Consider the surface given by $z = x^2 + y^2$. What is the (3d) tangent vector at $(1,1)$ that has an $\mathbf{i}$ component of 0 and a $\mathbf{j}$ component of 2?

Solution using Directional Derivatives

Let $\mathbf{v} = 0 \mathbf{i} + 2 \mathbf{j} + c \mathbf{j}$ be the vector we seek. We would like to find the value of $c$ such that $\mathbf{v}$ is tangent to $z(x,y) = x^2 + y^2$ at $(1,1)$.

The vector $\mathbf{v}$ has a rise of $c$ and a run in the $xy$ plane of $| 0\mathbf{i} + 2 \mathbf{j}| = 2$. Hence, the 'slope' of the vector $\mathbf{v}$ is $\frac{c}{2}$.