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🔬 Tutorial problems beta

🔬 Tutorial problems beta#

Note

This problems are designed to help you practice the concepts covered in the lectures. Not all problems may be covered in the tutorial, those left are for additional practice.

\(\beta\).1#

Find the (Euclidean) distances between the following pairs of points.

  1. \((1, 3)\) and \((2, 4)\)

  2. \((-1, 2)\) and \((3, 3)\)

  3. \((\tfrac{3}{2},-2)\) and \((-5,1)\)

  4. \((x,y)\) and \((2x,y+3)\)

  5. \((a, b)\) and \((-a, b)\)

  6. \((a,3)\) and \((2+a,5)\)

[Sydsæter, Hammond, Strøm, and Carvajal, 2016] Exercises for Section 5.5, Question 1

Recall that the distance between two points \((x_1,y_1)\) and \((x_2,y_2)\) is given by \(\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\).

  1. Distance between \((1, 3)\) and \((2, 4)\) is given by \(\sqrt{(1-2)^2+(3-4)^2} = \sqrt{1+1} = \sqrt{2}\)

  2. Distance between \((-1, 2)\) and \((3, 3)\) is given by \(\sqrt{(-1-3)^2+(2-3)^2} = \sqrt{(4)^2+(1)^2} = \sqrt{17}\)

  3. Distance between \((\tfrac{3}{2},-2)\) and \((-5,1)\) is given by \(\sqrt{(\tfrac{3}{2}-(-5))^2+(-2-1)^2} = \sqrt{(\tfrac{13}{2})^2+(3)^2} = \sqrt{(\tfrac{13}{2})^2+(3)^2} = \sqrt{\tfrac{169}{4} + 9} = \sqrt{\tfrac{205}{4}} = \tfrac{\sqrt{205}}{2}\)

  4. Distance between \((x,y)\) and \((2x,y+3)\) is given by \(\sqrt{(x-2x)^2+(y-(y+3))^2} = \sqrt{x^2+3^2} = \sqrt{x^2+9}\)

  5. Distance between \((a, b)\) and \((-a, b)\) is given by \(\sqrt{(a-(-a))^2+(b-b)^2} = \sqrt{(2a)^2+0^2} = \sqrt{4a^2} = 2|a|\). We need absolute value because distance is always non-negative.

  6. Distance between \((a,3)\) and \((2+a,5)\) is given by \(\sqrt{(a-(2+a))^2+(3-5)^2} = \sqrt{(2)^2+(2)^2} = 2\sqrt{2}\)

\(\beta\).2#

Suppose that the cost of producing \(Q\) units of a commodity is given by \(C(Q) = 1, 000 + 300Q + Q^2\).

  1. Compute \(C(0)\), \(C(100)\), and \(C(101) - C(100)\).

  2. Compute \(C(Q + 1) - C(Q)\) and explain the meaning of this expression.

[Sydsæter, Hammond, Strøm, and Carvajal, 2016] Exercises for Section 4.2, Question 6

Suppose that the cost of producing \(Q\) units of a commodity is given by \(C(Q)=1,000+300Q+Q2\). We have

\[ C(0)=1,000+300(0)+02 =1,000+0+0 = 1,000 \]
\[\begin{split} \begin{array}{c} C(100) = 1,000+300(100)+(100)^2 = 1,000+30,000+10,000 = 41,000 \\ C(101) = 1,000+300(101)+(101)^2 = 1,000+30,300+10,201 = 41,501 \\ C(101) - C(100) = 41, 501 - 41, 000 = 501 \end{array} \end{split}\]
\[\begin{split} \begin{array}{c} C(Q + 1) = 1,000 + 300(Q+1) + (Q+1)^2 = \\ = 1,000+300Q+300+Q2 +2Q+1 = 1,301 + 302Q + Q^2,\\ C(Q+1)-C(Q) = 1,301 + 302Q+ Q^2 - (1,000+300Q+Q^2) = \\ = 1,301+302Q+Q^2 -1,000-300 Q-Q^2 =301+2Q. \end{array} \end{split}\]

Note that \(C(Q+1)-C(Q)\) is the difference between the total cost of producing \(Q+1\) units of a commodity and the total cost of producing \(Q\) units of that commodity. As such, it is the incremental cost that is incurred by producing the \((Q + 1)\)-th unit of the commodity. Since the incremental increase is a one unit increase in this case, this incremental cost is known as the marginal cost of the \((Q + 1)\)-th unit of the commodity.

