2012 期中考

這裡包含了 2012 年臺北醫學大學牙醫系公衛系保健系醫管系的微積分期中考。因題目大同小異,故合併為一篇文章。公衛、保健二系採用同一份考卷。

第一題考線性回歸,同 2013 小考的第一題

,串場的文件是〈指數函數〉。

導函數

第 2 題

Given f(x) = x3 − 3x2 + 3, use the definition to find f′(x).

代入題目給的定義。

為了簡化計算,讓同次項分別相減,最後再總和。

(x + h)3x3 = 3x2h + 3xh2 + h3

(x + h)2x2 = 2xh + h2

切線與法線

公、保、管第 3 題

Find the equation to the tangent of the curve (x2 + 3) (x − 3)1/2 at x = 4.

牙醫系第 3 題

… at x = 5.

f(x) = (x2 + 3) (x − 3)1/2,則

我們知道 f(4) = 19f′(4) = 35/2,所以此處的切線方程式為

同理,,所以此處的切線方程式為

牙醫系第 7 題

Find the equation of the tangent line to the curve x2 + 3xy + y2 = 5 at (1, 1).

公、保、管第 7 題

… the normal line ….

首先,(1, 1) 的確滿足 x2 + 3xy + y2 = 5。我們對 x 隱微分。

2x + 3y + 3xy′ + 2yy′ = 0

代入 (1, 1)。

5 + 5y′ = 0

所以圖形在此處的斜率為 -1。切線方程式為

y − 1 = -(x − 1)

法線的斜率為 1,所以方程式為

y − 1 = x − 1

導數、二階導數

公、保、管第 4 題

Consider a curve . Find f′(5).

牙醫系第 4 題

… and f″(5).

公、保、管第 6 題

Given f(x) = 2x2+1, find f′(2).

y = x2 + 1,由鏈式法則

因為

所以

f′(x) = (ln 2) x 2y+1 = (ln 2) x 2x2+2

f′(2) = 128 ln 2

牙醫系第 6 題

Given f(x) = ln(sec4(x) tan2(x)), find f′(π/4).

f′(x) = y′ / y

再由乘法法則

y′ = 4 sec4(x) tan3(x) + 2 sec6(x) tan(x)

f′(π/4) = 8

函數圖形

第 5 題

Sketch the graph of and also find the relative extreme points and inflection points at the interval of [-1, 1].

多項式是光滑函數,局部極值只會出現在平坦處,即導數為 0 之處。因此首先我們找出所有平坦處以簡化問題。此外,這個函數是奇函數,省去了一些繪圖的麻煩。

平坦處出現在 0 與 ±2 三處,只有 0 在 [1, -1] 內。我們再進行高階導數測試

由於 f″(0) = 0f‴(0) < 0,所以此處為嚴格遞減的反曲點。

多項式的圖形的反曲點出現在導函數的極值,故只會出現在二階導數為 0 之處。因此我們找出所有這些地方,即 x = 0 或 ,只有 0 在 [1, -1] 內且這個點已討論過是反曲點。

的圖形

線性近似

公、保、管第 8 題

Use the differentials to approximate the quantity to two decimal points places.

牙醫系第 8 題

….

這附近看來沒什麼有希望的點,只好先遷就 嘍!設 ,則

線性近似告訴我們

f(x) ≈ f(a) + f′(a) (xa)

代入題目的函數。

f(5.6) ≈ 2 + f′(4) (5.6 − 4)

小數點後三位已無變化,所以

同理,

小數點後三位已無變化,所以

應用題

牙、公、保第 9 題

When a person coughs, the trachea (windpope windpipe) contracts, allowing air to be expelled at a maximum velocity. It can shown that during a cough the velocity v of airflow is given by the function v = f(r) = kr2(Rr) , where r is the trachea’s radius (in centimeters) during a cough, R is the trachea’s normal radius (in centimeters), and k is a positive constant that depends on the length of the trachea. Find the radius r for which the velocity of airflow is greatest.

