Quick question about linearization using the small angle method

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Jayalk97
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Hey guys, when you're linearizing a function that has a constant, what do you do to it?

An example would be y = x^2 + 3, would you just linearize it using its derivative and get rid of the constant?
 
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Hey guy, no.

The derivative gives you the slope of a linear approximation. (in this case 2x).
A constant is still needed for the approximation in a particular point; e.g. if x = 3 the slope is 6 and the line has to go through the point (3,12), so a linear approximation there is y = 6x - 6 (or, if you want: (y-12) = 6 (x-3) )
 
BvU said:
Hey guy, no.

The derivative gives you the slope of a linear approximation. (in this case 2x).
A constant is still needed for the approximation in a particular point; e.g. if x = 3 the slope is 6 and the line has to go through the point (3,12), so a linear approximation there is y = 6x - 6 (or, if you want: (y-12) = 6 (x-3) )
I see, thank you. I was just looking at the method of linearizing where you take the partial derivative of every variable and noticed that you wouldn't have any constants left doing this.