JEE Maths›Application of Derivatives›Visual Solution
Visual SolutionTricky
Question
A ladder of length 55 m is leaning against a wall. The bottom slides away at 22 cm/s. When the bottom is 44 m from the wall, the rate at which the area of the triangle formed by the ladder, wall, and floor is changing (in cm2^2/s) is:
Solution Path
Pythagorean constraint gives dy/dtdy/dt, product rule on A=xy/2A = xy/2 gives dA/dt=7dA/dt = 7 cm2^2/s.
01Question Setup
1/4
A ladder slides along a wall. Given rates, find dAdt\frac{dA}{dt} of the triangle formed.
dAdt=  ?\frac{dA}{dt} = \;?
02Pythagorean Setup
2/4
x2+y2=L2x^2 + y^2 = L^2. Differentiate: 2xdxdt+2ydydt=02x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0. Find dydt\frac{dy}{dt}.
dydt=−xy⋅dxdt\frac{dy}{dt} = -\frac{x}{y} \cdot \frac{dx}{dt}
03Differentiate AreaKEY INSIGHT
3/4
A=12xyA = \frac{1}{2}xy. By product rule: dAdt=12(xdydt+ydxdt)\frac{dA}{dt} = \frac{1}{2}\left(x\frac{dy}{dt} + y\frac{dx}{dt}\right). Substitute and simplify.
dAdt=12(xdydt+ydxdt)\frac{dA}{dt} = \frac{1}{2}(x\frac{dy}{dt} + y\frac{dx}{dt})
04Final Answer
4/4
Substituting the values gives dAdt=7\frac{dA}{dt} = 7 cm2^2/s.
7\boxed{7}
Concepts from this question3 concepts unlocked
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First Derivative Test

EASY

The sign change of f'(x) around a critical point determines whether it is a local maximum or minimum. If f'(x) changes from positive to negative, the point is a local max; if negative to positive, it is a local min.

f′(x) changes +→−  ⟹  local max,f′(x) changes −→+  ⟹  local minf'(x) \text{ changes } + \to - \implies \text{local max}, \quad f'(x) \text{ changes } - \to + \implies \text{local min}

JEE problems frequently ask to classify critical points of polynomial and trigonometric functions, and the first derivative test works even when the second derivative is zero or hard to compute

Local extrema classificationPolynomial optimizationCurve sketching
Practice (14 Qs) →

Tangent and Normal Geometry

EASY

The slope of the tangent to y = f(x) at a point (x_1, y_1) is f'(x_1), and the slope of the normal is -1/f'(x_1). The tangent line equation is y - y_1 = f'(x_1)(x - x_1).

y−y1=f′(x1)(x−x1)(tangent),y−y1=−1f′(x1)(x−x1)(normal)y - y_1 = f'(x_1)(x - x_1) \quad \text{(tangent)}, \quad y - y_1 = -\frac{1}{f'(x_1)}(x - x_1) \quad \text{(normal)}

Tangent and normal equations appear directly in JEE problems involving curves, and are also used to find subtangent, subnormal lengths and angle of intersection between curves

Tangent and normal equationsCurve intersection anglesLength of subtangent/subnormal
Practice (12 Qs) →

Optimization Strategy

TRICKY

Express the quantity to be optimized as a single-variable function using given constraints, differentiate to find critical points where the derivative is zero, then verify using the first or second derivative test whether the critical point gives a maximum or minimum.

dAdx=0 at extremum,verify with d2Adx2≶0\frac{dA}{dx} = 0 \text{ at extremum}, \quad \text{verify with } \frac{d^2A}{dx^2} \lessgtr 0

Word problems on maximizing area, minimizing cost, or optimizing geometric quantities are high-weightage in JEE and follow this systematic approach

Geometric optimizationApplied max/min word problemsRate of change applications
Practice (8 Qs) →
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