## anonymous 5 years ago the surface z=(x,y) has this line tangent to it: x=2+2t, y=1, z=-2+3t at the point (2,1,-2). Find f_x(2,1)

1. anonymous

What is f_x(2,1)

2. anonymous

Is it the derivative?

3. nowhereman

Do you mean the surface is $$(x,y, f(x,y))$$ and $$f_x = \frac{∂f}{∂x}$$?

4. anonymous

i meant f sub x , yes I believe it has to do with partial derivatives

5. nowhereman

Then it is $$\frac{3}{2}$$ because the tangent line is in x-direction.

6. anonymous

are you just taking coefficient in front of the t variable?

7. nowhereman

Yes.

8. anonymous

Okay, I'm still kind of lost as to the problem. I believe the first deriv the tangent line is 0 ? or something of that nature. I'm confused to what the answer is actually telling me.

9. anonymous

at the point$F_x (2,1)$ the tangent line is 3,2?

10. nowhereman

You see, the function is $$f: ℝ×ℝ→ℝ$$ and the tangent lines of the graph at a point $$(x,y)$$ are of the form $(x + d_xt,y + d_yt,f(x,y) + f_x(x,y)d_xt + f_y(x,y)d_yt)$ where $$(d_x, d_y)$$ is the direction of the argument. Here you can see, that $$d_x = 2$$ and $$d_y = 0$$ so you have $(x + 2t, y, f(x,y) + 2f_x(x,y)t)$ Then you only have to plug in the values for of the base coordinates and the function at that point.

11. amistre64

it gives you a point and a tangent line; determine the equation of the surface

12. amistre64

nowheres looks more technical :)