A function $\,f:\reel^2\rightarrow\reel\,$ is given by the expression
$$\,f(x,y)=x^2+y^2\,.$$
A
Describe the level curves given by $\,f(x,y)=c\,$ for the values $\,c \in\lbrace 1,2,3,4,5\rbrace\,.$
hint
Remember the equation of a circle: $\,(x-a)^2+(x-b)^2=r^2\,.$
answer
The level curves are circles all centered at $\,(0,0)\,.$ Their radii are $\,1,\,\sqrt 2,\,\sqrt 3,\,2,\,\sqrt 5\,$, respectively.
B
Determine the gradient of $\,f\,$ in the point $\,(1,1)\,$ and determine the directional derivative of $\,f\,$ in the point $\,(1,1)\,$ in the direction determined by the unit directional vector $\,\mathbf e=(1,0)\,.$
answer
$\,\nabla f(1,1)=(2,2)\,.$ The directional derivative is the dot product of the gradient and the given directional derivative, that is, $\,2\,.$
A function $\,f:\reel^2\rightarrow\reel\,$ is given by the expression
$$\,f(x,y)=x^2-4x+y^2\,.$$
C
Describe the level curves given by $\,f(x,y)=c\,$ for the values $\,c \in\lbrace -3,-2,-1,0,1\rbrace\,.$
hint
Remember the equation of a circle: $\,(x-a)^2+(x-b)^2=r^2\,.$
answer
We give the result for the first level curve: Since
$$\,x^2-4x+y^2=-3\Leftrightarrow (x-2)^2+y^2=1$$
is the level curve for a circle centered in $\,(2,0)\,$ and radius 1. The other level curves are also circles with the same center, but different radii.
D
Determine the gradient of $\,f\,$ in the point $\,(1,2)\,$ and determine the directional derivative of $\,f\,$ in the point $\,(1,2)\,$ in the direction towards the origin.
answer
We start with the gradient:
$$\,\nabla f(1,2)=(-2,4)\,.$$
hint
We shall use a unit directional vector that points from $\,(1,2)\,$ towards the origin.
hint
We can use the directional vector $\,(-1,-2)\,,$ but it must be normed that is it must have the length 1.
answer
The wanted unit directional vector is found by dividing the proposed directional vector by its length, that is:
When you then dot $\,\mathbf e\,$ with the gradient, you get the directional derivative
$$\,-\frac{6}{\sqrt 5}\,.$$
Exercise 2: Approximation of First Degree
For $\,(x,y)\in \reel^2\,$ we consider the function
$$\,f(x,y)=\exp(-x+\sin(y))\,.$$
A
Determine the approximating first-degree polynomial for $\,f\,$ with the development point $\,(x,y)=(0,0)\,.$
answer
$\,P_1(x,y)=1-x+y\,.$
B
Determine an equation for the tangent plane to the graph for $\,f\,$ in the point of tangency $\,(x,y,z)=\big(0,0,f(0,0)\big)\,.$ And determine a normal vector for the tangent plane.
answer
The tangent plane:
$$z=1-x+y\,\Leftrightarrow x-y+z=1\,.$$
From this a normal vector can immediately be read (highschool stuff):
$$\,\mathbf N=(1,-1,1)\,.$$
Exercise 3: Description of Regions in the (x,y)-plane
A
Draw in each of the four cases below a sketch of the given set of points $\,A\,$, det interior $\,A^{\circ}\,$, the boundary $\,\partial A\,$ and the closure $\,\bar{A}\,$. Further investigate whether $\,A\,$ is open, closed or neither. Finally state whether $\,A\,$ is bounded or not.
First focus on the axes: How are the set of points in question bounded in the direction of the $\,x\,$-axis and the $y$-axis, respectively?
The variables can also be mutually dependent, but only curve forms that you already know is used.
answer
$\lbrace(x,y)\,\,\vert xy\neq 0\rbrace\,$ is the real number plane ($\reel^2$), but without the coordinate axes. This region is also the interior of the set, while the boundary is the coordinate axes. The closure is the real number plane. The set is open and not bounded.
$\lbrace(x,y)\,\,\vert 0<x<1\wedge 1\leq y\leq 3\rbrace\,$ constitutes the rectangle, bounded by the straight lines $x=0$, $x=1$, $y=1$ and $y=3$, where $x=0$ and $x=1$ do not belong to the set, while $y=1$ and $y=3$ belong to the set. The interior of the set is the rectangle excluding the line segments, the boundary is all four line segments and the closure is the rectangle including the line segments. The set is neither open not closed, but it is bounded.
