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. AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
   Q.A point is moving on the curve $y=x^3-3x^2+2x-1$ and the y-coordinate of the point is increasing at the rate of 6 units per second. When the point is at $(2,-1)$, the rate of change of x-coordinate of the point is (A) 3 (B) $\dfrac{1}{2}$ (C) $-\dfrac{1}{2}$ (D) $-3$

   ›Reveal solutionSolution

   A related-rates problem: knowing $dy/dt$ and the curve's slope at the given point lets us solve directly for $dx/dt$. The answer is $3$.

   Concept and Intuition

   For a point moving along $y=f(x)$, the rates of change of $x$ and $y$ with respect to time are linked through the chain rule $\dfrac{dy}{dt}=f'(x)\dfrac{dx}{dt}$ — the curve's local slope is exactly the conversion factor between the two rates at that instant.

   Step-by-Step Solution

   1. $y=x^3-3x^2+2x-1\Rightarrow\dfrac{dy}{dx}=3x^2-6x+2$.
   2. At the point $(2,-1)$: $\dfrac{dy}{dx}=3(4)-6(2)+2=12-12+2=2$. …

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2. AP EAPCET 2024Set eng-2024-05-23-FN1 markMCQ
   Q.A is a point on the circle with radius 8 and centre at O. A particle P is moving on the circumference of the circle starting from A. M is the foot of the perpendicular from P on OA and $\angle POM=\theta$. When $OM=4$ and $\dfrac{d\theta}{dt}=6$ radians/sec, then the rate of change of PM is (in units/sec) (A) $24\sqrt{3}$ (B) $24$ (C) $15\sqrt{3}$ (D) $48\sqrt{3}$

   ›Reveal solutionSolution

   $M$ is the foot of the perpendicular from $P$ to line $OA$, so $OM=OP\cos\theta$ and $PM=OP\sin\theta$ in the right triangle $OMP$; differentiate $PM$ with respect to time.

   Concept and Intuition

   As $P$ moves around the circle, the right triangle $OMP$ (right-angled at $M$) has hypotenuse $OP=8$ (the radius) fixed, and angle $\theta=\angle POM$ varying with time. So $OM$ and $PM$ are both simple trig functions of $\theta$, and their time-rates follow directly by the chain rule using $\dot\theta$.

   Step-by-Step Solution

   1. In right triangle $OMP$ (right angle at $M$): $OM = OP\cos\theta = 8\cos\theta$ and $PM = OP\sin\theta = 8\sin\theta$.
   2. Given $OM=4$: $8\cos\theta=4 \Rightarrow \cos\theta=\tfrac12 \Rightarrow \theta=60^\circ$, and $\sin\theta=\tfrac{\sqrt3}{2}$.
   3. Differentiate $PM=8\sin\theta$ with respect to time: $\dfrac{d(PM)}{dt}=8\cos\theta\cdot\dfrac{d\theta}{dt}$. …

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3. AP EAPCET 2021Set eng-2021-08-19-AN1 markMCQ
   Q.Let '$x$' and '$y$' be the sides of two squares such that $y=x-x^2$. The rate of change of area of the second square with respect to area of the first square is ______. (A) $1-3x+2x^2$ (B) $1+3x-2x^2$ (C) $2x$ (D) $x+2x^3-3x^2$

   ›Reveal solutionSolution

   Use the chain rule to compute a derivative of one area with respect to another, both parametrized by $x$. Answer: $1-3x+2x^2$.

   Concept and Intuition

   "Rate of change of Area2 w.r.t. Area1" means $\dfrac{d(\text{Area2})}{d(\text{Area1})}$, which by the chain rule equals $\dfrac{d(\text{Area2})/dx}{d(\text{Area1})/dx}$ since both areas are functions of the common parameter $x$.

