A function $\,f:\reel^2\rightarrow\reel\,$ is given by the expression
$$\,f(x,y)=x^2+y^2\,.$$
A
Describe the level curves $\,f(x,y)=c\,$ for the values $\,c \in\lbrace 1,2,3,4,5\rbrace\,.$
hint
To describe a level curve, you must describe its shape, size, location and other defining features until it is fully defined for the reader. This is often possible in just one sentence.
hint
Remember that the circle equation is
$$\,(x-a)^2+(x-b)^2=r^2\,.$$
Anything on this same form is a circle.
answer
The level curves are circles all centered at $\,(0,0)\,.$ Their radiuses are, respectively, $\,1,\,\sqrt 2,\,\sqrt 3,\,2,$ and $\sqrt 5\,$.
B
Determine the gradient of $\,f\,$ at the point $\,(1,1)\,$ and determine the directional derivative of $\,f\,$ at this same point $\,(1,1)\,$ in the direction determined by the direction vector $\,\mathbf e=(1,0)\,.$
hint
A directional derivative is computed as the dot product of the gradient and the unit direction vector in the point.
hint
Luckily the given direction vector is already of unit length, so no need to divide it with its own length first.
answer
Gradient:
$$\,\nabla f(1,1)=(2,2)\,$$
Directional derivative:
$$f'\left((1,1);(1,0)\right)=2$$
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 $\,f(x,y)=c\,$ for the values $\,c \in\lbrace -3,-2,-1,0,1\rbrace\,.$
hint
Again, they are circles. But slightly more rewriting is needed this time before it is brought to the form of the circle equation.
This level curve is a circle centred at $\,(2,0)\,$ with radius 1. The other level curves are likewise circles with the same centre but different radiuses.
D
Determine the gradient of $\,f\,$ at the point $\,(1,2)\,$ and determine the directional derivative of $\,f\,$ at $\,(1,2)\,$ in the direction towards the origin.
hint
This time we need to first construct a direction vector. You can do that by subtracted end point from starting point.
For $\,(x,y)\in \reel^2\,$ we consider the function
$$\,f(x,y)=\exp(-x+\sin(y))\,.$$
A
Determine the approximating first-degree polynomial of $\,f\,$ with expansion point $\,(x,y)=(0,0)\,.$
answer
$$P_1(x,y)=1-x+y$$
B
Determine an equation for the tangent plane to the graph of $\,f\,$ in the point $\,(x,y,z)=\big(0,0,f(0,0)\big)\,.$ Determine a normal vector of the tangent plane.
answer
The tangent plane:
$$z=1-x+y\quad\Leftrightarrow\quad x-y+z=1\,.$$
A normal vector is read from this plane equation as the coefficients (highschool stuff):
$$\mathbf N=(1,-1,1)\,.$$
Exercise 3: Description of Regions in the (x,y) plane
Use proper sketching notation, such as highlighting for included regions, solid vs. dashed lines for included vs. excluded lines of points and use filled-in dots or empty dots for included vs. excluded single points.
hint
If you struggle to get started, it is often a good idea to first figure out where the boundary should be drawn. Then consider which other regions on either side of this boundary that do or do not belong to the set.
is the real number plane $\reel^2$ without the coordinate axes. This region is also the interior of the set, while the boundary is the coordinate axes. The closure is the entire real number plane. The set is open and not bounded.
covers the rectangle that is 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 or closed, but it is bounded.
covers 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 line segments and the segment of the parabola. The set is neither open or closed, and it is unbounded.
covers the region inside the circle centred at $(1,-3)$ and with a radius of $5$. 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 Elevation Function
We will now 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 of $\,f\,$ and characterize the domain using terms such as open, closed, bounded, unbounded etc.
hint
Logarithms are only defined for positive values, therefore
Determine the gradient of $\,f\,$ at the point $\,(1,1\,)$, and compute the directional derivative of $\,f\,$ at this point in the direction given by the vector $\,\mathbf s=(1,-1)\,.$
