Calculus Early Transcendentals: Differential & Multi-Variable Calculus for Social Sciences

Find the equations and describe the shapes of the cross-sections when \(x=0\text{,}\) \(y=0\) and \(x=y\text{.}\) Plot the surface with a three-dimensional graphing tool.

1. \(\displaystyle f(x,y)=(x-y)^2\)

   Answer
   \(z=y^2\) , \(z=x^2 \text{,}\) \(z=0\text{.}\)
   Solution

   Let \(z=f(x,y)=(x-y)^2\text{,}\) and consider the following cross-sections. Taking a slice along the \(y\)-axis, we have

   \begin{equation*} z = f(0,y) = (0-y)^2 = (-y)^2 = y^2\text{.} \end{equation*}

   This cross-section will then look like:

   

   Taking a slice along the \(x\)-axis, we have

   \begin{equation*} z = f(x,0) = x^2\text{,} \end{equation*}

   and graphically,

   

   Finally, along the line \(x=y\text{,}\)

   \begin{equation*} z=f(x,y) = (0)^2 = 0\text{,} \end{equation*}

   and so \(f\) is constant (and zero) along this cross-section. The surface \(z=f(x,y)\) is plotted below for \((x,y) \in [-5,5]\times[-5,5]\text{.}\)

   

2. \(\displaystyle f(x,y)=|x|+|y|\)

   Answer
   \(z= |y|\text{,}\) \(z= |x| \text{,}\) \(z=2|y|\text{.}\)
   Solution

   Let \(z=f(x,y)=|x|+|y|\text{,}\) and consider the following cross-sections. Taking a slice along the \(y\)-axis, we have

   \begin{equation*} z = f(0,y) = |y|\text{.} \end{equation*}

   This cross-section will then look like:

   

   Taking a slice along the \(x\)-axis, we have

   \begin{equation*} z = f(x,0) = |x|\text{,} \end{equation*}

   and graphically,

   

   Finally, along the line \(x=y\text{,}\)

   \begin{equation*} z=f(x,y) = |y|+|y| = 2|y|\text{,} \end{equation*}

   and graphically,

   

   The surface \(z=f(x,y)\) is plotted below for \((x,y) \in [-5,5]\times[-5,5]\text{.}\)

   

3. \(\displaystyle f(x,y)=e^{-(x^2+y^2)}\sin(x^2+y^2)\)

   Answer
   \(z=e^{-(y^2)}\sin(y^2) \text{,}\) \(z= e^{-(x^2)}\sin(x^2) \text{,}\) \(z= e^{-(2y^2)}\sin(2y^2)\text{.}\)
   Solution

   Let \(z=f(x,y)=e^{-(x^2+y^2)}\sin(x^2+y^2)\text{,}\) and consider the following cross-sections. Taking a slice along the \(y\)-axis, we have

   \begin{equation*} z = f(0,y) = e^{-(y^2)}\sin(y^2)\text{.} \end{equation*}

   This cross-section will then look like:

   

   Taking a slice along the \(x\)-axis, we have

   \begin{equation*} z = f(x,0) = e^{-(x^2)}\sin(x^2)\text{,} \end{equation*}

   and graphically,

   

   Finally, along the line \(x=y\text{,}\)

   \begin{equation*} z=f(x,y) = e^{-(y^2+y^2)}\sin(y^2+y^2) = e^{-(2y^2)}\sin(2y^2)\text{,} \end{equation*}

   and graphically,

   

   The surface \(z=f(x,y)\) is plotted below for \((x,y) \in [-5,5]\times[-5,5]\text{.}\)

   

4. \(\displaystyle f(x,y)=\sin(x-y)\)

   Answer
   \(z=-\sin(y) \text{,}\) \(z= \sin(x)\text{,}\) \(z=0 \text{.}\)
   Solution

   Let \(z=f(x,y)=\sin(x-y)\text{,}\) and consider the following cross-sections. Taking a slice along the \(y\)-axis, we have

   \begin{equation*} z = f(0,y) = \sin(-y) = -\sin(y)\text{,} \end{equation*}

   where we used the fact that sine is an even function. This cross-section will then look like:

   

   Taking a slice along the \(x\)-axis, we have

   \begin{equation*} z = f(x,0) = \sin(x)\text{,} \end{equation*}

   and graphically,

   

   Finally, along the line \(x=y\text{,}\)

   \begin{equation*} z=f(x,y) = \sin(0) = 0\text{,} \end{equation*}

   and so \(f\) is constant (and zero) along this cross-section. The surface \(z=f(x,y)\) is plotted below for \((x,y) \in [-2\pi,2\pi]\times[-2\pi,2\pi]\text{.}\)