   Step-by-Step Solution

   1. Area1 $=x^2\Rightarrow\dfrac{d(\text{Area1})}{dx}=2x$.
   2. Area2 $=y^2=(x-x^2)^2\Rightarrow\dfrac{d(\text{Area2})}{dx}=2(x-x^2)(1-2x)$. …

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4. AP EAPCET 2021Set eng-2021-08-20-FN1 markMCQ
   Q.The diameter and altitude of a right circular cone, at a certain instant, were found to be $10\ cm$ and $20\ cm$ respectively. If its diameter is increasing at a rate of $2\ cm/s$, then at what rate must its altitude change, in order to keep its volume constant? (A) $4\ cm/s$ (B) $6\ cm/s$ (C) $-4\ cm/s$ (D) $-8\ cm/s$

   ›Reveal solutionSolution

   This tests related rates: differentiating the volume of a cone with respect to time and setting the total rate of change to zero to keep volume constant. Answer: $-8$ cm/s.

   Concept and Intuition

   When a quantity built from several changing variables must stay constant, differentiate the formula with respect to time and set the total derivative to zero — this links the individual rates of change together.

   Step-by-Step Solution

   1. Diameter $=10$ cm $\Rightarrow$ radius $r=5$ cm; height $h=20$ cm.
   2. Diameter increasing at $2$ cm/s $\Rightarrow \dfrac{d(2r)}{dt}=2 \Rightarrow \dfrac{dr}{dt}=1$ cm/s.
   3. Volume: $V = \dfrac13\pi r^2 h$. Differentiate with respect to $t$: $\dfrac{dV}{dt} = \dfrac13\pi\left(2rh\dfrac{dr}{dt} + r^2\dfrac{dh}{dt}\right)$.
   4. For constant volume, $\dfrac{dV}{dt}=0 \Rightarrow 2rh\dfrac{dr}{dt} + r^2\dfrac{dh}{dt} = 0$. …

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5. AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
   Q.If the surface area of a spherical bubble is increasing at the rate of 4 sq.cm/sec, then the rate of change in its volume (in cubic cm/sec) when its radius is 8 cms is (A) $8$ (B) $12$ (C) $15$ (D) $16$

   ›Reveal solutionSolution

   A related-rates problem: given $\dfrac{dS}{dt}$, find $\dfrac{dV}{dt}$ by eliminating $\dfrac{dr}{dt}$ through the common variable $r$. Answer: $16$ cubic cm/sec.

   Concept and Intuition

   Both surface area $S$ and volume $V$ of a sphere depend on the single variable $r$ (radius), which itself changes with time. The strategy in any related-rates problem is: express both quantities in terms of the shared variable, differentiate each with respect to time using the chain rule, and then eliminate the unknown rate ($\dfrac{dr}{dt}$ here) between the two equations.

   Step-by-Step Solution

   1. Surface area: $S=4\pi r^2$. Differentiating w.r.t. $t$: $\dfrac{dS}{dt}=8\pi r\,\dfrac{dr}{dt}$.
   2. We're given $\dfrac{dS}{dt}=4$, so $8\pi r\,\dfrac{dr}{dt}=4 \Rightarrow \dfrac{dr}{dt}=\dfrac{4}{8\pi r}=\dfrac{1}{2\pi r}$.
   3. Volume: $V=\dfrac43\pi r^3$. Differentiating w.r.t. $t$: $\dfrac{dV}{dt}=4\pi r^2\,\dfrac{dr}{dt}$.
   4. Substitute $\dfrac{dr}{dt}$ from step 2: …

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6. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
   Q.If the rate of increase in the surface area of a cube is $6\ sq.cm./sec$, then the rate of increase in its volume (in $c.c./sec$), when the length of its edge is $12\ cm$, is (A) $6$ (B) $12$ (C) $18$ (D) $9$

   ›Reveal solutionSolution

   A related-rates problem: convert the given rate of change of surface area into the rate of change of the edge length, then into the rate of change of volume.

   Concept and Intuition

   Both surface area and volume of a cube are functions of the single variable $a$ (the edge length), so differentiating each with respect to time and using the chain rule links their rates through $da/dt$ — the one quantity actually changing independently.

