Q.A particle moves along the curve $6y = x^3 + 2$. Find the points on the curve at which the $y$-coordinate is changing $8$ times as fast as the $x$-coordinate.

Showing the 12 most recent of 17 on this concept.

1. 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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2. 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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3. 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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4. COMEDK 2024Set 2024-A1 markMCQ
   Q.The points on the ellipse $16 x^2+9 y^2=400$ at which the ordinate decreases at the same rate at which the abscissa increases are (A) $\left(3, \frac{16}{3}\right) \text { and }\left(-3,-\frac{16}{3}\right)$ (B) $\left(-3,-\frac{16}{3}\right) \text { and }\left(-3, \frac{16}{3}\right)$ (C) $\left(-3, \frac{16}{3}\right) \text { and }\left(3,-\frac{16}{3}\right)$ (D) $\left(3, \frac{16}{3}\right) \text { and }\left(-3, \frac{16}{3}\right)$

   ›Reveal solutionSolution

   "Ordinate decreases as fast as abscissa increases" means $\frac{dy}{dt}=-\frac{dx}{dt}$; differentiating the ellipse gives $y=\frac{16x}{9}$, and the points are $\left(3,\frac{16}{3}\right)$ and $\left(-3,-\frac{16}{3}\right)$. Option (A).

   The condition — ordinate $y$ decreasing at the same rate the abscissa $x$ increases — means

   $\frac{dy}{dt}=-\frac{dx}{dt}.$

   Differentiate the ellipse $16x^2+9y^2=400$ with respect to $t$:

   $32x\frac{dx}{dt}+18y\frac{dy}{dt}=0.$

   Substitute $\dfrac{dy}{dt}=-\dfrac{dx}{dt}$:

   $32x\frac{dx}{dt}-18y\frac{dx}{dt}=0\;\Rightarrow\;32x=18y\;\Rightarrow\;y=\frac{16x}{9}.$

   Put this into the ellipse:

   $16x^2+9\left(\frac{16x}{9}\right)^2=400\;\Rightarrow\;16x^2+\frac{256x^2}{9}=400.$ …

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5. KCET 2024Set A-11 markMCQ
   Q.The family of curves whose $x$ and $y$ intercepts of a tangent at any point are respectively double the $x$ and $y$ coordinates of that point is (A) $xy = C$ (B) $x^2 + y^2 = C$ (C) $x^2 - y^2 = C$ (D) $\frac{y}{x} = C$

   ›Reveal solutionSolution

   The condition that the $x$ and $y$ intercepts of the tangent are double the coordinates of the point of tangency leads to a differential equation whose solution is $xy = C$, so the correct option is (A).

   The key idea here is to translate a geometric condition about a tangent line into an algebraic relation between the coordinates of the point of tangency and the slope. Once you write the intercepts in terms of the slope and the point, the given condition becomes a differential equation that you can solve.

   Let the curve be $y = f(x)$, and let $P(x, y)$ be any point on it. The tangent at $P$ has slope $m = \frac{dy}{dx}$. The equation of the tangent line is

   $Y - y = m (X - x).$

   1. Find the $x$-intercept. Set $Y = 0$:

   $0 - y = m (X - x) \quad\Rightarrow\quad -y = m(X - x) \quad\Rightarrow\quad X = x - \frac{y}{m}.$

   So the $x$-intercept is $\left(x - \frac{y}{m}, 0\right)$.

   2. Find the $y$-intercept. Set $X = 0$:

   $Y - y = m (0 - x) \quad\Rightarrow\quad Y = y - mx.$

   So the $y$-intercept is $(0, y - mx)$.

   3. Apply the given condition. The $x$-intercept is double the $x$-coordinate of $P$, and the $y$-intercept is double the $y$-coordinate of $P$:

   $x - \frac{y}{m} = 2x \quad\text{and}\quad y - mx = 2y.$

   From the first equation:

   $x - \frac{y}{m} = 2x \quad\Rightarrow\quad -\frac{y}{m} = x \quad\Rightarrow\quad m = -\frac{y}{x}.$

   From the second equation:

   $y - mx = 2y \quad\Rightarrow\quad -mx = y \quad\Rightarrow\quad m = -\frac{y}{x}.$

   Both give the same condition, which is consistent. …

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6. 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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7. 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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8. 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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9. 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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10. COMEDK 2024Set 2024-E1 markMCQ
    Q.The side of a cube is equal to the diameter of a sphere. If the side and radius increase at the same rate then the ratio of the increase of their surface area is (A) $3: \pi$ (B) $\pi: 6$ (C) $2 \pi: 3$ (D) $3: 2 \pi$

    ›Reveal solutionSolution

    The problem asks for the ratio of the rates of increase of surface areas of a cube and a sphere when their side and radius increase at the same rate, given the side equals the sphere’s diameter. The answer is $3 : \pi$, which corresponds to option (A).

    We start by understanding the relationship: the cube’s side length $s$ equals the sphere’s diameter, so $s = 2r$, where $r$ is the sphere’s radius. Both $s$ and $r$ increase at the same rate, meaning $\frac{ds}{dt} = \frac{dr}{dt}$. We want the ratio of the rates of change of their surface areas.

    Concept and intuition:

    Surface area growth depends on both the current size and the rate of change of the linear dimension. Since the cube and sphere have different formulas for surface area, their rates of change will differ even when their linear dimensions grow at the same speed. The key is to differentiate each surface area with respect to time, then substitute the given relationship $s = 2r$ and the equal rate condition.

    ---

    1. Write the surface area formulas.

       • Cube surface area: $A_{\text{cube}} = 6s^2$
       • Sphere surface area: $A_{\text{sphere}} = 4\pi r^2$

    2. Differentiate both with respect to time $t$.

       • $\frac{dA_{\text{cube}}}{dt} = 12s \cdot \frac{ds}{dt}$
       • $\frac{dA_{\text{sphere}}}{dt} = 8\pi r \cdot \frac{dr}{dt}$

    3. Apply the given conditions.

       • The side equals the diameter: $s = 2r$.
       • The rates are equal: $\frac{ds}{dt} = \frac{dr}{dt}$. …

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11. 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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12. 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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