Use differentials to approximate square root calculator

Question

About "Use differentials to find an approximate value for the given number"

Use differentials to find an approximate value for the given number :

Here we are going to see how to use differentials to find an approximate value for the given number.

f(x + Δx) = f(x) + dy

To find approximate value of the given number using derivatives, we use the formula given above.

here "x" stands for the number for which we may find the exact value nearest to the given number.

"Δx" stands for the remainder part of the given number.

Let us look into some examples to understand the above concept.

Example 1 :

Use differentials to find the approximate value of ∛65

Solution :

f(x + Δx) = f(x) + dy

We cannot find cube root of the number 65, but we may find the cube root of the number 64 nearest 65.

So, x = 65 and Δx = dx = 1

Let y = f(x) = ∛x = x1/3

First let us find the value of "dy"

y = x1/3

dy/dx = (1/3)x1/3 - 1

dy = (1/3)x-2/3 dx

dy = (1/3)(64)-2/3 (1)

Applying the values of "x" and "dx".

= (1/3)4-2

= 1/3(16)

= 1/48

The value of dy is 0.208333...

f(x) = x1/3

f(64) = 641/3 = (43)1/3 = 4

The value of f(64) is 4.

f(x + Δx) = f(x) + dy

f(64 + 1) = f(64) + dy

f(65) = 4 + 0.2083333......

= 4.208333........

Hence the approximate value of ∛65 is 4.0283333.....

Example 2 :

Use differentials to find the approximate value of √36.1

Solution :

f(x + Δx) = f(x) + dy

We cannot find square root of the number 36.1, but we may find the cube root of the number 36 nearest 36.1.

So, x = 36 and Δx = dx = 0.1

Let y = f(x) = √x = x1/2

First let us find the value of "dy"

y = x1/3

dy/dx = (1/2)x1/2 - 1

dy = (1/2)x-1/2 dx

dy = (1/2)(36)-1/2 (0.1)

Applying the values of "x" and "dx".

= (1/2)6-1 (0.1)

= (1/2)(1/6)(0.1)

= 0.1/12

= 0.008333.......

The value of dy is 0.008333333...........

f(x) = x1/2

f(36) = 361/2 = (62)1/2 = 6

The value of f(36) is 6.

f(x + Δx) = f(x) + dy

f(36 + 0.1) = f(64) + dy

f(65) = 6 + 0.008333333......

= 6.00833333........

Hence the approximate value of √36.1 is 6.008333333....

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#1

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

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

Please see

Mathwords: Approximation by Differentials

#3

Approximate square roots

Hello Zabulius

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 \(\displaystyle \sqrt{110}\)?

Grandad

How do you find approximate value using differentiation?

General Form.

Find a point x0 near the point x′, at which the value of the function is known..

Differentiate the function with respect to x. ... .

Use the approximations i.e. the value of the change in x i.e. dx = Δx = x′ – x0 and calculate the derivative at x = x0 to get dy, which is approximated as Δy:.

differentials approximate square root calculator Differential approximation formula