📘 Calculus · Chapter MN13🎯 NDA Level : High Priority
Applications of Derivatives (AOD) translates calculus into geometry and optimisation. The derivative dy/dx is no longer just a calculation — it becomes the slope of a tangent, a sign chart that reveals where functions rise or fall, and a tool for finding the best (maximum or minimum) value of any quantity. NDA tests all three geometric, monotonicity, and extrema applications every year.
📌 What to expect in NDA (based on 2022–2024 papers): (1) Finding the slope of tangent and equation of tangent/normal at a given point on a curve; (2) Angle between two curves (using their slopes at the intersection); (3) Determining intervals of increase and decrease by sign of f′(x); (4) Finding critical points by solving f′(x) = 0; (5) Applying the first derivative test to classify maxima/minima; (6) Applying the second derivative test: f′′(a) > 0 → min, f′′(a) < 0 → max; (7) Finding global maximum/minimum on a closed interval [a, b]; (8) Optimisation problems — maximise area, minimise perimeter, etc.
Topics at a Glance
① Tangent & Normal
Slope, equation, perpendicularity condition
② Angle Between Curves
tan θ using slopes m₁, m₂
③ Monotonicity
Increasing (f′>0), decreasing (f′<0), sign chart
④ Critical Points
f′(x) = 0, stationary points, sign change
⑤ 1st & 2nd Derivative Tests
Local max/min classification
⑥ Optimisation
Global extrema, area/volume/perimeter problems
1. Tangent & Normal to a Curve
1.1
Slope, Equation of Tangent & Normal
The derivative gives the slope of the tangent; normal is perpendicular to it
The tangent at a point (x₀, y₀) on y = f(x) is the straight line that just touches the curve at that point. The normal is the line through the same point perpendicular to the tangent.
⚡ Tangent & Normal — Key Formulae
Slope of tangent at (x₀, y₀):
mₜ = [dy/dx]ₘ=(x₀,y₀) (evaluate the derivative at the given point)
Slope of normal at (x₀, y₀):
mₙ = −1 / mₜ (negative reciprocal, since tangent ⊥ normal)
Special: if mₜ = 0, normal is vertical (undefined slope)
if mₜ = ∞, normal is horizontal (slope 0)
Equation of tangent at (x₀, y₀):
y − y₀ = mₜ(x − x₀)
Equation of normal at (x₀, y₀):
y − y₀ = mₙ(x − x₀) = (−1/mₜ)(x − x₀)
Length of tangent, subtangent, normal, subnormal:
Length of tangent = y₀ √(1 + mₜ²) / |mₜ|
Length of normal = y₀ √(1 + mₜ²)
Length of subtangent = |y₀ / mₜ|
Length of subnormal = |y₀ mₜ|
For NDA: the most tested results are the slope formula and the equation of tangent/normal using point-slope form. The length formulas are occasionally asked in NDA as direct MCQs.
Fig 1: Tangent (green) and Normal (red) at point P on the curve. They are perpendicular to each other at P. The small square marks the 90° angle.
Worked Example — Equation of Tangent
Find the equation of the tangent to y = x² − 3x + 2 at x = 2.
The angle at the intersection is between their tangents at that point
⚡ Angle Between Curves — Formula
If two curves intersect at a point, and their tangent slopes there are m₁ and m₂:
tan θ = |(m₁ − m₂) / (1 + m₁m₂)| where θ is the angle between the curves
Orthogonal curves (intersect at right angles):
1 + m₁m₂ = 0 ⇔ m₁m₂ = −1
Curves touch each other (same tangent):
m₁ = m₂ (slopes equal at the intersection point)
This is identical to the angle between two straight lines formula from coordinate geometry. The trick: find the intersection point, compute dy/dx for each curve at that point, then substitute into the formula.
Worked Example — Orthogonal Curves
Show that the curves xy = 4 and x² − y² = 8 cut orthogonally.
Q1. The slope of the tangent to the curve y = x³ + 3x at x = 1 is:
(a) 3 (b) 6 (c) 4 (d) 9
Answer: (b) 6
dy/dx = 3x² + 3. At x = 1: slope = 3(1) + 3 = 6.
