Q.A cylindrical tank of radius $10$ m is being filled with wheat at the rate of $314$ cubic metre per hour. Then the depth of the wheat is increasing at the rate of (A) $1$ m/h (B) $0.1$ m/h (C) $1.1$ m/h (D) $0.5$ m/h

Showing the 12 most recent of 30 on this concept.

1. TG EAPCET 2024Set eng-2024-05-11-FN1 markMCQ
   Q.If Water is poured into a cylindrical tank of radius $3.5$ ft at the rate of $1$ cu ft/min, then the rate at which the level of the water in the tank increases (in ft/min) is (A) $\dfrac{1}{154}$ (B) $\dfrac{8}{77}$ (C) $\dfrac{2}{77}$ (D) $\dfrac{1}{11}$

   ›Reveal solutionSolution

   The water level rises at a constant rate because the tank’s cross‑sectional area is constant; the rate is the inflow divided by the area. The answer is $\frac{2}{77}$ ft/min, option (C).

   Concept & Intuition

   When you pour water into a cylinder, the volume added is directly proportional to the increase in height, because the cross‑sectional area doesn’t change with depth. So the rate of change of height is simply the volumetric flow rate divided by the area of the base. No calculus chain‑rule gymnastics needed — just a straightforward division.

   ---

   1. Identify the relationship The volume of water in a cylinder of radius $r$ and height $h$ is

   $V = \pi r^2 h.$

   Here $r = 3.5$ ft, so the base area is

   $A = \pi (3.5)^2 = \pi \times 12.25 = \frac{49\pi}{4} \ \text{ft}^2.$

   2. Differentiate with respect to time Since $r$ is constant,

   $\frac{dV}{dt} = \pi r^2 \frac{dh}{dt} = A \frac{dh}{dt}.$

   We are given $\frac{dV}{dt} = 1$ cu ft/min.

   3. Solve for $\frac{dh}{dt}$

   $\frac{dh}{dt} = \frac{1}{A} = \frac{1}{\frac{49\pi}{4}} = \frac{4}{49\pi}.$

   4. Simplify numerically Use $\pi \approx \frac{22}{7}$ (common in such problems): $$$$

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2. TG EAPCET 2021Set eng-2021-08-04-AN1 markMCQ
   Q.If the radius of a spherical balloon is increasing at the rate of 5 inch per minute, then the rate at which the volume increases (in cube inches per minute) when the radius is 10 inches is (A) $100\pi$ (B) $1000\pi$ (C) $2000\pi$ (D) $25000\pi$

   ›Reveal solutionSolution

   The rate of change of volume is found by differentiating the volume formula $V = \frac{4}{3}\pi r^3$ with respect to time, using the chain rule. When $r = 10$ inches and $\frac{dr}{dt} = 5$ in/min, the answer is $2000\pi$ cubic inches per minute.

   The core idea here is related rates — a classic application of the chain rule in calculus. When a quantity changes over time, and another quantity depends on it, their rates of change are linked through differentiation. For a sphere, volume depends on radius, so if the radius grows at a known speed, the volume’s growth speed follows directly.

   The trap many students fall into is forgetting that $\frac{dV}{dt}$ is not just the derivative of $V$ with respect to $r$ — you must multiply by $\frac{dr}{dt}$ because both are functions of time. Let’s walk through it cleanly.

   1. Write the relationship. The volume of a sphere of radius $r$ is

   $V = \frac{4}{3}\pi r^3.$

   2. Differentiate both sides with respect to time $t$. Since $r$ itself changes with $t$, use the chain rule:

   $\frac{dV}{dt} = \frac{d}{dt}\left(\frac{4}{3}\pi r^3\right) = \frac{4}{3}\pi \cdot 3r^2 \cdot \frac{dr}{dt} = 4\pi r^2 \frac{dr}{dt}.$

   Notice how the $3$ cancels with the $\frac{4}{3}$, leaving a clean $4\pi r^2$ — that’s the surface area of the sphere. Makes intuitive sense: the volume grows like the surface area times the radial speed.

