---
# System prepended metadata

title: 建築系微積分 Exercise 1 Solution
tags: [建築系, 微積分, Calculus, Arch]

---

# 建築系微積分 Exercise 1 Solution
###### tags: `Calculus` `微積分` `Arch` `建築系`

解答僅供參考，有問題請利用旁邊留言的功能、或者是和助教聯繫。


### Section 1.2
**58.** The demand and supply equations for an MP3 player are given by
\begin{eqnarray*}
\begin{array}{ll}
p=190-15x & \mbox{Demand equation}\\
p=75+8x & \mbox{Supply equation},
\end{array}
\end{eqnarray*}
where $p$ is the price (in dollars) and $x$ represents the number of units (in hundreds of thousands). Find the equilibrium point for this market.
***Solution:***
The equilibrium point means that the number of MP3 players in demand meets that in supply. So we solve the equation $190-15x=75+8x$ and we have $x=5$ and $p=115$.

**61.** A mathematical model for the numbers of associate's degrees conferred $y$ (in thousands) from 2008 to 2012 is given by the equation $y=7.79t^2-86.6t+941$, where $t$ represents the year, with $t=8$ corresponding to 2008.
a. Use the model to complete the table.
| Year     | 2008 | 2009 | 2010 | 2011 | 2012  | 2016  |
| -------- | ---- | ---- | ---- | ---- | ----  | ----  |
| Degrees  |747.76|792.59| 854.0|930.99|1023.56|1549.64|

b. This model was created using actual data from 2008 through 2012. How accurate do you think the model is in predicting the number of associate's degrees conferred in 2016? Explain your reasoning.

c. Using this model, what is the prediction for the number of associate's degrees conferred in 2020? Do you think this prediction is valid?
***Solution:***
Let $t=20$ and we have $y=2325$.

### Section 1.3
* Find the slope of the line passing through the pair of points.

**35.** $(-2,6)$, $(1,6)$
***Solution:***
The slope $m$ is computed by $m=\frac{6-6}{1-(-2)}=0$.

**40.** $(\frac{7}{8},\frac{3}{4})$, $(\frac{5}{4},-\frac{1}{4})$
***Solution:***
The slope $m$ is computed by $$m=\frac{\frac{3}{4}+\frac{1}{4}}{\frac{7}{8}-\frac{5}{4}}=-\frac{8}{3}.$$

* Find an equation of the line that passes through the given point and has the given slpoe. Then sketch the line. (Use the point-slope form)
 
**69.** $(-\frac{1}{3},1)$, $(-\frac{2}{3},\frac{5}{6})$
***Solution:***
We first find the slope by computing
$$m=\frac{1-\frac{5}{6}}{-\frac{1}{3}+\frac{2}{3}}=\frac{1}{2}.$$
So the equation of the line is therefore given by $y-1=\frac{1}{2}(x+\frac{1}{3})$. As we simplify, we obtain $3x-6y+7=0$. The graph is plotted below:
![80](https://i.imgur.com/nlPCILN.jpg)


* Find equations of the lines that are (a) parallel to the given line and (b) perpendicular to the given line. Then use a graphing utility to graph all three  equations in the same viewing window using a square setting.

**81.** Point: (1,1), Line: x-2=0.
***Solution:***
We find the line is vertical, so for (a), such line is vertical. So the equation of the line is $x-1=0$.  For (b), such line is horizonal. Therefore, the slope is zero and the equaltion of the line is $y=1$. The graph of these lines are plotted below:

![](https://i.imgur.com/raTbsFj.png)

**85.** You are driving on a road that has a 6% uphill grade. This means that the slope of the road is $\frac{6}{100}$. Approximate the amount of vertical change in your position when you drive 200 meters.
***Solution:***
We formulate the relationship between the vertical change $y$ and the distance $x$, which is given by $y=\frac{6}{100}x$. So the approximate vertical change is $\frac{6}{100}*200=12$ meters.

**88.** Personal income (in billions of dollars) in the United Stated was 12,430 in 2008 and 14,167 in 2013. Assume that the relationship  between the personal income $y$ and the time $t$ is linear. Let $t=0$ represent 2000.
a. Write a linear model for the data.
b. Estimate the personal incomes in 2011 and 2014.
***Solution:***
a. Denote the income by $y$, we formulate the relationship between $y$ and $t$:
$$y-12430=\frac{14167-12430}{13-8}(t-8)\Rightarrow y=347.4t+9650.8.$$
b. Substute t by 11 and 14 respectively, we have $y(11)=13472.2$ and $y(14)=14514.4$.

**90.** A hospital purchases a $500,000 magnetic resonance imaging (MRI) machine that has a useful life of 9 years. The salvage value at the end of 9 years is $77,000.
a. Write a linear equation that describes the value $y$ (in dollars) of the MRI machine in terms of the time $t$ (in years), $0\leq t\leq9$.
b. Find the value of the machine after 5 years.
c. Find the time when the value of the equipment will be $160,000.
***Solution:***
a. Suppose the linear equation is $y=mt+b$, then the linear equation can be computed by solving $500000=b$ and $77000=9m+b$. So $m=\frac{77000-500000}{9}=-\frac{423000}{9},b=500000$. Therefore, the linear equation is $y=500000-47000t$.
b. Let t=5, we have $y(5)=500000-47000^*5=265000$.
c. We solve $t$ from $16000=500000-47000t$ and have $t\approx7.2$.


### Section 1.4

* Decide whether the equation defines $y$ as a function of $x$.

**5.** $x^2+y=4$
***Solution:***
The equatuon defines $y$ as a function of $x$, since $y$ can be represented as $y=4-x^2$, that is, for each $x$ there is only one value of $y$.

**7.** $y=|x+2|$
***Solution:***
The equatuon defines $y$ as a function of $x$, since $y$ can be represented as $y=x+2$ for $x\geq -2$ and $y=-x-2$ for $x<-2$, that is, for each $x$ there is only one value of $y$.