Remark 4.2.1.
This section relies heavily on implicit differentiation, so referring back to Section 2.6 may help.
Section 4.2 Related Rates
Remark 4.2.1.
Example 4.2.2. Understanding related rates.
The radius of a circle is growing at a rate of 5 in⁄h. At what rate is the circumference growing?
Solution.
The circumference and radius of a circle are related by \(C = 2\pi r\text{.}\) We are given information about how the length of \(r\) changes with respect to time; that is, we are told \(\lz{r}{t}\) is 5 in⁄h. We want to know how the length of \(C\) changes with respect to time, i.e., we want to know \(\lz{C}{t}\text{.}\)
Implicitly differentiate both sides of \(C = 2\pi r\) with respect to \(t\text{:}\)
\begin{align*}
C \amp = 2\pi r\\
\lzoo{t}{C} \amp = \lzoo{t}{2\pi r}\\
\lz{C}{t} \amp =2\pi \lz{r}{t}\text{.}
\end{align*}
As we know \(\lz{r}{t}\) is 5 in⁄h, we know
\begin{equation*}
\lz{C}{t} = 2\pi 5 = 10\pi \approx 31.4\text{ in/hr }\text{.}
\end{equation*}
Key Idea 4.2.3. Related Rates.
Example 4.2.4. Finding related rates.
Water streams out of a faucet at a rate of 2 in3⁄s onto a flat surface at a constant rate, forming a circular puddle that is 1/8 in deep.
- At what rate is the area of the puddle growing?
- At what rate is the radius of the circle growing?
Solution.
-
We can answer this question two ways: using “common sense” or related rates. The common sense method states that the volume of the puddle is growing by 2 in3⁄s, where\begin{equation*} \text{volume of puddle}=\text{area of circle}\times\text{depth}\text{.} \end{equation*}Since the depth is constant at 1/8 in, the area must be growing by 16 in2⁄s since \(16\cdot \frac{1}{8}=2\text{.}\) This approach reveals the underlying related rates principle.Now let’s solve the problem using Key Idea 4.2.3. Based on the problem description, the quantities that change with time are the volume of water (the volume of the puddle), the area of the circular puddle and the radius of the circle. We don’t need a diagram for this problem. The important variables for this part of the problem are the volume and area.Let \(V\) and \(A\) represent the Volume and Area of the puddle. We know \(V= A\times \frac{1}{8}\text{.}\) Take the derivative of both sides with respect to \(t\text{,}\) employing implicit differentiation.\begin{align*} V \amp = \frac{1}{8}A\\ \lzoo{t}{V}\amp = \lzoo{t}{\frac{1}{8}A}\\ \lz{V}{t} \amp = \frac{1}{8}\lz{A}{t} \end{align*}We know the change in volume, \(\lz{V}{t} = 2\text{,}\) so we substitute this value into our related rates equation: \(2 = \frac{1}{8}\lz{A}{t}\text{,}\) and hence \(\lz{A}{t} = 16\text{.}\) Thus the area is growing by 16 in2⁄s.
-
We already identified the quantities that are changing in Part 1. The variables of interest in this problem are the radius and the volume. We need an equation that relates the volume of the circle to the radius. Since the puddle is a right circular cylinder, we will use a known volume formula, \(V=\pi r^2 h\) where \(V\) is the volume of the puddle (in in3, \(r\) is the radius (in inches) and \(h\) is the height (i.e. depth) of the puddle in inches. (Notice that this formula is equivalent to \(V=\text{area} \times \text{depth}\text{.}\)) We know that the height (depth) is a constant \(1/8\) inch. Since this quantity does not change in the problem, we can safely substitute this value now.Implicitly derive both sides of \(V=\pi r^2 \frac{1}{8}\) with respect to \(t\text{:}\)\begin{align*} V \amp = \frac{1}{8} \pi r^2\\ \frac{d}{dt}\big(V\big) \amp = \frac{d}{dt}\left(\frac{1}{8}\pi r^2\right)\\ \lz{V}{t} \amp = \frac{1}{8} 2\pi r\lz{r}{t}\\ \lz{V}{t} \amp = \frac{1}{4} \pi r\lz{r}{t} \end{align*}We know that \(\lz{V}{t}\) is 2 in3⁄s. So we have:\begin{equation*} 2 = \frac14 \pi r\lz{r}{t} \end{equation*}Solving for \(\lz{r}{t}\text{,}\) we have\begin{equation*} \lz{r}{t} = \frac{8}{\pi r}\text{.} \end{equation*}Note how our answer is not a number, but rather a function of \(r\text{.}\) In other words, the rate at which the radius is growing depends on how big the circle already is. If the circle is very large, adding 2 in3⁄s of water will not make the circle much bigger at all. If the circle is dime-sized, adding the same amount of water will make a radical change in the radius of the circle.In some ways, our problem was (intentionally) ill-posed. We need to specify a current (instantaneous) value of the radius in order to know a rate of change. When the puddle has a radius of 10 in, the radius is growing at a rate of\begin{equation*} \lz{r}{t} = \frac{8}{10\pi} = \frac{4}{5\pi} \approx 0.25\text{ in/s }\text{.} \end{equation*}
Example 4.2.5. Studying related rates.