   3. Plug in the given values. We know $\frac{dr}{dt} = 5$ inches per minute, and we want the rate when $r = 10$ inches: $\frac{dV}{dt} = 4\pi (10)^2 \cdot 5 = 4\pi \cdot 100 \cdot 5 = 2000\pi.$ …

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3. TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
   Q.The height of a cone with semi vertical angle $\dfrac{\pi}{3}$ is increasing at the rate of 2 units/min. The rate at which the radius of the cone is to be decreased so as to have a fixed volume always is (A) $\dfrac{1}{\sqrt{3}}$ (B) $\dfrac{1}{\sqrt{2}}$ (C) $\sqrt{3}$ (D) $\sqrt{2}$

   ›Reveal solutionSolution

   The problem uses related rates with the cone’s volume fixed. Differentiating $V = \frac13 \pi r^2 h$ and using $\frac{dh}{dt}=2$ gives $\frac{dr}{dt} = -\frac{r}{2h} \cdot 2$. With semi-vertical angle $\pi/3$, $\frac{r}{h} = \tan(\pi/3)=\sqrt{3}$, so $\frac{dr}{dt} = -\sqrt{3}$ units/min. The rate of decrease is $\sqrt{3}$.

   Concept & Intuition

   We have a cone whose height is increasing, but we want its volume to stay constant. That means the radius must shrink to compensate. The key is to relate the radius and height through the fixed semi-vertical angle — this gives a constant ratio $r/h = \tan(\pi/3) = \sqrt{3}$. Then we use calculus (related rates) to find how fast the radius must change when the height changes at 2 units/min.

   Step-by-step solution

   1. Volume of a cone

      The volume is $V = \frac13 \pi r^2 h$. Since the volume is fixed, $V$ is constant, so $\frac{dV}{dt} = 0$.

   2. Differentiate implicitly with respect to time

      Using the product rule:

   $\frac{dV}{dt} = \frac13 \pi \left( 2r \frac{dr}{dt} \cdot h + r^2 \frac{dh}{dt} \right) = 0.$

   Multiply through by $3/\pi$ (nonzero):

   $2rh \frac{dr}{dt} + r^2 \frac{dh}{dt} = 0.$

   3. Solve for $\frac{dr}{dt}$

   $2rh \frac{dr}{dt} = - r^2 \frac{dh}{dt} \quad\Rightarrow\quad \frac{dr}{dt} = -\frac{r}{2h} \frac{dh}{dt}.$

   4. Use the given rate and geometry We are told $\frac{dh}{dt} = 2$ units/min. The semi-vertical angle is $\pi/3$, so in a right triangle formed by the height, radius, and slant height: $$$$

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4. TG EAPCET 2025Set eng-2025-05-04-FN1 markMCQ
   Q.The height of a cone with semi vertical angle $\pi/3$ is increasing at the rate of 2 units/min. The rate at which the radius of the cone is to be decreased so as to have a fixed volume always is (A) $\sqrt{3}$ (B) $\dfrac{1}{\sqrt{2}}$ (C) $\dfrac{1}{\sqrt{3}}$ (D) $\sqrt{2}$

   ›Reveal solutionSolution

   For a cone of fixed volume, the radius must shrink at a rate that exactly compensates the growth in height. Using the relation $V = \frac13 \pi r^2 h$ and differentiating with respect to time gives $\frac{dr}{dt} = -\frac{r}{2h} \frac{dh}{dt}$. With semi-vertical angle $\pi/3$, we have $r/h = \tan(\pi/3) = \sqrt{3}$, so $\frac{dr}{dt} = -\frac{\sqrt{3}}{2} \cdot 2 = -\sqrt{3}$ units/min. The required rate of decrease is $\sqrt{3}$ units/min, so the correct option is (A).

   The key idea is that "fixed volume" ties the radius and height together through a constraint. When one changes, the other must change in a specific way to keep the product $r^2 h$ constant. The semi-vertical angle gives the instantaneous ratio of radius to height at the moment we are considering — that ratio is not constant over time (since the cone's shape changes), but at the instant we care about, it is fixed by the given angle.

   Let’s work through it step by step.

   1. Write the volume constraint. For a cone, $V = \frac13 \pi r^2 h$. Since the volume is fixed, $V$ is constant. Differentiating both sides with respect to time $t$:

   $\frac{dV}{dt} = \frac13 \pi \left( 2r \frac{dr}{dt} \cdot h + r^2 \frac{dh}{dt} \right) = 0.$

   Multiply through by $3/\pi$ (non-zero):

   $2rh \frac{dr}{dt} + r^2 \frac{dh}{dt} = 0.$

   2. Solve for $\frac{dr}{dt}$. Rearranging:

   $2rh \frac{dr}{dt} = - r^2 \frac{dh}{dt}.$

   Assuming $r \neq 0$, divide both sides by $r$:

   $2h \frac{dr}{dt} = - r \frac{dh}{dt}.$

   Hence:

   $\frac{dr}{dt} = -\frac{r}{2h} \frac{dh}{dt}.$

   The negative sign tells us that if height increases, radius must decrease — exactly what we expect.

