# Related Rates

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Objectives
• Understand the problem at hand and what related rates is trying to accomplish
• Learn the general approach to related rates questions
• Practice the half dozen or so very very common problem types teachers use for exams
Lesson Description

When observing an object moving in two dimensions, such as two trains moving away from one another, or a ladder falling down a wall, we can determine a relationship between the speeds of the objects in the system with some Calculus. This topic is many students' enemy, but while related rates problems are admittedly detailed, Mister Math has the step by step way to make these happen.

Practice Problems

Practice problems and worksheet coming soon!

## Multiple Movement Measures

Related Rates refers to calculus problems in which two or more related measurements are changing over time (usually distance, area, or volume). The rates at which the measurements are changing are represented as derivatives, and we'll be asked to solve for a number of unknowns. There are a handful of common question types, but any question can have unique aspects.There's no denying the truth - students collectively tend to dislike related rates problems at first. Part of the stigma is that these are word problems, and the other part is the challenge of setting up diagrams ourselves in conjunction with using in-context implicit differentiation.
Pro Tip
The keys to success in related rates problems are (1) a well-labeled diagram and (2) consistently tracking known and unknown information. Follow the four step method I preach for this topic and you'll be top banana on the test.
Related rates problems always involve several variables, each of which change over time. Therefore, in all of these problems, we will take implicit derivatives of relationships, but the variable of the derivative is always time ($t$).To get an idea of what these problems are like, let'a look at a classic related rates problem:

Example 1 (Setup Only)Two trains leave a station simultaneously. Train A travels due east at a constant speed of $60$ km/hr while Train B travels due north at a constant speed of $100$ km/hr.Find the rate at which the straight-line distance between the two trains is increasing after $30$ minutes.$\blacktriangleright$ This problem requires recognition of a few important details. First, that we have both given information and missing information, and that a diagram and variables would greatly help us keep track properly. Second, that the thing we're measuring probably isn't changing at a constant rate - otherwise, we wouldn't care about the time at which we observe the rate of change.Let's start with a diagram.The distance that Train A has traveled since leaving the station is $x$, and the distance Train B has since traveled is $y$. The direct distance between them is also a variable, which we have labeled $r$.Using these variables, we can better understand the given and missing information. We were given the rate at which each $x$ and $y$ is changing, and we want to know the rate at which $r$ is changing.
Warning!
Many students try to make a diagram using specific numbers, but this will not answer the question! We do not seek to know what $r$ is, but rather what the rate of change of $r$ is (i.e. the derivative of $r$). We will need the specified values of $x$, $y$, and $r$, but not until the end of the problem.
This example showcases the classic tasks that Related Rates problems ask of us - setting up and understanding the problem. Since the tasks are ones we'll need to repeat for each new problem we attempt, following my four-step approach is important if you want to succeed on this topic consistently.

## Related Rates - The Fail-safe Method

Though very problem is different, this approach will always work.
• 1. Draw and Label Everything
• 2. Write a relationship equation
• 3. Differentiate
• 4. Plug-in / Solve
Don't take Step 1 lightly - for many students it is the most important step!Let's execute Example 1 using this approach.