Q2. The equation of the normal to y = x² at (2, 4) is:
(a) x + 4y = 18 (b) x + 4y + 18 = 0 (c) 4x − y = 4 (d) x − 4y = 0
Answer: (a) x + 4y = 18
dy/dx = 2x. At (2,4): mₜ = 4. mₙ = −1/4.
Normal: y − 4 = (−1/4)(x − 2) → 4y − 16 = −x + 2 → x + 4y = 18.
Q3. The curves y = x² and y² = x intersect at the origin. The angle between them is:
(a) 0° (b) 45° (c) 90° (d) tan⁻¹(3/4)
Answer: (b) 45°
At (0,0): m₁ for y=x²: dy/dx=2x=0. m₂ for y²=x: 2y(dy/dx)=1, at (0,0) slope is ∞ (vertical).
Also at (1,1): m₁=2, m₂=1/2. tanθ=|(2−1/2)/(1+1)|=|(3/2)/2|=3/4 → θ = tan⁻¹(3/4).
At origin: one tangent is horizontal (m=0), other is vertical (∞) → angle = 90°. Option (c).
🔥 TRICKY QUESTIONS
Tangent & Normal — Classic NDA Traps
🤯 T1. Find the point on y = x³ where the tangent is parallel to y = 12x + 3.
Parallel tangent means same slope: dy/dx = 12.
dy/dx = 3x² = 12 → x² = 4 → x = ±2.
At x=2: y = 8. At x=−2: y = −8.
Points: (2, 8) and (−2, −8). Trap: Students forget x=−2. Always check both roots of x² = k when k > 0.
🤯 T2. The tangent to a curve at P is perpendicular to the x-axis. What does this tell you?
Perpendicular to x-axis means the tangent is vertical → slope = ∞.
dy/dx = ∞ at P means the denominator in dy/dx = 0 (if using implicit), or the derivative is undefined.
This occurs at a vertical tangent point — e.g., y² = x at x=0 has vertical tangent. Key: A vertical tangent does NOT mean a corner. The curve may still be smooth (like a cusp). But the function y=f(x) is NOT differentiable there in the usual sense since dy/dx is undefined.
Make a sign chart of f′(x) across its critical points
f′(x) > 0
Increasing Function rises as x increases. Graph slopes upward.
f′(x) < 0
Decreasing Function falls as x increases. Graph slopes downward.
f′(x) = 0
Critical Point Transition point. Could be max, min, or inflection.
⚡ Method for Finding Intervals of Monotonicity
Step 1: Find f′(x) by differentiating f(x).
Step 2: Solve f′(x) = 0 to find critical points x₁, x₂, x₃, …
Also note where f′(x) is undefined (if applicable).
Step 3: Arrange critical points on a number line.
Step 4: Test the sign of f′(x) in each interval:
• Pick any test point in the interval.
• Substitute into f′(x).
• Positive → increasing, Negative → decreasing.
Step 5: State the intervals:
f is increasing on (a, b) if f′(x) > 0 for all x in (a, b)
f is decreasing on (a, b) if f′(x) < 0 for all x in (a, b)
Always use open brackets for strict inequalities: f is “increasing on (a, b)” since at endpoints the sign may be zero. For NDA, the intervals are stated as open intervals unless the question specifies otherwise.
Worked Example — Sign Chart for Monotonicity
Find the intervals where f(x) = x³ − 3x² − 9x + 7 is increasing or decreasing.
Answer: (b) (π/4, 5π/4)
f′(x) = cos x + sin x = √2 sin(x + π/4).
f′ < 0 when sin(x+π/4) < 0 → π < x+π/4 < 2π → 3π/4 < x < 7π/4.
More precisely, f′ < 0 in (π/4 + interval). Standard result: decreasing on (π/4, 5π/4).
🔥 TRICKY QUESTIONS
Monotonicity — Interval & Sign-Chart Traps
🤯 T3. For what values of k is f(x) = kx³ − 9x² + 12x + 1 monotonically increasing on ℜ?
f′(x) = 3kx² − 18x + 12. For strictly increasing on ℜ: f′(x) ≥ 0 for all x.
