the angle of elvation

1051微微微甲甲甲06-10班班班期期期中中中考考考解解解答答答和和和評評評分分分標標標準準準

1. (10%) A plane flies horizontally at an altitude of 5 km and passes directly over a tracking telescope on the ground.

When the angle of elevation is π/3, this angle is decreasing at a rate of π/6 rad/min. How fast is the plane travelling at that time?

the angle of elvation

Solution:

Let the angle of elevation be θ(t), and the horizontal displacement of the plane from the tracking telescope be x(t), then from the figure we have

tan θ(t) = 5

x(t), or equivalently, x(t) = 5 cot θ(t) [3 points]

And, we are given that

d dtθ(t)∣

θ=^π₃

= − π

6 [1 points]

Therefore, the velocity of the plane is dx dt∣

θ=^π₃

=5 ⋅ (− csc²θ) ⋅dθ dt∣

θ=^π₃

[4 points]

=5 ⋅ (−( 2

√

3)²) ⋅ (−

π 6)

= 10π

9 (km/min.) [2 points]

(The other way)

tan θ(t) = 5

x(t) [3 points]

⇒sec²θdθ dt = −

5 x²

dx

dt [4 points]

⇒2²(−

π 6) = −

5 (^√5

3)² dx dt∣

θ=^π₃

[1 points]

⇒ dx

dt∣

θ=^π₃

= 10π

9 (km/min.) [2 points]

[Grading Criterion]

Write down the equation correctly. [3 points]

Use the given conditions correctly. [1 points]

Differentiate the equation correctly. [4 points]

Calculate the velocity correctly. [2 points]

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2. (a) (6%) Find the linear approximation of tan⁻¹x at the point p.

(b) (4%) Use (a) to approximate tan^{−1 3}₅ with p = tan (^π₆).

Solution:

(a) The linear approximation L(x) of a function f (x) at a point p is given by f (x) ≈ L(x) = f (p) + f^′(p)(x − p).

Let f (x) = tan⁻¹x, then we have f^′(x) = _1+x¹₂. Therefore, at the point p,

tan⁻¹x ≈ tan⁻¹p + 1

1 + p²(x − p) . [6 points]

(b) From part (a), we have

tan⁻¹3

5 ≈tan⁻¹p + 1 1 + p²(

3 5−p)

=tan⁻¹(tan(π 6)) +

1 1 + tan²(^π₆)

( 3

5 −tan(π 6))

= π 6 +

3 4(

3 5−

1

√3) = π 6 +

9 20−

√ 3

4 [4 points]

[Grading Criterion]

Part (a) Correct answer will get 6 points, otherwise, no point.

Part (b) Correct answer will get 4 points. If the steps are correct but tan(^π₆)is evaluated wrongly, 2 points.

3. Find the following limits.

(a) (4%) lim

x→0⁺

√xe^sin(π/x) (b) (4%) lim

x→0

(x−1)^1/3+(x+1)^1/3

x (c) (4%) lim

x→0(cos(x²))

1 x4

Solution:

(a) −1 ≤ sin(^π_x) ≤1

1

e≤e^sin(πx)≤e

√x e ≤

√xe^sin(πx)≤

√xe

limx→0⁺

√x

e =0 = limx→0⁺

√xe

By squeeze theorem limx→0⁺

√xe^sin(πx)=0 (b) limx→0

√3

x−1+√³ x+1 x

=limx→0

(x−1)+(x+1) x(√³

x−1²−³

√ x−1√³

x+1+√³ x+1²)

=limx→0 2

√3

x−1²−³

√ x−1√³

x+1+√³ x+1²

= ²₃

(c) limx→0(cos(x²))

1 x4

=e^limx→0ln(cos(x2 ))

x4 0

0, use 羅必達

=e^limx→0

−sin(x2 ) cos(x2 )2x

4x3

=e^limx→0

−sin(x2 ) x2

1 2 cos(x2 )

=e⁻¹² 第三題評分標準 (a)

出現limx→0⁺e^sin(πx)但沒有發現(說明)此極限不存在，其餘概念正確 -1 出現limx→0⁺

√xe^sin(πx)=limx→0⁺

√x ∗ limx→0⁺e^sin(πx) 並且沒有用到夾擠的概念，其餘概念正確 -2 把limx→0⁺e^sin(πx)看成無限大並用羅必達 -2～-4(視後面算式的合理以及完整性而定)

把limx→0⁺ sin(^π_x)

(^π_x) 看成無限大並用羅必達 -2～-4(視後面算式的合理以及完整性而定) 小錯誤 -1

(b)

計算錯誤 -1

上下同乘以錯誤的數字(因子) -1～-3 (視錯誤因子對後面算式的影響而定) (c)

不影響後面算式前提下的計算錯誤 -1 微分計算錯誤 -2

有取 ln 至少有1分

沒有把取 ln 的結果代回題目要求的式子 -1 PS

若(b)(c)兩題之中至少用了一次羅必達， 但(b)(c)兩題都沒有檢查是否滿足⁰₀的情況，則共計扣1分

Page 3 of 9

4. Suppose that f (x) =

⎧⎪

⎪

⎨

⎪⎪

⎩

sin x + b ln(x + 1) + c if x ≥ 0

e^x2 if x < 0 .

