## Problem on finding a tangent vector in the direction of steepest ascent

Consider the surface $z = 10 - x^2 - 2 y^2$. At $(1,-1,7)$, find a 3d tangent vector that points in the direction of steepest ascent.
• ## Solution

To devise a strategy for finding $\bfv$, we recall that
Our strategy will be: first, find the $x,y$ components of $\mathbf{v}$; then, find the $z$ component of $\mathbf{v}$.

#### Finding the $x,y$ components of $\bfv$

As the gradient provides the direction of steepest ascent, we compute it:
\begin{align}\nabla z(x,y) &= -2x \ \mathbf{i} - 4 y \ \mathbf{j} \\ \nabla z(1,-1) &= -2\bfi + 4 \bfj
\end{align}
Thus $a = -2$ and $b = 4$, and we are seeking a tangent vector $$\mathbf{v} = -2 \bfi + 4 \bfj + c \bfk.$$

#### Finding the $z$ component of $\bfv$

Recall that
To illustrate, we consider the cross-section of the surface at $x=1,y=-1$ in the direction of $-2 \mathbf{i} + 4 \mathbf{j}$:
The rise of $\mathbf{v}$ in $z$ is $c$. The run of $\mathbf{v}$ in $(x,y)$ is $\sqrt{a^2 + b^2}=\sqrt{20}$.
Thus, tangency requires \begin{align}\frac{c}{\sqrt{20}} &= \sqrt{20}\\ c &= 20. \end{align}
A tangent vector to the surface in the direction of maximal ascent is: $$\mathbf{v} = -2 \mathbf{i} -4 \mathbf{j} + 20 \mathbf{k}.$$