Velocity Vectors, Circles, and Calculus

Hello, everyone! While I have been drifting through my sea of homework, I came across a rather interesting problem involving circles and velocity vectors. The problem goes as follows:

“A particle moves along a curve so that its position vector and velocity vector are perpendicular at all times. If the particle passes through the point (4, 3), what is the equation of the curve?”

Building Intuition

Part of what makes the problem so interesting is our ability to come up with a general case, usable for any provided point of contact, and not just (4, 3)! First, let’s imagine what such a situation with perpendicular velocity and position vectors might look like.

One such possibility that fits the description would be a circle. As seen in the diagram, the velocity vectors v1, v2, and v3 (which happen to be tangent to the circle) are all perpendicular to their respective position vectors, r1, r2, and r3.

A circle with three velocity vectors pointing out as tangents from three position vectors on a circle.

However, as we can soon see, there are multiple different (and rather beautiful) ways that we can prove that the only possible general curve is a circle!

Solution

Let’s do this! First things first, let’s describe our particle’s motion by describing it as a vector, where the x and y positions and x(t) and y(t) respectively, where t is time. We can write our position vector as s(t) = (x(t), y(t)).

What’s more, we can take the derivative of our position vector with respect to time to get a velocity vector! 馃榾 To accomplish that, all we need to do is take the derivative of the x and y positions separately. (Taking the derivative of a single-variable function f(t) results in f'(t), pronounced f prime of t. One might visualize the derivative as being the “rate of change” of a function, or how fast the function is changing at time t.)

v(t) = (x'(t), y'(t)).

From here, we can employ a neat trick from linear algebra known as the dot product. Essentially, the dot product helps us find the angle between two vectors. In this case, we know that the angle between our position vector (x(t), y(t)) and our velocity vector (x'(t), y'(t)) must be 90 degrees.

This is where the fun begins! One important aspect of the dot product, as we see on the bottom left, is that the dot product of perpendicular vectors is always 0. Thus, we can say that in all cases, s(t) 路 v(t) = 0.

The dot product of two vectors is calculated by summing the products of the same components of both vectors. For example, (3, 4) 路 (5, 6) = 3 * 5 + 4 * 6 = 15+24 = 39. However, (0, 2) 路 (3, 0) = 0 * 3 + 2 * 0 = 0. This makes sense, because (0, 2) and (3, 0) are perpendicular.

If we expand out s(t) 路 v(t) = 0, we get (x(t), y(t)) 路 (x'(t), y'(t)) = 0. If we expand this out using the above definition, we get that x(t) * x'(t) + y(t) * y'(t) = 0. This is cool, but how can we proceed from here? ANTIDERIVATIVES!

Indeed, if you noticed that the original functions are multiplied by their derivatives and were reminded of the chain rule, you would be onto something important. Because we have x(t) * x'(t), we are provided with a very clean antiderivative.

Now we have some math steps:

(LaTeX converted at http://www.codecogs.com/latex/eqneditor.php?latex)

And we have it! Does the final equation look familiar? x2 + y2 = D, where D is the radius of the circle squared! In the above problem, we can plug in x(t) = 4 and y(t) = 3, and we get D = 9 + 16 = 25, meaning that the curve our particle is following is a circle with a radius of 5! 馃檪

This post is running longer than I was originally intended, so I’ll make sure to write a Part 2 regarding the two other interesting ways of solving this problem if you’re interested. Until then, best wishes and happy exploring!

– Ely


Many thanks to my Calculus instructor, Ms. Nguyen, for inspiring me to write this post, as well as providing students with many different opportunities to succeed this year! For further explanation in regards to calculus and linear algebra, I would highly suggest checking out the amazing YouTube channel 3Blue1Brown for detailed and immersive visuals to help understand those sorts of topics on a deeper level! Here is his channel: https://www.youtube.com/channel/UCYO_jab_esuFRV4b17AJtAw You can also subscribe; it’s completely free, WordPress handles all of privacy concerns, and you get notified whenever something new comes out!

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