Solving Related Rates - Calculus 1

1. A spherical balloon is inflated inside a cube so that it is always tangent to the cube's six faces. If the volume of the balloon is increasing at a constant rate of $10\pi$ cm³/s, what is the rate of increase of the volume of the cube at the instant the balloon's radius is 5 cm?
   Answer: $60$ cm³/s Let $r$ be the radius of the sphere and $s$ be the side length of the cube. Since the sphere is tangent to the faces of the cube, $s=2r$. The volume of the sphere is $V_s = \frac{4}{3}\pi r^3$ and the volume of the cube is $V_c = s^3 = (2r)^3 = 8r^3$. We are given $\frac{dV_s}{dt} = 10\pi$. We want to find $\frac{dV_c}{dt}$. First, differentiate both volume formulas with respect to $t$: $\frac{dV_s}{dt} = 4\pi r^2 \frac{dr}{dt}$ and $\frac{dV_c}{dt} = 24r^2 \frac{dr}{dt}$. From the first equation, we can find an expression for $\frac{dr}{dt} = \frac{1}{4\pi r^2}\frac{dV_s}{dt}$. Substitute this into the second equation: $\frac{dV_c}{dt} = 24r^2 \left( \frac{1}{4\pi r^2}\frac{dV_s}{dt} \right) = \frac{24}{4\pi}\frac{dV_s}{dt} = \frac{6}{\pi}\frac{dV_s}{dt}$. Plugging in the given rate $\frac{dV_s}{dt} = 10\pi$, we get $\frac{dV_c}{dt} = \frac{6}{\pi}(10\pi) = 60$ cm³/s. Note that the rate is constant and does not depend on the radius $r$.
2. A hemispherical bowl of radius 10 cm is being filled with water at a constant rate of $3\pi$ cm³/s. Using the formula provided, find the rate at which the water level is rising when the depth of the water is 5 cm.
   Answer: $1/25$ cm/s This is a related rates problem, where you need to find how fast one quantity changes based on how fast another quantity changes. When you see "rate of change" language with geometric shapes being filled, think about connecting the rates using the chain rule. Given the volume formula $V = \frac{\pi h^2}{3}(3R - h)$ for a spherical cap, you need to find $\frac{dh}{dt}$ when $h = 5$ cm, given that $\frac{dV}{dt} = 3\pi$ cm³/s and $R = 10$ cm. First, differentiate the volume formula with respect to time: $\frac{dV}{dt} = \frac{\pi}{3}\frac{d}{dt}[h^2(3R - h)] = \frac{\pi}{3}[2h(3R - h) + h^2(-1)] = \frac{\pi}{3}[6Rh - 3h^2]$ Substituting the known values ($R = 10$, $h = 5$, $\frac{dV}{dt} = 3\pi$): $3\pi = \frac{\pi}{3}[6(10)(5) - 3(5)^2] = \frac{\pi}{3}[300 - 75] = \frac{225\pi}{3} = 75\pi$ This gives us: $3\pi = 75\pi \cdot \frac{dh}{dt}$, so $\frac{dh}{dt} = \frac{3\pi}{75\pi} = \frac{1}{25}$ cm/s. Choice A ($\frac{1}{5}$) likely comes from incorrect differentiation or arithmetic errors. Choice B ($\frac{1}{50}$) might result from doubling the denominator incorrectly. Choice C ($\frac{3}{75}$) equals $\frac{1}{25}$ but wasn't simplified, or represents a calculation error. Study tip: In related rates problems, always differentiate the entire equation with respect to time, substitute known values only after differentiating, and double-check your product rule applications when dealing with composite expressions.
3. A pebble dropped in a pond creates a circular ripple. The area of the ripple increases at a constant rate of $6\pi$ cm²/s. What is the rate of increase of the circumference of the ripple when its area is $9\pi$ cm²?
   Answer: $2\pi$ cm/s Let $A$ be the area, $C$ be the circumference, and $r$ be the radius of the ripple. The formulas are $A = \pi r^2$ and $C = 2\pi r$. We are given $\frac{dA}{dt} = 6\pi$. We want to find $\frac{dC}{dt}$. First, let's find the radius when the area $A=9\pi$. From $9\pi = \pi r^2$, we get $r^2=9$, so $r=3$ cm. Next, we differentiate the area and circumference formulas with respect to time $t$: $\frac{dA}{dt} = 2\pi r \frac{dr}{dt}$ and $\frac{dC}{dt} = 2\pi \frac{dr}{dt}$. Using the first differentiated equation with the given values: $6\pi = 2\pi (3) \frac{dr}{dt}$, which gives $6\pi = 6\pi \frac{dr}{dt}$, so $\frac{dr}{dt} = 1$ cm/s. Now substitute this into the second differentiated equation: $\frac{dC}{dt} = 2\pi (1) = 2\pi$ cm/s.
