Marianopolis 201-SNU-MS tutoring

When finding $f_x$, hold $y$ fixed; when finding $f_y$, hold $x$ fixed. A direction vector gives an orientation, but the requested rate uses a unit vector.

1. Find the gradient at the specified point

The partial derivatives are $f_x=2x+y$ and $f_y=x+4y$. The gradient collects them: $\nabla f=(2x+y,x+4y)$, hence $\nabla f(P)=(5,6)$. These are the rates for the two coordinate directions; neither alone gives the requested slanted-direction rate.

2. Make the direction vector unit length

Since $\|\mathbf v\|=\sqrt{4^2+3^2}=5$, use $\mathbf u=\mathbf v/\|\mathbf v\|=(4/5,3/5)$. Its squared length is $16/25+9/25=1$. Doubling $\mathbf v$ doubles its length as well, leaving $\mathbf u$ unchanged.

3. Combine the coordinate rates

The polynomial is differentiable everywhere, so its directional derivative is the gradient dot product with the unit direction:

The positive result means $f$ increases initially in that direction, by about $(38/5)s$ for a small positive distance $s$ from $P$. This is a local linear estimate, not an exact change over every distance.

4. Check by following a unit-speed line

Along $(x,y)=(2+4s/5,1+3s/5)$, a positive $s$ is the distance travelled from $P$. Substitution gives $f=8+(38/5)s+(46/25)s^2$. Its derivative at $s=0$ is $38/5$, confirming the result without reusing the dot-product calculation.

Using $\mathbf v$ directly would give $38$. That is the rate with respect to a parameter on the line $P+t\mathbf v$, where each parameter unit moves five coordinate-distance units. It does not answer the requested rate per unit distance.

Here $\nabla g=(6x-2y,-2x+2y)$, so $\nabla g(Q)=(8,-4)$. Because $\|\mathbf w\|=5$, its unit direction is $\mathbf u=(-3/5,4/5)$. Thus

The function decreases initially in this direction. Doubling $\mathbf w$ leaves its unit direction and this derivative unchanged. For a separate check, substitute $(x,y)=(1-3s/5,-1+4s/5)$: $g=6-8s+(67/25)s^2$, whose derivative at zero is $-8$.