Q.The length $x$ of a rectangle is decreasing at the rate of $5 \text{ cm/minute}$ and the width $y$ is increasing at the rate of $4 \text{ cm/minute}$. When $x = 8 \text{cm}$ and $y = 6 \text{cm}$, find the rates of change of

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1. COMEDK 2026Set 2026-M1 markMCQ
   Q.A square plate is contracting at a uniform rate of $2 \mathrm{~cm}^2 / \mathrm{min}$. The rate at which the perimeter is decreasing when the side of the square is 16 cm is: (A) $\frac{1}{8} \mathrm{~cm} / \mathrm{min}$ (B) $\frac{1}{4} \mathrm{~cm} / \mathrm{min}$ (C) $16 \mathrm{~cm} / \mathrm{min}$ (D) $32 \mathrm{~cm} / \mathrm{min}$

   ›Reveal solutionSolution

   The area decreases at a constant rate; we relate the side length to area, differentiate with respect to time, and then find the perimeter’s rate of change. The perimeter decreases at $\frac{1}{4}$ cm/min when the side is 16 cm.

   We have a square plate whose area is shrinking at a steady rate of $2\ \text{cm}^2/\text{min}$. The question asks: at the moment the side length is 16 cm, how fast is the perimeter decreasing?

   The key idea is related rates: we connect the changing area to the changing side length, then connect the side length to the perimeter. Because the contraction is uniform, the side length shrinks at a rate that depends on the current side length.

   ---

   1. Define variables and given rate Let $s$ be the side length (in cm) and $A$ the area (in cm²). For a square:

   $A = s^2$

   We are told:

   $\frac{dA}{dt} = -2 \quad (\text{negative because area is decreasing})$

   2. Relate the rates of area and side Differentiate $A = s^2$ with respect to time $t$:

   $\frac{dA}{dt} = 2s \cdot \frac{ds}{dt}$

   Substitute the known rate:

   $-2 = 2s \cdot \frac{ds}{dt}$

   Solve for $\frac{ds}{dt}$:

   $\frac{ds}{dt} = -\frac{1}{s}$

   At the instant $s = 16$ cm:

   $\frac{ds}{dt} = -\frac{1}{16}\ \text{cm/min}$

   The negative sign confirms the side length is decreasing.

   3. Find the perimeter’s rate of change Perimeter $P = 4s$. Differentiate:

   $\frac{dP}{dt} = 4 \cdot \frac{ds}{dt}$

   Plug in $\frac{ds}{dt} = -\frac{1}{16}$:

   $$$$

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2. COMEDK 2025Set 2025-E1 markMCQ
   Q.If the length of the diagonal of a square is increasing at the rate of $0.1 \mathrm{~cm} / \mathrm{sec}$. What is the rate of increase of its area when the side is $\frac{15}{\sqrt{2}} \mathrm{~cm}$ ? (A) $3 \mathrm{~cm}^2 / \mathrm{sec}$ (B) $0.15 \mathrm{~cm}^2 / \mathrm{sec}$ (C) $1.5 \mathrm{~cm}^2 / \mathrm{sec}$ (D) $3 \sqrt{2} \mathrm{~cm}^2 / \mathrm{sec}$

   ›Reveal solutionSolution

   The key idea is to relate the side length and diagonal of a square, then use the chain rule to connect the rates of change of the diagonal and the area. The area increases at $1.5 \, \text{cm}^2/\text{sec}$ when the side is $\frac{15}{\sqrt{2}}$ cm, so the correct option is (C).

   We are told the diagonal of a square is increasing at a constant rate of $0.1 \, \text{cm/s}$. We need the rate of increase of the area when the side length is $\frac{15}{\sqrt{2}} \, \text{cm}$. The natural approach is to express the area in terms of the diagonal, then differentiate with respect to time.

   Concept and intuition:

   For a square, the diagonal $d$ and side $s$ are related by $d = s\sqrt{2}$. The area $A = s^2$. If we know how fast $d$ changes, we can find how fast $s$ changes, and then how fast $A$ changes. Alternatively, we can directly relate $A$ to $d$: since $s = d/\sqrt{2}$, then $A = (d/\sqrt{2})^2 = d^2/2$. Differentiating this gives $dA/dt = d \cdot (dd/dt)$. This is simpler because we don't need to find $ds/dt$ separately.

