## Useless

√♥ = ?

cos ♥ = ?

D ♥ = ?

I ♥ = ?

F{♥} = (1/√(2π)) ∫(-∞,∞)♥ e−iωt dt = ?

My normal approach is useless here.

## 線性回歸

### 第 1 題

In a study of five industrial areas, a researcher obtained these data relating the average number of units of a certain pollutant in the air and the incidence (per 100,000 people) of a certain disease:

 Units of pollutant Incidence of disease 3 4 5 8 10 48 52 58 70 96

Find the equation of the least-squares line y = Ax + B (to two decimal points places.)

Given that A = \frac {n \sum xy - \sum x \sum y} {n \sum x^2 - \left( \sum x \right)^2}$A = \frac {n \sum xy - \sum x \sum y} {n \sum x^2 - \left( \sum x \right)^2}$ and B = \frac {\sum y - A \sum x} n$B = \frac {\sum y - A \sum x} n$.

x y x2 xy
3 48 9 144
4 52 16 208
5 58 25 290
8 70 64 560
10 96100 960
303242142162


A = \frac {5 \cdot 2162 - 30 \cdot 324} {5 \cdot 214 - 30^2} = \frac{109}{17} \approx 6.411764705882353


B = \frac {324 - \left( 109/17 \right) \cdot 30} {5} = \frac{2238}{85} \approx 26.32941176470588


## 導函數

### 第 2 題，A 卷

Use the definition of the derivative \displaystyle f' \left( x \right) = \lim_{h \to 0} \frac {f \left( x + h \right) - f \left( x \right)} h $\displaystyle f' \left( x \right) = \lim_{h \to 0} \frac {f \left( x + h \right) - f \left( x \right)} h$ to find \displaystyle \frac d{dx} \sin x $\displaystyle \frac d{dx} \sin x$.


\frac d{dx} \sin x = \lim_{h \to 0} \frac {\sin \left( x + h \right) - \sin x} h



\sin \left( x + h \right) = \sin x \cos h + \cos x \sin h



\begin{align*}
\frac d{dx} \sin x &= \lim_{h \to 0} \frac {\sin x \cos h + \cos x \sin h - \sin x} h \\
&= \left( \sin x \right) \left( \lim_{h \to 0} \frac {\cos h - 1} h \right) + \left( \cos x \right) \lim_{h \to 0} \frac {\sin h} h
\end{align*}



\frac d{dx} \sin x = \cos x


## 切線與法線

### 第 3 題，A 卷

Find the equation of the tangent line to the curve y3xy2 + xy = 14 3 at (1, −2).

### 第 3 題，B 卷

… the normal line ….

y3xy2 + xy = 3

3y2y′ − (2xyy′ + y2) + (xy′ + y) = 0

(3y2 − 2xy + x) y′ = y2y

 y' = \frac{y^2 - y}{3y^2 - 2xy + x} 
 y' \left( 1, -2 \right) = \frac6{17} 

 y + 2 = \frac6{17} \left( x - 1 \right) 

 y + 2 = \frac{-17}6 \left( x - 1 \right) 

## 導數

### 第 4 題

Given \displaystyle f \left( x \right) = \frac {x^2 \left( 1-x \right)^3} {1+x} $\displaystyle f \left( x \right) = \frac {x^2 \left( 1-x \right)^3} {1+x}$, find f′(2)


f' \left( x \right) = \frac {-2x \left( x-1 \right)^3 - 3x^2 \left( x-1 \right)^2} {x+1} + \frac {x^2 \left( x-1 \right)^3} {\left( x+1 \right)^2}


f' \left( 2 \right) = \frac{-44}9


### 第 6 題，B 卷

Given f(x) = (x2 + 1)sin x, find f′(π/2)

f(x) = exp(sin x ln(x2 + 1))


\frac d{dx} \sin x \ln \left( x^2 + 1 \right) = \cos x \ln \left( x^2 + 1 \right) + \frac {2x \sin x} {x^2 + 1}


f' \left( x \right) = \left( x^2 + 1 \right)^{\sin x} \left( \cos x \ln \left( x^2 + 1 \right) + \frac {2x \sin x} {x^2 + 1} \right)


f' \left( \frac \pi 2 \right) = \left( \frac {\pi^2} 4 + 1 \right) \left( \frac \pi {\pi^2/4 + 1} \right) = \pi


## 微分

### 第 7 題

In the late 1830s, French physiologist Jean Poiseuille discovered the formula we use today to predict how much the radius of a particular clogged artery decreases the normal volume of flow. His formula,

V = kr4

says that volume of fluid flowing through a small pipe or tube in a unit of time at a fixed pressure is a constant times the fourth power of the tube’s radius r. How dose does a 10% decrease in r affect V?

#### 官方解法


\frac{dV}{dr} = 4kr^3


Δr = -0.1r


\Delta V \approx \frac{dV}{dr} \Delta r


ΔV ≈ -0.4kr4 = -0.4V

V 減少 40%

#### 直接解法

V0 = kr04

r = 0.9r0

V = k (0.9r0)4 = 0.94 V0


\frac {V - V_0} {V_0} = 0.9^4 - 1 = -0.3439


## 線性近似

### 第 8 題

Use the differentials to approximate the quantity \sqrt{4.6}$\sqrt{4.6}$ to four decimal points places.

f(x) ≈ f(a) + f′(a) (xa)


\sqrt x \approx \sqrt a + \frac {x - a} {2 \sqrt a}


x = 4 為起點，


\begin{align*}
\sqrt{4.6} &\approx 2 + \frac {4.6 - 4} {2 \cdot 2} = 2.15 \\
\sqrt{4.6} &\approx 2.15 + \frac {4.6 - 2.15^2} {2 \cdot 2.15} = \frac{3689}{1720} \approx 2.144767441860465 \\
\sqrt{4.6} &\approx \frac{3689}{1720} + \frac {4.6 - \left( 3689/1720 \right)^2} {2 \cdot \left( 3689/1720 \right)} \approx 2.144761058962219
\end{align*}



\sqrt{4.6} \approx 2.1448


## 應用題

### 第 9 題

Water boils at 212℉ at sea level and 200℉ at an elevation of 6000 ft. Assume that the boiling point B varies linearly with altitude α. Find the function B = f(α) that describes the dependence. Comment on whether a linear function gives a realistic model.

f(α) = mα + k

f(0) = k = 212

f(6000) = 6000m + 212 = 200

m 為 -0.002，所以

f(α) = -0.002α + 212