   3. Use the semi-vertical angle to find $r/h$. The semi-vertical angle is the angle between the axis and the slant height. In a right circular cone, $\tan(\text{semi-vertical angle}) = \frac{\text{radius}}{\text{height}}$. Given the angle is $\pi/3$:

   $\frac{r}{h} = \tan\frac{\pi}{3} = \sqrt{3}.$

   So $r = \sqrt{3}\,h$ at the instant under consideration.

   4. Plug in the given rate. …

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5. TG EAPCET 2022Set eng-2022-07-18-AN1 markMCQ
   Q.If an error of $0.02$ sq.cm is found in the surface area of a sphere when its radius is measured as $10$ cm, then the approximate error that occurs in the volume of the sphere, in cubic centimetres, is (A) $0.2$ (B) $0.01$ (C) $0.3$ (D) $0.1$

   ›Reveal solutionSolution

   The error in volume is found by relating differentials: $dV = \frac{r}{2} \, dS$. With $r = 10$ cm and $dS = 0.02$ sq.cm, the approximate error in volume is $0.1$ cubic cm.

   The key idea here is that when a small error is made in measuring a quantity (here, the radius), that error propagates into any other quantity calculated from it. We are not asked for the exact error — only an approximate error, which is exactly what differentials give us. The surface area and volume of a sphere are both functions of the radius, so a small change $\Delta r$ in radius produces small changes $\Delta S$ and $\Delta V$ that are well approximated by the differentials $dS$ and $dV$.

   We are told the error in surface area ($dS = 0.02$) and the measured radius ($r = 10$). We need the corresponding error in volume ($dV$). The direct link is through the radius: find $dr$ from $dS$, then use that $dr$ to find $dV$.

   1. Relate surface area error to radius error. Surface area of a sphere: $S = 4\pi r^2$. Differentiate: $dS = 8\pi r \, dr$. With $r = 10$ and $dS = 0.02$:

   $0.02 = 8\pi (10) \, dr = 80\pi \, dr$

   So

   $dr = \frac{0.02}{80\pi} = \frac{0.00025}{\pi}$

   This is the approximate error in the radius measurement.

   2. Relate volume error to the same radius error. Volume of a sphere: $V = \frac{4}{3}\pi r^3$. Differentiate: $dV = 4\pi r^2 \, dr$. Substitute $r = 10$ and the $dr$ we found:

   $dV = 4\pi (100) \cdot \frac{0.00025}{\pi} = 400\pi \cdot \frac{0.00025}{\pi}$

   The $\pi$ cancels:

   $dV = 400 \times 0.00025 = 0.1$ …

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6. TG EAPCET 2021Set eng-2021-08-05-AN1 markMCQ
   Q.The volume of a spherical balloon is increasing at the rate of $2\ \text{cm}^3/\text{sec}$. When its radius is $4\ \text{cm}$, the rate of change of its surface area (in $\text{cm}^2/\text{sec}$) is (A) 1 (B) 2 (C) 3 (D) 4

   ›Reveal solutionSolution

   We use related rates to connect the given rate of change of volume to the rate of change of surface area via the radius. The rate of change of the surface area is $\boxed{1\ \text{cm}^2/\text{sec}}$.

   This problem asks us to find the rate of change of the surface area of a spherical balloon, given the rate of change of its volume at a specific instant. This is a classic application of "related rates" in differential calculus. The core idea is that if two or more quantities are related by an equation, and they are all changing with respect to a common variable (usually time), then their rates of change are also related. We use the chain rule to establish these relationships.

   For a sphere, both its volume ($V$) and surface area ($S$) depend on its radius ($r$). If the radius changes over time, then both the volume and surface area will also change over time.

   • The volume of a sphere is given by $V = \frac{4}{3}\pi r^3$.
   • The surface area of a sphere is given by $S = 4\pi r^2$.

   We are given $\frac{dV}{dt}$ and need to find $\frac{dS}{dt}$. Both these rates depend on $\frac{dr}{dt}$, the rate at which the radius is changing. So, our strategy will be:

   1. Use the given rate of change of volume ($\frac{dV}{dt}$) and the volume formula to calculate $\frac{dr}{dt}$ at the specified radius.
   2. Use this calculated $\frac{dr}{dt}$ and the surface area formula to find $\frac{dS}{dt}$ at that same radius.

