How do you use linear approximation to the square root function to estimate square roots #sqrt 4.400#?

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Jim H Share
Feb 1, 2017

Answer:

The linear approximation is a form of (or a way of thinking about) the equation of a tangent line.

Explanation:

For function, #f#, the linear approximation at #a# is given by

#L(x) = f(a) + f'(a)(x-a)#.

Note that the equation of the line tangent to the graph at #(a,f(a))# has slope #m=f'(a)# and it has point-slope equation

#y-f(a) = f'(a)(x-a)# Solving for #a# gives us the linear approximation to #f# at #a#.

We are not told what to use for #a#, but we want to eventually use #x = 4.4#.

We want #a# to be a number "close to" 4.4 for which it is relatively easy to calculate #f(a)#.

In this case, we want a number close to #4.4# whose square root is relatively easy to find.

The 'obvious' (once you see it) choice is #a = 4#.

So,

#f(x) = sqrtx# and #f(a) = f(4) = 2#.

#f'(x) = 1/(2sqrttx)# so #f'(a) = f'(4) = 1/4#

#L(x) = 2 + 1/4(x-4)#.

And we finish with

#L(4.4) = 2+1/4(4.4-4) = 2.1#

If there is a typo in the question

If the question should ask us to approximate #sqrt(4,400)# we should find a different #a#.

#60^2 = 3600# and #70^2 = 4900# so, for a rough approximation, use #a=70#.

A bit more arithmetic will show that

#66^2 = 4356# and #67^2 = 4489#.

Since #4356# is close to #4400#, use #a=4356#.

OR

To get a rough estimate use #sqrt4400 = sqrt(4 xx 100 xx11) = 20sqrt11# and approximate #sqrt11# using #a=9#.

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Feb 1, 2017

Answer:

66.33, nearly.

Explanation:

Using binomial expansion,

# sqrt(a^2+b)=a(1+b/a^2)^(1/2)=a(1+1/2b/a^2)#, nearly.

Here, choose a = 66 and b = 44.

Now,

#sqrt4400=sqrt(66^2+44)=sqrt(66^2+44)=66(1+44/66^2)^(1/2)#

#66(1+1/2(44/66^2))=66(1+1/198)=66+1/3=66.33#, nearly.

Note that the error is of order 66(1/198)^2=O(.002), nearly.

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