Example 1 (Full)For clarity, let's restate the problem:Two trains leave a station simultaneously. Train A travels due east at a constant speed of $60$ km/hr while Train B travels due north at a constant speed of $100$ km/hr. Find the rate at which the straight-line distance between the two trains is increasing after $30$ minutes.Draw and Label EverythingIn the set up above, we already discussed drawing a figure to understand the situation. My advice for your figure for any problem is
• Stick to lines and points when you can - don't actually try to draw the objects.
• Don't make the common mistake of labelling your figure with specific moment information - if a measurement is changing over time, it needs a variable label, not a number.
• Generally speaking, I don't write rates of change on the figure, but rather off to the side. Keep your diagram as clean as possible!
Again, our diagram for this problem isMake sure you always write down given information on the side, such as rates of change and at what instances they occur at, if applicable. You should also jot down the rate you're looking for to solve the problem.Here, we know$$\frac{dx}{dt} = 60$$$$\frac{dy}{dt} = 100$$and we want to calculate $\displaystyle \frac{dr}{dt}$.Write a Relationship EquationMake a connection between the variables you know about and the one whose rate you want to solve for. This is often a Pythagorean, area, volume, or other geometry related relationship, but every problem is a little different.For this problem, we are looking for $dr/dt$, the rate of change of the variable $r$. We were given some information about the variables $x$ and $y$, so we should try to relate those three variables. Unsurprisingly, we end up with$$x^2 + y^2 = r^2$$DifferentiateTake the derivative of the relationship implicitly. Remember that this is implicit because the variables in your problem (in this case, $x$, $y$, and $r$) are each varying measurements that are changing over time, so for example, the derivative of $x$ with respect to time is $dx/dt$, not $1$.The derivative of the variable relationship for this problem gives us$$2x \cdot \frac{dx}{dt} + 2y \cdot \frac{dy}{dt} = 2r \cdot \frac{dr}{dt}$$
Remember!
When you take derivatives of $x$ in typical coursework while we learn derivatives, we are taking $x$ derivatives of $x$ expressions. That is why the derivative of $x$ is $1$ in those cases, and why the derivative of $x^3$ is $3x^2$. When $x$ is a variable that changes with time, we are taking $t$ derivatives of $x$, so the derivative of $x$ is $dx/dt$, and the derivative of something like $x^3$ would be$$\frac{d}{dt} \, x^3 = 3x^2 \cdot \frac{dx}{dt}$$
Plug-in / SolveHaving differentiated the generic relationship between the variables in the problem, we can plug in the measures at the specific moment we were asked to look at.This problem asked us to find $dr/dt$ at the instant when $30$ minutes have passed. Using the classic linear time-distance relationship (distance = speed $\times$ time), and the fact that Train A is travelling at a constant speed of $60$ km/hr and Train B is travelling at a constant speed of $100$ km/hr, we know that, at the moment when $30$ minutes (or $0.5$ hours) have elapsed, the positions of the trains are$$x = 30$$$$y = 50$$and the distance $r$ between the two trains at that instance is$$r = \sqrt{30^2 + 50^2} = 10\sqrt{34}$$We also already wrote down that$$\frac{dx}{dt} = 60$$$$\frac{dy}{dt} = 100$$and so, we can solve for the desired unknown in our equation: $dr/dt$.$$2x \cdot \frac{dx}{dt} + 2y \cdot \frac{dy}{dt} = 2r \cdot \frac{dr}{dt}$$$$2(30)(60) + 2(50)(100) = 2(10\sqrt{34}) \cdot \frac{dr}{dt}$$$$\longrightarrow \frac{dr}{dt} = \frac{680}{\sqrt{34}}$$$$\approx 116.62$$Therefore, the straight-line distance between the two trains is increasing at a rate of about $117$ km/hr, at the moment that $30$ minutes have elapsed. $\blacksquare$

## Common Problem Types

We already saw the moving vehicle problem, but here are a few other frequent fliers.Ladder ProblemsThese problems involve a ladder sliding up or down a wall, and are usually simpler problems because the ladder is a fixed length and the relationship we will use is the Pythagorean one.

Example 2A $16$ foot ladder is sliding down a wall such that the top of the ladder is moving at a constant rate of $3$ ft/sec.How fast is the bottom of the ladder moving along the ground when the top of the ladder is $6$ feet off the ground?$\blacktriangleright$ Let's start by marking up our figure, labelling the measurements that change over time.As usual, we should also write down what we know and what we seek. The top of the ladder is moving at a constant $3$ ft/sec, so we know$$\frac{dy}{dt} = -3$$Note that we must sign $dy/dt$ as negative because $y$ is shrinking as time progresses.The problem is asking us how fast the bottom of the ladder is moving when $y$ is $6$ so we want$$\frac{dx}{dt} \, , \mathrm{when} \:\: y=6$$Next we need a relationship. This diagram lends itself well to using the Pythagorean Theorem. From the figure, at any instant, it is true that$$x^2 + y^2 = 10^2$$Differentiating that relationship, we have$$2x \cdot \frac{dx}{dt} + 2y \cdot \frac{dy}{dt} = 0$$At the moment we're interested in, we know $y$ is $6$. Based on the Pythagorean relationship, that means$$x^2 = 10^2 - 6^2$$$$\longrightarrow \;\; x = 8$$And, since we already know $dy/dt$ from the given information, we are ready to solve for what we want.$$2x \cdot \frac{dx}{dt} + 2y \cdot \frac{dy}{dt} = 0$$$$\Rightarrow \;\; 2(8) \cdot \frac{dx}{dt} + 2(6) \cdot \left( -3 \right) = 0$$$$\frac{dx}{dt} = \frac{36}{16} = \frac{9}{4}$$Therefore, the bottom of the ladder is moving along the floor at a rate of $2.25$ ft/sec, at the instant that $y=6$ ft. $\blacksquare$

Volume ProblemsThese problems involve a geometry volume formula from your previous high school coursework as the key relationship.

