# Math Help - Partial Derivatives Problem

1. ## Partial Derivatives Problem

Hi all,

my professor posted a solution to a partial derivative in his class notes. I am not sure how he got to <0,1,5>. Any suggestions would be greatly appreciated!

Let f(x, y) = xy + y2. Find the tangent lines to the curves obtained by slicing f(x,y) using the planes x = 1,y = 2 at (1,2). When we slice f(x,y) at (1,2) with x = 1, what we are doing is finding the tangent line to the curve f(1,y) at (1,2). We see that fy(x,y) = x+2y, so fy(1,2) = 5. This line passes through (1,2,6), and has direction vector given by ⟨0, 1, 5⟩. Indeed, we see that the x-coordinate of this line never changes, which explains why the first coordinate equals 0. This means that the tangent line in question has equation x = 1, y = 2 + t, z = 6 + 5t.

2. Originally Posted by newman611
Hi all,

my professor posted a solution to a partial derivative in his class notes. I am not sure how he got to <0,1,5>. Any suggestions would be greatly appreciated!

Let f(x, y) = xy + y2. Find the tangent lines to the curves obtained by slicing f(x,y) using the planes x = 1,y = 2 at (1,2). When we slice f(x,y) at (1,2) with x = 1, what we are doing is finding the tangent line to the curve f(1,y) at (1,2). We see that fy(x,y) = x+2y, so fy(1,2) = 5. This line passes through (1,2,6), and has direction vector given by <0, 1, 5>.
Yes, "slicing" z= f(x,y) with the plane x= 1 gives $z= f(1, y)= y+ y^2$. The derivative of that is $dz/dy= 1+ 2y$ which is 1+ 4= 5 at y= 2. Of course, z(1, 2)= 2+ 4= 6. Ignoring the x-component, that tangent line would be z= 5(y- 2)+ 6= 5y- 4. Two points on that line are y= 2, z= 6, and y= 3, z= 15-4= 11 or, adding the x= 1 coordinate, (1, 2, 6) and (1, 3, 11). A vector from (1, 2, 6) to (1, 3, 11) which would point in the direction of that line is <1- 1, 3- 2, 11- 6>= <0, 1, 5>.

Indeed, we see that the x-coordinate of this line never changes, which explains why the first coordinate equals 0. This means that the tangent line in question has equation x = 1, y = 2 + t, z = 6 + 5t.
Notice that we could, as easily, have taken the second point to be at, say, y= 0 which would give z= 5(0)- 4= -4 so our two points would be (1, 2, 6) and (1, 0, -4). A vector from (1, 2, 6) to (1, 0, -4) which would also be pointing along the line (but in opposite direction) would be <1- 1, 0- 2, -4- 6>= <0, -2, -10> and parametric equations for the line would be x= 1, y= 2- 2t, z= 6- 10t. Do you see that these are parametric equations for exactly the same line as x= 1, y= 2+ t, z= 6+ 5t?

Note that when t= -1/2 these new equations give the point (1, 2+ 1, 6+ 5)= (1, 3, 11) as before. Since both lines contain (1, 2, 6) and (1, 3, 11) they are the same line.