Q.A kite is flying at a height of 3 metres and 5 metres of string is out. If the kite is moving away horizontally at the rate of 200 cm/s, find the rate at which the string is being released.

Showing the 12 most recent of 17 on this concept.

1. AP EAPCET 2021Set eng-2021-10-05-FN1 markMCQ
   Q.A ladder 5 m long is leaning against a wall. If the top of the ladder slides downwards at a rate of $10\ cm.sec^{-1}$, then the rate at which the angle between the floor and the ladder decreases, when the lower end of ladder is 2 m from the wall, is _____ $radian.sec^{-1}$ (A) $\dfrac{1}{10}$ (B) $\dfrac{1}{20}$ (C) $20$ (D) $10$

   ›Reveal solutionSolution

   This is a related-rates problem: relate $x,y,\theta$ via the ladder's fixed length, then differentiate twice (once for $x,y$, once for $\theta$) using the chain rule. The answer is (B).

   Concept and Intuition

   As the top of the ladder slides down, the angle it makes with the floor decreases. Using $x=5\cos\theta$ directly (with $x$ = distance of foot from wall) lets us relate $dy/dt$ to $d\theta/dt$ in one clean step.

   Step-by-Step Solution

   1. Let $x$ = horizontal distance of foot from wall, $y$ = height of top on the wall, $\theta$ = angle between floor and ladder. Then $x=5\cos\theta$, $y=5\sin\theta$.
   2. Given: $\dfrac{dy}{dt}=-10\text{ cm/s}=-0.1$ m/s (height decreasing), at the instant $x=2$ m.
   3. $\cos\theta = \dfrac x5 = \dfrac25$.
   4. From $y=5\sin\theta$: $\dfrac{dy}{dt}=5\cos\theta\,\dfrac{d\theta}{dt}$.
   5. Substitute: $-0.1 = 5\left(\dfrac25\right)\dfrac{d\theta}{dt} = 2\dfrac{d\theta}{dt} \Rightarrow \dfrac{d\theta}{dt}=-0.05 = -\dfrac1{20}\text{ rad/s}$.
   6. The negative sign confirms $\theta$ is decreasing; its rate of decrease is $\dfrac1{20}$ rad/s. …

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2. 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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3. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.If the vertical angle of a cone is $60^\circ$ and the rate of change of its total surface area is $2\sqrt{3}\ cm^2/sec$, then the rate of change of its volume (in $cm^3/sec$) when its radius is $5\ cm$, is (A) $15$ (B) $10$ (C) $5$ (D) $9$

   ›Reveal solutionSolution

   A related-rates problem: using the $60°$ vertical angle to fix $h$ and slant height $l$ in terms of $r$, then chaining $\dfrac{dS}{dt}\to\dfrac{dr}{dt}\to\dfrac{dV}{dt}$ gives $5\ \text{cm}^3/\text{sec}$.

   Concept and Intuition

   When a cone's vertical (apex) angle is fixed, its shape stays similar as it grows — radius and height stay in a fixed ratio determined by the semi-vertical angle. This lets us express both surface area and volume purely in terms of $r$, so a single related-rates chain (through $dr/dt$) connects the given rate of surface-area change to the unknown rate of volume change.

   Step-by-Step Solution

   1. Vertical angle $=60°$, so semi-vertical angle $\alpha=30°$. In the cone's cross-section, $\tan\alpha = r/h$, so $r = h\tan30° = h/\sqrt3$, i.e. $h = r\sqrt3$.
   2. Slant height: $l = \sqrt{r^2+h^2} = \sqrt{r^2+3r^2} = \sqrt{4r^2} = 2r$.
   3. Total surface area: $S = \pi r^2 + \pi r l = \pi r^2 + \pi r(2r) = 3\pi r^2$.
   4. Differentiate: $\dfrac{dS}{dt} = 6\pi r\dfrac{dr}{dt}$.
   5. Given $\dfrac{dS}{dt}=2\sqrt3$ and $r=5$: $2\sqrt3 = 6\pi(5)\dfrac{dr}{dt} = 30\pi\dfrac{dr}{dt}$, so $\dfrac{dr}{dt} = \dfrac{2\sqrt3}{30\pi} = \dfrac{\sqrt3}{15\pi}$.
   6. Volume: $V = \dfrac13\pi r^2 h = \dfrac13\pi r^2(r\sqrt3) = \dfrac{\pi r^3}{\sqrt3}$. …

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4. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
   Q.An aeroplane is flying at a constant speed, parallel to the horizontal ground at a height of 5 kms. A person on the ground observed that the angle of elevation of the plane is changed from 15° to 30° in the duration of 50 seconds, then the speed of the plane (in kmph) is (A) 100 (B) 720 (C) 360 (D) 540

   ›Reveal solutionSolution

   Convert both elevation angles to horizontal distances using $h\cot\theta$; the difference is the distance flown, giving a speed of 720 kmph.

   Concept and Intuition

   For a plane flying level at height $h$, if the angle of elevation from a fixed ground point is $\theta$, the horizontal distance from the point directly below the plane's original position (or rather, the ground projection) satisfies $\tan\theta = h/x \Rightarrow x=h\cot\theta$. As the plane approaches, $\theta$ increases and $x$ decreases — the plane has covered the difference in these horizontal distances.

   Step-by-Step Solution

   1. At $\theta_1=15^\circ$: horizontal distance $x_1=h\cot15^\circ = 5(2+\sqrt3)$ km.
   2. At $\theta_2=30^\circ$: horizontal distance $x_2=h\cot30^\circ=5\sqrt3$ km.
   3. Distance flown $=x_1-x_2=5(2+\sqrt3-\sqrt3)=5\times2=10$ km. …

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5. 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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6. 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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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 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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9. 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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10. AP EAPCET 2021Set eng-2021-08-20-AN1 markMCQ
    Q.A spherical iron ball $10\ cm$ in radius is coated with a layer of ice of uniform thickness, which melts at a rate of $50\ cm^3/min$. When the thickness of the ice is $15\ cm$, the rate at which the thickness of ice decreases is ______ $cm/min$ (A) $\dfrac{5}{6\pi}$ (B) $\dfrac{1}{54\pi}$ (C) $\dfrac{1}{18\pi}$ (D) $\dfrac{1}{36\pi}$

    ›Reveal solutionSolution

    This is a related-rates problem: only the outer radius of the ice matters for relating the rate of volume loss to the rate of thickness loss. The answer is $\dfrac{1}{18\pi}$ cm/min.

    Concept and Intuition

    The iron ball's own radius (10 cm) is fixed and drops out of the derivative — only the total outer radius $R=10+x$, where $x$ is the ice thickness, matters, since $V_{ice}=\tfrac43\pi R^3-\tfrac43\pi(10)^3$ and the constant term vanishes on differentiating. This reduces the problem to the standard "rate of change of a sphere's volume vs. its radius" relation, $dV/dt=4\pi R^2\,dR/dt$.

    Step-by-Step Solution

    1. Let $x(t)$ be the ice thickness at time $t$; outer radius $R=10+x$.
    2. $V_{ice}=\dfrac43\pi(10+x)^3-\dfrac43\pi(10)^3$.
    3. $\dfrac{dV_{ice}}{dt}=4\pi(10+x)^2\dfrac{dx}{dt}$ (the constant term differentiates to zero).
    4. The ice melts (volume decreases) at $50\text{ cm}^3/\text{min}$, so $\dfrac{dV_{ice}}{dt}=-50$. …

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11. 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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12. 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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