Example 3A conical sand pile, with a height that is four times its radius, is growing as more sand is being dropped on top of it, as shown in the figure.Sand is being added to the pile at a constant rate of $1.6$ m3/hr. As the pile grows, it retains its shape. Find the rate at which the height is changing at the instant that the radius is $5$ meters.$\blacktriangleright$ Write down what we know and what we want. Let's pick $V$ to represent the volume, since it is not directly part of the figure.$$\mathrm{Know:} \;\; \frac{dV}{dt} = 1.6$$$$\mathrm{Want:} \;\; \frac{dh}{dt} \;\; \mathrm{when} \;\; r=5$$Here we will see the need to use two relationships. Intuitively, we start with the one that relates what we have and what we want - the volume of a cone.$$V = \frac{\pi r^2 h}{3}$$Let's differentiate this relationship using the product rule:$$\frac{dV}{dt} = \frac{\pi}{3} \, \left[ r^2 \cdot \frac{dh}{dt} + 2r \cdot \frac{dr}{dt} \cdot h \right]$$We know $dV/dt$, we want to solve for $dh/dt$, but we weren't given any information about $dr/dt$.Instead, we have to use a second relationship specific to this problem that was described initially:$$h = 4r$$(as stated, the height is equal to four times the radius length, and the pile retains its shape over time)We can glean a relationship between $dr/dt$ and $dh/dt$ by taking the derivative of this relationship.$$\frac{dh}{dt} = 4 \, \frac{dr}{dt}$$or$$\frac{1}{4} \, \frac{dh}{dt} = \frac{dr}{dt}$$Now in our volume derivative equation, we can replace $dr/dt$.$$\frac{dV}{dt} = \frac{\pi}{3} \, \left[ r^2 \cdot \frac{dh}{dt} + 2r \cdot \frac{dr}{dt} \cdot h \right]$$$$\longrightarrow$$$$\frac{dV}{dt} = \frac{\pi}{3} \, \left[ r^2 \cdot \frac{dh}{dt} + 2r \cdot \frac{1}{4} \cdot \frac{dh}{dt} \cdot h \right]$$Finally, we can calculate the rate we seek using the specifics of the instant in question. When $r$ is $5$, we know $h$ must be $20$. Along with the other given info, we have$$1.6 = \frac{\pi \cdot 5^2}{3} \cdot \frac{dh}{dt} + \frac{2(5)\cdot 20}{4} \cdot \frac{dh}{dt}$$$$\frac{dh}{dt} = \frac{1.6}{\frac{25 \pi}{3} + 50} = \frac{4.8}{25 \pi + 150}$$$$\frac{dh}{dt} \approx 0.021 \mathrm{m/hr}$$If you show your work, you probably don't need to try and rationalize or clean up that second-to-last step. Whenever there is $\pi$ in the denominator, it's usually acceptable to turn the exact answer into a decimal approximation. $\blacksquare$

Shadow ProblemsShadow problems get a bad reputation because they require being setup in a particular way, and because there are two kinds that seem similar on the surface but require different approaches. However, you'll find that once you've done some practice, they will seem like routine procedures.This example is the more intuitive type of shadow problem. Make sure to try Example 13 below to see the other type.

Example 4A $20$ ft lamppost casts a shadow on a $7$ foot man who is walking directly away from the lamppost at a rate of $2$ feet per second. Find the rate at which his shadow is growing at the instance that he is $16$ feet from the lamppost.$\blacktriangleright$ The figure is especially important in these types of problems, because while there are a few distinct measurements that are changing size over time, you want to focus on giving labels to the pieces you care about - the ones you're asked about.Here's the figure before we add labels.As the man walks away, the lengths that are changing are
• the total length of the light beam (hypotenuse)
• the length of the light beam between the man and the ground
• the distance between the post and the man
• the distance between the post and the tip of the shadow
• the distance between the man and the tip of the shadow
Because this problem wants us to find the rate at which his shadow is growing, we should give his shadow a variable label.We should also label the distance between the post and the man with another variable, because we were given information about that distance - we were told that the rate at which he is walking away from the pole is 2 miles per hour.Our updated figure looks likeAnd our given information should be written using the variables we've chosen:$$\frac{dy}{dt} = 2$$In context of our selected variables, the problem is asking us to find $dx/dt$ at the instant that $y$ is $16$. Let's move on to the next step of finding a generic relationship.While the shape of the triangles in the problem is changing as the man walks away from the lamppost, the two triangles in the figure are similar at any instant due to the fixed heights of the post and the man. Therefore we know that the following proportion is always true:$$\frac{7}{20} = \frac{x}{x+y}$$Before we take derivatives, let's cross multiply and obtain a linear relationship.$$\longrightarrow \; 7(x+y) = 20x$$$$7x + 7y = 20x$$$$7y = 13x$$Now differentiate.$$7\cdot \frac{dy}{dt} = 13 \cdot \frac{dx}{dt}$$Finally, plug in what we know about $dy/dt$.$$7(2) = 13 \cdot \frac{dx}{dt}$$$$\frac{dx}{dt} = \frac{14}{13}$$Therefore, $x$ is changing at a rate of $14/13$ ft/sec. Notice that in this problem, it didn't matter what moment we were examining. $\blacksquare$

