CUET UG Mathematics: Calculus - Applications of Derivatives, Maxima/Minima, Tangent/Normal, Rate of Change

Must?Know (15–20 detailed bullets)

• The derivative of a function at a point gives the slope of the tangent to the curve at that point; for ( y = x^2 ), ( \frac{dy}{dx} = 2x ), so slope at ( x = 3 ) is 6.
• Equation of tangent to curve ( y = f(x) ) at ( (x_1, y_1) ) is ( y - y_1 = f'(x_1)(x - x_1) ); for ( y = x^3 ) at ( (1,1) ), ( f'(x) = 3x^2 ), so tangent: ( y - 1 = 3(x - 1) ).
• Equation of normal to curve ( y = f(x) ) at ( (x_1, y_1) ) is ( y - y_1 = -\frac{1}{f'(x_1)}(x - x_1) ), provided ( f'(x_1) \neq 0 ); for ( y = x^2 ) at ( (2,4) ), slope of tangent is 4, so normal: ( y - 4 = -\frac{1}{4}(x - 2) ).
• If ( f'(x) = 0 ) and ( f''(x) < 0 ), the point is a local maximum; for ( f(x) = -x^2 + 4x ), ( f'(x) = -2x + 4 = 0 ) at ( x = 2 ), ( f''(x) = -2 < 0 ), so max at ( x = 2 ).
• If ( f'(x) = 0 ) and ( f''(x) > 0 ), the point is a local minimum; for ( f(x) = x^2 - 6x ), ( f'(x) = 2x - 6 = 0 ) at ( x = 3 ), ( f''(x) = 2 > 0 ), so min at ( x = 3 ).
• Critical points occur where ( f'(x) = 0 ) or ( f'(x) ) does not exist; for ( f(x) = |x| ), critical point at ( x = 0 ) where derivative does not exist.
• First Derivative Test: if ( f'(x) ) changes from positive to negative at ( x = c ), then ( f(c) ) is a local maximum; if from negative to positive, local minimum.
• Second Derivative Test: if ( f'(c) = 0 ) and ( f''(c) < 0 ), then ( f(c) ) is local max; if ( f''(c) > 0 ), local min; if ( f''(c) = 0 ), test fails—use first derivative test.
• Absolute maxima/minima on a closed interval ([a,b]) occur either at critical points or endpoints; for ( f(x) = x^3 - 3x^2 + 1 ) on ([0,3]), critical points at ( x = 0,2 ), evaluate ( f(0)=1 ), ( f(2)=-3 ), ( f(3)=1 ), so absolute min is (-3) at ( x=2 ).
• For two curves to touch each other, they must have a common point and equal derivatives at that point; curves ( y = x^2 ) and ( y = 2x - 1 ) touch at ( (1,1) ) since both pass through it and ( \frac{dy}{dx} = 2 ) for both.
• Rate of change of quantity ( y ) with respect to time ( t ) is ( \frac{dy}{dt} ); if ( y = 4t^2 + 3t ), then ( \frac{dy}{dt} = 8t + 3 ), so at ( t = 2 ), rate is 19 units/sec.
• If a ladder 5 m long slides down a wall, with foot pulled away at 2 cm/s, then ( x^2 + y^2 = 25 ), differentiate: ( 2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0 ); when ( x = 4 ), ( y = 3 ), ( \frac{dx}{dt} = 0.02 ), so ( \frac{dy}{dt} = -\frac{x}{y}\frac{dx}{dt} = -\frac{4}{3}(0.02) \approx -0.0267 ) m/s.
• Marginal cost is the rate of change of total cost with respect to output; if ( C(x) = 0.005x^3 - 0.02x^2 + 30x + 5000 ), then marginal cost ( MC = \frac{dC}{dx} = 0.015x^2 - 0.04x + 30 ).
• Marginal revenue is rate of change of total revenue with respect to number of units sold; if ( R(x) = 50x - 0.5x^2 ), then ( MR = \frac{dR}{dx} = 50 - x ).
• For a function ( f(x) ), if ( f'(x) > 0 ) on an interval, function is increasing; if ( f'(x) < 0 ), decreasing; ( f(x) = 2x^3 - 3x^2 - 12x + 5 ), ( f'(x) = 6x^2 - 6x - 12 = 6(x-2)(x+1) ), so increasing on ( (-\infty, -1) \cup (2, \infty) ).
• A point of inflection occurs where ( f''(x) = 0 ) and sign of ( f''(x) ) changes; for ( f(x) = x^3 ), ( f''(x) = 6x = 0 ) at ( x = 0 ), and sign changes, so inflection at ( x = 0 ).
• The angle ( \theta ) between two curves at point of intersection is given by ( \tan\theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| ), where ( m_1, m_2 ) are slopes of tangents; for ( y = x^2 ) and ( y = x ) at ( (1,1) ), ( m_1 = 2 ), ( m_2 = 1 ), so ( \tan\theta = \left| \frac{2-1}{1+2} \right| = \frac{1}{3} ).
• For optimization problems, express quantity to be maximized/minimized as function of one variable, then find critical points; to maximize area of rectangle with perimeter 36 m: ( A = x(18 - x) = 18x - x^2 ), ( A' = 18 - 2x = 0 \Rightarrow x = 9 ), so max area 81 m².
• Approximate change in ( y ) due to small change ( \Delta x ) in ( x ) is ( \Delta y \approx \frac{dy}{dx} \Delta x ); for ( y = \sqrt{x} ), at ( x = 25 ), ( \Delta x = 0.1 ), ( \frac{dy}{dx} = \frac{1}{2\sqrt{x}} = \frac{1}{10} ), so ( \Delta y \approx 0.01 ), hence ( \sqrt{25.1} \approx 5.01 ).
• Verify from NCERT: exact year of first application of derivatives in economics models.