首先,變數 r 的自然限制是 0 ≤ rR。此外,vr 的多項式。所以 v[0, R] 的最大值只會出現在端點或平坦處。

我們先考慮端點。

f(0) = f(R) = 0

接著,我們考慮平坦處。為了方便計算,我們把 f(r) 展開。

f(r) = kRr2kr3

這樣微分就容易多了。

f′(r) = 2kRr − 3kr2

解得平坦處在 0 及 2R/3 二處。我們已經討論過 0 這個端點了,而

f(2R/3) = k (2R/3)2 (R/3) > 0

所以最大值發生在 r = 2R/3

醫管系第 9 題

Suppose that during a nationwide program to immunize the population against certain from of influenza, public health officials found that the cost of inoculating x% of the population was approximately million dollars.

  1. What was the cost of inoculating the first 50% of the population?
  2. What was the cost of inoculating the second 50% of the population?
  3. What percentage of the population had been inoculated by the time 37.5 million dollars had been spent?

本題是閱讀測驗,數學很簡單。

  1. C(50) = 50
  2. C(100) − C(50) = 100
  3. 這實質上是解 解得 x 為 40,答案為 40%。

牙醫系第 10 題

A rain gutter is made from sheets of metal 9 in wide. The gutters have a 3-in base and two 3-in sides, folded up at an angle θ (see figure). What angle θ maximizes the cross-sectional area of the gutter?

等腰梯形水槽,下底與兩腰皆三吋長。

[本圖受著作權保護,請勿轉載。]

我們知道 θ 的自然限制是 0 ≤ θ ≤ π。設此梯形的截面積為 A,則

A = 9 (sin θ) (1 + cos θ)

若代入端點,截面積均為 0。我們尋找導數為 0 的點。

解得當 cos θ 為 1/2 或 -1 時,導數為 0。由自然限制得 θ 分別為 π/3 與 π,其中 π 是端點,已經討論過了。我們確認當 θ 為 π/3,A > 0,大於端點的值。所以最大值發生在 θ = π/3

公、保第 10 題

Several mathematical stories originated with the second wedding of the mathematician and astronomer Johannes Kepler. Here is one: While shopping for wine for his wedding, Kepler noticed that the price of a barrel of wine (here assumed to be a cylinder) was determined solely by the length d of a dipstick that was inserted diagonally through a hole in the top of the barrel to the edge of the base of the barrel (see figure). Kepler realized that this measurement does not determine the volume of the barrel and that for a fixed value of d. The volume varies with the radius r and height h of the barrel. For a fixed value of d, what is the ratio r/h that maximizes the volume of the barrel?

圓柱形的酒桶

[本圖受著作權保護,請勿轉載。]

設體積為 V,則

V = πr2h

我們希望只留下單一變數,rh 其一。平方項好欺負!

r2 = d2h2

所以

V = π (d2hh3)

其中 h 的自然限制是 0 ≤ hd。代入端點會使得 V 為 0。接著我們對 h 微分得

導數為 0 的點有 ,但負不合。將 代入得

並且造成 V > 0,故此處確定是最大值。所以可以安心作答。

醫管系第 10 題

Based on a study conducted in 1997, the percentage of the U.S. population by age affilcted with Alzheimer’s disease is given by the function

P(x) = 0.0726x2 + 0.7902x + 4.9623 where 0 ≤ x ≤ 25

where x is measured in years, with x = 0 corresponding to age 65 yr. Show that P is an increasing function of x on the interval (0, 25). What does your result tell you about the relationship between Alzheimer’s disease and age for the population that is aged 65 year and or older?

P 在 [0, 25] 等於多項式,所以在 (0, 25) 可微。我們觀察它的導函數。

P′(x) = 0.1452x + 0.7902 where 0 ≤ x ≤ 25

顯然 P′(x) > 0,所以 P 在區間 (0, 25) 內遞增。這意味著 65 歲以上的人口,得到阿茲海默症的比例隨年齡提高。

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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