## Mr. Math Makes It Mean

Related Rates InceptionOne way in which questions become decidedly more mean is to be asked to find a rate that depends on another unknown rate, which you must first find.

Example 5A $10$ foot ladder is sliding down a wall because it is being pulled away from the wall at the bottom of the ladder at a constant rate of $3$ ft / sec. Find the rate at which the area of the triangle formed between the ladder, the wall, and the floor is changing at the instant that the top of the ladder is $6$ feet from the ground.$\blacktriangleright$ First let's get a figure drawn, with the appropriate variable labels.We'll call the triangular area formed between the wall, ladder, and floor $A$. Let's note also that we are given $dx/dt$, which is a constant positive $2$ (because $x$ is growing over time), and the fact that we seek $dA/dt$ to answer the question.Since we seek $dA/dt$, and $A$ is the area of a right triangle, the relationship we will use is$$A = \frac{xy}{2}$$Let's take the derivative of this relationship and see what we see. By the product rule:$$\longrightarrow \frac{dA}{dt} = \frac{x}{2}\cdot \frac{dy}{dt} + \frac{y}{2}\cdot \frac{dx}{dt}$$We know $dx/dt$ is $3$ but we don't know $dy/dt$. We need to determine its value at the instant we are examining, so it's like its own mini related rates problem within a related rates problem.Start with a relationship between $x$, whose rate we know about, and $y$, whose rate we don't. The ladder is a fixed length, so we know that no matter what the location of the latter is at any time during its fall, it is true that$$x^2 + y^2 = 100$$and so$$2x \; \frac{dx}{dt} + 2y \; \frac{dy}{dt} = 0$$Having found a generic relationship of their derivatives, we can plug in the information about the specific moment of interest. At the instant that $y$ is $6$, we know also that $x$ is $8$ (coming from the relationship $x^2 + y^2 = 100$). We also know that $dx/dt$ is $3$ regardless of instant, so we can now solve for $dy/dt$:$$2x \; \frac{dx}{dt} + 2y \; \frac{dy}{dt} = 0$$$$\longrightarrow 2(8)(3) + 2(6) \cdot \frac{dy}{dt} = 0$$$$\longrightarrow \frac{dy}{dt} = - 4$$As a quick check, it makes sense that $dy/dt$ is negative because $y$ is getting smaller as time progresses.Finally we can return to the question at hand about area, and use the fact that $dy/dt = -4$ at the instant that we are examining. Along the way, we also figured out the values of $x$ and $y$ at that moment.$$\frac{dA}{dt} = \frac{x}{2}\cdot \frac{dy}{dt} + \frac{y}{2}\cdot \frac{dx}{dt}$$$$\frac{dA}{dt} = \frac{8}{2} \cdot (-4) + \frac{6}{2} \cdot (3)$$$$= -16 + 9 = -7$$As always, be sure to turn in an answer with units.$\longrightarrow$ Full Answer: The area of the triangle formed between the ladder, the wall, and the floor is shrinking at a rate of $7$ ft2 / sec at the moment that the top of the ladder is $6$ ft from the floor. $\blacksquare$
Warning!
In a problem like this, it is tempting to assume that $dy/dt$ is the same as $dx/dt$, because the problem intentionally didn't tell us about $dy/dt$. Never make assumptions about rates! The only things you should take for granted in related rates problems are fixed lengths (such as the ladder in this example) and geometric formulas (such as the area of a triangle in this example).

Backward Related RatesBackward problems give you rates and ask you about what moment is happening. These are often confusing but also fairly rare, even for AP BC level problems.