   Let's work through step by step.

   1. Relate area to diagonal. For a square with side $s$, diagonal $d = s\sqrt{2}$ and area $A = s^2$. Substituting $s = d/\sqrt{2}$ gives:

   $A = \left(\frac{d}{\sqrt{2}}\right)^2 = \frac{d^2}{2}.$

   2. Differentiate with respect to time. Using the chain rule:

   $\frac{dA}{dt} = \frac{d}{dt}\left(\frac{d^2}{2}\right) = \frac{1}{2} \cdot 2d \cdot \frac{dd}{dt} = d \cdot \frac{dd}{dt}.$

   We are given $\frac{dd}{dt} = 0.1 \, \text{cm/s}$.

   3. Find the diagonal when the side is $\frac{15}{\sqrt{2}}$ cm. Since $d = s\sqrt{2}$, …

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3. COMEDK 2024Set 2024-M1 markMCQ
   Q.For a given curve $y=2 x-x^2$, when $x$ increases at the rate of 3 units/sec, then how does the slope of the curve change? (A) Decreasing at 3 units/sec (B) Increasing at 3 units/sec (C) Decreasing at 6 units/sec (D) Increasing at 6 units/sec

   ›Reveal solutionSolution

   The slope of the curve is given by the derivative $\frac{dy}{dx} = 2 - 2x$, and its rate of change with respect to time is $\frac{d}{dt}(\text{slope}) = -2 \cdot \frac{dx}{dt}$. With $\frac{dx}{dt} = 3$ units/sec, the slope decreases at 6 units/sec, so the answer is (C).

   ---

   The key idea is that we are not asked for the slope itself, but for how fast the slope changes over time. That means we need the time derivative of the slope, using the chain rule, because the slope depends on $x$, and $x$ itself changes with time.

   Concept & Intuition:

   Imagine a point moving along the parabola $y = 2x - x^2$. As $x$ increases steadily (3 units every second), the slope of the tangent line at the moving point changes. The slope is $m = 2 - 2x$, a linear function of $x$. If $x$ increases, $m$ decreases because of the $-2x$ term. The question is: how fast does $m$ decrease? That’s just the derivative of $m$ with respect to time: $\frac{dm}{dt} = \frac{dm}{dx} \cdot \frac{dx}{dt}$.

   ---

   Step-by-step solution:

   1. Find the slope of the curve as a function of $x$. The slope at any point is the derivative $\frac{dy}{dx}$:

   $y = 2x - x^2 \quad \Rightarrow \quad \frac{dy}{dx} = 2 - 2x.$

   So the slope $m(x) = 2 - 2x$.

   2. We need the rate of change of the slope with respect to time. That is $\frac{dm}{dt}$. Since $m$ depends on $x$, and $x$ depends on $t$, use the chain rule:

   $\frac{dm}{dt} = \frac{dm}{dx} \cdot \frac{dx}{dt}.$

   3. Compute $\frac{dm}{dx}$. From $m = 2 - 2x$, differentiate with respect to $x$:

   $\frac{dm}{dx} = -2.$

   This is constant — the slope of the curve changes linearly with $x$, so its rate of change per unit $x$ is always $-2$.

   4. Use the given $\frac{dx}{dt}$. …

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4. COMEDK 2026Set 2026-M1 markMCQ
   Q.If $\mathbf{3 ~ c m} / \mathbf{s}$ is the rate at which the side of an equilateral triangle increases, then the rate of change of area, when the side is $\mathbf{1 2 ~ c m}$ is: (A) $9 \sqrt{3} \mathrm{~cm}^2 / \mathrm{s}$ (B) $18 \mathrm{~cm}^2 / \mathrm{s}$ (C) $6 \sqrt{3} \mathrm{~cm}^2 / \mathrm{s}$ (D) $18 \sqrt{3} \mathrm{~cm}^2 / \mathrm{s}$

   ›Reveal solutionSolution

   The area of an equilateral triangle depends on its side length; by differentiating the area formula with respect to time, we find the rate of change of area when the side is 12 cm is $18\sqrt{3} \, \text{cm}^2/\text{s}$, which corresponds to option (D).

