Related Rates with Geometry - Calculus 1

1. The length $l$ of a rectangle is increasing at a rate of 3 cm/s while its width $w$ is decreasing at a rate of 4 cm/s. At the moment when $l = 8$ cm and $w = 6$ cm, what is the rate of change of the length of the rectangle's diagonal?
   Answer: 0 cm/s Let $D$ be the length of the diagonal. By the Pythagorean theorem, $D^2 = l^2 + w^2$. We are given $\frac{dl}{dt} = 3$ cm/s and $\frac{dw}{dt} = -4$ cm/s. We want to find $\frac{dD}{dt}$ when $l=8$ cm and $w=6$ cm. First, at this instant, the diagonal's length is $D = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10$ cm. Differentiating the Pythagorean equation with respect to time $t$ gives $2D \frac{dD}{dt} = 2l \frac{dl}{dt} + 2w \frac{dw}{dt}$, which simplifies to $D \frac{dD}{dt} = l \frac{dl}{dt} + w \frac{dw}{dt}$. Substituting the known values: $10 \frac{dD}{dt} = (8)(3) + (6)(-4)$. $10 \frac{dD}{dt} = 24 - 24 = 0$. Therefore, $\frac{dD}{dt} = 0$ cm/s. At this specific instant, the length of the diagonal is momentarily not changing. Distractor A, $-14$ cm/s, is the rate of change of the area of the rectangle: $\frac{dA}{dt} = l\frac{dw}{dt} + w\frac{dl}{dt} = 8(-4) + 6(3) = -32 + 18 = -14$. Distractor B, $-1$ cm/s, is the sum of the rates, $\frac{dl}{dt} + \frac{dw}{dt} = 3 + (-4) = -1$. Distractor D, 1 cm/s, is a sign error on the sum of the rates.
2. Water is being pumped into an inverted conical tank at a rate of 10 ft$^3$/min. At the same time, water is leaking out of the bottom at a rate of 2 ft$^3$/min. The tank has a height of 9 ft and a top radius of 3 ft. What is the rate at which the water level is rising when the water is 6 ft deep?
   Answer: $2 / (\pi)$ ft/min Let $V$ be the volume of water in the tank. The net rate of change of the volume is the rate in minus the rate out: $\frac{dV}{dt} = 10 - 2 = 8$ ft$^3$/min. The volume of the water, which is in the shape of a cone, is $V = \frac{1}{3}\pi r^2 h$. For the tank, the ratio of radius to height is $R/H = 3/9 = 1/3$. This ratio holds for the water at any level, so $r/h = 1/3$, which means $r = h/3$. Substitute this into the volume formula to express $V$ in terms of $h$ only: $V = \frac{1}{3}\pi (\frac{h}{3})^2 h = \frac{1}{3}\pi (\frac{h^2}{9}) h = \frac{\pi}{27}h^3$. Now, differentiate with respect to time $t$: $\frac{dV}{dt} = \frac{\pi}{27}(3h^2) \frac{dh}{dt} = \frac{\pi}{9}h^2 \frac{dh}{dt}$. We want to find $\frac{dh}{dt}$ when $h=6$ ft. We have $\frac{dV}{dt} = 8$. So, $8 = \frac{\pi}{9}(6^2) \frac{dh}{dt} = \frac{\pi}{9}(36) \frac{dh}{dt} = 4\pi \frac{dh}{dt}$. Solving for $\frac{dh}{dt}$ gives $\frac{dh}{dt} = \frac{8}{4\pi} = \frac{2}{\pi}$ ft/min. Distractor A, $1/\pi$, results from an arithmetic error or from using $r=h/6$. Distractor C, $3/(2\pi)$, might arise from a calculation error. Distractor D, $4/\pi$, results from using only the inflow rate (10 ft$^3$/min) instead of the net rate, which would give $\frac{dh}{dt} = 10/(4\pi) = 5/(2\pi)$, or from another calculation error leading to $4/\pi$.
3. The radius of a right circular cylinder is increasing at a rate of 2 in/min, and the height is decreasing at a rate of 5 in/min. At the instant when the radius is 10 inches and the height is 8 inches, what is the rate of change of the volume of the cylinder?
