2011 期中考

這裡包含了 2011 年臺北醫學大學醫學系、牙醫系的微積分期中考。學號為奇數者考 A 卷,偶數考 B 卷。

先前我寫了一篇爛尾的詳解,因為本文尚未完成,請湊合著看。

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 pollutant345810
Incidence of disease4852587096

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} and B = \frac {\sum y - A \sum x} n.
試算表
x y x2 xy
3 48 9 144
4 52 16 208
5 58 25 290
8 70 64 560
10 96100 960
303242142162

把這些總和代入公式。


			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 
		

為了避免捨入誤差,我先化簡分式。作答只須估算即可,畢竟題目只要求小數點後二位。

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 to find \displaystyle \frac d{dx} \sin x .

代入題目給的定義。


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

和角公式如下。


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

所以


		\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*}
		

這帶我們回到二個經典的極限問題。答題時不須證明,可直接代入。


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

第 2 題,B 卷

本題同 2013 期中小考的第 2 題

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 ….

我們已經知道切線通過 (1, −2) 了,所以知道斜率就可以代入點斜式,求出切線方程式。

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} 

所以切線方程式為

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

法線方程式為

 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} , 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,A 卷

本題同 2012 牙醫系期中考第 6 題

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 
		

函數圖形

第 5 題

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 
			

由題意知

Δr = -0.1r

我們利用


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

得出

ΔV ≈ -0.4kr4 = -0.4V

V 減少 40%

直接解法

設原有的半徑為 r0,原有的體積為 V0

V0 = kr04

因為半徑減少 10%,即

r = 0.9r0

所以

V = k (0.9r0)4 = 0.94 V0


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

體積減少 34.39%

討論

為何二組解差異這麼大?

線性近似

第 8 題

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

設函數 f \left( x \right) = \sqrt x 線性近似告訴我們

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*}
	

小數點下四位已經不動,所以


		\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.

因為題目要線性關係,所以設二數 mk 使得

f(α) = mα + k

已知

f(0) = k = 212

已求得 k。又

f(6000) = 6000m + 212 = 200

m 為 -0.002,所以

f(α) = -0.002α + 212

第 10 題

本題同 2012 牙醫系期中考第 10 題

作者: 何 震邦

我叫何震邦,目前就讀臺北醫學大學醫學系。在漢字不宜的場合,我也叫 Chen-Pang He。各位看倌可以在 FacebookGoogle+ 找到我。詳細資訊請洽本站的《關於》頁面。

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