use a linear approximation (or differentials) to estimate the given number?

Question

cos(29) degree's

Answer 1

Linear approximation works like this: we know the value of a function at a particular value of a variable and we are going to use it to approximate the value of that same function when the variable takes on a different number (provided the numbers are not too far apart)

 

To really derive the linearization of an equation, you need to be familiar with Taylor Series. If you are, then linearization is just the first-order Taylor expansion of the function about a particular point. If you are not, then no worries, I give you the equation below:

 

y(x) = y(a) + y^'(a)*(x - a), where you

 

1) know what a is (it looks like in your problem your instructor lets you pick an "a" value that you want--remember that it has to be close to the x value you are going to be approximating for!)

 

2) know what y(a) is

3) know y^'(a)  --Note that this is to be interpreted as "the derivative of the function evaluated at x=a" so if y = x² and a = 5, then y^'(x) = 2x and thus y^'(a) = 10

 

Having established that, let's begin:

 

For your case, we need an "a" value. Well, let's not make it too difficult on ourselves and recognize that from trig cos(30) = √3/2 (√ = square root). This works perfect since we now know y(a) and 30 is not too far from 29.

 

Using our linearization equation, then cos(29) = cos(30) - sin(30)*(29 - 30) = √3/2 - 1/2(-1) = √3/2 + 1/2

 

Now you can plug this in your calculator to get 1.366

 

If you type cos(29) into your calculator, you will get 0.8746

 

Now, you may be wondering if something went wrong. The answer is no. So what happened then? The cosine function is nonlinear and we used a linear equation to approximate it! That will not work! 

 

Now this isn't to say that linearization is not good. It will give you better and better results if you had taken a number that was much closer to 29 for "a" like 29.0001. This would work well, BUT it's pretty hard to do that without a calculator, so it's a safe bet that your teacher wanted to show you the dangers of using linear approximations of nonlinear equations when you are working with an "a" value that is not very close to the value of x that you want to get the approximation at.

 

-Hope this helps!