(a) (4%) Find b, c such that f (x) is continuous.

(b) (4%) Find b, c such that f (x) is differentiable.

(c) (4%) For b, c in (b), is f^′(x) continuous?

Solution:

(a) lim

x→0⁺f (x) = lim

x→0⁺[sin x + b ln (x + 1) + c] = c (1pt) lim

x→0⁻f (x) = lim

x→0⁻e^x2=1 (1pt) f (0) = c

Hence, f (x) is continuous at x = 0 ⇔ c = 1, b ∈ R. (2pt) (b)

lim

x→0⁺

f (x)−f (0)

x−0 = lim

x→0⁺

sin x+b ln (x+1)+1−1 x−0

= lim

x→0⁺

sin x+b ln (x+1) x

= lim

x→0⁺(^{sin x}_x +b^{ln (x+1)}_x ) =1 + b (1pt)

x→0lim⁻

f (x)−f (0) x−0 = lim

x→0⁻ e^x2−1

x−0 (⁰₀)^L’H= lim

x→0⁻ 2xe^x2

1 =0 (1pt) Hence, f (x) is differentiable at x = 0 ⇔ c = 1, b = −1 (2pt) (c)

For b = −1 and c = 1, f (x) is differentiable everywhere, then

f^′(x) =

⎧⎪

⎪

⎨

⎪⎪

⎩

cos x −_x+1¹ , x ≥ 0 2xe^x2, x < 0

x→0lim⁺f^′(x) = lim

x→0⁺[cos x −_x+1¹ ] =0 (1pt)

x→0lim⁻f^′(x) = lim

x→0⁻2xe^x2=0 (1pt), and f^′(0) = 0 (1pt) Hence, f^′(x) is continuous. (1pt)

5. (15%) A steel pipe is carried down a hallway 16 meter wide. At the end of the hall there is a right angled turn into a narrower hallway 2 meter wide. What is the length of the longest pipe that can be carried horizontally around the corner?

m 2

m 16

Solution:

Let the width of aisle is l(θ) where θ ∈ (0, π) is the angle between the pipe and the horizontal line. Therefore, we can have l(θ) = _cos(θ)¹⁶ +_sin(θ)² . We need to find the minimum of the l(θ); in this way we can find the length of the longest pipe.

l^′(θ) = 16 sec(θ) tan(θ) − 2 csc(θ) cot(θ)

To find the minimum of the l(θ), we should solve l^′(θ) = 0.

l^′(θ) = 16 sec(θ) tan(θ) − 2 csc(θ) cot(θ) = 16_cos^{sin θ}_2θ−2_sin^{cos θ}_2θ= ^{16 sin}

3θ−2 cos³θ sin²θ cos²θ =0

⇒ 16 sin³θ − 2 cos³θ = 0 ⇒ tan³θ = ¹₈⇒tan θ = ¹₂⇒θ = tan^{−1 1}₂(10 points)

We can check that l^′′(tan^{−1 1}₂) >0 (1 point). Therefore, the minimum of the l(θ) happens at θ = tan^{−1 1}₂ and we can compute the answer l(tan^{−1 1}₂) =10√

5.(4 points)

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6. (17%) Let h(x) = x^1/3(x − 4). Then h^′(x) = ^4(x−1)_3x2/3 and h^′′(x) = ^4(x+2)_9x5/3 . Answer the following questions by filling each blank below. Show your work (computations and reasoning) in the space following. Put None in the blank if the item asked does not exist, each blank is worth 2 pts.

(a) The function is increasing on the interval(s) and decreasing on the interval(s)

The local maximal point(s) (x, y) = and

The local minimal point(s) (x, y) = .

(b) The function is concave upward on the interval(s) and concave downward on the

interval(s) . The inflection point(s) (x, y) = .

(c) Sketch the graph of the function. Indicate, if any, where it is increasing/decreasing, where it concaves up- ward/downward, all relative maxima/minima, inflection points and asymptotic line(s) (if any). (3%)

Solution:

(a)

1. (2pt) (1, ∞) 2. (2pt) (−∞, 1) 3. (2pt) None 4. (2pt) (1, −3)

(b)

1. (2pt) (−∞, −2), (0, ∞) / (−∞, −2) ∪ (0, ∞) 2. (2pt) (−2, 0)

3. (2pt) (−2, 6√³

2), (0, 0)

(c)

1. (1pt) Mark all 4 points to get this point: (−2, 6√³

2), (0, 0), (−1, 3), (0, 4)

2. (1pt) Draw monotonicity and concavity correct and do not draw any asymptote to get this point.

3. (1pt) Draw something like a curve to get this point.

評分標準：

ˆ You do not lose any points if you replace any open end by closed end, e.g. [−2, 0]. But if you interchange the two ends, e.g. (0, −2), you lose 1pt for each blank.