4. A camera is positioned on the ground 2000 feet from the launchpad of a vertically ascending rocket. The rocket's velocity is a constant 500 ft/s. What is the rate of change of the camera's angle of elevation when the rocket is 4000 feet high?
   Answer: $1/20$ rad/s This is a classic related rates problem where you need to find how fast one quantity changes when you know how fast another quantity changes. When you see a setup involving changing distances and angles, think about forming a triangle and using trigonometry to relate the variables. Set up the problem by drawing a right triangle where the horizontal leg is the 2000-foot distance from camera to launchpad, the vertical leg is the rocket's height $h$, and the angle of elevation is $\theta$. From trigonometry: $\tan(\theta) = \frac{h}{2000}$. To find $\frac{d\theta}{dt}$, differentiate both sides with respect to time: $\sec^2(\theta) \cdot \frac{d\theta}{dt} = \frac{1}{2000} \cdot \frac{dh}{dt}$. Since the rocket rises at 500 ft/s, $\frac{dh}{dt} = 500$. When the rocket is 4000 feet high, you can find $\sec^2(\theta)$ using the triangle. At this moment, the hypotenuse has length $\sqrt{2000^2 + 4000^2} = \sqrt{20,000,000} = 2000\sqrt{5}$. Therefore, $\cos(\theta) = \frac{2000}{2000\sqrt{5}} = \frac{1}{\sqrt{5}}$, so $\sec^2(\theta) = 5$. Substituting: $5 \cdot \frac{d\theta}{dt} = \frac{500}{2000} = \frac{1}{4}$, giving $\frac{d\theta}{dt} = \frac{1}{20}$ rad/s. Answer B is correct. Answer A ($\frac{1}{10}$) likely comes from forgetting to square the secant. Answer C ($\frac{1}{5}$) might result from using $\cos^2(\theta)$ instead of $\sec^2(\theta)$. Answer D ($\frac{1}{4}$) occurs if you forget to account for the $\sec^2(\theta)$ factor entirely. Remember: in related rates problems, always identify your triangle, write the trigonometric relationship, then differentiate with respect to time.
5. Ship A leaves a port at noon and sails north at 15 knots. Ship B leaves the same port at 1:00 PM and sails east at 20 knots. At 3:00 PM, what is the rate of change of the angle $\theta$ of the line of sight from Ship A to Ship B, measured clockwise from the north direction?
   Answer: $12/145$ rad/hr Let $y$ be the distance of Ship A from the port and $x$ be the distance of Ship B. At 3:00 PM, Ship A has been sailing for 3 hours and Ship B for 2 hours. So, $y = 15 \times 3 = 45$ nautical miles, and $x = 20 \times 2 = 40$ nautical miles. We have $\frac{dy}{dt} = 15$ knots and $\frac{dx}{dt} = 20$ knots. The angle $\theta$ satisfies $\tan(\theta) = x/y$. Differentiating with respect to time $t$: $\sec^2(\theta) \frac{d\theta}{dt} = \frac{y(dx/dt) - x(dy/dt)}{y^2}$. We can find $\frac{d\theta}{dt} = \frac{y(dx/dt) - x(dy/dt)}{y^2 \sec^2(\theta)}$. The distance between the ships is $D = \sqrt{x^2+y^2}$, so $\sec(\theta) = D/y$. Thus, $y^2 \sec^2(\theta) = y^2(D^2/y^2) = D^2 = x^2+y^2$. So, $\frac{d\theta}{dt} = \frac{y(dx/dt) - x(dy/dt)}{x^2+y^2}$. Plugging in the values: $\frac{d\theta}{dt} = \frac{(45)(20) - (40)(15)}{40^2 + 45^2} = \frac{900 - 600}{1600 + 2025} = \frac{300}{3625} = \frac{12}{145}$ rad/hr.
6. A spherical snowball is melting such that its surface area decreases at a constant rate of $8\pi$ cm²/min. What is the rate at which the radius is decreasing at the moment when the snowball's volume is $288\pi$ cm³?