   Answer: $-180\pi$ in$^3$/min This is a related rates problem involving a cylinder whose dimensions are changing over time. When you see multiple quantities changing simultaneously and need to find how another quantity changes, you're dealing with related rates - use implicit differentiation with respect to time. The volume of a cylinder is $V = \pi r^2 h$. To find how volume changes with time, differentiate both sides with respect to time: $\frac{dV}{dt} = \pi \frac{d}{dt}(r^2 h)$. Using the product rule: $\frac{dV}{dt} = \pi(2r \frac{dr}{dt} \cdot h + r^2 \frac{dh}{dt})$. Now substitute the given values: $\frac{dr}{dt} = 2$ in/min, $\frac{dh}{dt} = -5$ in/min (negative because height is decreasing), $r = 10$ inches, and $h = 8$ inches. $\frac{dV}{dt} = \pi(2(10)(2)(8) + (10)^2(-5)) = \pi(320 - 500) = -180\pi$ cubic inches per minute. Choice A ($220\pi$) likely comes from adding instead of subtracting: $320 + 500 = 820$, then making an error. Choice B ($-120\pi$) might result from incorrectly using the derivative formula or arithmetic mistakes. Choice C ($180\pi$) gives the correct magnitude but wrong sign - this happens when you forget that decreasing height contributes negatively to the rate of change. The key insight for related rates problems: always pay careful attention to whether rates are positive (increasing) or negative (decreasing), and remember to use the product rule when variables multiply together in your formula.
4. A particle moves along the curve $y = \sqrt{x}$. As the particle passes through the point $(4, 2)$, its $x$-coordinate is increasing at a rate of 3 units/sec. How fast is the distance from the particle to the origin changing at this instant?
   Answer: $27\sqrt{5}/20$ units/sec This is a related rates problem where you need to find how fast the distance to the origin changes as the particle moves along a curve. When you see "how fast is changing," think derivatives with respect to time. Let's set up the problem systematically. The particle is at point $(x, y)$ where $y = \sqrt{x}$, and the distance from the origin is $d = \sqrt{x^2 + y^2}$. Since $y = \sqrt{x}$, we have $d = \sqrt{x^2 + x} = \sqrt{x(x+1)}$. To find $\frac{dd}{dt}$, differentiate both sides: $\frac{dd}{dt} = \frac{1}{2\sqrt{x(x+1)}} \cdot \frac{d}{dt}[x(x+1)] = \frac{1}{2\sqrt{x(x+1)}} \cdot (2x+1) \cdot \frac{dx}{dt}$. At point $(4,2)$, we have $x = 4$ and $\frac{dx}{dt} = 3$. Substituting: $\frac{dd}{dt} = \frac{1}{2\sqrt{4 \cdot 5}} \cdot (8+1) \cdot 3 = \frac{9 \cdot 3}{2\sqrt{20}} = \frac{27}{2 \cdot 2\sqrt{5}} = \frac{27}{4\sqrt{5}}$. Rationalizing the denominator: $\frac{27}{4\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = \frac{27\sqrt{5}}{20}$, which is choice D. Choice A ($9/2$) likely comes from forgetting to account for the curve constraint $y = \sqrt{x}$. Choice B ($9\sqrt{5}/4$) appears to miss the factor of $5$ in the denominator. Choice C ($27/4$) forgets to rationalize the denominator and incorrectly simplifies $\sqrt{20}$. Key strategy: In related rates problems involving distance, always write distance as a function of the given constraint, then differentiate the entire expression with respect to time.
5. The surface area of a spherical balloon is increasing at a constant rate of $12 \pi$ cm$^2$/s. What is the rate of change of the volume of the balloon when its radius is 3 cm?
   Answer: $18 \pi$ cm$^3$/s Let $A$ be the surface area, $V$ be the volume, and $r$ be the radius of the sphere. The formulas are $A = 4\pi r^2$ and $V = \frac{4}{3}\pi r^3$. We are given $\frac{dA}{dt} = 12\pi$ cm$^2$/s and we want to find $\frac{dV}{dt}$ when $r=3$ cm. This is a two-step problem. First, we find $\frac{dr}{dt}$ using the surface area information. Differentiate $A$ with respect to time $t$: $\frac{dA}{dt} = 8\pi r \frac{dr}{dt}$. Substitute the given values: $12\pi = 8\pi (3) \frac{dr}{dt}$, which gives $12\pi = 24\pi \frac{dr}{dt}$, so $\frac{dr}{dt} = \frac{12\pi}{24\pi} = \frac{1}{2}$ cm/s. Next, we find $\frac{dV}{dt}$. Differentiate $V$ with respect to time $t$: $\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$. Now substitute $r=3$ and $\frac{dr}{dt} = \frac{1}{2}$: $\frac{dV}{dt} = 4\pi (3)^2 (\frac{1}{2}) = 4\pi(9)(\frac{1}{2}) = 18\pi$ cm$^3$/s. Distractor A, $9\pi$ cm$^3$/s, might result from an algebraic error, perhaps dividing by 2 twice. Distractor C, $36\pi$ cm$^3$/s, is the value of the volume when $r=3$, not its rate of change. It could also result from forgetting the $\frac{1}{2}$ factor for $\frac{dr}{dt}$. Distractor D, $72\pi$ cm$^3$/s, might result from misremembering the derivative of surface area or volume.
