Formula for the tangent plane to a functional surface












0












$begingroup$


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?






multivariable-calculus







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  • $begingroup$
    Have you at least tried to compute the gradient of $z-f(x,y)$?
    $endgroup$
    – amd
    Jan 21 at 23:18












  • $begingroup$
    Essentially a duplicate of math.stackexchange.com/q/3073917/265466.
    $endgroup$
    – amd
    Jan 21 at 23:20
















0












$begingroup$


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?






multivariable-calculus







share|cite|improve this question









$endgroup$












  • $begingroup$
    Have you at least tried to compute the gradient of $z-f(x,y)$?
    $endgroup$
    – amd
    Jan 21 at 23:18












  • $begingroup$
    Essentially a duplicate of math.stackexchange.com/q/3073917/265466.
    $endgroup$
    – amd
    Jan 21 at 23:20














0












0








0





$begingroup$


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?






multivariable-calculus







share|cite|improve this question









$endgroup$




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?






multivariable-calculus




multivariable-calculus






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 21 at 22:16









Random StudentRandom Student

443




443












  • $begingroup$
    Have you at least tried to compute the gradient of $z-f(x,y)$?
    $endgroup$
    – amd
    Jan 21 at 23:18












  • $begingroup$
    Essentially a duplicate of math.stackexchange.com/q/3073917/265466.
    $endgroup$
    – amd
    Jan 21 at 23:20


















  • $begingroup$
    Have you at least tried to compute the gradient of $z-f(x,y)$?
    $endgroup$
    – amd
    Jan 21 at 23:18












  • $begingroup$
    Essentially a duplicate of math.stackexchange.com/q/3073917/265466.
    $endgroup$
    – amd
    Jan 21 at 23:20
















$begingroup$
Have you at least tried to compute the gradient of $z-f(x,y)$?
$endgroup$
– amd
Jan 21 at 23:18






$begingroup$
Have you at least tried to compute the gradient of $z-f(x,y)$?
$endgroup$
– amd
Jan 21 at 23:18














$begingroup$
Essentially a duplicate of math.stackexchange.com/q/3073917/265466.
$endgroup$
– amd
Jan 21 at 23:20




$begingroup$
Essentially a duplicate of math.stackexchange.com/q/3073917/265466.
$endgroup$
– amd
Jan 21 at 23:20










1 Answer
1






active

oldest

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1












$begingroup$

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 $1$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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 at 22:39






  • 1




    $begingroup$
    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 at 23:07










  • $begingroup$
    Ohh I see. Thanks for your help! I get it now!
    $endgroup$
    – Random Student
    Jan 21 at 23:30











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$begingroup$

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 $1$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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 at 22:39






  • 1




    $begingroup$
    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 at 23:07










  • $begingroup$
    Ohh I see. Thanks for your help! I get it now!
    $endgroup$
    – Random Student
    Jan 21 at 23:30
















1












$begingroup$

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 $1$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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 at 22:39






  • 1




    $begingroup$
    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 at 23:07










  • $begingroup$
    Ohh I see. Thanks for your help! I get it now!
    $endgroup$
    – Random Student
    Jan 21 at 23:30














1












1








1





$begingroup$

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 $1$.






share|cite|improve this answer









$endgroup$



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 $1$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 21 at 22:26









John DoumaJohn Douma

5,54211319




5,54211319












  • $begingroup$
    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 at 22:39






  • 1




    $begingroup$
    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 at 23:07










  • $begingroup$
    Ohh I see. Thanks for your help! I get it now!
    $endgroup$
    – Random Student
    Jan 21 at 23:30


















  • $begingroup$
    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 at 22:39






  • 1




    $begingroup$
    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 at 23:07










  • $begingroup$
    Ohh I see. Thanks for your help! I get it now!
    $endgroup$
    – Random Student
    Jan 21 at 23:30
















$begingroup$
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 at 22:39




$begingroup$
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 at 22:39




1




1




$begingroup$
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 at 23:07




$begingroup$
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 at 23:07












$begingroup$
Ohh I see. Thanks for your help! I get it now!
$endgroup$
– Random Student
Jan 21 at 23:30




$begingroup$
Ohh I see. Thanks for your help! I get it now!
$endgroup$
– Random Student
Jan 21 at 23:30


















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