   Answer: $1/6$ cm/min This is a related rates problem involving a melting sphere. When you see rates of change for different properties of the same object, you need to connect them using differentiation and the geometric formulas. Start by identifying what you know and need. The surface area decreases at $\frac{dA}{dt} = -8\pi$ cm²/min (negative because it's decreasing), and you need $\frac{dr}{dt}$ when the volume is $288\pi$ cm³. First, find the radius when $V = 288\pi$. Since $V = \frac{4}{3}\pi r^3$: $288\pi = \frac{4}{3}\pi r^3$ $288 = \frac{4}{3}r^3$ $r^3 = 216$ $r = 6$ cm Next, relate the surface area to the radius. For a sphere, $A = 4\pi r^2$. Differentiate both sides with respect to time: $\frac{dA}{dt} = 8\pi r \frac{dr}{dt}$ Substitute the known values: $-8\pi = 8\pi(6)\frac{dr}{dt}$ $-8\pi = 48\pi\frac{dr}{dt}$ $\frac{dr}{dt} = -\frac{8\pi}{48\pi} = -\frac{1}{6}$ cm/min The radius decreases at $\frac{1}{6}$ cm/min, making D correct. Choice A ($\frac{1}{9}$) likely comes from miscalculating $r^3 = 216$ as $r = 9$. Choice B ($\frac{1}{3}$) results from using the wrong surface area formula or arithmetic errors. Choice C ($24\pi$) probably comes from confusing which derivative formula to use or forgetting to solve for $\frac{dr}{dt}$. For related rates problems, always: identify all rates, find intermediate values using given conditions, write the relationship equation, then differentiate with respect to time.
7. A fixed amount of gas in a cylinder is being compressed. The relationship between pressure $P$ (in kPa), volume $V$ (in L), and temperature $T$ (in K) is given by the law $PV=kT$ for some constant $k$. At an instant when the volume is 50 L, the pressure is 400 kPa, and the temperature is 400 K, the volume is decreasing at 1 L/s and the temperature is increasing at 2 K/s. What is the rate of change of the pressure at this instant?
   Answer: It is increasing at 10 kPa/s. This is a related rates problem involving three variables connected by the ideal gas law. When you see multiple changing quantities governed by an equation, you need to differentiate implicitly with respect to time to find how their rates of change relate. Starting with $PV = kT$, differentiate both sides with respect to time: $\frac{d}{dt}(PV) = k\frac{dT}{dt}$. Using the product rule on the left side: $P\frac{dV}{dt} + V\frac{dP}{dt} = k\frac{dT}{dt}$. First, find the constant $k$ using the given conditions: $k = \frac{PV}{T} = \frac{400 \times 50}{400} = 50$. Now substitute all known values into the differentiated equation. Given: $V = 50$ L, $P = 400$ kPa, $\frac{dV}{dt} = -1$ L/s (negative because volume is decreasing), and $\frac{dT}{dt} = 2$ K/s. $400(-1) + 50\frac{dP}{dt} = 50(2)$ $-400 + 50\frac{dP}{dt} = 100$ $50\frac{dP}{dt} = 500$ $\frac{dP}{dt} = 10$ kPa/s The pressure is increasing at 10 kPa/s, which is answer C. Answer A (2 kPa/s) likely results from calculation errors or forgetting the product rule. Answer B (-6 kPa/s) suggests sign confusion with the decreasing volume. Answer D (136 kPa/s) probably comes from incorrectly using the original values instead of properly applying the differentiated equation. For related rates problems, always differentiate the constraint equation first, then substitute known values—never substitute before differentiating.
8. A water trough is 10 m long and has a cross-section in the shape of an isosceles trapezoid. The trapezoid is 2 m wide at the bottom, 4 m wide at the top, and has a height of 4 m. If the trough is being filled with water at a rate of 5 m³/min, how fast is the water level rising when the water is 2 m deep?
   Answer: $1/6$ m/min Let $V$ be the volume of water, $h$ be the water level, and $w$ be the width of the water surface. The volume is $V = \text{Area} \times \text{Length} = \frac{1}{2}(2+w)h \times 10$. We need to express $w$ in terms of $h$. The total change in width is $4-2=2$ m over a height of $4$ m. The trapezoid has slanted sides. By similar triangles formed by the slanted sides, the extra width on each side is $x$. So $x/h = (1)/4$, which means $x = h/4$. The total width of the water surface is $w = 2 + 2x = 2 + 2(h/4) = 2 + h/2$. Substitute this into the volume formula: $V = 5(2 + (2+h/2))h = 5(4+h/2)h = 20h + \frac{5}{2}h^2$. Now, differentiate with respect to $t$: $\frac{dV}{dt} = 20\frac{dh}{dt} + 5h\frac{dh}{dt} = (20+5h)\frac{dh}{dt}$. We are given $\frac{dV}{dt} = 5$ m³/min and we want to find $\frac{dh}{dt}$ when $h=2$ m. So, $5 = (20 + 5(2))\frac{dh}{dt} = 30\frac{dh}{dt}$. Thus, $\frac{dh}{dt} = \frac{5}{30} = \frac{1}{6}$ m/min.

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