   

5. \(\displaystyle f(x,y)=(x^2-y^2)^2\)

   Answer
   \(z =y^4 \text{,}\) \(z =x^4 \text{,}\) \(z =0 \text{.}\)
   Solution

   Let \(z=f(x,y)=(x^2-y^2)^2\text{,}\) and consider the following cross-sections. Taking a slice along the \(y\)-axis, we have

   \begin{equation*} z = f(0,y) = (0-y^2)^2 = (-y^2)^2=y^4\text{.} \end{equation*}

   This cross-section will then look like:

   

   Taking a slice along the \(x\)-axis, we have

   \begin{equation*} z = f(x,0) = x^4\text{,} \end{equation*}

   and graphically,

   

   Finally, along the line \(x=y\text{,}\)

   \begin{equation*} z=f(x,y) = (0)^2 = 0\text{,} \end{equation*}

   and so \(f\) is constant (and zero) along this cross-section. The surface \(z=f(x,y)\) is plotted below for \((x,y) \in [-5,5]\times[-5,5]\text{.}\)

   

Find the domain of each of the following functions:

1. \(\displaystyle f(x,y) = \ds\sqrt{9-x^2}+\sqrt{y^2-4}\)

   Answer
   \(\{(x,y)\mid |x|\le3\ \hbox{and}\ |y|\ge2\}\)
   Solution

   The domain of \(f(x,y)=\sqrt{9-x^2}+\sqrt{y^2-4}\) is given by

   \begin{equation*} 9-x^2 \geq 0 \ \ \text{ and } \ \ y^2-4 \geq 0\text{.} \end{equation*}

   This will be the area of the \(x\)-\(y\)-plane such that \(x \in [-3,3]\) and \(y \in (-\infty,-2]\cup[2,\infty)\text{.}\) This area is given by the two infinite rectangles graphed below.

   

2. \(\displaystyle h(u,v)=\sqrt{2-u^2-v^2}\)

   Answer
   \(\{(u,v)\mid u^2+v^2 \leq 2\}\)
   Solution

   The domain of \(h(u,v)=\sqrt{2-u^2-v^2}\) is given by

   \begin{equation*} u^2+v^2 \leq 2\text{.} \end{equation*}

   This will be the area of the circle of radius \(\sqrt{2}\text{,}\) including the boundary.

   

3. \(\displaystyle f(x,y) = \ds\sqrt{16-x^2-4y^2}\)

   Answer
   \(\{(x,y)\mid x^2+4y^2\le16\}\)
   Solution

   The domain of \(f(x,y)=\sqrt{16-x^2-4y^2}\) is given by

   \begin{equation*} \dfrac{x^2}{4^2}+\dfrac{y^2}{2^2} \leq 1\text{.} \end{equation*}

   This will be the area of the ellipse with major radius equal to \(4\) and minor radius equal to \(2\text{,}\) including the boundary.

   

   The domain of the function \(f(x,y)=\sqrt{16-x^2-y^2}\) is all points \((x,y)\) such that \(x^2+y^2 \leq 16\text{.}\) This gives a disk in the \(x\)-\(y\)-plane:

   

4. \(\displaystyle f(x,y)= x+4y\)

   Answer
   All real \(x\) and \(y\)
   Solution

   The domain of \(f(x,y)=x+4y\) is all \((x,y) \in \R \times \R\text{.}\)

5. \(\displaystyle g(x,y,z)=x^2+y^2+z^2\)

   Answer
   All real \(x\text{,}\)\(y\text{,}\) and \(z\)
   Solution

   The domain of \(g(x,y,z)=x^2+y^2+z^2\) is all \((x,y,z) \in \R \times \R \times \R\text{.}\)

6. \(\displaystyle f(x,y) = \arcsin(x^2+y^2-2)\)

   Answer
   \(\{(x,y)\mid 1\le x^2+y^2\le3\}\)
   Solution

   The domain of \(f(x,y)=\arcsin(x^2+y^2-2)\) is given by

   \begin{equation*} -1 \leq x^2+y^2-2 \leq 1 \implies 1 \leq x^2 + y^2 \leq 3\text{.} \end{equation*}

   This is the area between the two circles \(x^2+y^2 =1\) and \(x^2+y^2=3\) (inclusive of the boundaries):

   

7. \(\displaystyle h(u,v)= \dfrac{uv}{u-v}\)

