squenshl
- 468
- 4
- Homework Statement
- Let ##U := \left\{(x,y)\in \mathbb{R}^2: xy\neq 0\right\}## and let ##f: U\mapsto \mathbb{R}## be defined by
$$f(x,y) := (\log_{e}{(|x|)})^2+(\log_{e}{(|y|)})^2.$$
1. Calculate ##\nabla f(x,y)## at each point of ##U##.
2. Let ##\mathbf{r}: (0,1)\mapsto \mathbb{R}^2## be defined by ##\mathbf{r}(t) := \left(e^{\sin{(t)}},e^{\cos{(t)}}\right).##
Calculate the derivative of ##\mathbf{r}## at each point of ##(0,1).##
3. Justify whether you can use the chain rule to calculate the derivative of ##f\circ \mathbf{r}.##
If it is justifiable, calculate the derivative of ##f\circ \mathbf{r}## using the chain rule.
- Relevant Equations
- None
1. We find the partial derivatives of ##f## with respect to ##x## and ##y## to get ##f_x = \frac{2\ln{(x)}}{x}## and ##f_y = \frac{2\ln{(y)}}{y}.## This makes the gradient vector
$$\nabla{f} = \begin{bmatrix}
f_x \\
f_y
\end{bmatrix} = \begin{bmatrix}
\frac{2\ln{(x)}}{x} \\
\frac{2\ln{(y)}}{y}
\end{bmatrix}.$$
2. We have
$$\mathbf{r}'(t) = \left(\cos{(t)}e^{\sin{(t)}},-\sin{(t)}e^{\cos{(t)}}\right).$$
After this I'm a little confused. Any help is appreciated.
$$\nabla{f} = \begin{bmatrix}
f_x \\
f_y
\end{bmatrix} = \begin{bmatrix}
\frac{2\ln{(x)}}{x} \\
\frac{2\ln{(y)}}{y}
\end{bmatrix}.$$
2. We have
$$\mathbf{r}'(t) = \left(\cos{(t)}e^{\sin{(t)}},-\sin{(t)}e^{\cos{(t)}}\right).$$
After this I'm a little confused. Any help is appreciated.