For $f ( x , y ) = x ^ { 2 } y$ find a line in the x direction tangent to the surface defined by f at (1,2).
A) $\vec { r } ( t ) = \langle 1,2,2 \rangle + t \langle 0,0,4 \rangle$
B) $\vec { r } ( t ) = \langle 1,2,2 \rangle + t \langle 0,0,2 \rangle$
C) $\vec { r } ( t ) = \langle 1,2,2 \rangle + t \langle 0,0,1 \rangle$
D) $\vec { r } ( t ) = \langle 1,2,2 \rangle + t \langle 4,1,0 \rangle$

Question

Quizplus · Accepted Answer

To find the tangent line in the x direction, we first calculate the partial derivative of $f$ with respect to $x$, which gives us $f_x = 2xy$. At the point (1,2), $f_x = 2(1)(2) = 4$, indicating the slope of the tangent in the x direction. The point on the surface is found by substituting $x=1$ and $y=2$ into $f$, giving $f(1,2) = 1^2(2) = 2$. Thus, the tangent line is $\vec{r}(t) = \langle 1,2,2 \rangle + t\langle 0,0,4 \rangle$, where the direction vector $\langle 0,0,4 \rangle$ reflects movement only in the z direction (consistent with the slope of the tangent line in the x direction) and no movement in x or y, as the line is tangent in the x direction at the point (1,2).

For f(x,y)=x2yf ( x , y ) = x ^ { 2 } yf(x,y)=x2y

For $f ( x , y ) = x ^ { 2 } y$