Example 6 [calculator required]Two trains left the same station. Train X is traveling due east and Train Y is traveling due north. The position of Train X is given by the function$$x(t) = 3t^2 + 2$$while the position of Train Y is given by the function$$y(t) = \sin(t) + 1$$where $t$ is measured in hours and $x$ and $y$ are measured in miles.At the moment that the distance between them is changing at a rate of $5$ mi/hr, find the position of each train.$\blacktriangleright$ We'll still want to tackle this problem by following the four steps. The only real oddity here is that the given information consists of all rates, and the thing we want to solve for isn't a rate for once.Let's start with a diagram, including variable labels.Off to the side, let's write the given information:$\displaystyle x(t) = 3t^2 + 2$$\displaystyle y(t) = \sin(t) + 1$$\displaystyle D'(t) = 5$ at the moment we're interested inNext, identify the generic relationship among the variables in the problem. Unsurprisingly,$$\left( D(t) \right)^2 = \left( x(t) \right)^2 + \left( y(t) \right)^2$$or, using implicit shorthand,$$D^2 = x^2 + y^2$$and so, we can take the derivative of this relationship with respect to time and obtain$$2DD' = 2xx' + 2yy'$$At the moment we are looking at, we know $D'$. Additionally, we can use the explicit relationships we have to ultimately end up with an equation that only has one variable in it - $t$.$$D = \sqrt{x^2 + y^2}$$$$x = 3t^2 + 2$$$$y = \sin(t) + 1$$$$\therefore$$$$x' = 6t$$$$y' = \cos(t)$$Putting all of this together, $\displaystyle 2D D' = 2xx' + 2yy'$ becomes$$2 \left( \sqrt{\left( 3t^2 + 2 \right)^2 + \left( \sin(t) + 1 \right)^2} \right) (5) = 2 \left( 3t^2 + 2 \right) (6t) + 2 \left( \sin(t) + 1 \right) \left(\cos(t) \right)$$This is where the calculator requirement becomes apparent. This isn't solvable algebraically but the value of $t$ that makes this true can be obtained fairly quickly from your calculator, either by using a "solver" function or a graphing approach (graph each side of the equation and find where they intersect).With some quick button mashing, you'll find that when $t \approx 0.9335$, the equation has a solution.Therefore, plug this value of $t$ into each position function to answer the question.$$x(0.9335) \approx 4.6143$$$$y(0.9335) \approx 1.8037$$Therefore Train X is about $4.61$ miles east of the station, and Train Y is about $1.8$ miles north of the station. $\blacksquare$
Remember!
The calculator solving trick we saw in Example 6 is a commonly needed approach in many AP problems, both for multiple choice and written response questions, among all topics on the syllabus. This is why the "calculator required" sections of AP exams are not "calculator optional" - they are truly necessary to answer the question.

## Put It To The Test

Example 7A $3$ cm by $5$ cm rectangle is growing. The initially short side length is increasing at a rate of $4$ cm per minute, while the initially longer side is increasing at a rate of $1$ cm per minute. Find the rate at which the area of the rectangle is changing $10$ minites later.
Show solution
$\blacktriangleright$ First draw a figure, keeping in mind that measurements that change over time should have variables as labels, not numbers.I have labelled the side that was initially $3$ cm as $y$, and the initially longer side as $x$.Write down what we know and what we want to know in context of the drawn figure:$$\frac{dy}{dt} = 4$$$$\frac{dx}{dt} = 1$$$$\frac{dA}{dt} = ?$$Again, note that this given information depends on how we drew and labelled our figure. Next, we need a relationship involving the variables we have and the area, since that's what we're asked about.$$A = xy$$Finally, we'll take the derivative of this relationship and then plug in the specific values at the moment we're interested in. Using the product rule:$$\frac{dA}{dt} = x \cdot \frac{dy}{dt} + y \cdot \frac{dx}{dt}$$And, when we're looking at $10$ minutes after the initial start of the problem, $x$ is $5 + 10(1) = 15$ (initially it was $5$ long and grew at a rate of $1$ per minute). Similarly, at this moment, $y$ is $3 + 10(4) = 43$.Putting this all together,$$\frac{dA}{dt} = (15)(4) + (43)(1) = 103$$Therefore, $10$ minutes later, this rectangle is growing at a rate of $103$ cm2 / min. $\blacksquare$

