Intro to Related Rates - Calculus 1

A 10-foot ladder is leaning against a vertical wall. The base of the ladder is pulled away from the wall at a constant rate of 2 ft/s. At what rate is the angle between the ladder and the ground changing when the base of the ladder is 6 feet from the wall?
Answer: The angle is decreasing at a rate of $1/4$ rad/s. Let $x$ be the distance from the base of the ladder to the wall, and let $\theta$ be the angle between the ladder and the ground. The relationship between $x$ and $\theta$ is given by $\cos(\theta) = x/10$ . Differentiating both sides with respect to time $t$ , we get $-\sin(\theta) \frac{d\theta}{dt} = \frac{1}{10} \frac{dx}{dt}$ . We are given $\frac{dx}{dt} = 2$ ft/s. We need to find $\frac{d\theta}{dt}$ when $x=6$ . At this instant, the ladder forms a right triangle with sides 6 and $y = \sqrt{10^2 - 6^2} = \sqrt{64} = 8$ . Thus, $\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{y}{10} = \frac{8}{10} = \frac{4}{5}$ . Substituting the known values into the differentiated equation: $-\left(\frac{4}{5}\right) \frac{d\theta}{dt} = \frac{1}{10}(2)$ . This simplifies to $-\frac{4}{5} \frac{d\theta}{dt} = \frac{1}{5}$ . Solving for $\frac{d\theta}{dt}$ gives $\frac{d\theta}{dt} = -\frac{1}{4}$ rad/s. The negative sign indicates that the angle is decreasing.
Two cars start from the same point. Car A travels east at 60 mph and Car B travels northeast (45° east of north) at 50 mph. At what rate is the distance between the cars increasing after 2 hours?
Answer: $\sqrt{6100 - 3000\sqrt{2}}$ mph Let $a(t)$ be the distance of Car A from the start and $b(t)$ be the distance of Car B. We have $\frac{da}{dt} = 60$ and $\frac{db}{dt} = 50$ . The angle $\theta$ between their paths is $45^\circ$ . Let $c(t)$ be the distance between them. By the Law of Cosines, $c^2 = a^2 + b^2 - 2ab\cos(45^\circ) = a^2 + b^2 - \sqrt{2}ab$ . Differentiate with respect to $t$ : $2c\frac{dc}{dt} = 2a\frac{da}{dt} + 2b\frac{db}{dt} - \sqrt{2}\left(\frac{da}{dt}b + a\frac{db}{dt}\right)$ . After $t=2$ hours, $a = 60(2) = 120$ miles and $b = 50(2) = 100$ miles. At this time, $c^2 = 120^2 + 100^2 - \sqrt{2}(120)(100) = 14400 + 10000 - 12000\sqrt{2} = 24400 - 12000\sqrt{2}$ . So $c = \sqrt{24400 - 12000\sqrt{2}}$ . Now substitute into the derivative equation: $2c\frac{dc}{dt} = 2(120)(60) + 2(100)(50) - \sqrt{2}(60 \cdot 100 + 120 \cdot 50) = 14400 + 10000 - \sqrt{2}(6000 + 6000) = 24400 - 12000\sqrt{2}$ . So, $2c\frac{dc}{dt} = c^2$ , which implies $\frac{dc}{dt} = \frac{c}{2} = \frac{1}{2}\sqrt{24400 - 12000\sqrt{2}} = \sqrt{\frac{24400 - 12000\sqrt{2}}{4}} = \sqrt{6100 - 3000\sqrt{2}}$ mph.
Water is pumped into an inverted conical tank at a rate of 2 cubic meters per minute. The tank has a height of 6 meters and the radius at the top is 3 meters. How fast is the water level rising when the water is 4 meters deep?
Answer: $\frac{1}{2\pi}$ m/min The volume of a cone is $V = \frac{1}{3}\pi r^2 h$ . The ratio of the radius to the height of the water in the cone is constant and equal to the ratio for the tank itself. So, $\frac{r}{h} = \frac{3}{6} = \frac{1}{2}$ , which means $r = \frac{h}{2}$ . Substituting this into the volume formula gives $V = \frac{1}{3}\pi \left(\frac{h}{2}\right)^2 h = \frac{1}{3}\pi \left(\frac{h^2}{4}\right) h = \frac{\pi}{12}h^3$ . Differentiating with respect to time $t$ , we get $\frac{dV}{dt} = \frac{\pi}{12}(3h^2) \frac{dh}{dt} = \frac{\pi}{4}h^2 \frac{dh}{dt}$ . We are given $\frac{dV}{dt} = 2$ m³/min and we want to find $\frac{dh}{dt}$ when $h=4$ m. Plugging in the values: $2 = \frac{\pi}{4}(4^2) \frac{dh}{dt} \Rightarrow 2 = \frac{\pi}{4}(16) \frac{dh}{dt} \Rightarrow 2 = 4\pi \frac{dh}{dt}$ . Solving for $\frac{dh}{dt}$ gives $\frac{dh}{dt} = \frac{2}{4\pi} = \frac{1}{2\pi}$ m/min.