$\lbrace(x,y)\,\,\vert y\geq x^2\wedge \vert x\vert<2 \rbrace\,$ constitutes the section of the region that lies above the parabola given by the equation $\,y=x^2\,$ and the region between the lines $\,x=-2\,$ and $\,x=2\,$. The line does not belong to the set but the segment of the parabola does. The interior of the set is the region excluding the lines and the segment of the parabola, the boundary consists of the two line segments and the segment of the parabola, while the closure is the region including the linesegments and segment of the parabola. The set is neither open nor closed, and it is unbounded.
$\lbrace(x,y)\,\,\vert x^2+y^2-2x+6y\leq 15 \rbrace\,$ constitutes the region inside the circle with center at $(1,-3)$ and radius 5. The interior is the region excluding the circumference, the boundary is the circumference and the closure is the region including the circumference. Thus the closure is the set itself. The set is closed and bounded.
Exercise 4: An Altitude Function
We consider a real function of two real variables given by the expression
$$f(x,y)=\ln(9-x^2-y^2)\,.$$
A
Determine the domain for $\,f\,,$ and characterize the domain using concepts such as open, closed, bounded, unbounded.
answer
It is only possible to take the logarithm of positive values, therefore
$$9-x^2-y^2>0\Leftrightarrow x^2+y^2<3^2\,.$$
Dm$(f)=\left{(x,y)\,|\,x^2+y^2<3^2\right}\,.$ this is a circular disk with center at the origin and radius 3, but excluding the circumference. The set is open and bounded.
Now we consider a parametrized curve $\,\mathbf r\,$ in the $\,(x,y)$-plane given by
Which curve are we talking about (you already know its equation)?
answer
It is the graph for the third-degree polynomial $\,y=x^3\,,\,\,x\in \left[-1.2\,,\,1.2\right]\,.$
Now we consider the composite function
$$\,h(u)=f(\mathbf r(u))\,.$$
C
Why is it fair to call $\,h\,$ an altitude function?
D
Determine $\,h\,’(1)\,$ by two different methods:
1) Determine a functional expression for $\,h(u)\,$ and differentiate in the ordinary fashion.
2) Use Theorem $\,19.49\,$ in eNote 19: The chain rule along curves.
answer
Method 1: We get $\,h(u)=ln(-u^6-u^2+9)\,$ and $\,\displaystyle{h’(1)=-\frac{8}{7}}\,.$
Methode 2: The tangent vector is determined: $\,\mathbf r’(u)=(1,3\,u^2)\Rightarrow \mathbf r’(1)=(1,3)\,.$ The gradient of $\,\nabla f(x,y)\,$ is found, whereafter $\,\displaystyle{\nabla f(\mathbf r(1))=\nabla f(1,1)=(-\frac{2}{7},-\frac{2}{7})\,.}$ The dot product of the vectors found is $\,\displaystyle{-\frac{8}{7}\,.}$
Exercise 5: Summary Exercise
A real function $f$ of two real variables are given by:
$$f(x,y)=\frac {\mathrm e^x}y\,.$$
A
Determine the domain for $f\,$.
answer
$\,Dm(f)=\left{(x,y)\in\reel^2\,|\,y\neq 0\,\right}.$ Geometrically that is all point in the $\,(x,y)$-plane except the $\,x$-axis.
B
Compute the functional value of $f\,$ in the following three points: $\,A(1,1),\,\,B(0,1)\,\,\,\mathrm{and}\,\,\,C(-1,\frac 1{\mathrm e}\,)\,.$
Two out of the three points are on the same level curve for $\,f\,$. Describe this level curve.
answer
Since $\,f(B)=f(C)=1\,,$$B$ and $C$ lie on the same level curve. It is given by
Determine the gradient of $\,f\,$ in the point $\,(1,1\,)$, and find the directional derivative of $\,f\,$ in the direction that is determined by the vector $\,\mathbf s=(1,-1)\,.$
answer
The gradient: $\,\nabla f(1,1)=(\e,-\e)\,.$
The directional derivative: $f’\left(\,(1,1),\mathbf e\,\right)=\sqrt 2\,\e\,.$
For $\,u>0\,$ and in the $\,(x,y)$-plane is given the parametrized curve $\,\mathbf r(u)=(u,u)\,$. Furthermore we are given the composite function
$$\,h(u)=f\big(\mathbf r(u)\big)\,.$$
D
Determine the point $\,\mathbf r(u_\texttt{o})\,$ in the $\,(x,y)$-plane, for which $\,h\,’(u_\texttt{o})=0\,$.
answer
By computation you get $u_o=1\,.$ Therefore the wanted point on the curve is the point $\mathbf r(1)=(1,1)\,.$