Example 8A ship loses a barrel of oil in the ocean, and it begins to leak, slowly at first and then worsening, spreading a circle-shaped spill as shown in the figure.Find the rate at which the area of the spill is changing when the radius of the spill is $8$ ft, if the radius is increasing at a rate of $2$ ft/min at that instant.
Show solution
$\blacktriangleright$ We already have a figure provided to us, so let's write down the given and wanted information. Let $r$ represent the radius of the spill at a given time, and let $A$ represent the area of the spill at a given time.$$\frac{dr}{dt} = 2 \;\; \mathrm{when} \;\; r = 8$$$$\frac{dA}{dt} = \mathrm{?}$$Next, we need a relationship between the variables $r$ and $A$. This is one you're definitely expected to remember from geometry!$$A = \pi r^2$$Next, we'll take the derivative of this relationship implicitly with respect to time:$$\frac{dA}{dt} = 2 \pi r \, \frac{dr}{dt}$$Now that we have a generic relationship between the derivatives of the quantities in the problem, we can use this relationship to inspect the specific instance at which the problem is asking us about. At the instant that $r=8$, we were also told that $dr/dt$ is 2, so we have$$\frac{dA}{dt} = 2 \pi (8) \cdot (2)$$$$= 32\pi$$Therefore, at the instant that $r$ is $8$, the area of the spill is changing at a rate of $32\pi$ ft2 / min (make sure you write down the units!). $\blacksquare$

Example 9A boat is tied to a dock, as shown in the figure. The top of the boat is $2$ feet from the surface of the water, and the top of the dock is $8$ feet from the surface of the water.A child comes along and begins to pull the rope in at a constant rate of $4$ ft / sec, causing the boat to move along the water toward the dock.When there is $20$ feet of rope left between the child and the boat, how fast is the boat moving along the water?
Show solution
$\blacktriangleright$ Let's update our figure with labels. The rope length and the horizontal distance from boat to dock are our two variable measures. Note also that the vertical distance between the boat and the dock is $6$ feet.Write down what we know and what we want to know:$$\frac{dr}{dt} = -4 \;\; \mathrm{ft / sec}$$$$\frac{dx}{dt} = ? \;\; \mathrm{ft / sec}$$Note that $dr/dt$ is negative because the rope length is getting shorter as the boat is being pulled in. Relate the variables in the figure with a relationship, and then differentiate. Here, a simply pythagorean relationship will suffice.$$6^2 + x^2 = r^2$$$$\frac{d}{dt} \left[ 6^2 + x^2 \right] = \frac{d}{dt} \left[ r^2 \right]$$$$2x \; \frac{dx}{dt} = 2r \; \frac{dr}{dt}$$At the moment we are examining, we know $r$ is $20$, and $dr/dt$ is $-4$. Since we are seeking to solve for $dx/dt$, we need to find the value of $x$ at that instant.Looking back at the relationship between $x$ and $r$:$$36 + x^2 = r^2$$$$x^2 = r^2 - 36$$$$x^2 = 20^2 - 36 = 364$$$$x = \sqrt{364} = 2\sqrt{91}$$We can now finish by solving for $dx/dt$ in our equation above.$$2x \; \frac{dx}{dt} = 2r \; \frac{dr}{dt}$$$$2\left( 2\sqrt{91} \right) \cdot \frac{dx}{dt} = 2(20) \cdot (-4)$$$$\frac{dx}{dt} = \frac{-160}{4\sqrt{91}}$$$$= -\frac{40}{\sqrt{91}}$$Therefore, the boat is moving toward the dock at a rate of $40/\sqrt{91}$ ft / sec at the instant that there is $20$ feet of rope left. Note that our answer is appropriately negative because the measurement $x$ in our diagram is shrinking as time goes on. $\blacksquare$

