1.1Compute a Gradient
For $f(x,y)=x^2y+3y$, compute $\nabla f$ at $(2,-1)$.
Solution
$f_x=2xy$ and $f_y=x^2+3$, so $\nabla f(2,-1)=\langle-4,7\rangle$.
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Difficulty
For $f(x,y)=x^2y+3y$, compute $\nabla f$ at $(2,-1)$.
Solution
$f_x=2xy$ and $f_y=x^2+3$, so $\nabla f(2,-1)=\langle-4,7\rangle$.
For $f(x,y)=x^2+y^2$, find the directional derivative at $(1,2)$ toward $(4,6)$.
Solution
$\nabla f(1,2)=\langle2,4\rangle$ and the unit direction is $\langle3,4\rangle/5$. Thus $D_uf=22/5$.
Difficulty
Use the linearization of $f(x,y)=\sqrt{x^2+y^2}$ at $(3,4)$ to estimate $f(3.1,3.9)$.
Solution
$f(3,4)=5$, $f_x=3/5$, and $f_y=4/5$. Therefore $L=5+(3/5)(0.1)+(4/5)(-0.1)=4.98$.
Classify the critical point $(0,0)$ of $f(x,y)=x^2+4y^2$.
Solution
The Hessian is $\operatorname{diag}(2,8)$, which is positive definite. The point is a strict local minimum.
Difficulty
Set up, but do not solve, the Lagrange equations for maximizing $f(x,y)=xy$ subject to $x^2+y^2=1$.
Solution
Set $g=x^2+y^2-1$. The system is $\nabla f=\lambda\nabla g$, namely $\langle y,x\rangle=\lambda\langle2x,2y\rangle$ together with $x^2+y^2=1$.
Difficulty
For $F(x,y)=(x+y,x-y)$, write its Jacobian and explain what it maps locally.
Solution
$J_F=\begin{bmatrix}1&1\\1&-1\end{bmatrix}$. It maps a small input displacement to the corresponding first-order output displacement.