I'm having a hard time grasping what they're saying in my textbook for this chapter.

If $z=f(x,y)$ defines a surface, then the tangent plane to the surface at (a,b) may be obtained as $$(-f_x(a,b),-f_y(a,b),1) * (x-a,y-b,z-f(a,b))=0$$ This is the result of applying the gradient-of-a-surface formula to the surface $z-f(x,y)=0$ at the point $(a,b,f(a,b))$.

How is the formula comparable to the formula for finding tangent planes to a surface which is $$(f_x(a,b,c),f_y(a,b,c),f_z(a,b,c))*(x-a,y-b,z-c)=0$$ and why are the $f_x$ and $f_y$ components negative in the first formula?

  • $\begingroup$ Have you at least tried to compute the gradient of $z-f(x,y)$? $\endgroup$ – amd Jan 21 '19 at 23:18
  • $\begingroup$ Essentially a duplicate of math.stackexchange.com/q/3073917/265466. $\endgroup$ – amd Jan 21 '19 at 23:20

The two formulations are the same. We are taking $F(x,y,z)=z-f(x,y)$ and setting $F=0$. Taking partial derivatives of $F$ gives $$F_x(a,b,c)=-f_x(a,b)$$ $$F_y(a,b,c)=-f_y(a,b)$$ $$F_z(a,b,c)=1$$

The partial derivatives with respect to $x$ and $y$ have negative coefficients because we are taking the gradient of $z-f(x,y)$. They would have positive coefficients if we defined $F(x,y,z)=f(x,y)-z$ but then $F_z(a,b,c)$ would be $-1$ instead of $1$.

  • $\begingroup$ Alright this may be a stupid question but if the formulations are the same, can't we use the second formula (formula of a tangent plane to a surface) for everything rather than this? What's the significance of this formula? $\endgroup$ – Random Student Jan 21 '19 at 22:39
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    $\begingroup$ You need a level surface so that the gradient is normal to the surface. You are given a function of two variables. To use the second formulation you need a function of three variables equal to a constant, i.e. a level surface. $\endgroup$ – John Douma Jan 21 '19 at 23:07
  • $\begingroup$ Ohh I see. Thanks for your help! I get it now! $\endgroup$ – Random Student Jan 21 '19 at 23:30

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