Q.Find an angle $\theta$, $0 < \theta < \dfrac{\pi}{2}$, which increases twice as fast as its sine.

Showing the 12 most recent of 17 on this concept.

1. 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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2. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.The surface area of a sphere is $49\pi\ sq.cm$. If it is increased by $0.016\ sq.cm$, then the approximate increase in its volume (in $c.c.$) is (A) $0.07$ (B) $0.04$ (C) $0.032$ (D) $0.028$

   ›Reveal solutionSolution

   Using differentials to connect a small change in surface area to the corresponding small change in volume, via the shared variable $r$, gives an approximate volume increase of $0.028$ c.c.

   Concept and Intuition

   When a small change in one geometric quantity (surface area) causes a small change in another (volume), and both depend on a common variable ($r$), we can relate their differentials directly: $dS = 8\pi r\,dr$ and $dV = 4\pi r^2\,dr$ share the same $dr$, so we solve for $dr$ from the given $dS$ and substitute into the $dV$ formula — this is the standard "approximate change" technique using derivatives.

   Step-by-Step Solution

   1. Surface area of sphere: $S=4\pi r^2 = 49\pi \Rightarrow r^2 = \dfrac{49}{4} \Rightarrow r = \dfrac72 = 3.5\ \text{cm}$.
   2. Differentiate $S=4\pi r^2$: $dS = 8\pi r\, dr$.
   3. Given $dS = 0.016$, and $r=3.5$: $dr = \dfrac{0.016}{8\pi(3.5)} = \dfrac{0.016}{28\pi}$.
   4. Volume: $V = \dfrac43\pi r^3$. Differentiate: $dV = 4\pi r^2\, dr$.
   5. $r^2 = 12.25$, so $dV = 4\pi(12.25)\,dr = 49\pi\, dr$. …

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3. 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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4. 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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5. 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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6. 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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7. 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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8. 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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9. 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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10. 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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11. 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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12. 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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