# Linear Approximations

## Related Calculator: Linear Approximation Calculator

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.