A person who is 6 feet tall is walking away from a 15-foot tall lamppost at a constant rate of 3 ft/s. What is the rate at which the length of the person's shadow is increasing?
Answer: 2 ft/s Let $x$ be the distance of the person from the lamppost and $s$ be the length of the shadow. Let $H=15$ be the height of the lamppost and $h=6$ be the height of the person. By similar triangles, we have the proportion $\frac{s}{h} = \frac{x+s}{H}$ . Substituting the heights, we get $\frac{s}{6} = \frac{x+s}{15}$ . Cross-multiplying gives $15s = 6(x+s) = 6x + 6s$ . Solving for $s$ , we get $9s = 6x$ , or $s = \frac{2}{3}x$ . Differentiating with respect to time $t$ , we get $\frac{ds}{dt} = \frac{2}{3}\frac{dx}{dt}$ . Given that the person is walking at 3 ft/s, $\frac{dx}{dt} = 3$ . Therefore, $\frac{ds}{dt} = \frac{2}{3}(3) = 2$ ft/s. Distractor D represents the rate at which the tip of the shadow is moving, which is $\frac{d}{dt}(x+s) = \frac{dx}{dt} + \frac{ds}{dt} = 3 + 2 = 5$ ft/s, a common point of confusion.
A particle moves along the curve $y = \sqrt{x^2+9}$ . The x-coordinate of the particle is increasing at a rate of 8 units per second. What is the rate of change of the particle's y-coordinate when the particle is at the point $(4, 5)$ ?
Answer: 6.4 units/s This is a related rates problem, where you need to find how fast one quantity changes when you know how another is changing. When you see "rate of change" language with connected variables, think about using implicit differentiation with respect to time. Start with the given curve equation $y = \sqrt{x^2+9}$ and differentiate both sides with respect to time $t$ . Using the chain rule: $\frac{dy}{dt} = \frac{1}{2\sqrt{x^2+9}} \cdot 2x \cdot \frac{dx}{dt} = \frac{x}{\sqrt{x^2+9}} \cdot \frac{dx}{dt}$ You're told that $\frac{dx}{dt} = 8$ units/s and need $\frac{dy}{dt}$ when the particle is at $(4,5)$ . Substituting these values: $\frac{dy}{dt} = \frac{4}{\sqrt{4^2+9}} \cdot 8 = \frac{4}{\sqrt{25}} \cdot 8 = \frac{4}{5} \cdot 8 = 6.4$ units/s. Answer choice A (5 units/s) likely comes from incorrectly using just the y-coordinate value. Answer choice B (10 units/s) might result from miscalculating $\frac{4}{5} \cdot 8$ as $\frac{5}{4} \cdot 8$ . Answer choice C (8 units/s) is the trap of assuming $\frac{dy}{dt} = \frac{dx}{dt}$ , ignoring that the curve's slope affects how changes in x translate to changes in y. For related rates problems, always: (1) write down the constraint equation, (2) differentiate implicitly with respect to time, (3) substitute known values including the specific point, and (4) solve for the unknown rate. The key insight is that rates of change along a curve depend on both the given rate and the curve's geometry at that point.
The length $l$ of a rectangle is increasing at a rate of 8 cm/s and its width $w$ is decreasing at a rate of 3 cm/s. When $l=20$ cm and $w=15$ cm, at what rate is the length of the diagonal changing?
Answer: $23/5$ cm/s Let $D$ be the length of the diagonal. By the Pythagorean theorem, $D^2 = l^2 + w^2$ . We are given $\frac{dl}{dt} = 8$ and $\frac{dw}{dt} = -3$ (since the width is decreasing). We want to find $\frac{dD}{dt}$ when $l=20$ and $w=15$ . At this instant, the diagonal is $D = \sqrt{20^2 + 15^2} = \sqrt{400 + 225} = \sqrt{625} = 25$ cm. Differentiating the relationship with respect to time $t$ : $2D\frac{dD}{dt} = 2l\frac{dl}{dt} + 2w\frac{dw}{dt}$ . Divide by 2: $D\frac{dD}{dt} = l\frac{dl}{dt} + w\frac{dw}{dt}$ . Substitute the known values: $25\frac{dD}{dt} = 20(8) + 15(-3)$ . This gives $25\frac{dD}{dt} = 160 - 45 = 115$ . Solving for $\frac{dD}{dt}$ yields $\frac{dD}{dt} = \frac{115}{25} = \frac{23}{5}$ cm/s. Distractor C is the result of using $\frac{dw}{dt}=+3$ , a common sign error. Distractors A and D are the rates of change of the perimeter and area, respectively.
