Linear Approximations

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After studying differentials we know that if `Delta y=f(a+Delta x)-f(a)` and `dy=f'(x)Delta x` then making `Delta x` very small, i.e. if we let `Delta x->0` we can write that `dy~~Delta y`.

This can be rewritten as `f(a+Delta x)-f(a)~~f'(x)Delta x`.

If we now let `x=a+Delta x` then we obtain that `f(x)-f(a)~~f'(x)(x-a)`.

Definition. Linearization of function `y=f(x)` at point `a` is `f(x)~~f(a)+f'(x)(x-a)`.

We need this linearization because it might be easy to calculate a value of a function `f(a)`, but difficult (or even impossible) to compute nearby values of `f`. So we settle for the easily computed values of the linear function (tangent line).

Example 1. Find a linear approximation of `y=sqrt(x)` at `(4,2)`. Then approximate `sqrt(3.99)`, `sqrt(4.01)` and `sqrt(4.05)`.

Since `f'(x)=1/(2sqrt(x))` then `f'(4)=1/4`.

Therefore linear approximation `L(2)=1/4(x-4)+2` or `sqrt(x)~~1/4x+1`.

`sqrt(3.99)~~1/4 3.99+1=1.9975`. True value is given for comparison `sqrt(3.99)~~1.997498435544`.

`sqrt(4.01)~~1/4 4.01+1=2.0025`. True value is given for comparison `sqrt(4.01)~~2.00249843945`.

`sqrt(4.05)~~1/4 4.05+1=2.0125`. True value is given for comparison `sqrt(4.05)~~2.01246117974981`.

As can be seen approximate values are overestimated. That's because tangent line at `x=4` lies above `f(x)=sqrt(x)` .

Example 2. Suppose that temperature of tea is 95°C. After one minute its temperature is 85°C. Find temperature of tea in two minutes.

If `T(t)` represents the temperature of the tea after `t` minutes, then `T(0)=95`, `T(1)=85`.

In order to make linear approximation with `a=1` we need to approximate `T'(1)`.

Since `T'(1)=lim_(t->1)(T(t)-T(1))/(t-1)` we can't find exact value of derivative, but we could estimate `T'(1)` by taking `t=0`: `T'(1)~~(T(0)-T(1))/(0-1)=(95-85)/(0-1)=-10`.

Thus we approximated instantaneous rate of change by average rate of change between `t =0` and `t =1`, which is -10°C per minute. Thus, `T(2)~~T'(1)(2-1)+T(1)=-10(3-2)+85=75^0C`.