6. A spherical snowball is melting in such a way that its surface area decreases at a rate of 2 cm$^2$/min. What is the rate of change of the radius, in cm/min, at the instant the volume is $288\pi$ cm$^3$?
   Answer: $-1/(24\pi)$ Let $A$ be the surface area, $V$ be the volume, and $r$ be the radius of the snowball. The formulas are $A = 4\pi r^2$ and $V = \frac{4}{3}\pi r^3$. We are given $\frac{dA}{dt} = -2$ cm$^2$/min (negative because it's decreasing). We need to find $\frac{dr}{dt}$ at the instant $V = 288\pi$ cm$^3$. First, we must find the radius $r$ at this instant. Using the volume formula: $288\pi = \frac{4}{3}\pi r^3$. Dividing by $\pi$ and multiplying by $3/4$ gives $r^3 = 288 \times \frac{3}{4} = 72 \times 3 = 216$. Thus, $r = \sqrt[3]{216} = 6$ cm. Now, we use the surface area formula's derivative to find $\frac{dr}{dt}$. Differentiating $A = 4\pi r^2$ with respect to time $t$ gives $\frac{dA}{dt} = 8\pi r \frac{dr}{dt}$. Substitute the known values: $-2 = 8\pi (6) \frac{dr}{dt}$, which is $-2 = 48\pi \frac{dr}{dt}$. Solving for $\frac{dr}{dt}$ gives $\frac{dr}{dt} = -\frac{2}{48\pi} = -\frac{1}{24\pi}$ cm/min. Distractor B, $-1/(12\pi)$, results from an error in finding the radius, possibly using $r=3$. Distractor C, $-1/(6\pi)$, may result from forgetting the radius in the derivative, computing $\frac{dr}{dt} = \frac{dA}{dt} / (8\pi)$. Distractor D, $-1/(4\pi)$, might arise from a mistake in the surface area formula or its derivative.
7. Two cars start moving from the same point. One travels south at 48 mi/h, and the other travels west at 36 mi/h. At what rate is the distance between the cars increasing two hours later?
   Answer: 60 mi/h Let $y$ be the distance the first car has traveled south, and $x$ be the distance the second car has traveled west. Let $D$ be the distance between the cars. By the Pythagorean theorem, $D^2 = x^2 + y^2$. We are given $\frac{dy}{dt} = 48$ mi/h and $\frac{dx}{dt} = 36$ mi/h. We want to find $\frac{dD}{dt}$ after $t=2$ hours. Differentiating the equation with respect to time $t$, we get $2D \frac{dD}{dt} = 2x \frac{dx}{dt} + 2y \frac{dy}{dt}$, which simplifies to $D \frac{dD}{dt} = x \frac{dx}{dt} + y \frac{dy}{dt}$. After 2 hours, the distances are $y = 48 \times 2 = 96$ miles and $x = 36 \times 2 = 72$ miles. The distance $D$ between them at this time is $D = \sqrt{x^2 + y^2} = \sqrt{72^2 + 96^2} = \sqrt{5184 + 9216} = \sqrt{14400} = 120$ miles. Now substitute all values into the differentiated equation: $120 \frac{dD}{dt} = (72)(36) + (96)(48)$. $120 \frac{dD}{dt} = 2592 + 4608 = 7200$. Solving for $\frac{dD}{dt}$ gives $\frac{dD}{dt} = \frac{7200}{120} = 60$ mi/h. A common shortcut is to notice that if the velocities are constant, the rate of separation is constant and equal to $\sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} = \sqrt{36^2+48^2} = \sqrt{1296+2304} = \sqrt{3600} = 60$ mi/h. This problem is designed so that this shortcut works, but the full method is required for more complex scenarios. Distractor B, $84$ mi/h, is the sum of the speeds, $48 + 36$. Distractor C, $96$ mi/h, is one of the car's positions at $t=2$. Distractor D, $120$ mi/h, is the distance between the cars at $t=2$, not the rate of change of the distance.

Start this quiz