Consider the following problem:

problem. A traffic enforcement camera lies thirty feet straight from the road. A vehicle approaches the camera, and the camera's radar bounces off the vehicle. The camera reads the vehicle as fifty feet away. The camera also reads vehicle as travelling at eighty feet per second. The speed limit is ninety feet per second (or sixty-five miles per hour). Question: Is the vehicle speeding?

As usual, we begin by drawing a diagram. We know that the camera lies thirty feet straight from the road, so it forms a right angle. And since the camera reads the vehicle as fifty feet away, we have a Heronian triangle,¹ specifically a ${3:4:5}$ triangle:

The next step is to name our quantities. We'll use ${t}$ for time, measured in seconds.

t \coloneqq time~~(\text{s})

Next, we want to denote some quantities directly on our diagram:

In our diagram above, we made the following definitions:

${x = 40}$ (the ${x}$ distance from the camera)
${D = 50}$ (the ${x}$ distance between the camera and vehicle)
${\text{SL} = 90 \frac{\text{ft}}{\text{s}}}$ (the speed limit)
${\dfrac{dD}{dt} = 80 \frac{\text{ft}}{\text{s}}}$ (the approach speed)

Why did we denote ${x = 40?}$ Because the problem boils down to determining whether this proposition is true:

\dfrac{dx}{dt} > (\text{SL} = 95)

We also denoted ${D = 50}$ because the dstiance between the camera and the vehicle varies. This is all to show that whenever we deal with variable quantities, we always want to give them variable names. Constants, on the other hand, come down to personal style. In some problems, we might want to give constants variable names to avoid numeric distractions. In others, we might not want to give constants separate names because numeric quantities can provide intuition.

Now, the first thing to note is that we're moving towards the camera. This means that the approach speed is actually:

\dfrac{dD}{dt} = -80 \frac{\text{ft}}{\text{s}}

In other words, the speed is decreasing at ${80 \frac{\text{ft}}{\text{s}}.}$ Returning to our diagram, we want to express a relationship between the triangle's lengths. Here, we rely on the pythagorean theorem:

x^2 + 30^2 = D^2

At this point, we'll want to differentiate. Here, the easiest approach is to use implicit differentiation. We'll differentiate the entire equation with respect to time. In doing so, we want to keep mind not to just substitute ${40}$ into ${x.}$ Why? Because ${x}$ is variable. Performing the implicit differentiation:

\begin{aligned} \dfrac{d}{dx} x^2 + \dfrac{d}{dx} 30^2 &= \dfrac{d}{dx} D^2 \\[1em] 2x \dfrac{dx}{dt} + 0 &= 2D \dfrac{dD}{dt} \\[1em] 2x \dfrac{dx}{dt} &= 2D \dfrac{dD}{dt} \end{aligned}

With the derivative in place, now we can substitute ${x}$ for ${40:}$

\begin{aligned} 2(40) \dfrac{dx}{dt} &= 2(50)(-80) \\[1em] 80 \dfrac{dx}{dt} &= 2(50)(-80) \\[1em] \cancel{80} \dfrac{dx}{dt} &= 2(50)(-1)\cancel{(80)} \\[1em] \dfrac{dx}{dt} &= -100 \end{aligned}

This gives us our answer:

\left(\left\lvert \dfrac{dx}{dt} \right\rvert = 100 \frac{\text{ft}}{\text{s}}\right) > 90 \frac{\text{ft}}{\text{s}}

So yes, the vehicle was, in fact, speeding. Assuming, of course, that there were no calibration issues with the camera and all the necessary elements of the traffic offense are met.

Let's consider another problem:

problem. A conical tank with a top radius of four feet and a depth of ten feet is being filled with water at a rate of two cubic feet per minute. How fast is the water rising when the water reaches a depth of five feet?

Once again, we begin by diagramming:

Looking at our diagram, we see that the cross-section depicting the water presents similar triangles. Thus:

\dfrac{r}{h} = \dfrac{4}{10}

Next, we know from geometry that the volume of a cone is given by the formula:

V = \dfrac{1}{3}(\pi r^2)h

Next, we're told that the tank is being filled at a rate of two cubic feet per minute, so we have:

\dfrac{dV}{dt} = 2 ~\frac{\text{ft}^3}{\text{min}}

Finally, we want to formulate our question:

What is ${\dfrac{dh}{dt}}$ when ${h = 5?}$

From our ratio ${r/h = 4/10,}$ we can easily express ${r}$ as a function of ${h:}$

r = \dfrac{2}{5}h

Substitute the expression into ${V:}$

\begin{aligned} V &= \dfrac{1}{3} \pi \left(\dfrac{2}{5}h\right)^2 h \end{aligned}

At this point, we want to just implicitly differentiate. We don't want to try and solve for ${h,}$ because just looking at that equation, we can tell that we'll end up with a cubic root, and that can easily turn nasty.

\begin{aligned} \dfrac{d}{dt} V &= \dfrac{d}{dt} \dfrac{1}{3} \pi \left(\dfrac{2}{5}h\right)^2 h \\[1em] \dfrac{dV}{dt} &= \dfrac{\pi}{3} \left( \dfrac{2}{5} \right)^2 \dfrac{d}{dt} h^3 \\[1em] \dfrac{dV}{dt} &= \dfrac{\pi}{3} \left( \dfrac{2}{5} \right)^2 3h^2 \dfrac{dh}{dt} \end{aligned}

Now we can plug in the numbers:

\begin{aligned} 2 &= \dfrac{\pi}{3} \left( \dfrac{2}{5} \right)^2 (3)(5)^2 \dfrac{dh}{dt} \\[1em] 2 &= \dfrac{\pi}{3} \dfrac{2}{5} \dfrac{2}{5} (3)(5)(5) \dfrac{dh}{dt} \\[1em] \cancel{2} &= \dfrac{\pi}{\cancel{3}} \dfrac{\cancel{2}}{\cancel{5}} \dfrac{2}{\cancel{5}} (\cancel{3})(\cancel{5})(\cancel{5}) \dfrac{dh}{dt} \\[1em] 1 &= 2 \pi \dfrac{dh}{dt} \\[1em] \dfrac{1}{2 \pi} &= \dfrac{dh}{dt} \end{aligned}

So, when ${h = 5,}$ we have:

\dfrac{dh}{dt} = \dfrac{1}{2\pi} ~\frac{\text{ft}}{\text{min}}

A Heronian triangle is a triangle whose side lengths and area are all integers, named after the Greek mathematician Hero of Alexandria. ↩

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Footnotes