Linear Approximation: Find & Use for f(2.28,8.22)

In summary, the linear approximation equation for f(x,y) = 3 sqrt((x y)/4) at the point (2,8,6) is L(x,y) = 6 + 0.75(x-2) + 0.75(y-8), and using it to approximate f(2.28,8.22) gives a value of approximately 6.43. The correct derivatives for fx(x,y) and fy(x,y) are 0.75(y/4) and 0.75(x/4), respectively. The linear approximation equation can be thought of as the tangent line equation.
  • #1
andyk23
26
0
Find the linear approximation to the equation f(x,y) = 3 sqrt((x y)/4) at the point (2,8,6), and use it to approximate f(2.28,8.22). I know you take the derivative of fx(x,y) and fy(x,y), I think I'm taking the derivative wrong. Then after that you put x and y in the equation and solve for fx(2,8) and fy(2,8). Then take f(a,b)+fx(a,b)(x-a)+fy(a,b)(y-b). For fx(x,y) I'm coming up with .75(y/4) and for fy(x,y) I'm getting .75(x/4)
 
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  • #2
Your derivatives are indeed wrong. Let's start with a simpler function of just one variable.

[tex]g(x) = 3 \sqrt{x / 4}[/tex]What is [itex]g'(x)[/itex]?
 
  • #3
g(x)=3(x/4)^-.5
g'(x)= [3(x)^-.5]/4
 
  • #4
Want to know an easy way to derive a linear approximation equation? Well, I'll give you some intuition: it's basically the same thing as the tangent line equation in so many words.
 
  • #5
L(x)= f(x) + f'(a)(x-a)

(2.8,6) is the point...and so on
 

Related to Linear Approximation: Find & Use for f(2.28,8.22)

1. What is linear approximation?

Linear approximation is a method used in calculus to estimate the value of a function at a certain point by using the value of the function and its derivative at a nearby point. It involves creating a tangent line to the curve at the nearby point and using its slope to approximate the value of the function at the desired point.

2. How do I find the linear approximation of a function?

To find the linear approximation of a function, you will need to know the value of the function and its derivative at a nearby point. Then, you can use the formula y = f(a) + f'(a)(x-a) where a is the nearby point and x is the desired point, to find the approximate value of the function at x.

3. Why is linear approximation useful?

Linear approximation is useful because it allows us to estimate the value of a function at a certain point without having to use more complex methods. It is especially helpful when dealing with non-linear functions that are difficult to evaluate directly.

4. Can linear approximation be used for any function?

No, linear approximation can only be used for functions that are differentiable. This means that the function must have a well-defined derivative at the nearby point. If the function is not differentiable, then linear approximation cannot be used.

5. How accurate is linear approximation?

The accuracy of linear approximation depends on how close the nearby point is to the desired point. The closer the two points are, the more accurate the approximation will be. However, as the distance between the two points increases, the accuracy decreases. It is important to choose a nearby point that is as close as possible to the desired point for the most accurate approximation.

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