# Dealing with Undefined Values in Calculus: A Lesson in Limits

• archaic
In summary, the conversation discusses a graph showing a system of two points with constant velocities. The goal is to find the rate of change of the distance between them and the time of collision. The conversation also delves into the concept of derivatives and limits in order to address potential issues with the solution. Ultimately, the "correct" approach is to cancel inside the limit.
archaic
Homework Statement
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Relevant Equations
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This is not to ask for a solution, rather I want comments on the rigor of the step. Thank you for your time!

The graph shows a system of two points at ##t## and ##t+dt##, it is a bit exaggerated of course.
The upper point is moving strictly in the ##x## direction and has a constant velocity ##u##, while the lower point is always moving towards the upper point with constant velocity ##v>u##. I want to find the rate of change of the distance ##\mathcal{L}(t)## which separates them (to find the time of collision ##\tau##).

Keeping in mind that ##\cos(\pi-\theta)=-\cos(\theta)## and the cosines' law:
\begin{align*} \mathcal{L}^2(t+\Delta t)&\approx(u\Delta t)^2+(\mathcal{L}(t)-v\Delta t)^2+2(u\Delta t)(\mathcal{L}(t)-v\Delta t)\cos(\theta(t))\\ &\approx(u\Delta t)^2+\mathcal{L}^2(t)-2v\mathcal{L}(t)\Delta t+(v\Delta t)^2+2u\mathcal{L}(t)\cos(\theta(t))\Delta t-2uv\cos(\theta(t))(\Delta t)^2 \end{align*}
\begin{align*} \mathcal{L}^2(t+\Delta t)-\mathcal{L}^2(t)&\approx(u\Delta t)^2-2v\mathcal{L}(t)\Delta t+(v\Delta t)^2+2u\mathcal{L}(t)\cos(\theta(t))\Delta t-2uv\cos(\theta(t))(\Delta t)^2\\ \frac{\mathcal{L}^2(t+\Delta t)-\mathcal{L}^2(t)}{\Delta t}&\approx u^2\Delta t-2v\mathcal{L}(t)+v^2\Delta t+2u\mathcal{L}(t)\cos(\theta(t))-2uv\cos(\theta(t))\Delta t \end{align*}
Now let ##\Delta t\to 0##, we have:
\begin{align*} \frac{d}{dt}\mathcal{L}^2(t)&=-2v\mathcal{L}(t)+2u\mathcal{L}(t)\cos(\theta(t))\\ \Leftrightarrow 2\mathcal{L}(t)\mathcal{L}'(t)&=-2v\mathcal{L}(t)+2u\mathcal{L}(t)\cos(\theta(t))\\ \Leftrightarrow \mathcal{L}'(t)&=-v+u\cos(\theta(t)) \end{align*}
My problem is here, when ##t=\tau##, ##\mathcal{L}(\tau)=\mathcal{L}'(\tau)=0## and the division is then undefined. How can I go around this?

archaic said:
... and the division is then undefined
I don't see this as a problem. In the definition of the derivative of f at a point x0, you have this limit:
$$\lim_{h \to 0}\frac{f(x_0 + h) - f(x_0)} h$$
When h = 0, the fraction is undefined, but that is the whole idea of defining the derivative in terms of a limit.

Mark44 said:
I don't see this as a problem. In the definition of the derivative of f at a point x0, you have this limit:
$$\lim_{h \to 0}\frac{f(x_0 + h) - f(x_0)} h$$
When h = 0, the fraction is undefined, but that is the whole idea of defining the derivative in terms of a limit.
Oh no, not that. I'am talking about the division by the length in the last step.

Your equation will be valid as ##\mathcal L (t)## approaches 0.

Mark44 said:
Your equation will be valid as ##\mathcal L (t)## approaches 0.
How can I have a definite answer then? If I consider the simplification true when ##t=\tau##, the answer is correct.

Your question seems to me to be similar to this:
If ##f(x) = \frac{x^2 - 1}{x - 1}##, what is f(1)?
Well, f(1) is undefined, because both numerator and denominator are zero when x = 1.
But, by using limits, we can determine that ##\lim_{x \to 1} = \lim_{x \to 1}\frac{x^2 - 1}{x - 1} = \lim_{x \to 1} \frac{(x - 1)(x + 1)}{x - 1} = \lim_{x \to 1}x + 1 = 2##.
Clearly if x = 1, cancalling x - 1 from numerator and denominator isn't valid, but for any other value of x, ##\frac{(x - 1)(x + 1)}{x - 1} = x + 1##

archaic
Mark44 said:
Your question seems to me to be similar to this:
If ##f(x) = \frac{x^2 - 1}{x - 1}##, what is f(1)?
Well, f(1) is undefined, because both numerator and denominator are zero when x = 1.
But, by using limits, we can determine that ##\lim_{x \to 1} = \lim_{x \to 1}\frac{x^2 - 1}{x - 1} = \lim_{x \to 1} \frac{(x - 1)(x + 1)}{x - 1} = \lim_{x \to 1}x + 1 = 2##.
Clearly if x = 1, cancalling x - 1 from numerator and denominator isn't valid, but for any other value of x, ##\frac{(x - 1)(x + 1)}{x - 1} = x + 1##
i.e the "correct" way to answer the problem is by canceling inside the limit.

## What is a rate of change problem?

A rate of change problem is a type of mathematical problem that involves finding the relationship between two quantities that are changing at different rates. It is often used in calculus and physics to analyze the behavior of changing variables over time.

## How do you find the rate of change?

The rate of change is typically calculated by finding the slope of a line or curve on a graph. This can be done by dividing the change in the dependent variable by the change in the independent variable.

## What is the difference between average rate of change and instantaneous rate of change?

Average rate of change is the overall change in a quantity over a specific interval, while instantaneous rate of change is the change at a specific point in time. Average rate of change is calculated by finding the slope of a secant line, while instantaneous rate of change is calculated by finding the slope of a tangent line.

## Why is rate of change important?

Rate of change is important because it helps us understand the relationship between two variables and how they are changing over time. It is often used to make predictions and analyze patterns in data, and is a fundamental concept in many fields of science and engineering.

## What are some real-world applications of rate of change?

Rate of change is used in many real-world applications, such as calculating the speed and acceleration of objects in motion, analyzing population growth and decay, and predicting stock market trends. It is also used in fields such as biology, economics, and engineering to understand and model changing systems.

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