   Here's the step-by-step solution:

   1. Identify the given information and what needs to be found.

      We are given:

      • The rate at which the volume of the spherical balloon is increasing: $\frac{dV}{dt} = 2\ \text{cm}^3/\text{sec}$.
      • The radius of the balloon at the specific instant we are interested in: $r = 4\ \text{cm}$. We need to find:
      • The rate of change of its surface area, $\frac{dS}{dt}$, at that instant.

   2. Write down the formulas for the volume and surface area of a sphere.

      The volume of a sphere with radius $r$ is $V = \frac{4}{3}\pi r^3$.

      The surface area of a sphere with radius $r$ is $S = 4\pi r^2$.

   3. Differentiate the volume formula with respect to time ($t$) to find $\frac{dr}{dt}$.

      Since $V$ is a function of $r$, and $r$ is a function of $t$, we apply the chain rule to differentiate $V$ with respect to $t$:

   $\frac{dV}{dt} = \frac{d}{dt}\left(\frac{4}{3}\pi r^3\right)$

   $\frac{dV}{dt} = \frac{4}{3}\pi \cdot (3r^2) \cdot \frac{dr}{dt}$

   $\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$

   Now, substitute the given values: $\frac{dV}{dt} = 2\ \text{cm}^3/\text{sec}$ and $r = 4\ \text{cm}$.
   

   $2 = 4\pi (4)^2 \frac{dr}{dt}$

   $2 = 4\pi (16) \frac{dr}{dt}$

   $2 = 64\pi \frac{dr}{dt}$

   Solving for $\frac{dr}{dt}$:
   

   $\frac{dr}{dt} = \frac{2}{64\pi} = \frac{1}{32\pi}\ \text{cm/sec}$

   This is the rate at which the radius is increasing at the instant when $r=4\ \text{cm}$.
   

   4. Differentiate the surface area formula with respect to time ($t$) to find $\frac{dS}{dt}$.

   Similarly, $S$ is a function of $r$, and $r$ is a function of $t$. We use the chain rule to differentiate $S$ with respect to $t$:

   $\frac{dS}{dt} = \frac{d}{dt}(4\pi r^2)$ …

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7. TG EAPCET 2023Set eng-2023-05-14-AN1 markMCQ
   Q.A ladder of length 13 mts has one end resting against a vertical wall and the other on the ground. If the lower end moves away from the wall at a speed of 2 mts/minute, then the speed (in mts/min) at which upper end falls when the bottom is 5 mts away from the wall is (A) $\dfrac{6}{5}$ (B) $\dfrac{12}{5}$ (C) $\dfrac{5}{6}$ (D) $\dfrac{5}{12}$

   ›Reveal solutionSolution

   This is a classic related-rates problem: use the Pythagorean theorem to relate the ladder’s height and base distance, then differentiate with respect to time. The upper end falls at $\frac{5}{6}$ m/min when the bottom is 5 m from the wall.

   We have a ladder of fixed length 13 m leaning against a vertical wall. The bottom slides away from the wall at a constant speed of 2 m/min. We need the speed at which the top slides down the wall at the instant the bottom is 5 m from the wall.

   Concept & Intuition

   The ladder, wall, and ground form a right triangle: the ladder is the hypotenuse (always 13 m), the distance from the wall to the bottom is one leg, and the height of the top along the wall is the other leg. As the bottom moves, both legs change, but the hypotenuse stays fixed. This gives a relationship between the rates of change of the two legs — a classic related rates problem. Differentiating the Pythagorean relation with respect to time lets us connect the known speed (bottom moving away) to the unknown speed (top moving down).

   ---

   1. Set up variables and the fixed relation Let $x$ = distance from the wall to the bottom of the ladder (in m). Let $y$ = height of the top of the ladder on the wall (in m). The ladder length is constant:

   $x^2 + y^2 = 13^2 = 169.$

   2. Differentiate with respect to time Both $x$ and $y$ change with time $t$. Differentiate implicitly:

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

   Divide by 2:

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

   3. Identify known and unknown rates

      We are given $\frac{dx}{dt} = 2$ m/min (positive because $x$ increases).

      We want $\frac{dy}{dt}$ when $x = 5$ m.

      Note: $\frac{dy}{dt}$ will be negative because $y$ decreases (top falls). The problem asks for the speed (magnitude), so we will take the absolute value at the end.