   We are told the side length $s$ of an equilateral triangle increases at a constant rate: $\frac{ds}{dt} = 3$ cm/s. We need the rate of change of the area $A$ when $s = 12$ cm. The key idea is to relate $A$ to $s$ using geometry, then differentiate with respect to time $t$ using the chain rule — this turns a static formula into a dynamic relationship.

   1. Area of an equilateral triangle in terms of its side For an equilateral triangle of side $s$, the height is $\frac{\sqrt{3}}{2}s$ (by splitting it into two 30-60-90 right triangles). The area is

   $A = \frac{1}{2} \cdot \text{base} \cdot \text{height} = \frac{1}{2} \cdot s \cdot \frac{\sqrt{3}}{2}s = \frac{\sqrt{3}}{4} s^2.$

   2. Differentiate with respect to time Both $A$ and $s$ are functions of time $t$. Using the chain rule:

   $\frac{dA}{dt} = \frac{d}{dt}\left( \frac{\sqrt{3}}{4} s^2 \right) = \frac{\sqrt{3}}{4} \cdot 2s \cdot \frac{ds}{dt} = \frac{\sqrt{3}}{2} s \cdot \frac{ds}{dt}.$

   3. Substitute the given values We have $\frac{ds}{dt} = 3$ cm/s and $s = 12$ cm at the moment of interest. $$$$

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5. COMEDK 2024Set 2024-M1 markMCQ
   Q.The side of an equilateral triangle expands at the rate of $\sqrt{3} \mathrm{~cm} / \mathrm{sec}$. When the side is $12 \mathrm{~cm}$, the rate of increase of its area is (A) $18 \mathrm{~cm}^2 / \mathrm{sec}$ (B) $12 \mathrm{~cm}^2 / \mathrm{sec}$ (C) $10 \mathrm{~cm}^2 / \mathrm{sec}$ (D) $3 \sqrt{3} \mathrm{~cm}^2 / \mathrm{sec}$

   ›Reveal solutionSolution

   The area of an equilateral triangle is $A = \frac{\sqrt{3}}{4} s^2$. Differentiating with respect to time gives $\frac{dA}{dt} = \frac{\sqrt{3}}{2} s \frac{ds}{dt}$. Substituting $s = 12$ cm and $\frac{ds}{dt} = \sqrt{3}$ cm/s yields $\frac{dA}{dt} = 18$ cm²/s, so the correct option is (A).

   Concept & Intuition

   This is a classic related rates problem. The key idea: when a geometric shape changes size, its area changes at a rate that depends on both its current dimensions and how fast those dimensions are changing. Here, the side length grows at a constant speed, but the area grows faster as the side gets longer because area depends on the square of the side. We connect the rates using calculus — specifically, implicit differentiation with respect to time.

   Step-by-step solution

   1. Write the formula for the area of an equilateral triangle. For an equilateral triangle of side $s$, the area is

   $A = \frac{\sqrt{3}}{4} s^2.$

   (Derivation: height $= \frac{\sqrt{3}}{2}s$, so area $= \frac12 \cdot \text{base} \cdot \text{height} = \frac12 \cdot s \cdot \frac{\sqrt{3}}{2}s = \frac{\sqrt{3}}{4}s^2$.)

   2. Differentiate both sides with respect to time $t$. Since $s$ changes with time, $A$ also changes. Using the chain rule:

   $\frac{dA}{dt} = \frac{\sqrt{3}}{4} \cdot 2s \cdot \frac{ds}{dt} = \frac{\sqrt{3}}{2} s \frac{ds}{dt}.$

   This equation tells us the rate of change of area at any instant, given the side length $s$ and its rate of change $\frac{ds}{dt}$.

