Problem on finding a tangent vector in the direction of steepest ascent
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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
Level curves and surfaces
Let our tangent vector be $\mathbf{v} = a \mathbf{i} + b \mathbf{j} + c \mathbf{k}$. First, we sketch the surface and the tangent vector in a couple ways: in perspective, and relative to the level curves of $z$.
To devise a strategy for finding $\bfv$, we recall that
Gradient is in direction of steepest ascent
The direction of steepest ascent of $z(x,y)$ is given by the two-dimensional vector $\nabla z$.
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 $\bfv$
Tangent vectors to surfaces
We have reduced the problem to one of finding the tangent vector with specified values of the $x$ and $y$ components.
Recall that
Using directional derivatives to find a tangent vector to a surface
A tangent vector $a\bfi + b\bfj + c\bfk$ to a surface has a slope (rise in $z$ over run in $xy$) equal to the directional derivative of the surface height $z(x,y)$ in the direction of $a \bfi + b \bfj$.
To illustrate, we consider the cross-section of the surface at $x=1,y=-1$ in the direction of $-2 \mathbf{i} + 4 \mathbf{j}$:
(see details)
We can compute that the directional derivative of $f$ in the direction of $-2 \mathbf{i} + 4 \mathbf{j}$ is $\sqrt{20}$.
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: $$\mathbf{v} = -2 \mathbf{i} -4 \mathbf{j} + 20 \mathbf{k}.$$
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