   4. Find $y$ when $x = 5$

      From $x^2 + y^2 = 169$:

      $$$$

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8. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
   Q.A man of 5 feet height is walking away from a light fixed at a height of 15 feet at the rate of K miles/hour. If the rate of increase of his shadow is $\dfrac{11}{5}$ feet/sec, then K = (Take 1 mile = 5280 feet) (A) 2 (B) 3 (C) 4 (D) 5

   ›Reveal solutionSolution

   Using similar triangles the shadow length is $s=\tfrac{x}{2}$, so $\tfrac{ds}{dt}=\tfrac12\tfrac{dx}{dt}$. From $\tfrac{ds}{dt}=\tfrac{11}{5}$ ft/s the man's speed is $\tfrac{22}{5}$ ft/s $=3$ mph. Answer: (B) 3.

   Setup (similar triangles). Let the lamp be at height $15$ ft, the man ($5$ ft tall) at distance $x$ ft from the lamp post, and $s$ the length of his shadow. The lamp-ground-shadow-tip triangle and the man-feet-shadow-tip triangle are similar:

   $\frac{15}{x+s}=\frac{5}{s}\;\Rightarrow\;15s=5x+5s\;\Rightarrow\;10s=5x\;\Rightarrow\;s=\frac{x}{2}.$

   Differentiate.

   $\frac{ds}{dt}=\frac12\,\frac{dx}{dt}.$

   Solve for the man's speed. Given $\dfrac{ds}{dt}=\dfrac{11}{5}$ ft/s,

   $\frac{dx}{dt}=2\cdot\frac{11}{5}=\frac{22}{5}\ \text{ft/s}.$ …

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9. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
   Q.A man of 5 feet height is walking away from a light fixed at a height of 15 feet at the rate of K miles/hour. If the rate of increase of his shadow is $\dfrac{11}{5}$ feet/sec, then K = (Take 1 mile = 5280 feet) (A) 2 (B) 3 (C) 5 (D) 4

   ›Reveal solutionSolution

   $K = 3$ miles/hour — option (B).

   By similar triangles, if $x$ is the man's distance from the pole and $s$ his shadow's length, the tip of the shadow, the top of the lamp and the man's head are collinear:

   $\frac{x+s}{15} = \frac{s}{5} \;\Rightarrow\; 5(x+s) = 15s \;\Rightarrow\; 5x = 10s \;\Rightarrow\; s = \frac{x}{2}.$

   Differentiating with respect to time:

   $\frac{ds}{dt} = \frac{1}{2}\,\frac{dx}{dt}.$

   Given $\dfrac{ds}{dt} = \dfrac{11}{5}$ ft/sec, …

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10. TG EAPCET 2021Set eng-2021-08-06-FN1 markMCQ
    Q.The semi vertical angle of a right circular cone is $30^\circ$. If the height of the cone is $6.125$ cm, then the approximate value of the volume of the cone (in cubic cm) is (A) $(23.5)\pi$ (B) $(76.5)\pi$ (C) $48\pi$ (D) $(25.5)\pi$

    ›Reveal solutionSolution

    With semi-vertical angle $30^\circ$, $r=h\tan30^\circ$, so $V=\tfrac13\pi h^3\tan^2 30^\circ\approx (25.5)\pi$.

    Radius from the semi-vertical angle. The semi-vertical angle $\alpha$ satisfies $\tan\alpha=\dfrac{r}{h}$, so

    $r=h\tan 30^\circ=\frac{h}{\sqrt3},\qquad r^2=\frac{h^2}{3}.$

    Volume. With $h=6.125$ cm, …

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11. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
    Q.There is a possible error of 0.03 cm in a scale of length 1 foot with which the height of a closed right circular cylinder and the diameter of a sphere are measured as 3.5 feet each. If the radii of both cylinder and sphere are same, then the approximate error in the sum of the surface areas of both cylinder and sphere is (in square feet) (A) 0.385 (B) 0.0962 (C) 0.77 (D) 0.1925

    ›Reveal solutionSolution

    With equal radii, the total surface area is $S=\pi d h+\tfrac32\pi d^2$; propagating the scale error by differentials gives $\Delta S=17.5\pi\,\Delta x\approx0.1925$ ft$^2$. Answer: (D) 0.1925.

    Surface areas (radii equal, cylinder closed). Let the common radius be $r=\tfrac{d}{2}$ with measured diameter $d=3.5$ ft and height $h=3.5$ ft.