   Answer
   \(\{(u,v)\mid u \neq v\}\)
   Solution

   The domain of \(h(u,v) = \dfrac{uv}{u-v}\) is the set of all points \((u,v) \in \mathbb{R} \times \mathbb{R}\) such that \(u \neq v\text{:}\)

   

8. \(\displaystyle f(s,t)=\sqrt{s^2+t^2}\)

   Answer
   All real \(s\) and \(t\)
   Solution

   The domain of \(f(s,t) = \sqrt{s^2+t^2}\) is all \((s,t) \in \mathbb{R} \times \mathbb{R}\text{.}\)

9. \(\displaystyle g(r,s)=\sqrt{rs}\)

   Answer
   \(\{(r,s)\mid rs > 0\}\)
   Solution

   The domain of \(g(r,s)=\sqrt{rs}\) is given by

   \begin{equation*} rs \geq 0\text{.} \end{equation*}

   This is given by

   \begin{equation*} r \geq 0 \ \text{ and } \ s \geq 0 \end{equation*}

   or

   \begin{equation*} r \leq 0 \ \text{ and } \ s \leq 0 \end{equation*}

10. \(\displaystyle f(x,y)=e^{-xy}\)

    Answer
    All real \(x\) and \(y\)
    Solution

    The domain of \(f(x,y)=e^{-xy}\) is all \((x,y) \in \R \times \R\text{.}\)

11. \(\displaystyle h(x,y)=\ln(x+y-7)\)

    Answer
    \(\{(x,y)\mid x+y > 7 \}\)
    Solution

    The domain of \(h(x,y)=\ln(x+y-7)\) is given by

    \begin{equation*} x+y-7 > 0 \ \text{ or } \ y > 7-x\text{.} \end{equation*}

Below are two sets of level curves at a sequence of equally-spaced heights \(z\text{.}\) One is for a cone, one is for a paraboloid. Which is which? Explain.

Sketch the level curves of the function corresponding to the given values of \(z\text{.}\)

1. \(f(x,y)=3x+2y\text{,}\) \(z=-2,-1,0,1,2\)

   Solution

   The level curves of \(z=f(x,y)=3x+2y\) are

   \begin{equation*} k=3x+2y \implies y=\frac{1}{2}(k-3x) \end{equation*}

   for some constant \(k\text{.}\) We sketch below the level curves corresponding to \(k=-2,-1,0,1,2\text{.}\)

   

2. \(f(x,y)=x^2-y\text{,}\) \(z=-2,-1,0,1,2\)

   Solution

   The level curves of \(z=f(x,y)=x^2-y\) are

   \begin{equation*} k=x^2-y \implies y=x^2-k \end{equation*}

   for some constant \(k\text{.}\) We sketch below the level curves corresponding to \(k=-2,-1,0,1,2\text{.}\)

   

3. \(f(x,y)=2y+x^2\text{,}\) \(z=-2,-1,0,1,2\)

   Solution

   The level curves of \(z=f(x,y)=2y+x^2\) are

   \begin{equation*} k=2y+x^2 \implies y=\frac{1}{2}(k-x^2) \end{equation*}

   for some constant \(k\text{.}\) We sketch below the level curves corresponding to \(k=-2,-1,0,1,2\text{.}\)

   

4. \(f(x,y)=2xy\text{,}\) \(z=-4,-2,2,4\)

   Solution

   The level curves of \(z=f(x,y)=2xy\) are

   \begin{equation*} k=2xy \implies y=\dfrac{k}{2x} \end{equation*}

   for some constant \(k\text{.}\) We sketch below the level curves corresponding to \(k=-4,-2,2,4\text{.}\)

   

5. \(f(x,y)=\sqrt{25-x^2-y^2}\text{,}\) \(z=0,1,3,5\)

   Solution

   The level curves of \(z=f(x,y)=\sqrt{25-x^2-y^2}\) are

   \begin{equation*} k=\sqrt{25-x^2-y^2} \implies x^2+y^2=25-k^2 \end{equation*}

   for some constant \(k\text{.}\) We sketch below the level curves corresponding to \(k=0,1,3,5\text{.}\)

   

6. \(f(x,y)=e^y-x^2\text{,}\) \(z=1,2,3,4\)

   Solution

   The level curves of \(z=f(x,y)=e^y-x^2\) are

   \begin{equation*} k=e^y-x^2 \implies y=\ln(x^2+k) \end{equation*}

   for some constant \(k\text{.}\) For \(k=1,2,3,4\text{,}\) we therefore have

   

A clothing retailer sells casual and business jackets. Let \(q_c\) denote the quantity of casual jackets demanded monthly and \(q_b\) denote the quantity of business jackets demanded monthly. It is determined that the unit price of casual jackets is approximately

and that the unit price of business jackets is approximately

dollars, respectively.