   3. Plug in the known values. We are given:
      • $\frac{ds}{dt} = \sqrt{3}$ cm/s (the side expands at this rate),
      • $s = 12$ cm (the side length at the moment we care about). Substituting: …

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6. COMEDK 2026Set 2026-A1 markMCQ
   Q.An open hemispherical storage tank has radius 13 m . Oil flows into the tank such that the depth ' $\boldsymbol{h}$ ' of oil in the tank changes at the rate of $3 \mathrm{~m} / \mathrm{hr}$. When the depth $\boldsymbol{h}=1 \mathrm{~m}$, the rate of change of the area of the top surface of the oil is (A) $72 \pi \mathrm{~m}^2 / \mathrm{hr}$ (B) $75 \pi \mathrm{~m}^2 / \mathrm{hr}$ (C) $24 \pi \mathrm{~m}^2 / \mathrm{hr}$ (D) $26 \pi \mathrm{~m}^2 / \mathrm{hr}$

   ›Reveal solutionSolution

   The oil surface is a circle of radius $r$ with $r^2=2Rh-h^2$ for a bowl of radius $R$. So area $A=\pi(2Rh-h^2)$ and $\frac{dA}{dt}=\pi(2R-2h)\frac{dh}{dt}$. At $R=13,\ h=1,\ \frac{dh}{dt}=3$ this is $72\pi\ \text{m}^2/\text{hr}$ — option (A).

   Concept & Intuition

   In a hemispherical bowl of radius $R$, the free surface at oil depth $h$ (measured from the lowest point) is a horizontal circle. Its radius $r$ comes from the sphere geometry: the surface is a distance $R-h$ from the centre, so $r^2=R^2-(R-h)^2=2Rh-h^2$. Differentiate the surface area with respect to time and use the given $\frac{dh}{dt}$.

   Step-by-step solution

   1. Radius of the top circle at depth $h$:

   $r^2=R^2-(R-h)^2=2Rh-h^2.$

   2. Area of the top surface: …

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7. COMEDK 2025Set 2025-M1 markMCQ
   Q.Oil from a conical funnel is dripping at the rate of $5 \mathrm{~cm}^3 / \mathrm{s}$. If the radius and height of the funnel are 10 cm and 20 cm respectively, then the rate at which the oil level drops when it is 5 cm from the top is (A) $\frac{8}{45 \pi} \mathrm{~cm} / \mathrm{s}$ (B) $-\frac{2 \pi}{45} \mathrm{~cm} / \mathrm{s}$ (C) $-\frac{4 \pi}{45} \mathrm{~cm} / \mathrm{s}$ (D) $-\frac{4}{45 \pi} \mathrm{~cm} / \mathrm{s}$

   ›Reveal solutionSolution

   The rate at which the oil level drops is found by relating the volume of a cone to its height using similar triangles, then differentiating with respect to time. The answer is $-\frac{4}{45\pi} \text{ cm/s}$, which corresponds to option (D).

   ---

   Concept & Intuition

   This is a classic related rates problem. Oil is draining from a conical funnel, so the volume is decreasing at a known rate ($dV/dt = -5$ cm³/s). We want the rate at which the height of the oil changes ($dh/dt$) when the oil is at a particular depth. The key twist: as the oil level drops, the radius of the oil's surface also shrinks, because the funnel is conical. The radius and height of the oil are not independent — they are linked by the geometry of the cone (similar triangles). So we first express volume purely in terms of height, then differentiate.

   ---

   Step-by-step solution

   1. Set up the geometry. The funnel is a right circular cone with radius $R = 10$ cm and height $H = 20$ cm. At any moment, the oil forms a smaller cone of height $h$ (measured from the tip) and radius $r$. By similar triangles:

   $\frac{r}{h} = \frac{R}{H} = \frac{10}{20} = \frac{1}{2}$

   So $r = \frac{h}{2}$.

   2. Write the volume of oil in terms of $h$. Volume of a cone: $V = \frac{1}{3}\pi r^2 h$. Substitute $r = h/2$:

   $V = \frac{1}{3}\pi \left(\frac{h}{2}\right)^2 h = \frac{1}{3}\pi \cdot \frac{h^2}{4} \cdot h = \frac{\pi}{12} h^3$

   3. Differentiate with respect to time $t$. Using the chain rule:

   $\frac{dV}{dt} = \frac{\pi}{12} \cdot 3h^2 \cdot \frac{dh}{dt} = \frac{\pi}{4} h^2 \frac{dh}{dt}$

   4. Plug in known values. We are told the oil is dripping out at $5$ cm³/s, so $dV/dt = -5$ (negative because volume is decreasing). The oil is 5 cm from the top of the funnel. Since the funnel is 20 cm tall, the oil height from the tip is $h = 20 - 5 = 15$ cm. …