    $S=\underbrace{(2\pi r h+2\pi r^2)}_{\text{closed cylinder}}+\underbrace{4\pi r^2}_{\text{sphere}}=2\pi r h+6\pi r^2.$

    In terms of the measured quantities $d$ and $h$ (with $r=d/2$):

    $S=\pi d h+\frac{3}{2}\pi d^2.$

    Differential error.

    $dS=\frac{\partial S}{\partial d}\,\Delta d+\frac{\partial S}{\partial h}\,\Delta h,\qquad \frac{\partial S}{\partial d}=\pi h+3\pi d,\quad \frac{\partial S}{\partial h}=\pi d.$

    At $d=h=3.5$:

    $\frac{\partial S}{\partial d}=\pi(3.5)+3\pi(3.5)=14\pi,\qquad \frac{\partial S}{\partial h}=3.5\pi.$ …

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12. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
    Q.If the base of an isosceles triangle is $3\sqrt{2}$ feet and the two equal sides of it are increasing at the rate of $1$ ft/s, then the rate of increase of its area (in sq.ft/sec) when the angle between the equal sides is a right angle is (A) $3\sqrt{3}$ (B) $\sqrt{3}$ (C) $9$ (D) $3$

    ›Reveal solutionSolution

    The area of an isosceles triangle is expressed in terms of the equal side length and the included angle. Using the given rate of change of the side and the fact that the angle is fixed at the instant of interest, the rate of increase of area is found to be $3$ sq.ft/sec.

    The problem gives an isosceles triangle with base $3\sqrt{2}$ feet and equal sides that are increasing at $1$ ft/s. We need the rate of increase of its area at the moment when the angle between the equal sides is a right angle.

    The key is to choose a formula for area that directly involves the changing quantity (the equal side length) and the angle. For any triangle, area is $\frac12 ab \sin C$. Here, the two equal sides are the ones forming the included angle, so that formula is perfect.

    1. Set up the variables. Let the equal sides each have length $s$ feet, and let $\theta$ be the angle between them. The area $A$ of the triangle is

    $A = \frac12 \cdot s \cdot s \cdot \sin\theta = \frac12 s^2 \sin\theta.$

    2. What is given and what is wanted? We know $\frac{ds}{dt} = 1$ ft/s. We want $\frac{dA}{dt}$ at the instant when $\theta = 90^\circ = \frac{\pi}{2}$ radians. But note: the base is fixed at $3\sqrt{2}$ feet. Does that give a relation between $s$ and $\theta$? Yes — by the law of cosines, the base $b$ satisfies

    $b^2 = s^2 + s^2 - 2s^2 \cos\theta = 2s^2(1 - \cos\theta).$

    So $b = 3\sqrt{2}$ is constant, meaning $s$ and $\theta$ are not independent — as $s$ increases, $\theta$ must change to keep the base fixed. However, we only need the rate at a specific instant, not a full functional relation.

    3. Differentiate the area with respect to time. Since both $s$ and $\theta$ can change with time,

    $\frac{dA}{dt} = \frac12 \left( 2s \frac{ds}{dt} \sin\theta + s^2 \cos\theta \cdot \frac{d\theta}{dt} \right) = s \frac{ds}{dt} \sin\theta + \frac12 s^2 \cos\theta \frac{d\theta}{dt}.$

    4. Find $s$ and $\frac{d\theta}{dt}$ at the required instant. At $\theta = \frac{\pi}{2}$, $\sin\theta = 1$, $\cos\theta = 0$. The law of cosines gives

    $(3\sqrt{2})^2 = 2s^2 (1 - \cos\frac{\pi}{2}) = 2s^2 (1 - 0) = 2s^2.$

    So $18 = 2s^2$, hence $s^2 = 9$ and $s = 3$ feet (positive length).

    Now we need $\frac{d\theta}{dt}$ at that instant. Differentiate the law of cosines relation with respect to time. From $b^2 = 2s^2(1 - \cos\theta)$, since $b$ is constant,

    $0 = \frac{d}{dt}\left[2s^2(1 - \cos\theta)\right] = 4s \frac{ds}{dt}(1 - \cos\theta) + 2s^2 \sin\theta \frac{d\theta}{dt}.$

    At $\theta = \frac{\pi}{2}$, $\cos\theta = 0$, $\sin\theta = 1$, $s = 3$, $\frac{ds}{dt} = 1$: …

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