UQAM MAT1115 tutoring

Find the gradient at the starting point

The gradient $\nabla f$ is the vector of partial derivatives, each found while holding the other coordinate fixed: $$\begin{aligned}&\nabla f(x,y)\\&=(2x+y,x+2y)\end{aligned}$$. At the starting point, $\nabla f(1,2)=(4,5)$.

Normalize the direction vector to length 1, then calculate the rate per unit distance

The length of $(3,4)$ is $\sqrt{3^2+4^2}=5$, so the unit direction vector is $u=(3/5,4/5)$.

Let $D_u f(1,2)$ denote the rate of change per unit distance in this direction. It is the dot product $$\begin{aligned}&D_u f(1,2)\\&=\nabla f(1,2)\cdot u\\&=4(3/5)+5(4/5)\\&=32/5=6.4\end{aligned}$$.

Estimate the value after moving a distance of 0.01

Let $s=0.01$ be the distance moved. The displacement is $su=(0.006,0.008)$, so the new point is $(1.006,2.008)$.

Since $f(1,2)=7$, the estimate is $$\begin{aligned}&f(1.006,2.008)\\&\approx f(1,2)+sD_u f(1,2)\\&=7+0.01(6.4)\\&=7.064\end{aligned}$$. This is an approximation because it keeps only the starting value and the linear change.

Check the error and compare the two direction vectors

Direct substitution gives $f(1.006,2.008)=7.064148$. The omitted quadratic terms are $$\begin{aligned}&(0.006)^2+(0.006)(0.008)\\&\quad +(0.008)^2=0.000148\end{aligned}$$, exactly the difference.

Using $(3,4)$ without normalizing would give a rate of $32$ per unit of its path parameter, not per unit distance. The vector $(6,8)$ has length $10$ and normalizes to the same $u$, so the requested directional derivative and estimate do not change.