29 - 30 find the gradient vector field $\nabla f$ of $f$ and sketch it. 29. $f(x,y)=\frac{1}{2}(x - y)^2$…

29.

Explanation:

Step1: Recall gradient formula

The gradient of a function $f(x,y)$ is $\nabla f=\left(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y}\right)$.

Step2: Calculate $\frac{\partial f}{\partial x}$

Given $f(x,y)=\frac{1}{2}(x - y)^2$. Using the chain - rule, if $u=x - y$, then $f=\frac{1}{2}u^{2}$. So $\frac{\partial f}{\partial x}=\frac{\partial f}{\partial u}\cdot\frac{\partial u}{\partial x}$. We have $\frac{\partial f}{\partial u}=u=x - y$ and $\frac{\partial u}{\partial x}=1$. Thus, $\frac{\partial f}{\partial x}=x - y$.

Step3: Calculate $\frac{\partial f}{\partial y}$

Again, using the chain - rule. $\frac{\partial f}{\partial y}=\frac{\partial f}{\partial u}\cdot\frac{\partial u}{\partial y}$. Since $\frac{\partial f}{\partial u}=u=x - y$ and $\frac{\partial u}{\partial y}=-1$, then $\frac{\partial f}{\partial y}=-(x - y)=y - x$.

Step4: Write the gradient vector field

$\nabla f=(x - y,y - x)$.

Answer:

$\nabla f=(x - y,y - x)$

30.

Explanation:

Step1: Recall gradient formula

The gradient of a function $f(x,y)$ is $\nabla f=\left(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y}\right)$.

Step2: Calculate $\frac{\partial f}{\partial x}$

Given $f(x,y)=\frac{1}{2}(x^{2}-y^{2})$. Differentiating with respect to $x$ while treating $y$ as a constant, we get $\frac{\partial f}{\partial x}=\frac{1}{2}\cdot2x=x$.

Step3: Calculate $\frac{\partial f}{\partial y}$

Differentiating $f(x,y)=\frac{1}{2}(x^{2}-y^{2})$ with respect to $y$ while treating $x$ as a constant. We have $\frac{\partial f}{\partial y}=\frac{1}{2}\cdot(- 2y)=-y$.

Step4: Write the gradient vector field

$\nabla f=(x,-y)$.

Answer:

$\nabla f=(x,-y)$