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 \(\theta\). The distance between the tether point and launch point remains constant, hence the side labeled as \(400\). 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.
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 |
\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.
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 \(300\). |
\(\theta\) | \(\fe{\tan}{\theta}=\frac{3}{4}\), which implies that \(\fe{\cos}{\theta}=\frac{4}{5}\). |
\(\lz{h}{t}\) | The value of this variable is \(0.88\). |
\(\lz{\theta}{t}\) | 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}&=\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}.