For a quadratic 3kx²−18x+12 to be ≥ 0 everywhere:
1. Coefficient of x² must be positive: 3k > 0 → k > 0.
2. Discriminant must be ≤ 0: (18)² − 4(3k)(12) ≤ 0 → 324 ≤ 144k → k ≥ 9/4.
Therefore: k ≥ 9/4. Trap: Students only check k > 0 and forget the discriminant condition. Both are needed.
3. Maxima & Minima — Extreme Values
3.1
Critical Points, First Derivative Test & Second Derivative Test
Find where f′ = 0, then classify using sign change or f′′
⚡ First Derivative Test (FDT)
Step 1: Find all critical points by solving f′(x) = 0.
Step 2: Check the sign of f′(x) on either side of each critical point:
f′ changes + → − (pos to neg): LOCAL MAXIMUM at x = c
f′ changes − → + (neg to pos): LOCAL MINIMUM at x = c
f′ does NOT change sign: INFLECTION POINT (not an extremum)
Visual memory:
Maximum: f′ > 0 on left of c, f′ < 0 on right (“rises then falls”)
Minimum: f′ < 0 on left of c, f′ > 0 on right (“falls then rises”)
The first derivative test always works and never gives wrong answers. The second derivative test is faster but can be inconclusive when f′′(c) = 0.
⚡ Second Derivative Test (SDT)
Step 1: Solve f′(x) = 0 to find critical points.
Step 2: Compute f′′(x) and evaluate at each critical point x = c:
f′′(c) < 0 → LOCAL MAXIMUM at x = c (curve concave down = “cap”)
f′′(c) > 0 → LOCAL MINIMUM at x = c (curve concave up = “cup”)
f′′(c) = 0 → INCONCLUSIVE; use the first derivative test instead
Physical meanings:
f′′ > 0 → concave up (like a valley, holds water)
f′′ < 0 → concave down (like a hill, water flows off)
Point of inflection: where f′′(x) changes sign (concavity changes)
Concavity mnemonic: f′′ > 0 looks like a CUP (holds water, minimum). f′′ < 0 looks like a CAP (sheds water, maximum). “Cup = positive, Cap = negative”.
Fig 2: f′ > 0 (increasing, green), f′ < 0 (decreasing, red). Local max at x₁ where f′ changes + to −. Local min at x₂ where f′ changes − to +. Inflection at x where concavity changes.
Worked Example — First & Second Derivative Tests
Find the local maxima and minima of f(x) = x³ − 3x² − 9x + 7.
f′(x) = 3x² − 6x − 9 = 3(x−3)(x+1). Critical points: x = −1 and x = 3.
f′′(x) = 6x − 6.
At x = −1: f′′(−1) = −12 < 0 → Local Maximum. f(−1) = −1−3+9+7 = 12.
At x = 3: f′′(3) = 12 > 0 → Local Minimum. f(3) = 27−27−27+7 = −20.
📝 TOPIC-WISE PYQ
Maxima & Minima — NDA-Pattern Questions
Q6. The function f(x) = x² − 6x + 5 has a minimum at x = ?
(a) 3 (b) −3 (c) 6 (d) 0
Answer: (a) 3
f′(x) = 2x − 6 = 0 → x = 3.
f′′(x) = 2 > 0 → local minimum at x = 3. f(3) = 9−18+5 = −4.
Q7. The maximum value of sin x + cos x is:
(a) 1 (b) √2 (c) 2 (d) 1/√2
Answer: (b) √2
f(x) = sin x + cos x = √2 sin(x + π/4). Maximum of sin = 1.
Maximum value = √2 ⋅ 1 = √2.
Q8. If f(x) = ax³ + bx² + cx + d has a local max at x = 0 and local min at x = 1, then:
(a) a > 0, b > 0 (b) a < 0, b > 0 (c) a > 0, b < 0 (d) a < 0, b < 0
Answer: (a) a > 0, b > 0 ... rechecking
f′(x) = 3ax² + 2bx + c. f′(0) = 0 → c = 0.
f′′(x) = 6ax + 2b. Local max at x=0 → f′′(0) = 2b < 0 → b < 0.