   Step-by-Step Solution

   1. Surface area: $S=6a^2\Rightarrow\dfrac{dS}{dt}=12a\dfrac{da}{dt}$.
   2. Given $\dfrac{dS}{dt}=6$: $12a\dfrac{da}{dt}=6\Rightarrow\dfrac{da}{dt}=\dfrac{1}{2a}$.
   3. Volume: $V=a^3\Rightarrow\dfrac{dV}{dt}=3a^2\dfrac{da}{dt}$. …

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7. AP EAPCET 2021Set eng-2021-08-20-AN1 markMCQ
   Q.The volume of a spherical balloon is increasing at the rate of $30\ cc$ per minute. Find the rate of change of surface area of the balloon, when its radius is $6\ cm$. (A) $5\ cm^2.min^{-1}$ (B) $30\ cm^2.min^{-1}$ (C) $10\ cm^2.min^{-1}$ (D) $20\ cm^2.min^{-1}$

   ›Reveal solutionSolution

   A standard related-rates chain: volume rate → radius rate → surface-area rate. The answer is $10\ cm^2/\text{min}$.

   Concept and Intuition

   Both $V$ and $S$ of a sphere depend only on $r$, so their rates of change are linked through $dr/dt$ via the chain rule. First use the given $dV/dt$ to find $dr/dt$ at the specified radius, then plug that into $dS/dt$.

   Step-by-Step Solution

   1. $V=\dfrac43\pi r^3\Rightarrow\dfrac{dV}{dt}=4\pi r^2\dfrac{dr}{dt}$.
   2. Given $\dfrac{dV}{dt}=30\ cc/min$ and $r=6$ cm: $30=4\pi(6)^2\dfrac{dr}{dt}=144\pi\dfrac{dr}{dt}$.
   3. $\dfrac{dr}{dt}=\dfrac{30}{144\pi}=\dfrac{5}{24\pi}$ cm/min.
   4. $S=4\pi r^2\Rightarrow\dfrac{dS}{dt}=8\pi r\dfrac{dr}{dt}$. …

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8. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
   Q.If the velocity of a particle moving on a straight line is proportional to the cube root of its displacement, then its acceleration is (A) constant (B) inversely proportional to its velocity (C) proportional to its velocity (D) proportional to its displacement

   ›Reveal solutionSolution

   Express acceleration as $v\,dv/dx$, substitute the given proportionality, and eliminate $x$ in favour of $v$. Answer: acceleration is inversely proportional to velocity.

   Concept and Intuition

   For motion along a straight line, acceleration can be written as $a = \dfrac{dv}{dt} = v\dfrac{dv}{dx}$ (chain rule through position). Given a relation between $v$ and $x$, this lets us compute $a$ purely as a function of $x$, and then re-express it in terms of $v$ using the same original relation.

   Step-by-Step Solution

   1. Given $v \propto x^{1/3}$, write $v = kx^{1/3}$ for some constant $k$.
   2. $\dfrac{dv}{dx} = \dfrac{k}{3}x^{-2/3}$.
   3. $a = v\dfrac{dv}{dx} = kx^{1/3}\cdot\dfrac{k}{3}x^{-2/3} = \dfrac{k^2}{3}x^{-1/3}$.
   4. From $v=kx^{1/3}$, cube both sides: $v^3 = k^3x \Rightarrow x = \dfrac{v^3}{k^3}$, so $x^{-1/3} = \dfrac{k}{v}$.
   5. Substitute: $a = \dfrac{k^2}{3}\cdot\dfrac{k}{v} = \dfrac{k^3}{3v}$. …

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9. AP EAPCET 2024Set eng-2024-05-22-AN1 markMCQ
   Q.If a man of height 1.8 mt. is walking away from the foot of a light pole of height 6 mt. with a speed of 7 km per hour on a straight horizontal road opposite to the pole, then the rate of change of the length of his shadow is (in kmph) (A) $7$ (B) $5$ (C) $3$ (D) $2$

   ›Reveal solutionSolution

   Similar triangles link shadow length to distance walked; the shadow grows at $3$ kmph.

   Concept and Intuition

   The tip of the shadow, the top of the pole, and the top of the man's head are collinear (that's what casts the shadow). This gives a similar-triangles relationship between the man's distance from the pole and his shadow's length, which can be differentiated with respect to time (related rates).