Radar guns measure the rate of distance change between the gun and the object it is measuring. For instance, a reading of “55 mph” means the object is moving away from the gun at a rate of \(55\) miles per hour, whereas a measurement of “-25 mph” would mean that the object is approaching the gun at a rate of \(25\) miles per hour.
If the radar gun is moving (say, attached to a police car) then radar readouts are only immediately understandable if the gun and the object are moving along the same line. If a police officer is traveling 60 mph and gets a readout of 15 mph, he knows that the car ahead of him is moving away at a rate of \(15\) miles an hour, meaning the car is traveling 75 mph. (This straight-line principle is one reason officers park on the side of the highway and try to shoot straight back down the road. It gives the most accurate reading.)
Suppose an officer is driving due north at 30 mph and sees a car moving due east, as shown in Figure 4.2.6. Using his radar gun, he measures a reading of 20 mph. By using landmarks, he believes both he and the other car are about \(1/2\) mile from the intersection of their two roads.
The passenger car is on an eastward road, while the police car is on a northward road, and these roads are perpendicular to each other. The police car is \(1/2\) units south of the intersection point, and the passenger car is \(1/2\) units east of the same intersection.
A straight line, labeled as \(C\text{,}\) connects the current positions of both the police car and the passenger car.
If the speed limit on the other road is 55 mph, is the other driver speeding?
Solution.
The important quantities that are changing are: the distance of the officer to the intersection, the distance of the car to the intersection, and the distance of the officer to the car. (There are other quantities that are changing as well such as the angles and area of the triangle, but these are not important to this problem.)
Using the diagram in Figure 4.2.6, let’s label what we know about the situation. As both the police officer and other driver are \(1/2\) mile from the intersection, we have \(A = 1/2\text{,}\) \(B = 1/2\text{,}\) and through the Pythagorean Theorem, \(C = 1/\sqrt{2}\approx 0.707\text{.}\) These values are “instantaneous” values for our variables, so we won’t use them until the end of the problem. Instead, we will use the variables \(A\text{,}\) \(B\text{,}\) and \(C\text{.}\)
We need an equation that relates \(A\text{,}\) \(B\text{,}\) and \(C\text{.}\) The Pythagorean Theorem is a good choice: \(A^2+B^2 = C^2\text{.}\) Differentiate both sides with respect to \(t\text{:}\)
\begin{align*}
A^2 + B^2 \amp = C^2\\
\lzoo{t}{A^2+B^2} \amp = \lzoo{t}{C^2}\\
2A\lz{A}{t} + 2B\lz{B}{t} \amp = 2C\lz{C}{t}
\end{align*}
We know the police officer is traveling at 30 mph; that is, \(\lz{A}{t} = -30\text{.}\) The reason this rate of change is negative is that \(A\) is getting smaller; the distance between the officer and the intersection is shrinking. The radar measurement is \(\lz{C}{t} = 20\text{.}\) We want to find \(\lz{B}{t}\text{.}\)
We have values for everything except \(\lz{B}{t}\text{.}\) Solving for this we have:
\begin{equation*}
\lz{B}{t} = \frac{C\lz{C}{t}- A\lz{A}{t}}{B}\text{.}
\end{equation*}
Now we substitue in our known rates and instantaneous values of our variables:
\begin{align*}
\lz{B}{t} \amp \approx \frac{0.707(20)- 0.5(-30)}{(0.5)}\\
\amp= 58.28 \text{ mph }\text{.}
\end{align*}
The other driver appears to be speeding slightly.
