Partial differential (multivariable calculus)

[Poetria]

Homework Statement

Recall that when we take a partial derivative, at a point (a,b), we fix the x-coordinate at a , and differentiate with respect to the y variable.

Which of the following best describes fixing the x-coordinate at the value a?

Relevant Equations

E.g. f(x,y)=x^2+y^2
f(y)= 2*y

Intersecting the graph of the surface z=f(x,y) with the yz -plane.

This is the option I have chosen, but it's wrong. I don't understand why. x is fixed so I thought the coordinates: y and z are left.

I thought this source may be helpful: https://www.whitman.edu/mathematics/calculus_online/section14.03.html

 

[fresh_42]

You fixed the value of $x$ to $x=0$. Why? You should consider $g(y)=f(c,y)$ with an arbitrary constant $c$. Now that we are interested in the derivative at $(x,y)=(a,b)$ we get $c=a$. Your solution is only correct if $a=0.$ E.g. if we have $f(x,y)=xy^2$ then we look for $g'(y)=(ay^2)'$ at $y=b.$ What would we get in the $y-z-$plane?

 

[Poetria]

Well, g'(y)=2*a*y
If y=b, then g'(y)=2*a*b

Is it a line?
So the correct answer is: Intersecting the graph of the surface z=f(x,y) with the plane y=b ?

 

[fresh_42]

Well, g'(y)=2*a*y
If y=b, then g'(y)=2*a*b

Is it a line?
So the correct answer is: Intersecting the graph of the surface z=f(x,y) with the plane y=b ?

Are you sure the partial derivative according to $y$ was meant? I'd say we consider the intersection with the plane $x=a$ since we want to know changes in the $y-$direction with constant $x=a.$ The plane with $y=b$ doesn't change anything in the $y-$direction.

A partial derivative is only a directional derivative in a coordinate direction, either $\partial_x$ or $\partial_y.$ Evaluating it at $(a,b)$ means to put the origin of the tangent space in $(a,b)$, but the directions remain the same.

 

[Poetria]

I got it. The plane a is parallel to the yz plane.
Many thanks.