Activity8.2Diagrams for Related Rates¶ permalink

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!

Example8.2.1

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:

“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 increase…”

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 the helps. That said, you need a clean variable diagram in your solution, so leave Figure 8.2.2 as it is.

Figure8.2.2Variable Diagram

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

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

Table8.2.3Variable table for the drone problem

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

This is our related rates equation.

Figure8.2.4Variable Diagram

At the moment the drone is \(300\) ft above the ground and rising at a rate of \(0.88\) ^ft⁄_s.

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.

Table8.2.5What we know about the variables at the described moment

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

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

Subsection8.2.1Exercises

4

Go ahead and finish the three problems stated in Exercises 8.2.1.1–3. Remember to complete each step in the problems solving process.

Draw a properly labeled variable diagram. (D) Note that you completed this step while working Exercises 8.2.1.1–3.
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.

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.

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.

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.

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!

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.

Figure8.2.6Slushy Cone

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 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 \(h\text{.}\)

Go ahead and solve the related rates problem.