Find the slope of the tangent line that results when intersecting the following surfaces: $z=x^3y+5y^2$ with the plane $x=2$ at $y=1$
Attempt:
The surfaces must intersect, so I plug in $x=2$ into the first equation and get:
$$z=8y+5y^2$$
Then I take the derivative and evaluate it at 1:
$$ z'=8+10y \implies z'(1)=18 $$
So the answer is $18$, right?
This confused me in a few points: First I have $z=f(x,y)$, then when I plug in $x=2$ my $z$ becomes exclusively a function of $y$, right? So instead $z'$ would be equal to $z_y$?
Also, if I wanted to right out the equation of the tangent line whose slope I found, how would I procceed?
At first, z(x,y) is a surface, and after you plug in the value of x, you get a curve, which uses y as its parameter. So when you take the partial derivative, you actually get the tangent vector of the curve, and the result you got is the slope.
More concretely, you can think of the curve as defined as $(y,8y+5y^2)$. And then the tangent vector is $(1,8+10y)=(1,18)$ and the slope is 18/1=18.