Activity 8.2 Diagrams for Related Rates

A particular challenge when working most related rates problems is identifying the quantity-variables. For example, when working a problem that involves a right triangle, there are six potential quantity-variables (the lengths of the three sides, the measurements of the two acute angles, and the area). In each related rates problem involving a right triangle, at least two of these measurements must be quantity-variables, or you will not end up with two (or more) rates to relate!

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Example 8.2.1.

A small drone is flying straight upward at the rate of 0.88 ^ft⁄_s. The drone is tethered to the ground (with a lot of slack) at a point that lies 400 ft west of the launch point of the drone. Assuming that the ground is perfectly level, determine the rate at which the angle of elevation from the tether point to the drone increases at the moment the drone is 300 ft above ground.

When searching for your quantity-variables, you need to identify sentences or phrases in the problem statement that contain the word “rate” or something similar (hence the R in DREDS). There are two such phrases in this problem:

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“A small drone is flying straight upward at the rate of…”
“…determine the rate at which the angle of elevation from the tether point to the drone increases…”

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The pieces of the right triangle that correspond to those phrases have been marked in Figure 8.2.2 as, respectively, \(h\) and \(\theta\text{.}\) The distance between the tether point and launch point remains constant, hence the side labeled as \(400\text{.}\) Please note that there should be no other annotations of any kind on this diagram. If writing other numbers down helps you think through the problem, draw a second triangle on your scratch paper and annotate that in any way that helps. That said, you need a clean variable diagram in your solution, so leave Figure 8.2.2 as it is.

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Now we can finish using the rest of the DREDS process.

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Table 8.2.3. Variable table for the drone problem

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Variable	Variable Description	Unit
\(h\)	The elevation of the drone \(t\) seconds after the drone begins to rise.	ft
\(\theta\)	The angle of elevation from the tether point to the drone \(t\) seconds after the drone begins to rise.	rad
\(\lz{h}{t}\)	The rate of change in elevation of the drone \(t\) seconds after the drone begins to rise.	^ft⁄_s
\(\lz{\theta}{t}\)	The rate of change in the angle of elevation from the tether point to the drone \(t\) seconds after the drone begins to rise.	^rad⁄_s
\(t\)	The amount of time that elapses after the drone begins to rise.	s

\begin{align*} \fe{\tan}{\theta} \amp= \frac{h}{400}\\ \lzoo{t}{\fe{\tan}{\theta}} \amp= \lzoo{t}{\frac{h}{400}}\\ \fe{\sec^2}{\theta}\lz{\theta}{t} \amp= \frac{1}{400}\lz{h}{t} \end{align*}

This is our related rates equation.

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At the moment the drone is 300 ft above the ground and rising at a rate of 0.88 ^ft⁄_s:

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Table 8.2.5. What we know about the variables at the described moment

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Variable	What we know
\(h\)	The value of this variable is \(300\text{.}\)
\(\theta\)	\(\fe{\tan}{\theta} = \frac{3}{4}\text{,}\) which implies that \(\fe{\cos}{\theta} = \frac{4}{5}\text{.}\)
\(\lz{h}{t}\)	The value of this variable is \(0.88\text{.}\)
\(\lz{\theta}{t}\)	This is the variable for which we need to solve.

From our related rates equation and the values in Table 8.2.5,

\begin{align*} \lz{\theta}{t} \amp= \frac{1}{400}\fe{\cos^2}{\theta}\lz{h}{t}\\ \lzoa{\theta}{t}{\fe{\cos}{\theta} = \frac{4}{5}, \lz{h}{t} = 0.88 } \amp= \frac{1}{400}\left(\frac{4}{5}\right)^2(0.88)\\ \amp= 0.001408\text{.} \end{align*}

When the elevation of the drone is 300 ft and the drone is rising at a rate of 0.88 ^ft⁄_s, the angle of elevation from the tether point to the drone is increasing at a rate of (about) 0.0014 ^rad⁄_s.

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Exercises Exercises

Exercise Group.

Three related rates problems are stated below. Each of the problems can be modeled using a right triangle. The sole task in these problems is to identify which pieces of the triangle correspond to quantity-variables and which piece of the triangle (if any) remains constant as the other pieces change.

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For each problem, you need to identify which two or three pieces of the triangle correspond to quantity-variables and which pieces remain constant or are not relevant. For each problem, draw a right triangle and label the pieces appropriately (finally a D for Diagram). When you are done, any given triangle should either have two pieces labeled as variables and one piece labeled with a constant value or three pieces labeled as variables.

