I'm having a little difficulty understanding how to do the .05 using differentials. I'm just hoping someone can walk me through, step by step, and explain why they are during it that way.

$$\sqrt[3] {27.05}$$

Edit Apologies to all, I wasn't very clear on what I was asking.

I was looking for someone to explain how to find an approximate answer to the equation above using differentials. (such as @Arturo has done)

@Sasha @Ross I apologize for wasting your time when answering the question using series.

  • $\begingroup$ I have voted to close. I do not understand what it is that is being asked, and why the differentials are involved. Do you need to compute the above expression approximately, using series? $\endgroup$ – Sasha May 3 '12 at 16:18
  • $\begingroup$ @Sasha I'm looking to approximate the value of 27.05^1/3 using differentials. Knowing that the cubed root of 27 is 3, but I have to figure out the .05 as well. (without a calculator). Also, it isn't homework. It is a practice problem as I'm preparing for an exam $\endgroup$ – Math_Phase May 3 '12 at 16:22
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    $\begingroup$ @MathPhase: Then please add that information to the post itself. $\endgroup$ – Arturo Magidin May 3 '12 at 16:24
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    $\begingroup$ Questions should be asked in such a way that readers can understand what the question is. $\endgroup$ – Siminore May 3 '12 at 16:28

The general philosophy when using differentials:

You are trying to find the value of a function at a difficult point, but there is a point, very close to the difficult point, where the function is easy to evaluate. Then one can use the formula $$f(a+\Delta x) = f(a)+\Delta y \approx f(a)+dy = f(a)+f'(a)dx=f(a)+f'(a)(x-a)\quad \text{ for }x\text{ near }a,$$ where $\Delta x = dx$ is the change in $x$ (how far the difficult point is from $a$, which is $x-a$), $\Delta y$ is the change in value (which is $f(x)-f(a)$), and $dy$ is the differential of $y$ (which is $dy = f'(x)dx$). If $\Delta x=dx$ is small, then $\Delta y$ is very close to $dy$, so we can use $dy$ to approximate.

So you want to do the following:

  1. Identify the function you are trying to evaluate. This will be your $f$.

  2. Find a point near the point where you are trying to evaluate where the function is easy: this will be your $a$.

  3. Compute $f(a)$ (which is supposed to be easy) and compute $f'(a)$ (which hopefully will be easy as well).

  4. Plug into the formula.

In this case, $f(x) = \sqrt[3]{x}$, and the "hard point" is $x=27.05$. Is there a point $a$ near $x$ where $f$ is easy to evaluate? Yes! $f(27)$ is easy. So take $a=27$, $\Delta x = dx = .05$, and use the approximation. You'll have to compute $f'(x)$ and $f'(27)$.

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    $\begingroup$ Thank you for the explanation. I had to read through it a few times to understand what was going on. My take from it is, I need to take the 'easy point' and add it to the derivative multiplied by the 'hard point'. (3 + (1/27*.05)) $\endgroup$ – Math_Phase May 3 '12 at 20:44

You are basically expanding in a Taylor series around a convenient point. $\sqrt[3]{27.05}=\sqrt[3]{27(1+\frac {.05}{27})}=3\sqrt[3]{1+\frac{.05}{27}}\approx 3 (1+\frac{.05}{81})$. The last step comes from $(1+x)^n\approx 1+nx$ for $x \ll 1$ and is the first term in the Taylor series.

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    $\begingroup$ Thank you for the reply, I apologize if I wasted your time by not mentioning that I haven't reached Taylor series yet. But I do appreciate you taking the time to provide an answer $\endgroup$ – Math_Phase May 3 '12 at 20:46

Here is an answer that starts from the tangent line approximation, in order to visualize the differentials approach better. Let $f(x)=x^{1/3}$. Then $f'(x)=\frac{1}{3}x^{-2/3}$.

The point $x=27$ is very near to $27.05$, and our function is easy to evaluate at $27$, since $27^{1/3}=3$. Let us find the equation of the tangent line to $y=x^{1/3}$ at $x=27$. The slope is $(1/3)27^{-2/3}=1/27$. So the equation of the tangent line is $$y-3=(1/27)(x-27),\tag{$\ast$}$$ or equivalently $$y=\frac{x}{27}+2.$$ Recall that the tangent line at $x=27$ kisses the curve at $x=27$. So near $x=27$, the tangent line is very close to the curve. Now let $x=27.05$. The $y$-coordinate of the point on the curve with $x$-coordinate $27.05$ is given approximately by $$\frac{27.05}{27}+2=3+\frac{0.5}{27}.$$ That is the desired approximation.

For differentials, it is best to go back to $(\ast)$.
Let $\Delta x$ be the change in $x$. In our case, $\Delta x=0.05$. Let $\Delta y$ be the corresponding change in $y=f(x)$ as we move from $x=27$ to $x=27+\Delta x$. Then from $(\ast)$ we have $\Delta y\approx \frac{1}{27}\Delta x$.

More casually, $y+dy\approx y +\frac{dy}{dx}dx$. We take $dx=0.05$. We found that at $x=27$, $\frac{dy}{dx}=\frac{1}{27}$. So $y+dy\approx 3+\frac{1}{27}(0.05)$.

Exactly how you should write out the calculation depends on the notation that your instructor uses.

  • $\begingroup$ Thank you for your explanation, it helped clarify the steps needed to solve the problem! $\endgroup$ – Math_Phase May 3 '12 at 20:45

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