# Finding linear approximation

1. Nov 6, 2015

### catch22

1. The problem statement, all variables and given/known data
Find linear approximation of the surface z2 = xy + y + 3 at the point (0,6, -3) and use it to approximate f(-0.01, 6.01, -2.98)

2. Relevant equations

3. The attempt at a solution

so this means the total surface has decrease by -0.17?

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2. Nov 6, 2015

### qq545282501

its not about the surface thats decreased, its about how accurate your approximation is. in this case, you used the newly found equation of the tangent plane to approximate f(-0.01, 6.01, -2.98). you want to compare using the original function to see how accurate it is.

3. Nov 6, 2015

### catch22

so you comparing it to the actual change?
which should be final - initial :
F(-0.01, 6.01, -2.98) - F(0,6, -3)

4. Nov 6, 2015

### qq545282501

but, if you want to see how close your approximation is, use z^2 = xy + y + 3 as the initial function to find the exact value of the point that you are trying to approximate with newly found tangent plane.

5. Nov 6, 2015

### qq545282501

hm..if you look at this way: the tangent plane and the original function are like twin brothers at a specific time[ point in our case] . we are testing to see how close the brothers are at that time[point] , but as we move away from that time[point], they become not so similar anymore, and if we go far enough, they are completely different from each other.

6. Nov 6, 2015

### HallsofIvy

You are asked to do two different problems but seem to have mixed them together! You have, as your final answer "L(x, y, z)= -0.17". That is certainly not true since the left side, L(x, y, z), the linear approximation to the function, is NOT a constant.

You need two different answers to the two different problems. First you need to write the equation for the tangent approzimation, then evaluate that at (-0.01, 6.01, -2.98).

Yes, given $F(x, y, z)= xy+ y- z^2+ 3$, then $F_x= -y$, $F_y= -x- 1$, and $F_z= 2z$.
At (0, 6, -3) we have $F_x=$

..... etc, see how you go with this now. [mod edit]

Last edited by a moderator: Nov 6, 2015