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8. COMEDK 2025Set 2025-M1 markMCQ
   Q.A spherical snowball is melting such that its volume is decreasing at the rate of $1 \mathrm{~cm}^3 / \mathrm{min}$. The rate at which the diameter is decreasing when the diameter is 10 cm is (A) $\frac{11}{75 \pi} \mathrm{~cm} / \mathrm{min}$ (B) $\frac{1}{50 \pi} \mathrm{~cm} / \mathrm{min}$ (C) $\frac{2}{75 \pi} \mathrm{~cm} / \mathrm{min}$ (D) $\frac{1}{25 \pi} \mathrm{~cm} / \mathrm{min}$

   ›Reveal solutionSolution

   We relate the rate of change of volume to the rate of change of diameter using the formula for the volume of a sphere and implicit differentiation. The diameter decreases at $\frac{1}{50\pi}$ cm/min when the diameter is 10 cm, so the correct option is (B).

   ---

   Concept and intuition:

   This is a classic related rates problem. The snowball’s volume shrinks at a known constant rate, and we want how fast its diameter shrinks at a particular instant. The key is to connect volume $V$ and diameter $D$ through the sphere’s volume formula, then differentiate both sides with respect to time $t$. Because we know $\frac{dV}{dt}$ and want $\frac{dD}{dt}$, we just substitute the given diameter and solve.

   ---

   Step-by-step solution:

   1. Write the volume in terms of diameter. The volume of a sphere of radius $r$ is $V = \frac{4}{3}\pi r^3$. Since the diameter $D = 2r$, we have $r = D/2$. Substituting:

   $V = \frac{4}{3}\pi \left(\frac{D}{2}\right)^3 = \frac{4}{3}\pi \cdot \frac{D^3}{8} = \frac{\pi}{6} D^3.$

   This expresses $V$ directly as a function of $D$, which is convenient because we want $\frac{dD}{dt}$.

   2. Differentiate with respect to time $t$. Both $V$ and $D$ depend on $t$, so we use implicit differentiation:

   $\frac{dV}{dt} = \frac{\pi}{6} \cdot 3D^2 \cdot \frac{dD}{dt} = \frac{\pi}{2} D^2 \frac{dD}{dt}.$

   3. Plug in known values. We are told $\frac{dV}{dt} = -1$ cm³/min (negative because volume is decreasing). At the instant of interest, $D = 10$ cm. Substitute: $-1 = \frac{\pi}{2} (10)^2 \cdot \frac{dD}{dt} = \frac{\pi}{2} \cdot 100 \cdot \frac{dD}{dt} = 50\pi \cdot \frac{dD}{dt}.$ …

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9. KCET 2023Set A-21 markMCQ
   Q.A particle moves along the curve $\frac{x^2}{16} + \frac{y^2}{4} = 1$. When the rate of change of abscissa is 4 times that of its ordinate, then the quadrant in which the particle lies is (A) II or IV (B) III or IV (C) II or III (D) I or III

   ›Reveal solutionSolution

   The problem uses implicit differentiation on an ellipse to relate $\frac{dx}{dt}$ and $\frac{dy}{dt}$. Setting $\frac{dx}{dt} = 4\frac{dy}{dt}$ leads to a relation between $x$ and $y$, which together with the ellipse equation gives the possible quadrants. The particle lies in quadrants II or IV.

   The curve is an ellipse centered at the origin. The abscissa is $x$, the ordinate is $y$. We are told that the rate of change of $x$ is 4 times the rate of change of $y$ — that is, $\frac{dx}{dt} = 4\frac{dy}{dt}$. The question asks: in which quadrants can the particle be when this condition holds?

   The key is to connect the rates through the geometry of the ellipse. Differentiating the ellipse equation with respect to time gives a relation between $x$, $y$, $\frac{dx}{dt}$, and $\frac{dy}{dt}$. Substituting the given rate condition then yields a simple linear relation between $x$ and $y$. The quadrants are determined by the signs of $x$ and $y$ that satisfy both this relation and the ellipse equation.