1. Determine the monthly revenue, \(R(q_c,q_b)\text{.}\)

   Answer
   \(R(q_c,q_b) = \frac{1}{420}\left(-105q_c^2-102q_cq_b+42000q_c-140q_b^2+63000q_b\right)\)
   Solution

   The monthly revenue is a two-variable function given by the sum of the revenue from selling casual jackets and the revenue from selling the business jackets:

   \begin{equation*} \begin{split} R(q_c,q_b) \amp = p_cq_c + p_bq_b\\ \amp = \bigl(100-\frac{1}{4}q_c - \frac{1}{7}q_b\bigr)q_c + \bigl(150-\frac{1}{10}q_c- \frac{1}{3}q_b\bigr)q_b\\ \amp = \frac{1}{420}\left(-105q_c^2-102q_cq_b+42000q_c-140q_b^2+63000q_b\right) \end{split} \end{equation*}

2. Determine the domain of \(R\text{.}\)

   Answer
   The region where all of \(p_c,p_b,q_c,q_b \geq 0\text{.}\)
   Solution

   The domain of \(R\) is the region for which all of \(q_c\text{,}\) \(q_b\text{,}\) \(p_c\) and \(p_b\) are positive:

   \begin{equation*} p_c \geq 0 \implies q_c \leq 400 - \frac{4}{7}q_b \end{equation*}

   \begin{equation*} p_b \geq 0 \implies q_c \leq 1500 - \frac{10}{3}q_b \end{equation*}

   The domain is the shaded region graphed below.

   

A certain manufacturer produces basic and enhanced versions of their product. Let \(q_b\) denote the quantity of basic units daily demanded and \(q_e\) denote the quantity of enhanced units demanded daily. It is determined that the unit price of the basic units is approximately

and that the unit price of the enhanced units is approximately

dollars, respectively.

1. Determine the daily revenue, \(R(q_b,q_e)\text{.}\)

   Answer
   \(R = -0.1q_b^2-0.9q_bq_c+10q_b-q_c^2+30q_c\)
   Solution

   The price \(p_b\) per unit of \(q_b\) is given by

   \begin{equation*} p_b=10-0.1q_b-0.5q_e\text{,} \end{equation*}

   and the price \(p_e\) per unit of \(q_e\) is given by

   \begin{equation*} p_e=30-0.4q_b-q_e\text{.} \end{equation*}

   Therefore, the revenue \(R\) as a function of \(q_b\) and \(q_e\) is

   \begin{equation*} \begin{split} R(q_b,q_e) \amp = p_bq_b + p_eq_e \\ \amp = \left(10-0.1q_b-0.5q_e\right)q_b+\left(30-0.4q_b-q_e\right)q_e\\ \amp = -0.1q_b^2-0.9q_bq_e+10q_b-q_e^2+30q_e \end{split} \end{equation*}

2. Determine the domain of \(R\text{.}\)

   Answer
   The region where all of \(p_b,p_c,q_b,q_c \geq 0\text{.}\)
   Solution

   The domain of \(R\) is constrained by \(p_b \geq 0\) and \(p_e \geq 0\text{:}\)

   \begin{equation*} p_b=10-0.1q_b-0.5q_e\geq 0 \implies q_e \leq 20-0.2q_b \ \ \text{ and } \end{equation*}

   \begin{equation*} p_e=30-0.4q_b-q_e \geq 0 \implies q_e \leq 30-0.4q_b\text{,} \end{equation*}

   such that both \(q_b,q_e \geq 0\text{.}\) The domain is the shaded region graphed below.

   

We can calculate the outstanding principal on a loan at the end of \(i\) months by the formula

where \(A\) is the principal loan, \(r\) is the annual interest rate, and \(t\) is the amortization period in years. Suppose the original amount borrowed is $100,000, and the interest rate charged by the bank is 3%. What is the amount owed to the bank after 2 years if the loan is to be repaid in equal instalments over 25 years? What is the amount owed after 20 years?

In economics, the given optimal quantity \(Q\) of goods for a store to order is given by the Wilson lot-size formula:

where \(C\) is the cost of placing an order, \(N\) is the number of items the store sells per day, and \(H\) is the daily holding cost for each item. What is the most economical quantity of winter tires to order if the store pays $25 for placing an order, pays $2 for holding a tire per day, and expects to sell 60 tires a day?