Related rates

related rates are actually completely made up. "related rates" describes a type of proble, not some special new technique. related rates as a thing does not exist. you just rearrange the problem algebraically so you can find the derivative of some function [footnote: "source: original research motherfucker"]. it is a lie peddled by big school to weed out students.

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Madeleine Sydney Ślaga, in the first version of this file.

In calculus, related rates are related differentials of related quantities.[cite:@enwiki:1267043375] Often, there will be several rates of change known with respect to time.

Solving related rates problems

Textbook related rates problems will often look like implicit differentiation problems. And just as in implicit differentiation, the problems don't require any new techniques of calculus as long as you write dependencies explicitly. Consider the following example problem:

Find $\frac{dx}{dt}$ at $x=4$ if $y=5x^2-4$ and $\frac{dy}{dt}=-5$.

As this problem is phrased, the values of variables $x$ and $y$ are implicitly dependent on $t$. We the people, of functional programming heritage, shall acknowledge this as what it is: bullshit. To solve this problem sanely,Calculus teachers will not do this, making it possible to — in the words of my own teacher — "substitute variables too early;" breaking the entire concept of equality. Amazing! start by rewriting $x$ and $y$ as functions of $t$:

Find $x^{\prime }\left(t\right)$ at $x\left(t\right)=4$ if $y\left(t\right)=5x^2-4$ and $y^{\prime }\left(t\right)=-5$.

Now it should be obvious that this is just a bit of algebra and an obvious application of the chain rule: $$\begin{array}{cc}\frac{d}{dt}[5{\left(xt\right)}^2-4]& =-5\\ 10\left(xt\right)\left(x^{\prime }t\right)& =-5\\ x^{\prime }t& =-\frac{5}{10\left(xt\right)}.\end{array}$$