   1. Differentiate the ellipse equation implicitly. The curve is $\frac{x^2}{16} + \frac{y^2}{4} = 1$. Differentiating both sides with respect to $t$:

   $\frac{2x}{16}\frac{dx}{dt} + \frac{2y}{4}\frac{dy}{dt} = 0$

   Simplify:

   $\frac{x}{8}\frac{dx}{dt} + \frac{y}{2}\frac{dy}{dt} = 0$

   2. Substitute the given rate condition. We have $\frac{dx}{dt} = 4\frac{dy}{dt}$. Replace $\frac{dx}{dt}$:

   $\frac{x}{8}\left(4\frac{dy}{dt}\right) + \frac{y}{2}\frac{dy}{dt} = 0$

   $\frac{x}{2}\frac{dy}{dt} + \frac{y}{2}\frac{dy}{dt} = 0$

   Factor $\frac{1}{2}\frac{dy}{dt}$:

   $\frac{1}{2}\frac{dy}{dt}(x + y) = 0$

   3. Interpret the equation. Since the particle is moving, $\frac{dy}{dt}$ is not zero (otherwise the condition $\frac{dx}{dt}=4\frac{dy}{dt}$ would force both rates to zero, meaning the particle is stationary — a trivial case not intended). Therefore:

   $x + y = 0 \quad \Rightarrow \quad y = -x$

   So the particle lies on the line $y = -x$ at the moment the condition holds.

   4. Find the intersection points of the line and the ellipse. Substitute $y = -x$ into $\frac{x^2}{16} + \frac{y^2}{4} = 1$:

   $\frac{x^2}{16} + \frac{(-x)^2}{4} = 1$

   $\frac{x^2}{16} + \frac{x^2}{4} = 1$

   Common denominator 16:

   $$$$

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10. COMEDK 2025Set 2025-E1 markMCQ
    Q.A man is moving away from a tower 41.6 m high at a rate of $2 \mathrm{~m} / \mathrm{s}$. If the eyelevel of the man is 1.6 m above the ground, then the rate at which the angle of elevation of the top of the tower changes, when he is at a distance of 30 m from the foot of the tower is : (A) $-\frac{4}{125} \mathrm{rad} / \mathrm{sec}$ (B) $\frac{4}{625} \mathrm{rad} / \mathrm{sec}$ (C) $-\frac{2}{125} \mathrm{rad} / \mathrm{sec}$ (D) $\frac{1}{625} \mathrm{rad} / \mathrm{sec}$

    ›Reveal solutionSolution

    The angle of elevation decreases as the man moves away; using related rates and the tangent function, the rate is found to be $-\frac{4}{125}$ rad/s, so the correct option is (A).

    We have a tower of height 41.6 m, and the man’s eye level is 1.6 m above ground. So the effective height of the tower above the man’s eye level is $41.6 - 1.6 = 40$ m. The man moves away from the tower at 2 m/s. We need the rate of change of the angle of elevation $\theta$ when his horizontal distance from the foot is 30 m.

    Concept and intuition:

    The angle of elevation $\theta$ satisfies $\tan\theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{40}{x}$, where $x$ is the horizontal distance from the man to the tower. As $x$ increases, $\theta$ decreases, so we expect a negative rate. Differentiating with respect to time gives a relation between $\frac{d\theta}{dt}$ and $\frac{dx}{dt}$. This is a classic related-rates problem: we know $\frac{dx}{dt} = 2$ m/s, and we want $\frac{d\theta}{dt}$ at $x = 30$.

    Step-by-step solution:

    1. Set up the relationship. Let $x$ be the distance from the man to the foot of the tower. The effective height above eye level is $H = 40$ m. Then

    $\tan\theta = \frac{40}{x}.$

    2. Differentiate implicitly with respect to time $t$. Using the chain rule:

    $\sec^2\theta \cdot \frac{d\theta}{dt} = -\frac{40}{x^2} \cdot \frac{dx}{dt}.$

    Here $\frac{dx}{dt} = 2$ m/s (positive because distance increases).

    3. Find $\sec^2\theta$ at the given instant. When $x = 30$ m,

    $\tan\theta = \frac{40}{30} = \frac{4}{3}.$

    Recall $\sec^2\theta = 1 + \tan^2\theta = 1 + \left(\frac{4}{3}\right)^2 = 1 + \frac{16}{9} = \frac{25}{9}$.