Example 4.2.7. Studying related rates.
A camera is placed on a tripod 10 ft from the side of a road. The camera is to turn to track a car that is to drive by at 100 mph for a promotional video. The video’s planners want to know what kind of motor the tripod should be equipped with in order to properly track the car as it passes by. Figure 4.2.8 shows the proposed setup.
A car is shown to the left of the image, driving at a 100 mph east, with a tripod placed 10 ft from the side of the road, set to track the car.
A right triangle is drawn, with the hypotenuse connecting the car to the tripod, and the other two sides parallel to the road and perpendicular to the road. The side parallel to the road has length \(x\text{,}\) and the angle between the hypotenuse and the side perpendicular to the road is labeled as \(\theta\text{.}\) The side perpendicular to the road has a length of 10 ft.
How fast must the camera be able to turn to track the car?
Solution.
The quantities that changing are \(x\) and \(\theta\) as drawn on Figure 4.2.8. (The hypotenuse of the triangle is also changing, but this isn’t important to the problem). We seek information about how fast the camera is to turn; therefore, we need an equation that will relate an angle \(\theta\) to the position of the camera and the speed and position of the car.
Figure 4.2.8 suggests we use a trigonometric equation. Letting \(x\) represent the distance the car is from the point on the road directly in front of the camera, we have
\begin{equation}
\tan(\theta) = \frac{x}{10}\text{.}\tag{4.2.1}
\end{equation}
Now take the derivative of both sides of Equation (4.2.1) using implicit differentiation:
\begin{align*}
\tan(\theta) \amp = \frac{x}{10}\\
\lzoo{t}{\tan(\theta)} \amp = \lzoo{t}{\frac{x}{10}}\\
\sec^2(\theta)\lz{\theta}{t} \amp = \frac{1}{10}\lz{x}{t}
\end{align*}
Now we solve for \(\lz{\theta}{t}\text{:}\)
\begin{equation}
\lz{\theta}{t} = \frac{\cos^2(\theta)}{10}\lz{x}{t}\tag{4.2.2}
\end{equation}
As the car is moving at 100 mph, we have that \(\lz{x}{t}\) is -100 mph (as in the last example, since \(x\) is getting smaller as the car travels, \(\lz{x}{t}\) is negative). We need to convert the measurements so they use the same units (we chose ft); rewrite -100 mph in terms of ft⁄s:
\begin{align*}
\lz{x}{t} \amp = -100\frac{\text{mi} }{\text{ hr } }\\
\amp = -100\frac{\text{mi} }{\text{ hr } }\cdot5280\frac{\text{ ft } }{\text{mi} }\cdot\frac{1}{3600}\frac{\text{ hr } }{\text{s} }\\
\amp =-146.\overline{6}\text{ ft/s }\text{.}
\end{align*}
We want to know the fastest the camera has to turn. Common sense tells us this is when the car is directly in front of the camera (i.e., when \(\theta = 0\)). Our mathematics bears this out. In Equation (4.2.2) we see this is when \(\cos^2(\theta)\) is largest; this is when \(\cos(\theta) = 1\text{,}\) or when \(\theta = 0\text{.}\) We also know that we should get an answer that is in rad⁄s. Since \(\cos(\theta)\) is a “dimensionless” measure, it won’t contribute to the units. However, radians are also dimensionless. This means we can write (or erase) the word “radian” without any unit consequences. (The same is not true of degrees — always convert degress to radians).
With \(\lz{x}{t}\) approximately -146.7 ft⁄s, we have
\begin{align*}
\lz{\theta}{t} \amp \approx -\frac{1}{10\text{ ft } }146.67\text{ ft/s }\\
\amp = -14.667\text{ radians/s }
\end{align*}
We find that \(\lz{\theta}{t}\) is negative; this matches our diagram in Figure 4.2.8 for \(\theta\) is getting smaller as the car approaches the camera.
What is the practical meaning of -14.667 rad⁄s? Recall that \(1\) circular revolution goes through \(2\pi\) radians, thus 14.667 rad⁄s means \(14.667/(2\pi)\approx 2.33\) revolutions per second. The negative sign indicates the camera is rotating in a clockwise fashion.
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