# Thread: Using differentials to approximate

1. ## Using differentials to approximate

I'm going to write this problem word for word because I'm not sure how else to describe it.

Q1. For y=f(x), differentials are used to approximate the change in y for a given change in x. Suppose you have misplaced your calculator and need a decent approximation of the square root of 27. The yknow 25^1/2 = 5 so you assume that 27^1/2 must be slightly larger than 5.

A. Use differentials to approximate the square root of 27.

Step 1. f(x) = x^1/2, x=25, dx=2.

Calculate dy = f '(x) dx and evaluate it.

__________________________________________________ ___________

In addition to doing it for 27, I have to do it for 21, 39, and 110. The main problem is I don't know which values to put where. I think the x is always 25 and only dx changes? So for 27, dy = .5(25^.50)*2 which is .2 I believe. So my approximation would be 5.2? Any help or insight is appreciated. Thanks

2. Originally Posted by Zabulius
I'm going to write this problem word for word because I'm not sure how else to describe it.

Q1. For y=f(x), differentials are used to approximate the change in y for a given change in x. Suppose you have misplaced your calculator and need a decent approximation of the square root of 27. The yknow 25^1/2 = 5 so you assume that 27^1/2 must be slightly larger than 5.

A. Use differentials to approximate the square root of 27.

Step 1. f(x) = x^1/2, x=25, dx=2.

Calculate dy = f '(x) dx and evaluate it.

__________________________________________________ ___________

In addition to doing it for 27, I have to do it for 21, 39, and 110. The main problem is I don't know which values to put where. I think the x is always 25 and only dx changes? So for 27, dy = .5(25^.50)*2 which is .2 I believe. So my approximation would be 5.2? Any help or insight is appreciated. Thanks
yes, sqrt(27) = 5.2

Mathwords: Approximation by Differentials

3. ## Approximate square roots

Hello Zabulius
Originally Posted by Zabulius
dy = .5(25^.50)*2
I don't think your formula is quite right. You need to divide by 25^0.5

$\displaystyle y = \sqrt{x} \Rightarrow \frac{dy}{dx} = \frac{1}{2\sqrt{x}}$

$\displaystyle \Rightarrow \delta y \approx \frac{1}{2\sqrt{x}}\delta x$

So to find $\displaystyle \sqrt{27}$, put $\displaystyle x = 25$ and $\displaystyle \delta x = 2$:

$\displaystyle \delta y \approx \frac{1}{2 \cdot 5}\cdot 2 = 0.2$

So $\displaystyle y+ \delta y = 5.2 \approx \sqrt {27}$

To find $\displaystyle \sqrt{21}$, use the fact that $\displaystyle 21 = 25 - 4$. So you can again use $\displaystyle x = 25$, but this time $\displaystyle \delta x = -4$.

The nearest perfect square to $\displaystyle 39$ is $\displaystyle 36$. So to find $\displaystyle \sqrt{39}$, use $\displaystyle x = 36$. And since $\displaystyle 39 = 36 + 3$, use $\displaystyle \delta x = 3$.

Can you see what values to use for $\displaystyle x$ and $\displaystyle \delta x$ to find $\displaystyle \sqrt{110}$?

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# using differential find the approximate value of β 26

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