    4. Substitute into the differentiated equation. $$$$

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11. COMEDK 2023Set 2023-E1 markMCQ
    Q.The altitude of a cone is $20 \mathrm{~cm}$ and its semi vertical angle is $30^{\circ}$. If the semi vertical angle is increasing at the rate of $2^0$ per second, then the radius of the base is increasing at the rate of (A) $160 \mathrm{~cm} / \mathrm{sec}$ (B) $10 \mathrm{~cm} / \mathrm{sec}$ (C) $\frac{160}{3} \mathrm{~cm} / \mathrm{sec}$ (D) $30 \mathrm{~cm} / \mathrm{sec}$

    ›Reveal solutionSolution

    (Note on units: the paper quotes the angular rate as '2 degrees per second' but the options are only consistent with treating that rate as 2 in the same angular unit used for the derivative - i.e. the option set is built on dr/dt = h sec^2 a * 2 = 160/3. Converting 2 degrees to radians would give 20*(4/3)*(pi/90) ~ 0.93 cm/s, which matches none of the four options. The intended and only available answer is 160/3 cm/sec.)

    Concept: related rates on a cone. With altitude h fixed and semi-vertical angle alpha varying, r = h tan alpha, so dr/dt = h sec^2(alpha) * d(alpha)/dt.

    Given h = 20 cm, alpha = 30 degrees, d(alpha)/dt = 2 per second.

    sec^2(30) = 1/cos^2(30) = 1/(3/4) = 4/3.

    dr/dt = 20 * (4/3) * 2 = 160/3 cm/sec. …

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12. COMEDK 2025Set 2025-A1 markMCQ
    Q.$x=a(\theta+\sin \theta)$ and $y=a(1-\cos \theta)$ represents the equation of a curve. If $\theta$ changes at a constant rate $k$ then the rate of change of the slope of the tangent to the curve at $\theta=\frac{\pi}{3}$ is (A) $2 k$ (B) $\frac{k}{3}$ (C) $\frac{2 k}{\sqrt{3}}$ (D) $\frac{2 k}{3}$

    ›Reveal solutionSolution

    The problem asks for the rate of change of the slope of the tangent, not the slope itself. We find $\frac{dy}{dx}$ in terms of $\theta$, then differentiate with respect to time using the chain rule, using $\frac{d\theta}{dt}=k$. At $\theta=\frac{\pi}{3}$, the result simplifies to $\frac{2k}{3}$, so the correct option is (D).

    We are given a cycloid-like parametric curve:

    $x = a(\theta + \sin\theta)$, $y = a(1 - \cos\theta)$, with $\theta$ increasing at constant rate $k$, i.e. $\frac{d\theta}{dt} = k$.

    The slope of the tangent is $\frac{dy}{dx}$. But the question asks for the rate of change of that slope with respect to time — that is $\frac{d}{dt}\left(\frac{dy}{dx}\right)$. This is a classic related-rates problem in parametric form.

    ---

    1. Find the slope in terms of $\theta$

    First compute derivatives with respect to $\theta$:

    $\frac{dx}{d\theta} = a(1 + \cos\theta), \quad \frac{dy}{d\theta} = a\sin\theta$

    Then the slope is:

    $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{a\sin\theta}{a(1+\cos\theta)} = \frac{\sin\theta}{1+\cos\theta}$

    Using the identity $\sin\theta = 2\sin(\theta/2)\cos(\theta/2)$ and $1+\cos\theta = 2\cos^2(\theta/2)$, this simplifies to:

    $\frac{dy}{dx} = \frac{2\sin(\theta/2)\cos(\theta/2)}{2\cos^2(\theta/2)} = \tan\frac{\theta}{2}$

    So the slope at any $\theta$ is simply $\tan(\theta/2)$.

    ---

    2. Differentiate the slope with respect to time

    We want $\frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{d}{dt}\left(\tan\frac{\theta}{2}\right)$.

    By the chain rule:

    $\frac{d}{dt}\left(\tan\frac{\theta}{2}\right) = \sec^2\left(\frac{\theta}{2}\right) \cdot \frac{1}{2} \cdot \frac{d\theta}{dt}$

    Given $\frac{d\theta}{dt} = k$, we have:

    $$$$

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