A right cylindrical tank with a radius of 5 meters is being filled with water at a rate of $4\pi$ cubic meters per minute. However, the tank is also expanding in such a way that its radius is increasing at a rate of 0.1 meters per minute. How fast is the water level, $h$ , rising when the height of the water is 2 meters?
Answer: 0.08 m/min The volume of a cylinder is given by $V = \pi r^2 h$ . Since both the radius $r$ and height $h$ are changing with time, we must use the product rule when differentiating: $\frac{dV}{dt} = \pi \left( 2r \frac{dr}{dt} h + r^2 \frac{dh}{dt} \right)$ . We are given $\frac{dV}{dt} = 4\pi$ m³/min, $\frac{dr}{dt} = 0.1$ m/min. We want to find $\frac{dh}{dt}$ at the instant when $r=5$ m and $h=2$ m. Substitute the known values into the equation: $4\pi = \pi \left( 2(5)(0.1)(2) + (5^2) \frac{dh}{dt} \right)$ . Dividing by $\pi$ gives $4 = 2(5)(0.1)(2) + 25 \frac{dh}{dt}$ . Simplify: $4 = 2 + 25 \frac{dh}{dt}$ . This leads to $2 = 25 \frac{dh}{dt}$ , so $\frac{dh}{dt} = \frac{2}{25} = 0.08$ m/min.
A rocket is launched vertically and is tracked by a radar station located on the ground 5 miles from the launch pad. What is the rate of change of the rocket's angle of elevation $\theta$ from the radar station when the rocket is 10 miles high and traveling at 600 miles per second?
Answer: 24 rad/s This is a classic related rates problem where you need to find how one quantity changes with respect to time when given information about another quantity's rate of change. When you see a problem involving angles and moving objects, set up the geometric relationship first, then differentiate with respect to time. Let's establish the setup: the rocket moves vertically upward at height $h$ , the radar station is 5 miles away horizontally, and the angle of elevation is $\theta$ . From basic trigonometry: $\tan \theta = \frac{h}{5}$ To find $\frac{d\theta}{dt}$ , differentiate both sides with respect to time: $\sec^2 \theta \cdot \frac{d\theta}{dt} = \frac{1}{5} \cdot \frac{dh}{dt}$ When the rocket is 10 miles high, we can find $\sec^2 \theta$ . Since $\tan \theta = \frac{10}{5} = 2$ , we have $\sec^2 \theta = 1 + \tan^2 \theta = 1 + 4 = 5$ . Substituting the known values ( $\frac{dh}{dt} = 600$ miles/s): $5 \cdot \frac{d\theta}{dt} = \frac{1}{5} \cdot 600 = 120$ Therefore: $\frac{d\theta}{dt} = \frac{120}{5} = 24 \text{ rad/s}$ The answer is D) 24 rad/s. A) 12 rad/s results from incorrectly using $\sec^2 \theta = 10$ instead of 5. B) 120 rad/s comes from forgetting to divide by $\sec^2 \theta$ . C) 60 rad/s occurs from using the wrong geometric relationship or calculation error. Strategy tip: In related rates problems, always identify the geometric relationship first, differentiate with respect to time, then substitute known values in the correct order.
The volume of a cube is increasing at a constant rate of 150 cm³/s. What is the rate of change of the cube's surface area at the instant when the edge length is 5 cm?
Answer: 120 cm²/s Let $s$ be the edge length of the cube. The volume is $V = s^3$ and the surface area is $A = 6s^2$ . We are given $\frac{dV}{dt} = 150$ . We want to find $\frac{dA}{dt}$ when $s=5$ . First, use the volume information to find $\frac{ds}{dt}$ . Differentiate $V$ with respect to time $t$ : $\frac{dV}{dt} = 3s^2 \frac{ds}{dt}$ . Substitute the known values: $150 = 3(5^2)\frac{ds}{dt} = 3(25)\frac{ds}{dt} = 75\frac{ds}{dt}$ . Solving for $\frac{ds}{dt}$ gives $\frac{ds}{dt} = \frac{150}{75} = 2$ cm/s. Next, differentiate $A$ with respect to time $t$ : $\frac{dA}{dt} = 12s \frac{ds}{dt}$ . Substitute the values for the specific instant: $\frac{dA}{dt} = 12(5)(2) = 120$ cm²/s. Distractor B results from forgetting the chain rule factor $\frac{ds}{dt}$ (or assuming it is 1). Distractor C results from confusing the formulas for volume and surface area.
