - 10t = 0 se t = 2. h'' = -10 < 0, isliye maximum. Height = 20(2) - 5(4) = 20 metres.

Q8. Use the first derivative test to classify the critical point of f(x) = x cubed.

Answer (English): The derivative is 3x squared, which is zero at x equal to 0 and positive on both sides. Since the derivative does not change sign at the critical point, the point is neither a local maximum nor a local minimum. The curve flattens there and continues to rise, and such a point is called a point of inflection.

Hinglish: f'(x) = 3x² zero par zero hai lekin dono taraf positive hai, isliye neither maximum nor minimum. Aise point ko point of inflection kehte hain.

Q9. What is the slope of the normal to the curve y = x squared at the point (1, 1)?

Answer (English): The derivative is 2x, so the tangent slope at x equal to 1 is 2. The normal is perpendicular to the tangent, so its slope is the negative reciprocal, which is minus 1 by 2. Writing the tangent slope first and then taking the negative reciprocal is the standard two-step answer.

Hinglish: Tangent slope f'(1) = 2. Normal perpendicular hota hai, isliye slope -1/2 hoga.

Q10. Find the maximum area of a rectangle whose perimeter is 20 cm.

Answer (English): Let the sides be x and y with 2x plus 2y equal to 20, so y equals 10 minus x. The area is x(10 minus x) equal to 10x minus x squared. Differentiating gives 10 minus 2x, which is zero at x equal to 5, and the second derivative is minus 2, so the area is maximum. Both sides are 5 cm and the maximum area is 25 square centimetres, which is a square.

Hinglish: Perimeter se y = 10 - x. Area = x(10 - x). Derivative zero se x = 5, aur f'' = -2 < 0 isliye maximum. Area = 25 cm², shape square.

A last habit: for a related rate answer, write the unit and say whether the quantity is growing or shrinking. For an extrema answer, write the value and the point where it occurs. Both are part of the answer, not extra detail.

Quiz below: five multiple choice questions on related rates, monotonicity, extrema, tangents and optimisation. Attempt them after the formula sheet and the three topic pages.