Linear Approximation of z2 = xy + y + 3 at (0,6,-3)

[catch22]

Homework Statement

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)

Homework Equations

The Attempt at a Solution

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

 

[qq545282501]

its not about the surface that's 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.

 

[catch22]

its not about the surface that's 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.

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

 

[qq545282501]

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

this is not what the question is asking though, you have already answered the question.
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.

 

[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.

 

[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]