Local min at x=1 → f′′(1) = 6a+2b > 0 → 6a > −2b > 0 → a > 0.
Answer: (c) a > 0, b < 0.
Q9. Find the global maximum of f(x) = x³ − 3x on [−2, 2].
(a) 2 (b) −2 (c) 4 (d) −4
Answer: (a) 2
f′(x) = 3x²−3 = 0 → x = ±1 (both in [−2,2]).
Evaluate at critical points and endpoints:
f(−2)=−8+6=−2; f(−1)=−1+3=2; f(1)=1−3=−2; f(2)=8−6=2.
Global maximum = 2, achieved at x = −1 and x = 2.
🔥 TRICKY QUESTIONS
Maxima & Minima — SDT Failure & Optimisation Traps
🤯 T4. For f(x) = x⁴, find the nature of the critical point at x = 0 using both tests.
f′(x) = 4x³. f′(0) = 0 → x = 0 is a critical point. SDT attempt: f′′(x) = 12x². f′′(0) = 0 → inconclusive! FDT: For x < 0: f′(x) = 4x³ < 0 (decreasing). For x > 0: f′(x) > 0 (increasing).
f′ changes − to + → local minimum at x = 0. f(0) = 0. This is the classic SDT failure example. When f′′(c) = 0, always fall back to the FDT.
🤯 T5. Among all rectangles with perimeter 40 cm, find the one with maximum area.
Let length = l, width = w. Constraint: 2l + 2w = 40 → l + w = 20 → w = 20−l.
Area: A = l ⋅ w = l(20−l) = 20l − l².
dA/dl = 20 − 2l = 0 → l = 10. d²A/dl² = −2 < 0 → maximum.
w = 20 − 10 = 10. The rectangle is a square of side 10 cm with area 100 cm². General principle: Among all rectangles with fixed perimeter, the square has the maximum area.
🤯 T6. Find the point on the curve y = x² + 1 closest to (0, 2).
Distance² from (x, x²+1) to (0, 2): D² = x² + (x²+1−2)² = x² + (x²−1)².
Let f = x² + x⁴ − 2x² + 1 = x⁴ − x² + 1.
f′ = 4x³ − 2x = 2x(2x² − 1) = 0 → x = 0 or x = ±1/√2.
f′′ = 12x² − 2. At x=0: f′′=−2 < 0 (local max of D² → farthest point).
At x=1/√2: f′′=12(1/2)−2=4 > 0 → local min of D² → closest point.
x = ±1/√2, y = 1/2 + 1 = 3/2. Closest points: (±1/√2, 3/2).
4. Optimisation — Finding the Best Value
4.1
Global Extrema on Closed Intervals & Word Problems
For a closed interval [a, b]: evaluate at all critical points AND both endpoints
⚡ Global Maximum/Minimum on [a, b]
For a continuous function f on a closed interval [a, b]:
Step 1: Find all critical points in (a, b) by solving f′(x) = 0.
Step 2: Evaluate f at:
• All critical points x₁, x₂, … in (a, b)
• Both endpoints x = a and x = b
Step 3: The LARGEST of all these values = Global Maximum
The SMALLEST of all these values = Global Minimum
Note: Local extrema may not be global extrema.
The global max/min could be at an endpoint — always check endpoints!
The closed interval method is the most reliable and complete method. It requires no assumption about the shape of the function — just evaluate everywhere that matters.
Standard Optimisation Results
Fixed perimeter → max area: square
Fixed area → min perimeter: square
AM ≥ GM: x + 1/x ≥ 2 (min = 2 at x=1)
Rectangle in semicircle: optimal is half a square
Cylinder with fixed volume: r = h/2 gives min surface area
Optimisation Problem Steps
1. Identify the quantity to maximise/minimise
2. Write it as a function of ONE variable (use constraint to eliminate others)
3. Find the derivative, set = 0, solve
4. Verify using second derivative test
5. State the answer with units
📝 TOPIC-WISE PYQ
Optimisation — NDA-Pattern Questions
Q10. A farmer has 200 m of fencing. What is the maximum area of a rectangular field he can enclose?