   Step-by-Step Solution

   1. Let $x$ = distance of the man from the pole, $s$ = length of his shadow. The tip of the shadow is at distance $x+s$ from the pole.
   2. Similar triangles (pole-to-shadow-tip vs man-to-shadow-tip): $\dfrac{\text{pole height}}{x+s}=\dfrac{\text{man height}}{s}\Rightarrow \dfrac{6}{x+s}=\dfrac{1.8}{s}$.
   3. Cross-multiply: $6s=1.8(x+s)=1.8x+1.8s\Rightarrow 4.2s=1.8x\Rightarrow s=\dfrac{1.8}{4.2}x=\dfrac37x$. …

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10. AP EAPCET 2021Set eng-2021-08-23-AN1 markMCQ
    Q.The volume of a spherical ball is increasing at a rate of $4\pi \text{ cm}^3.\text{s}^{-1}$. The rate at which its radius increases, when its volume is $288\pi \text{ cm}^3$, is ______ $\text{cm}.\text{s}^{-1}$. (A) $\dfrac{1}{6}$ (B) $\dfrac{1}{36}$ (C) $\dfrac{1}{9}$ (D) $\dfrac{1}{24}$

    ›Reveal solutionSolution

    Related rates on $V=\tfrac43\pi r^3$ give $\dfrac{dr}{dt}=\dfrac{1}{r^2}$, and at the given volume $r=6$, so $\dfrac{dr}{dt}=\dfrac1{36}$.

    Concept and Intuition

    This is a classic related-rates problem: differentiate the volume formula with respect to time using the chain rule, then substitute the known rate and the radius at the instant of interest (found from the given volume).

    Step-by-Step Solution

    1. $V=\dfrac43\pi r^3$.
    2. Differentiate w.r.t. time: $\dfrac{dV}{dt}=4\pi r^2\dfrac{dr}{dt}$.
    3. Given $\dfrac{dV}{dt}=4\pi$: $4\pi=4\pi r^2\dfrac{dr}{dt} \Rightarrow \dfrac{dr}{dt}=\dfrac{1}{r^2}$.
    4. Find $r$ when $V=288\pi$: $\dfrac43\pi r^3=288\pi \Rightarrow r^3=216 \Rightarrow r=6$. …

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11. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
    Q.If a cylindrical tank of radius 3 m is filled with water at the rate of $\dfrac{3}{2}$ $m^3$/sec, then the rate of change of its water level in (m/sec) is (A) $\dfrac{1}{3\pi}$ (B) $\dfrac{1}{2\pi}$ (C) $\dfrac{1}{\pi}$ (D) $\dfrac{1}{6\pi}$

    ›Reveal solutionSolution

    This tests related rates for a cylinder of fixed radius; the water level rises at $\dfrac{1}{6\pi}$ m/s.

    Concept and Intuition

    Since the radius doesn't change as the tank fills, the volume $V=\pi r^2h$ is a function of $h$ alone (with $r$ a constant), so differentiating with respect to time directly links $dV/dt$ to $dh/dt$ through the constant cross-sectional area $\pi r^2$.

    Step-by-Step Solution

    1. $V = \pi r^2 h$, with $r=3$ constant, so $V=9\pi h$.
    2. Differentiate w.r.t. time: $\dfrac{dV}{dt} = 9\pi\dfrac{dh}{dt}$. …

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12. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
    Q.If the volume of a sphere is increasing at the rate of 12 c.c./sec, then the rate (in sq. cm/sec) at which its surface area is increasing, when the diameter of the sphere is 12 cm is (A) 2 (B) 3 (C) 4 (D) 6

    ›Reveal solutionSolution

    This is a related-rates problem: given how fast volume grows, find how fast surface area grows at a specific radius; the answer is $4\text{ cm}^2/\text{s}$.

    Concept and Intuition

    Both $V$ and $S$ of a sphere depend only on $r$, so differentiating each with respect to time and eliminating $\frac{dr}{dt}$ (found from the volume-rate condition) gives the surface-area rate.

    Step-by-Step Solution

    1. $V=\dfrac43\pi r^3$, so $\dfrac{dV}{dt}=4\pi r^2\dfrac{dr}{dt}$.
    2. Given $\dfrac{dV}{dt}=12$ and diameter $=12\Rightarrow r=6$: $12=4\pi(36)\dfrac{dr}{dt}\Rightarrow \dfrac{dr}{dt}=\dfrac{12}{144\pi}=\dfrac{1}{12\pi}$.
    3. $S=4\pi r^2\Rightarrow \dfrac{dS}{dt}=8\pi r\dfrac{dr}{dt}$. …

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