1
$\begingroup$

I have a question below but I missed this day of class maybe someone can show me how to approach?

Find $dy$ and evaluate $dy$ for the given values of $x$ and $dx$

  1. $\displaystyle y=e^\frac{x}{10}$
  2. $\displaystyle x = 0$ and $dx = 0.1$
  • 1
    Well, it seems like you should recall $$ \mathrm{d}y=\frac{\mathrm{d}y}{\mathrm{d}x}\mathrm{d}x$$ From there, it should be straightforward to solve...2012-11-23

1 Answers 1

3

With differentials, it's an abuse of notation, but the answer is that $$ \mathrm{d}y=\frac{\mathrm{d}y}{\mathrm{d}x}\mathrm{d}x$$ Since $\mathrm{d}x$ is impossible to use, we make the approximation $\Delta x\approx\mathrm{d}x$.

For us, $\Delta x=0.1$. Now, we find the derivative $$\frac{\mathrm{d}}{\mathrm{d}x}e^{x/10}=\frac{1}{10}e^{x/10}.$$ Then we plug in the data: $$\mathrm{d}y\approx \left.\left(\frac{1}{10}e^{x/10}\right)\Delta x\right|_{x=0}$$ which means we are evaluating the parenthetic term when $x=0$, and multiply by $\Delta x=0.1$.

Now we have to just plug these things in to find: $$\begin{align}\left.\left(\frac{1}{10}e^{x/10}\right)\Delta x\right|_{x=0}&=\left(\frac{1}{10}e^{0}\right)0.1\\ &=0.01\end{align}$$ Thus $\mathrm{d}y\approx 0.01$.

  • 0
    Precisely how is it an abuse of notation? I recall reading about this, but it has now escaped me. Oh, and wouldn't it be best to write: $$ \begin{align} dy&\approx\left. \left(\frac{1}{10}e^{\frac{x}{10}}\right)\Delta x\right|_{x=0}\\ \left. \left(\frac{1}{10}e^{\frac{x}{10}}\right)\Delta x\right|_{x=0}&=\left(\frac{1}{10}e^{0}\right)0.1\\ \left(\frac{1}{10}e^{0}\right)0.1&=1. \end{align} $$ ?2012-11-23
  • 2
    Because the notation treats differentials like $\mathrm{d}x$ "as if" they were numbers, and could have $$\frac{\mathrm{d}y}{\color{red}{\mathrm{d}x}}\color{red}{\mathrm{d}x}$$ which is illegal mathematics. The result is right, the reasoning fallacious :(2012-11-23
  • 2
    Oh, right. They're not even algebraic quantities since they're simply a notational convention just like $\int$. Am I following you?2012-11-23
  • 0
    Exactly right! :)2012-11-23
  • 1
    Isn't the answer $\frac1{10}0.1=0.01$?2012-11-23
  • 0
    @robjohn: one day, I will remember arithmetic ;) Thanks!2012-11-23
  • 0
    @Limitless: thanks for the advice on typesetting :) I agree, having exactly one relational operator per equation is good; but once I got started, I just got too lazy to stop and sort it out :p2012-11-23
  • 0
    @AlexNelson, you're welcome. Also, thanks to robjohn for understanding arithmetic. It constantly eludes me, and I find myself running to fields, rings, and groups for comfort. :P2012-11-23