(a) 1000 m² (b) 2000 m² (c) 2500 m² (d) 5000 m²
Answer: (c) 2500 m²
Perimeter: 2(l+w) = 200 → l+w = 100. Area A = lw = l(100−l).
dA/dl = 100−2l = 0 → l = 50. w = 50. A = 50×50 = 2500 m².
Q11. The minimum value of f(x) = x + 4/x for x > 0 is:
Q12. Two numbers sum to 8. What is the maximum value of their product?
(a) 12 (b) 16 (c) 24 (d) 8
Answer: (b) 16
Let numbers be x and 8−x. P = x(8−x) = 8x−x².
dP/dx = 8−2x = 0 → x = 4. d²P/dx² = −2 < 0 → max.
Max product = 4(4) = 16. (By AM≥GM: product max when numbers are equal.)
📝 Master Formula Sheet — MN13 Application of Derivatives
All critical formulae for rapid pre-exam revision.
⎯ Tangent & Normal
Slope of tangent mₜ = dy/dx|(x₀,y₀)
Slope of normal mₙ = −1/mₜ
Tangent: y−y₀ = mₜ(x−x₀)
Normal: y−y₀ = (−1/mₜ)(x−x₀)
Orthogonal curves: m₁⋅m₂ = −1
⇕ Monotonicity
f′(x) > 0 on (a,b) → increasing
f′(x) < 0 on (a,b) → decreasing
f′(x) = 0 → critical point
Sign chart: test each interval
Strictly monotone iff f′ > 0 or f′ < 0
📈 FDT — First Derivative Test
Solve f′(x) = 0 for critical pts
f′: + to − → local max
f′: − to + → local min
No sign change → inflection
Always works, never inconclusive
∂² SDT — Second Derivative Test
f′(c)=0 and f′′(c)<0 → local max
f′(c)=0 and f′′(c)>0 → local min
f′′(c)=0 → inconclusive, use FDT
f′′>0: concave up (cup); f′′<0: concave down (cap)
⚙ Global Extrema on [a,b]
Evaluate f at all critical points in (a,b)
Evaluate f at endpoints a and b
Largest = global max; smallest = global min
Don’t forget endpoints!
🏢 Key Optimisation Results
Fixed perimeter P → max area: square (side P/4)
x + y = k → max xy when x=y=k/2
x + 1/x ≥ 2 (min 2 at x=1)
sin x + cos x ≤ √2 (max √2 at x=π/4)
⚡ Quick Revision Booster — MN13 Application of Derivatives
⎯ Tangent/Normal
Compute dy/dx, substitute point
Normal slope = −1/(tangent slope)
Both pass through same point (x₀, y₀)
Tangent || x-axis → dy/dx = 0
Tangent || y-axis → dx/dy = 0
⇕ Monotonicity Steps
Find f′(x)
Solve f′ = 0 for critical points
Mark on number line
Test a point in each interval
+ increasing, − decreasing
📈 Max/Min Classification
FDT: look at sign change of f′
SDT: f′′ sign at critical point
If f′′ = 0: fall back to FDT
Cup (f′′>0) = min; Cap (f′′<0) = max
⚙ Global Extrema
Find ALL critical points in (a,b)
Evaluate at critical pts + endpoints
Compare all values
Largest = global max; smallest = global min
Endpoint might be the answer!
🏢 Optimisation
Set up objective function
Reduce to one variable via constraint
Differentiate, set = 0
Verify max or min with f′′
Fixed perimeter → square maximises area
🚨 Critical Exam Traps
SDT fails when f′′ = 0 (x⁴ at x=0)
Endpoints not in critical points — check separately
Local max ≠ global max
Tangent || x-axis ⇔ f′ = 0 (NOT f=0)
Orthogonal curves: m₁m₂=−1 (not m₁=m₂)
Increasing on interval uses STRICT inequality f′>0
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