Example 10Bored of merely dropping them, Galileo is throwing objects straight down off of the leaning tower of Piza, and an observer is wondering how fast they are falling. The observer records his angle of elevation on the line of sight of objects as they fall.When the angle of elevation is $60^{\circ}$, the observer notes that the line of sight angle of elevation is changing at a rate of $5^{\circ}$ per second. The observer is $100$ feet from the drop line.Find the rate at which the object is falling at that instant.
Show solution
$\blacktriangleright$ Update the figure with the variable measurements. In this case, the angle of the observer's sight is changing over time as well as the distance between the falling object and the ground. Let's label the angle of sight $\theta$, and label the height of the falling object $y$.Now let's jot down what we know and what we want to know:$$\frac{d \theta}{dt} = -5^{\circ} \;\; \mathrm{per} \;\; \mathrm{sec}$$$$\frac{dy}{dt} = ?$$However, whenever we work with angles in Calculus, they must be measured in radians, so let's adjust our $d \theta / dt$ rate:$$\frac{d \theta}{dt} = -\frac{5 \pi}{180} = - \frac{\pi}{36}$$As usual, we'll need a relationship between the variables in our figure. In this case, the clearest right triangle relationship between $\theta$ and $y$ is via SOH-CAH-TOA.$$\tan(\theta) = \frac{y}{100}$$Now we can proceed as we often do by taking the derivative of both sides to obtain the relationship of derivatives, and then apply the measurements of the specific moment we want to know about.$$\frac{d}{dt} \; \tan(\theta) = \frac{d}{dt} \frac{y}{100}$$$$\sec^2 (\theta) \cdot \frac{d \theta}{dt} = \frac{1}{100} \cdot \frac{dy}{dt}$$At the moment that $\theta$ is $60^{\circ}$, or, in radians, $\pi / 3$, we have$$\sec^2 \left( \frac{\pi}{3} \right) \cdot \frac{\pi}{36} = \frac{1}{100} \cdot \frac{dy}{dt}$$Solving for $\displaystyle \frac{dy}{dt}$:$$\frac{dy}{dt} = \frac{400 \pi}{36} = \frac{100 \pi}{9}$$Therefore, the rate at which the object is falling at the instant that the observer's line of sight is at a $60^{\circ}$ angle of elevation is $\displaystyle \frac{100 \pi}{9}$ feet per second (try and follow where the units come from!). $\blacksquare$

Example 11A baseball diamond is a square with side lengths of $90$ feet. A runner at first base is running to second base, as shown in the diagram.If the runner can move at $20$ ft / sec, how fast is the distance between the runner and home plate changing when the runner is $10$ feet from reaching second base?
Show solution
$\blacktriangleright$ As always, start with a well-labelled diagram and remember to mark measurements that change over time as variables.Now, using our variable labels, let's write down our information.$$\frac{dy}{dt} = 20$$$$\frac{dr}{dt} = ?$$Hopefully the relationship between the variables is clear: a pythagorean relationship.$$r^2 = 90^2 + y^2$$Differentiating with respect to time,$$\frac{d}{dt} \; r^2 = \frac{d}{dt} \left[ 90^2 + y^2 \right]$$$$2r \; \frac{dr}{dt} = 2y \; \frac{dy}{dt}$$We seek to find the value of $dr/dt$ at the specified moment. Looking at the diagram, when the runner is $10$ feet from reaching second base, that means that $y$ is $80$, since the problem told us that the bases are $90$ feet apart. We also need to find out what $r$ is at that moment. Using the relationship between $r$ and $y$:$$r^2 = 90^2 + y^2$$$$r^2 = 90^2 + 80^2$$$$r = \sqrt{14500} = 10\sqrt{145}$$Finally, we will plug in these moment-specific measurements into the derivatives relationship of $r$ and $y$.$$2r \; \frac{dr}{dt} = 2y \; \frac{dy}{dt}$$$$2 \left( 10\sqrt{145} \right) \cdot \frac{dr}{dt} = 2(80) \cdot 20$$$$\frac{dr}{dt} = \frac{160}{\sqrt{145}}$$Answer: at the moment that the runner is $10$ feet from second base, the distance between the runner and home plate is changing at a rate of $160/\sqrt{145}$ feet per second. $\blacksquare$
Remember!
If you're still having trouble understanding why we say differentiate "with respect to time" or "with respect to $t$", and the variable $t$ shows up, even though we don't explicitly see the variable $t$ in the problem, remember that technically the variables that we use in related rates problems are actually functions. For example, in Example 11, $r$ represented the distance between home plate and the runner, but $r$ changes as time passes, so technically, we could have called that distance $r(t)$, not simply $r$. However the shorthand practice of writing a variable only is common because we will not explicitly define what $r(t)$ is as a written function.

