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- Apr 14, 2013

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Hey!!

We consider the function $f:\mathbb{R}^2\rightarrow \mathbb{R}$ with $$f(x,y)=\frac{x^2-1}{y^2+1}$$

(a) Describe and draw the level lines $N_c$ of $f$ for all $c\in \mathbb{R}$. Determine for each connected component of each non-empty level lines $N_c$ a parametrization $\gamma_c:I\rightarrow \mathbb{R}$.

(b) Calculate the gradient $\nabla f(\gamma_c(t))$ of $f$ in the point $\gamma_c(t)\in \mathbb{R}^2$. Show directly that $\nabla f(\gamma_c(t))$ is orthogonal to the level lines $N_c$ at the point $\gamma_c(t)$.

(c) Let $x\in \mathbb{R}^2$, such that $\nabla f(x)\neq 0$. Determine the direction with maximum directional derivative of $f$ in point $x$.

For (a) :

We have that $$f(x,y)=c \Rightarrow \frac{x^2-1}{y^2+1}=c \Rightarrow x^2-1=(y^2+1)c \Rightarrow x^2-cy^2-(c+1)=0$$

If $c=0$ then we have $x^2=1 \Rightarrow x\pm 1$, so we get the lines $x=1$ and $x=-1$.

If $c>0$ we have then $c=m^2>0$. So $x^2-m^2y^2-(m^2+1)=0 \Rightarrow x^2-m^2y^2=(m^2+1)$. We have $B^2-4AC=0-4\cdot 1\cdot (-m^2)=m^2>0$. So it is the general equation of an hyperbola with axes parallel to the coordinate axes.

If $c<0$ we have then $c=-m^2<0$. So $x^2+m^2y^2-(-m^2+1)=0 \Rightarrow x^2+m^2y^2=(-m^2+1)$. The left side is always non-negative, so also the right side must be. So $1-m^2\geq0 \Rightarrow m^2\leq1 \Rightarrow -1\leq -m^2 \Rightarrow -1\leq c$. That means that is $c$ is negative then $c$ must be greater or equal to $-1$. If $-1<c<0$ then we have the general formula of an ellipse. If $c=-1$ then we get $x^2+y^2=0$ which is the point $(0,0)$.

Is that correct?

How do we determine for each connected component of each non-empty level lines $N_c$ a parametrization $\gamma_c:I\rightarrow \mathbb{R}$ ?

For (b) :

We have that $$\nabla f(x,y)=\frac{2x}{y^2+1}\hat{i}+\frac{-2(x^2-1)y}{(y^2+1)^2}\hat{j}$$ But how do we get $\nabla f(\gamma_c(t))$ ?

For (c) :

The maximum value of the directional derivative occurs when $\nabla f$ and the unit vector point in the same direction.

But how can we find that? Dowenot need aspecific point here?

We consider the function $f:\mathbb{R}^2\rightarrow \mathbb{R}$ with $$f(x,y)=\frac{x^2-1}{y^2+1}$$

(a) Describe and draw the level lines $N_c$ of $f$ for all $c\in \mathbb{R}$. Determine for each connected component of each non-empty level lines $N_c$ a parametrization $\gamma_c:I\rightarrow \mathbb{R}$.

(b) Calculate the gradient $\nabla f(\gamma_c(t))$ of $f$ in the point $\gamma_c(t)\in \mathbb{R}^2$. Show directly that $\nabla f(\gamma_c(t))$ is orthogonal to the level lines $N_c$ at the point $\gamma_c(t)$.

(c) Let $x\in \mathbb{R}^2$, such that $\nabla f(x)\neq 0$. Determine the direction with maximum directional derivative of $f$ in point $x$.

For (a) :

We have that $$f(x,y)=c \Rightarrow \frac{x^2-1}{y^2+1}=c \Rightarrow x^2-1=(y^2+1)c \Rightarrow x^2-cy^2-(c+1)=0$$

If $c=0$ then we have $x^2=1 \Rightarrow x\pm 1$, so we get the lines $x=1$ and $x=-1$.

If $c>0$ we have then $c=m^2>0$. So $x^2-m^2y^2-(m^2+1)=0 \Rightarrow x^2-m^2y^2=(m^2+1)$. We have $B^2-4AC=0-4\cdot 1\cdot (-m^2)=m^2>0$. So it is the general equation of an hyperbola with axes parallel to the coordinate axes.

If $c<0$ we have then $c=-m^2<0$. So $x^2+m^2y^2-(-m^2+1)=0 \Rightarrow x^2+m^2y^2=(-m^2+1)$. The left side is always non-negative, so also the right side must be. So $1-m^2\geq0 \Rightarrow m^2\leq1 \Rightarrow -1\leq -m^2 \Rightarrow -1\leq c$. That means that is $c$ is negative then $c$ must be greater or equal to $-1$. If $-1<c<0$ then we have the general formula of an ellipse. If $c=-1$ then we get $x^2+y^2=0$ which is the point $(0,0)$.

Is that correct?

How do we determine for each connected component of each non-empty level lines $N_c$ a parametrization $\gamma_c:I\rightarrow \mathbb{R}$ ?

For (b) :

We have that $$\nabla f(x,y)=\frac{2x}{y^2+1}\hat{i}+\frac{-2(x^2-1)y}{(y^2+1)^2}\hat{j}$$ But how do we get $\nabla f(\gamma_c(t))$ ?

For (c) :

The maximum value of the directional derivative occurs when $\nabla f$ and the unit vector point in the same direction.

But how can we find that? Dowenot need aspecific point here?

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