\(\beta\).3#

Consider the function \(f(x) = \frac{3x+6}{x-2}\).

  1. Find the domain of this function

  2. Show that \(5\) belongs to the range of this function. Hint: Find a value for \(x\) such that \(f(x) = 5\).

  3. Show that \(3\) does not belong to the range of this function.

[Sydsæter, Hammond, Strøm, and Carvajal, 2016] Exercises for Section 4.2, Question 14

Note that \(f(x)\) will be defined whenever the denominator \((x-2)\) does not equal zero.Thus the only restriction on the domain of this function is that \(x \ne 2\). Therefore the domain for this function is the set \(D = \{x \in \mathbb{R} : x \ne 2\} = \mathbb{R} \setminus \{2\}\).

Suppose that \(f(x) = 5\). This requires that \(f^{-1}(5)\) is defined:

\[\begin{split} \begin{array}{c} \frac{3x+6}{x-2} = 5 \\ 3x+6 = 5(x-2) \\ 6+10 = 5x-3x \\ x = 8 \end{array} \end{split}\]

Thus, for \(x=8\) \(f(x) = 5\) and therefore \(5\) belongs to the range of \(f\).

Suppose instead that f(x) = 3. This requires that

\[\begin{split} \begin{array}{c} \frac{3x+6}{x-2} = 3 \\ 3x+6 = 3(x-2) \\ 6+6 = 3x-3x \\ 12 = 0 \end{array} \end{split}\]

This equation does not have any solutions. Therefore, \(f^{-1}(3)\) is undefined, and so \(3\) does not belong to the range of \(f\).

\(\beta\).4#

Suppose that the (Euclidean) distance between \((2,4)\) and \((5,y)\) is \(\sqrt{13}\).

  1. Find all possible values for \(y\).

  2. Explain geometrically why there must be exactly two distinct values for \(y\).

[Sydsæter, Hammond, Strøm, and Carvajal, 2016] Exercises for Section 5.5, Question 2

Let \(a = (a_1,a_2) = (2,4)\) and \(b = (b_1,b_2) = (5,y)\). The distance between these two points is

\[ \begin{array}{c} d(a,b)=\sqrt{(b_1 -a_1)^2 +(b_2 -a_2)^2} =\sqrt{(5-2)^2 +(y-4)^2} =\sqrt{32 +(y-4)^2} =\sqrt{9+(y-4)^2}. \end{array} \]

Given the distance \(d(a,b)=\sqrt{13}\), we have

\[\begin{split} \begin{array}{c} \sqrt{9+(y-4)^2} = \sqrt{13}\\ 9+(y-4)^2 = 13\\ (y-4)^2 = 4\\ y-4 = \pm \sqrt{4}\\ y = 4 \pm 2 \end{array} \end{split}\]

Thus we know that either (i) y = 6 or (ii) y = 2.

The set of points that are \(\sqrt{13}\) away from \((2,5)\) is a circle centered at \((2,4)\) with radius \(\sqrt{13}\). The set of points whose x-coordinate is 5 form a vertical straight line parallel to the y–axis. There are two points of the form \((5,y)\) that are exactly \(\sqrt{13}\) units distance from \((2,4)\) because the vertical line given by the equation \(x = 5\) cuts the circle of radius \(\sqrt{13}\) that is centred on the point \((2,4)\) in exactly two places.

\(\beta\).5#

To show that \(x^2+y^2-10x+14y+58 = 0\) is a circle, we argue like this: first, rearrange the equation to read \((x^2-10x) + (y^2 +14y) = -58\). Completing the two squares gives: \((x^2-10x+5^2) + (y^2 +14y+7^2) = -58 + 25 + 49\). This equation becomes \((x-5)^2 + (y+7)^2 = 16\), whose graph is a circle with center \((5,-7)\) and radius \(\sqrt{16} = 4\).

