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Calculus help


a.) Suppose that we want to approximate sqrt(x) for x near 81…?

 …Find the linear approximation to sqrt(x) at x=81. L(x)=? 
b.) Use the linear approximation at x=81 that you found in part a to approximate the square root of 76.
Let f(x) = x^(1/2). The idea.

Whats “linear approximation”? Basically, you interpolate linearly. How? As follows :
[ f(y) -f(x) ] / ( y – x ) = ‘the gradient value at y ; the basis for linear approximation used’. [ an approximate equation ] … ( 1)

Now, in our case, y =81 because we are using ’81’ to ‘approximate’ the ‘square root of ’76’. Thus, x = 76.

We can readily ‘compute’ the following :
f(x) = f(76) =76^(1/2).
f(y) = f(81) = 81^(1/2) = 9.
x = 76 & y = 81.
( y – x ) = 81 – 76 = 5.

The trick is that ‘calculus’ helps us with the ”the gradient value at y ; the basis for linear approximation used’ part”.
”the gradient value at y” = f ‘(y) or value of ‘dy/dx’ at y.
f ‘(y)
= dy/dx
= (1/2 ) x^(1/2 -1)
=(1/2) x^( -1/2 )

”the gradient value at y = 81 the basis for linear approximation used’ part”
is =(1/2) 81^( -1/2 )
= (1/2) * ( 1/9)
= 1/18

Hence, ‘plugging’ all those values computed into the (1)…we get,

[ f(y) -f(x) ] / ( y – x ) = 1/18
[ 9 – 76^(1/2) ] / 5 = 1/18
76^(1/2) = 9 – [ 5/18 ] = 8.722 ( approximately)

Note that The Linear approximation is NOT THE EXACT value. It is derived based on the assumption that since the points are close together, hence their gradient of the curve sqrt(x) between these two points is approximately linear! If you didnt understand what I said in this paragraph… its ok. Because you dont need to know it but its good to know it.

Also, L(x) = f(x) above. The reason they introduce the term ‘L(x)’ is because our L(x) is NOT the exact value of f(x) but is an approximation for f(x) using the Linear Interpolation Idea. However, we have “loosely” used ‘L(x)’ to ‘equal’ f(x) in the above because it does NOT thwart our purpose.

I just wanted to clarify as much things as possible so that you may now not just know “how to do it” but “understand the idea behind this method and how it related to calculus as well”.

If you noticed, you could have “approximated” sqrt 76 by the method above without a calculator since we are just dividing and multiplying integers and fractions! The method above is “one way” where one can “estimate” the “square root” of a “integer” without using a calculator… Go ahead and impress your friends!

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