Linear Approximations and Differentials

• loadsy
In summary, to estimate the value of cos31.5*, a linear approximation is used. Choosing a value of a=30* and using the formula dy=f'(x)dx, the value of cos(31.5*) is estimated to be approximately 0.853. The steps used in solving this question are correct and can be verified by evaluating the expression using a calculator.
loadsy
Alright I just did the following question, and was hoping I did it right:

Use differentials(or, equivalently, a linear approximation) to estimate the given number)

cos 31.5* (* meaning degrees)

f(x) = cosx
f(31.5*) = ?

a is chosen to be the closest number to the number evaluating in the function of, such that the equation can be easily evaluated.
a= 30*, then dx = 31.5*-30*= 1.5*

delta y is approximately equal to dy
delta y = f(31.5*)-f(30*)
= cos31.5* - cos30*
= cos31.5* - root3/2

dy = f'(30*)dx

f'(x) = -sinx -> f'(30*) = -sin(30*) = -0.5

dy = (-1/2)(1.5*) = -1/2(pi/120) = -(pi)/240

Hence: cos31.5* - root3/2 is approximately equal to -(pi)/240 or:
cos(31.5*) = root3/2 - (pi)/240

I'm just checking to see if I followed the correct steps in solving this question. Thanks guys.

loadsy said:
Hence: cos31.5* - root3/2 is approximately equal to -(pi)/240 or:
cos(31.5*) = root3/2 - (pi)/240
At the time of writing, this is a 15+ year old question. However the following may help someone who comes across it.

The question specifically asks for an estimate. Giving the answer in terms of √3 and π (which are exact) is not appropriate. A numerical answer to 3 significant figures is needed. (Be guided by the fact that ‘31.5º' has 3 significant figures.)

The answer is correct though working is a bit untidy. The answer can easily be checked: find cos(31.5º) and then evaluate the expression ##\frac {\sqrt 3}{2} - \frac {\pi}{240}##. They agree - both give 0.853

Delta2
Yes, to add some detail, this is the tangent line approximation, which follows from the differentiability of cosx. For those interested in the Math side.

What is a linear approximation?

A linear approximation is an estimation of the value of a function at a certain point by using the tangent line to the function at that point.

Why is linear approximation useful?

Linear approximation is useful because it allows us to approximate the value of a function at a certain point without having to know the exact value. It can also be used to estimate the behavior of a function near a specific point.

How is linear approximation calculated?

To calculate a linear approximation, we first find the slope of the tangent line to the function at the given point. Then, we use this slope and the point to construct the equation of the tangent line, which can be used to approximate the value of the function at that point.

What is the difference between linear approximation and linearization?

Linear approximation and linearization are often used interchangeably, but they are actually slightly different concepts. Linear approximation refers to the process of estimating the value of a function at a specific point, while linearization refers to the process of transforming a nonlinear function into a linear one near a specific point.

How can linear approximation be used in real-life situations?

Linear approximation can be used in a variety of real-life situations, such as in physics to approximate the behavior of a system near a certain point, in economics to estimate the impact of small changes in variables, and in engineering to approximate the performance of a system near a specific operating point.

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