Use this method to find the center and the radius of the two circles with equations:

  • \(x^2+y^2+10x-6y+30 = 0\)

  • \(3x^2+3y^2+18x-24y = -39\)

[Sydsæter, Hammond, Strøm, and Carvajal, 2016] Exercises for Section 5.5, Question 5

To see that a circle has the formula \(\sqrt{(x-a)^2+(y-b)^2}=r\) it is sufficient to realize that the left hand side of the expression is nothing else but the Euclidean distance between the point \((x,y)\) and the point \((a,b)\), which is set equal to \(r\).

\[\begin{split} \begin{array}{c} x^2+y^2+10x-6y+30 = 0 \\ x^2+10x+25 +y^2-6y+9 = 25 + 9 -30 \\ (x+5)^2+(y-3)^2 = 2^2 \end{array} \end{split}\]

The center of the circle is \((-5,3)\) and the radius is \(2\).

\[\begin{split} \begin{array}{c} 3x^2+3y^2+18x-24y = -39 \\ x^2+y^2+6x-8y = -13 \\ x^2+6x+9 + y^2-8y+16 = 9+16-13 \\ (x+3)^2+(y-4)^2 = 12 \end{array} \end{split}\]

The center of the circle is \((-3,4)\) and the radius is \(2\sqrt{3}\).

\(\beta\).6#

Find the domains of the following functions and then illustrate those domains in the \((x, y)\)–coordinate-plane.

  1. \(f(x,y)= \frac{x^2+y^3}{y-x+2}\)

  2. \(f(x,y)=\sqrt{2-(x^2 +y^2)}\)

  3. \(f(x,y)=\sqrt{(4-x^2 -y^2)(x^2 +y^2 -1)}\)

[Sydsæter, Hammond, Strøm, and Carvajal, 2016] Exercises for Section 11.1, Question 6

Part 1.

Consider the function \(f(x,y) = \frac{x^2+y^3}{y-x+2}\). The only restriction on the domain of this function is the impossibility of division by zero. Thus the only restriction on the domain of this function is that \(y - x + 2 \ne 0\). In other words, the only restriction on the domain of this function is that it not include any points on the line \(y = x - 2\). Thus the domain of this function is

\[ D = \left\{ (x,y) \in \mathbb{R}^2 : y \ne x-2 \right\}. \]

This is the entire two-dimensional Euclidean coordinate-plane excluding the points on the line \(y = x - 2\).

Note

Because the question explicitly asks to illustrate, full marks for the question like this will be awarded for the answer that includes both the derivation of this result and a sketch of the domain in a two-dimensional coordinate plane.

Part 2.

Consider the function \(f(x,y)=\sqrt{2-(x^2 +y^2)}\). Restriction on the domain of this function stems from the fact that the function \(\sqrt{x}\) is defined only for non-negative \(x\). Thus the only restriction on the domain of this function is that \(2 - (x^2 + y^2) \ge 0\). This requires that \(x^2 + y^2 \le 2\). As such, the domain for this function is the set

\[ D = \left\{ (x,y) \in \mathbb{R}^2 : x^2 + y^2 \le 2 \right\}. \]

This region consists of both the boundary of, and area inside, a circle of radius \(\sqrt{2}\) that is centred on the origin (that is, the point \((0, 0)\).

Part 3.

Consider the function \(f(x,y)=\sqrt{(4-x^2 -y^2)(x^2 +y^2 -1)}\). As in the previous part, the only restriction on the domain of this function stems from the fact that the function \(\sqrt{x}\) is defined only for non-negative \(x\).

This might occur in any of the following ways.

  1. Both \(4-x^2 -y^2 >0\) and \(x^2 +y^2 -1 >0\), or

  2. Both \(4-x^2 -y^2 <0\) and \(x^2 +y^2 -1<0\), or

  3. \(4 - x^2 - y^2\) = 0, or

  4. \(x^2 +y^2 -1=0\).

The first of these cases requires that \(1 < x^2 + y^2 < 4\). This is the area between the circle centred on the origin with radius one and the circle centred on the origin with radius two, not including either of the boundaries.

The second of these cases requires that \(4 < x^2 + y^2 < 1\), which is impossible because \(4>1\). Thus we know that this case cannot occur.

The third of these cases requires that \(x^2 + y^2 = 4\). This is the set of points on the boundary of a circle of radius two that is centred on the origin.

The third of these cases requires that \(x^2 + y^2 = 1\). This is the set of points on the boundary of a circle of radius one that is centred on the origin.

After combining cases one, three, and four, we conclude that the domain for this function is

\[ D = \left\{ (x,y) \in \mathbb{R}^2 : 1 \le x^2 + y^2 \le 4 \right\}. \]

This is the area between the circle centred on the origin with radius one and the circle centred on the origin with radius two, including both of the boundaries. It looks like a dough-nut or a simple washer when viewed from side-on.

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