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1.

A jet is climbing at a constant airspeed of 430 mph and at a constant angle of elevation of 12 °. At what rate is the elevation of the jet changing when the jet is 2640 ft above the ground? (Assume that the ground elevation below the jet’s path does not change as the jet makes its climb.)

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2.

A 17 ft ladder is propped against a very tall wall but the top of the ladder is slipping down the wall. At the moment the top of the ladder is 12 ft above the ground it is slipping at the rate of 0.2 ^ft⁄_s. At what rate is the distance between the foot of the ladder and the wall changing at that moment?

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3.

In Kansas, Highway A runs due north-south for 125 miles and Highway B runs due east-west for 222 miles. The elevation of both highways is a constant 725 feet along their entire lengths. The highways intersect at a point exactly halfway through their runs. One Tuesday a car is traveling east on Highway B at the constant speed of 76 mph and a truck is traveling north on Highway A at a constant speed of 84 mph. The car passes through the point of intersection at 4:12 pm and the truck passes through the point of intersection at 4:17 pm. At what rate is the distance between the two vehicles changing at 4:27 that afternoon? Assume that the speeds of the car and truck do not vary at all between 4:10 pm and 4:30 pm.

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4.

Go ahead and finish the three problems stated in Exercises 8.2.1–3. Remember to complete each step in the problem-solving process.

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Draw a properly labeled variable diagram. (D) Note that you completed this step while working Exercises 8.2.1–3.
Make a variable definition table. Include each variable type (quantity, rate, and time). Include the unit for each variable. (R)
Write down an equation that relates the two quantity-variables. (E)
Differentiate both sides of the equation you just wrote with respect to \(t\text{.}\) Simplify the result. This equation is called the related rates equation. (D)
Write an orientation sentence or phrase and create a summary table for your quantity and rate variables. (preparing for S)
Substitute the known values into the related rates equation and solve for the unknown rate. (S)
Write a conclusion.

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5.

When working with “common knowledge” formulas, it is important to identify which variables in the formula are really relevant to the stated problem and to then eliminate any irrelevant variables from the formula. Consider the following problem.

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Slushy is flowing out of the bottom of a cup at the constant rate of 0.9 ^cm³⁄_s. The cup is the shape of a right circular cone. The height of the cup is 12 cm and the radius at the top of the cup is 3 cm. Determine the rate at which the depth of the remaining slushy changes at the instant there are exactly 8\(\pi\) cm³ of slushy remaining in the cup.

The presence of the cubic-centimeter unit indicates that the problem is rooted in a volume formula. Specifically, we need the volume formula for a right circular cone which is \(V = \frac{\pi}{3}r^2h\) where \(r\) is the radius at the top of the cone and \(h\) is the height of the cone.

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In the stated problem, we are told the rate at which the volume of the remaining slushy changes and we are asked to determine the rate at which the depth of the remaining slushy changes. The rates correspond to \(\lz{V}{t}\) and \(\lz{h}{t}\) where \(V\) and \(h\) are, respectively, the volume (cm³) and depth (cm) of the slushy that remains in the cup \(t\) seconds after the slushy begins to flow out of the bottom of the cup.

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Since the value of \(\lz{r}{t}\) is neither given nor desired, we need to eliminate the variable \(r\) from the formula \(V = \frac{\pi}{3}r^2h\text{.}\) We do this by exploiting a property of similar triangles. Two triangles are said to be similar if and only if the three angular measurements of the first triangle are identical to the angular measurements of the second triangle. One thing that is always true about similar triangles is that their corresponding sides are proportional. This latter fact is the only reason we are able to define trigonometric functions in terms of the ratios of the sides of a right triangle!

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A diagram of the cup is shown in Figure 8.2.6. The radius of the cone at the top of the remaining slushy is labeled as \(r\) and the depth of the remaining slushy is labeled as \(h\text{.}\) The radius at the top of the cup as well as the height of the cup are also shown.

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Use the fact that the two right triangles in Figure 8.2.6 are similar to write a proportion that relates \(r\text{,}\) \(h\text{,}\) 3 cm, and 12 cm. Solve that equation for \(r\) and substitute the resultant formula into \(V = \frac{\pi}{3}r^2h\text{.}\) Simplify the resultant formula in terms of \(h\text{;}\) that’s the actual volume formula you want to use while solving the related rates problem—the formula that relates \(V\) and \(h\text{.}\)

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Go ahead and solve the related rates problem.

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