Difficulty Level

Intermediate — because it combines conceptual understanding of derivatives with application in real-world contexts and requires algebraic manipulation.

Common CUET Traps

• Trap: Assuming every point where ( f'(x) = 0 ) is a maxima or minima.
  Avoid: Use the second derivative test or first derivative test to confirm nature; e.g., ( f(x) = x^3 ) has ( f'(0) = 0 ) but no max/min at ( x = 0 ).

• Trap: Using ( \frac{dy}{dx} ) directly as rate of change without chain rule when both variables depend on time.
  Avoid: Apply chain rule: ( \frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt} ); e.g., in circular motion, ( A = \pi r^2 ), so ( \frac{dA}{dt} = 2\pi r \frac{dr}{dt} ).

• Trap: Confusing marginal cost with average cost.
  Avoid: Marginal cost = derivative of total cost; average cost = total cost divided by quantity; for ( C(x) = 100 + 5x ), MC = 5, AC = ( \frac{100}{x} + 5 ).

Practice MCQs

Q1. The slope of the tangent to the curve ( y = 3x^2 - 4x + 7 ) at ( x = 1 ) is:
A. 1
B. 2
C. 3
D. 4
Answer: B
Explanation: ( \frac{dy}{dx} = 6x - 4 ), at ( x = 1 ), slope = 2.
Why others fail: Option A (1) may come from incorrect differentiation or substitution error.

Q2. For the function ( f(x) = x^3 - 6x^2 + 9x + 15 ), the local minimum occurs at:
A. ( x = 1 )
B. ( x = 3 )
C. ( x = 2 )
D. ( x = 0 )
Answer: B
Explanation: ( f'(x) = 3x^2 - 12x + 9 = 3(x-1)(x-3) ), critical points at ( x = 1, 3 ); ( f''(x) = 6x - 12 ), ( f''(3) = 6 > 0 ), so min at ( x = 3 ).
Why others fail: Option A is local max (( f''(1) = -6 < 0 )), often mistaken due to sign error.

Q3. The marginal revenue when ( R(x) = 60x - 0.5x^2 ) at ( x = 10 ) is:
A. 50
B. 55
C. 60
D. 45
Answer: A
Explanation: ( MR = \frac{dR}{dx} = 60 - x ), at ( x = 10 ), MR = 50.
Why others fail: Option B (55) may result from using average revenue ( \frac{R(x)}{x} = 60 - 0.5x = 55 ), confusing MR with AR.

Q4. A balloon is being inflated at 900 cm³/s. The rate at which the radius increases when radius is 15 cm is:
A. ( \frac{1}{\pi} ) cm/s
B. ( \frac{2}{\pi} ) cm/s
C. ( \frac{3}{\pi} ) cm/s
D. ( \frac{1}{3\pi} ) cm/s
Answer: A
Explanation: ( V = \frac{4}{3}\pi r^3 ), ( \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} ), so ( 900 = 4\pi (225) \frac{dr}{dt} \Rightarrow \frac{dr}{dt} = \frac{900}{900\pi} = \frac{1}{\pi} ).
Why others fail: Option D comes from misusing formula, e.g., using surface area instead of volume.

Q5. The equation of the normal to the curve ( y = \log x ) at ( x = 1 ) is:
A. ( y = x - 1 )
B. ( y = 1 - x )
C. ( y = x + 1 )
D. ( y = -x + 1 )
Answer: D
Explanation: At ( x = 1 ), ( y = \log 1 = 0 ); ( \frac{dy}{dx} = \frac{1}{x} = 1 ), so slope of normal = -1; equation: ( y - 0 = -1(x - 1) \Rightarrow y = -x + 1 ).
Why others fail: Option B is tempting due to sign confusion but doesn’t pass through (1,0).

Last?Minute Revision (15–20 one?liners)

• Slope of tangent = ( f'(x) ); slope of normal = ( -\frac{1}{f'(x)} ) (if ( f'(x) \neq 0 )).
• Local max/min occur at critical points where ( f'(x) = 0 ) or undefined.
• First Derivative Test: sign change in ( f'(x) ) determines max/min.
• Second Derivative Test: ( f''(c) < 0 )-max; ( f''(c) > 0 )-min.
• Absolute extrema on [a,b]: check endpoints and critical points.
• Marginal Cost = ( \frac{dC}{dx} ); Marginal Revenue = ( \frac{dR}{dx} ).
• Rate of change with time: use chain rule ( \frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt} ).
• For approximation: ( f(x + \Delta x) \approx f(x) + f'(x)\Delta x ).
• Two curves touch if same point and same derivative.
• Angle between curves: ( \tan\theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| ).
• Increasing function: ( f'(x) > 0 ); decreasing: ( f'(x) < 0 ).
• Point of inflection: ( f''(x) = 0 ) and sign change in ( f''(x) ).
• For max/min in geometry: express area/volume as function of one variable.
• ( \frac{d}{dx}(\log x) = \frac{1}{x} ); ( \frac{d}{dx}(e^x) = e^x ).
• Normal is perpendicular to tangent-product of slopes = -1.
• If ( f'(x) = 0 ) and ( f''(x) = 0 ), use first derivative test.
• Volume of sphere: ( \frac{4}{3}\pi r^3 ); surface area: ( 4\pi r^2 ).
• Chain rule is essential in related rates problems.
• Remember: derivative = instantaneous rate of change.
• Mnemonic: "MAX" — when derivative changes + to –.