Gravel is falling from a conveyor belt onto a conical pile at a rate of 10 cubic feet per minute. The diameter of the base of the pile is always three times its height. How fast is the height of the pile increasing when the pile is 4 feet high?
Answer: $\frac{5}{18\pi}$ ft/min Let $V$ , $r$ , and $h$ be the volume, radius, and height of the conical pile. We are given $\frac{dV}{dt} = 10$ . The relationship between diameter $D$ and height is $D = 3h$ . Since the radius is half the diameter, $r = \frac{D}{2} = \frac{3h}{2}$ . The volume of a cone is $V = \frac{1}{3}\pi r^2 h$ . Substituting for $r$ : $V = \frac{1}{3}\pi \left(\frac{3h}{2}\right)^2 h = \frac{1}{3}\pi \left(\frac{9h^2}{4}\right)h = \frac{3\pi}{4}h^3$ . Now, differentiate with respect to time $t$ : $\frac{dV}{dt} = \frac{3\pi}{4} (3h^2) \frac{dh}{dt} = \frac{9\pi}{4}h^2 \frac{dh}{dt}$ . We want to find $\frac{dh}{dt}$ when $h=4$ . Substitute the known values: $10 = \frac{9\pi}{4}(4^2) \frac{dh}{dt} = \frac{9\pi}{4}(16) \frac{dh}{dt} = 36\pi \frac{dh}{dt}$ . Solving for $\frac{dh}{dt}$ gives $\frac{dh}{dt} = \frac{10}{36\pi} = \frac{5}{18\pi}$ ft/min.
Air is being pumped into a spherical balloon so that its volume increases at a rate of 100 cm³/s. How fast is the surface area of the balloon increasing when its radius is 5 cm?
Answer: 40 cm²/s Let $V$ be the volume and $A$ be the surface area of the sphere with radius $r$ . The formulas are $V = \frac{4}{3}\pi r^3$ and $A = 4\pi r^2$ . We are given $\frac{dV}{dt} = 100$ and want to find $\frac{dA}{dt}$ when $r=5$ . First, we find the rate of change of the radius, $\frac{dr}{dt}$ . Differentiate the volume formula: $\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$ . Substitute the given values: $100 = 4\pi (5^2) \frac{dr}{dt} = 100\pi \frac{dr}{dt}$ . So, $\frac{dr}{dt} = \frac{1}{\pi}$ cm/s. Next, differentiate the surface area formula: $\frac{dA}{dt} = 8\pi r \frac{dr}{dt}$ . Now substitute $r=5$ and $\frac{dr}{dt} = \frac{1}{\pi}$ : $\frac{dA}{dt} = 8\pi (5) \left(\frac{1}{\pi}\right) = 40$ cm²/s. Alternatively, note that $\frac{dA}{dt} = 8\pi r \frac{dr}{dt}$ and $\frac{dr}{dt} = \frac{1}{4\pi r^2}\frac{dV}{dt}$ . Substituting gives $\frac{dA}{dt} = 8\pi r \left(\frac{1}{4\pi r^2}\frac{dV}{dt}\right) = \frac{2}{r}\frac{dV}{dt}$ . At $r=5$ , $\frac{dA}{dt} = \frac{2}{5}(100) = 40$ cm²/s.
A stone is dropped into a calm pond, causing circular ripples. The radius of the outer ripple is increasing at a rate given by the equation $\frac{dr}{dt} = \frac{2}{r}$ cm/s, where $r$ is the radius in cm. At what rate is the area of the disturbed water increasing when the radius is 10 cm?
Answer: $4\pi$ cm²/s The area of a circle is given by $A = \pi r^2$ . To find the rate at which the area is increasing, we differentiate this equation with respect to time $t$ : $\frac{dA}{dt} = 2\pi r \frac{dr}{dt}$ . We are given that $\frac{dr}{dt} = \frac{2}{r}$ . Substituting this expression into our equation for $\frac{dA}{dt}$ gives $\frac{dA}{dt} = 2\pi r \left(\frac{2}{r}\right) = 4\pi$ . The rate of change of the area is a constant $4\pi$ cm²/s, independent of the radius. The information that the radius is 10 cm is not needed to find the rate, which is a key feature of this problem.

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