Use differentials to approximate the value of 7.8^3??

Should be easy, but canīt find the correct answer. How do you do this... help....

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- Nov 30th 2006, 07:57 AMmath34Use differentials to approximate?
Use differentials to approximate the value of 7.8^3??

Should be easy, but canīt find the correct answer. How do you do this... help.... - Nov 30th 2006, 08:26 AMSoroban
Hello, math34!

Quote:

Use differentials to approximate the value of $\displaystyle 7.8^3$

Evidently, we have the function: .$\displaystyle f(x) \:=\:x^3$

Then: .$\displaystyle df \:=\:3x^2dx$

We know that: .$\displaystyle f(8) \,= \,8^3 \,= \,512$

Let $\displaystyle x = 8,\;dx = -0.2$

Then: .$\displaystyle dy \:=\:3(8^2)(-0.2)\:=\:-38.4$

Hence: .$\displaystyle 7.8^3\:\approx\:512 - 38.4 \:=\:\boxed{473.6}$

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How good is this approximation?

The actual value is: .$\displaystyle 7.8^3\:=\:474.552$

Our answer is short by only: .$\displaystyle 474.552 - 473.6 \:=\:0.952$

As a percent: .$\displaystyle \frac{0.952}{474.552}\:=\:0.002066103 \:\approx\:0.2\%$

We were off by only two-tenths of a percent (2 parts in 1000) . . . not bad!

- Nov 30th 2006, 08:30 AMCaptainBlack
- Nov 30th 2006, 09:02 AMmath34How did you get -0.2 for dx?
Thanks

- Nov 30th 2006, 09:13 AMmath34How did you get -0.2 for dx? ,,
- Nov 30th 2006, 09:29 AMThePerfectHacker
Because that was the decrease in the value. You went from 8 to 7.8 thus the change was -.2