Example 12A perfectly spherical balloon is being inflated for a birthday party at a rate of $50$ cm3 / sec. How fast is the surface area of the balloon increasing at the moment that the diameter of the balloon is $20$ cm?
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$\blacktriangleright$ You may or may not need to rely on a diagram for this problem given that the only variable that matters in measuring a sphere is its radius, which we'll call $r$. You can still draw a circle if you prefer, but the real challenge here is understanding the given and needed information, and remembering the formulas for the volume and surface area of a sphere.And how do we know that volume is a required formula for this problem when all they are asking for is surface area? Because 1) the rate at which the radius is changing is not given, and 2) the given information tells us about the rate of volume increase, which is also indicated by the units of measure ($50$ cm3).The volume of a sphere is given by the formula$$V = \frac{4\pi r^3}{3}$$While the given information did not give us the volume, it did give us the rate of change of the volume, which is $dV/dt$. Let's differentiate the volume formula:$$\frac{dV}{dt} = \frac{12 \pi r^2 }{3} \cdot \frac{dr}{dt}$$$$\frac{dV}{dt} = 4 \pi r^2 \cdot \frac{dr}{dt}$$At the moment at which we are asked to examine, the radius is $10$ cm. Careful not to incorrectly use the given diameter length of $20$ cm in place of the radius (a common trick teachers use)! Now we can solve for $dr/dt$.$$50 = 4 \pi (10)^2 \cdot \frac{dr}{dt}$$$$\frac{dr}{dt} = \frac{1}{8 \pi}$$Now that we know how fast the radius is changing at the moment we want to know about, we can use that information to answer the question that the problem is asking us.To find out how fast the surface area is changing, we'll start with the formula for the surface area of a sphere.$$S = 4 \pi r^2$$Let's differentiate:$$\frac{dS}{dt} = 8 \pi r \cdot \frac{dr}{dt}$$Because we used the given information of the problem to discover the value of $dr/dt$ at the moment that the diameter is $20$ cm, we can finish the problem quickly.$$\frac{dS}{dt} = 8 \pi (10) \cdot \frac{1}{8 \pi}$$$$\frac{dS}{dt} = 10$$Therefore, at the moment that this inflating spherical balloon has a diameter of $20$ cm, the surface area is changing at a rate of $10$ cm2 / sec. $\blacksquare$

Example 13A $2$ m tall man is walking away from a $8$ m tall lamppost at a rate of $1$ m/s. How fast is the tip of the man's shadow moving when he is $10$ m away from the base of the post? Assume the light source is at the exact top of the lamppost.
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$\blacktriangleright$ Let's start with a diagram.Many students have trouble setting up moving shadow problems, but the key to success is to focus on what measurements are changing over time, and to zone in on the measurements we care about. The problem wants us to find how fast the tip of the shadow is changing. Based on our diagram and labels, that is the same as asking how fast $x$ is increasing. Even though $x$ is the entire distance from the lamp post to the end of the shadow, the rate at which that measurement is growing is also the rate at which the shadow tip is moving, because the other end of that length is the base of the lamp post, which is stationary.Now that we understand that the answer to our problem is to find $dx/dt$, let's write down what we know as well.$$\frac{dx}{dt} = ?$$$$\frac{dy}{dt} = 1$$The instant at which we are asked to find this rate of change is when $y$ is $10$ m. Let's try to relate $x$ and $y$.Looking at the diagram, we have two similar triangles to work with. Using proportionality:$$\frac{8}{x} = \frac{6}{z}$$However, we don't want to work with $z$, since it is not part of our given information or desired rate. Using the diagram, it makes sense that we can represent $z$ in terms of the two variables that are part of our information: $z = x-y$. Now we have$$\frac{8}{x} = \frac{6}{x-y}$$Before differentiating a proportion like this, cross multiply to obtain a linear style relationship.$$\longrightarrow 8(x-y) = 6x$$$$8x - 8y = 6x$$$$2x = 8y$$$$x = 4y$$Now, differentiate with respect to time.$$\frac{dx}{dt} = 4 \cdot \frac{dy}{dt}$$At the moment we want to know about, $dy/dt$ is $1$, so we have$$\frac{dx}{dt} = 4 \cdot 1 = 4$$Therefore, the tip of the man's shadow is moving at a rate of $4$ m/s when he is $10$ m away from the base of the post.It's fair to point out that the $dx/dt$ relationship only depended on $dy/dt$, so it didn't actually matter that he was $10$ m from the post - the shadow tip in this situation moves at a rate of $4$ m/s no matter how far he is from the lamp. $\blacksquare$
Pro Tip
As you practice more related rates problems, continue to focus on which parts of the diagram are important. When you have choices to make about which parts of the diagram to use, focus on the measurements that are involved with the question being asked. As always, practice makes perfect!

Lesson Takeaways
• Understand the premise of Related Rates problems
• Become a master of figure drawing and labelling
• Know the four step process to handle any related rates problem
• Become very familiar with the usual kinds of problems you'll encounter
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