Problem on directional derivatives
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Let $f(x,y) = x^2 + y^3 $ and $P=(1,1)$.
Find a direction at $P$ along which $f$ is not changing.
Solution
Directional derivative definition
We are after a direction along which a function is not changing. Using technical terms, this means we are after a $\bfu$ such that the directional derivative of $f$ in the direction of $\bfu$ is zero. $$ D_\bfu f = 0.$$
Recall that
Directional derivative and gradient
The directional derivative of $f(x,y)$ in the direction $\mathbf{u}$ is $$ D_\mathbf{u} f = \nabla f \cdot \mathbf{u} \textrm{, if } |\mathbf{u}|=1$$
Hence, we are looking for a unit vector $\mathbf{u}$ such that $$ \nabla f(1,1) \cdot \mathbf{u} = 0$$
Definition of the gradient
We can simplify this expression by evaluating the gradient of $f$ at $(1,1)$: $$\begin{align}\nabla f(x,y) &= 2x \ \mathbf{i} + 3y \ \mathbf{j} \\ \nabla f(1,1) &= 2 \mathbf{i} + 3 \mathbf{j} \end{align}$$
Hence, we are looking for a unit vector $\mathbf{u}$ such that $$ \langle 2, 3 \rangle \cdot \mathbf{u} = 0$$
Recall that
Dot product of perpendicular vectors is zero
Two vectors are perpendicular if their dot product is zero.
Finding a perpendicular vector
A vector perpendicular to $\langle a,b \rangle$ is $\langle -b, a \rangle$.
(see details)
Using these two facts, we get that $\langle 2,3 \rangle \cdot \langle -3, 2\rangle = 0$.
We can not say $\bfu = \langle -3, 2\rangle$ because this vector does not have unit length. We are after the multiple of this vector that has unit length.
Recall that
Direction of a vector
A multiple of the vector $\bfx$ that has unit length is given by $\text{dir } \bfx = \frac{\bfx}{| \bfx|}$.
Length of a vector
Thus $\bfu = \frac{\langle -3, 2 \rangle}{|\langle -3, 2 \rangle|}$, and the direction in which $w$ is not changing is $ \frac{\langle -3, 2 \rangle}{\sqrt{13}}$.
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