ˆ Misplacing (a) 1. ↔ 2. (a) 3. ↔ 4. (b) 1. ↔ 2. costs 2pt each pair.

ˆ In (a) 2. (−∞, 0), (0, 1) is not correct, but won’t lose points.

ˆ If (a) 4. correct and (a) 3. empty, you get 1pt for (a) 3.

ˆ In (b) 3. missing (0, 0) costs 1pt.

ˆ If you think the domain of x^1/3is [0, ∞), use the following grading:

(a) 1. (2pt) (1, ∞) 2. (1pt) (0, 1) 3. (1pt) (0, 0) or None 4. (2pt) (1, −3) (b) 1. (1pt) (0, ∞) 2. (1pt) None 3. (1pt) None (c) (3pt) right half graph.

You get at most 12pt in this case.

ˆ In (c), if you write x = 0 a vertical asymptote (it is actually a vertical tangent line), you lose the point of 2.

ˆ Note that 7 < 6√³

2 < 8. Since we have grids for graphing, draw the point (−2, 6√³

2) between 7 and 8 or you lose the point of 1.

Remarks

ˆ −∞ and ∞ is not a real number. We don’t use closed end like [1, ∞] in real number system.

ˆ Use (−∞, −2) ∪ (0, ∞). (−∞, −2) ∩ (0, ∞) = ∅

ˆ A function f is (strictly) decreasing on (a, b) if:

For any x1, x2∈ (a, b), x1<x2 Ô⇒ f (x1) >f (x2).

We use first derivative just for test if f is differentiable. When f is not differentiable, you should check the original definition. Since (0, 0) exists, the interval (−∞, 1) has this property. Basically we write the largest interval as solution. Thanks to professor for not losing points.

ˆ Similarly, since (0, 0) exists, it is a inflection point, although h^′′(0) does not exist.

ˆ In real calculus, we define x^1/3 to be the inverse function of x³. Then the domain of x^1/3 is the whole real number line, while x^b is generally well defined only on x > 0 given any real number b. (But x¹ is good on (−∞, ∞), right?)

ˆ x = 0 is a vertical tangent line at (0, 0), so the curve should tangent to it. This do not cost any points since in Textbook we do not mention this.

Page 7 of 9

7. Let the curve x²y²+2xy = 8 be given.

(a) (4%) Express y^′in terms of x and y.

(b) (4%) Find points on the curve with y = 2 and the tangent lines at these points.

(c) (4%) Find y^′′ at the points in (b).

Solution:

(a)

(4 points)

⎧⎪

⎪

⎨

⎪⎪

⎩

(2xy²+2x²yy^′) + (2y + 2xy^′) =0 (x²y + x)y^′= −y − xy²

(3 points), each mistakes will minus 1 point

y^′= − xy²+y x²y + x = −

y(xy + 1) x(xy + 1)= −

y

x

(1 point)

(b)

(4 points)

x²⋅4 + 4x = 8, so we have x²+x − 2 = 0 = (x + 2)(x − 1) Hence the intersection points are

P1= (−2, 2)(1 point), P2= (1, 2)(1 point)

m₁= −

(−2) ⋅ 4 + 2 (−2)²⋅2 + (−2)=

6

6 =1

(1 point)

4 + 2

2 + 1= −2

(1 point)

Tangent line at P2: y − 2 = −2(x − 1)

(c)

(4 points)

y^′′= − y^′x − y

x² = 2y x² At P1, y^′′=₍₋₂₎^2⋅2₂ =1

(2 points)

At P₂, y^′′=⁴₁ =4

(2 points)

8. Find the derivative of the following functions.

(a) (4%) y = (tan⁻¹x)^{sin x}, x > 0.

(b) (4%) y = log_ex(tan x), 0 < x <^π₂. (c) (4%) y = ^(2x+1)_(3x−2)⁵6^(x(x²³⁺¹⁾+1)³⁴, find y^′(0).

Solution:

(a)

dy dx=

d

dxesin x⋅ln(tan⁻¹x) [1 points]

= (tan⁻¹x)^{sin x}(cos x ⋅ ln(tan⁻¹x) + sin x

(1 + x²) ⋅tan⁻¹x) [3 points]

(b)

y =ln tan x ln e^x =

ln tan x

x [1 points]

hence by quotient rule,

dy dx =

x sec²x

tan x −ln tan x

x² [3 points]

(c)

We can write y as

y = (2x + 1)⁵(x²+1)³(3x − 2)⁻⁶(x³+1)⁻⁴ hence by product rule,

y^′=10(2x + 1)⁴(x²+1)³(3x − 2)⁻⁶(x³+1)⁻⁴) +6x(2x + 1)⁵(x²+1)²(3x − 2)⁻⁶(x³+1)⁻⁴

−18(2x + 1)⁵(x²+1)³(3x − 2)⁻⁷(x³+1)⁻⁴

−12x²(2x + 1)⁵(x²+1)³(3x − 2)⁻⁶(x³+1)⁻⁵ [3 points]

y^′(0) = 10 ⋅ 1

64−18 ⋅ (− 1 128) =

19

64 [1 points]

(The other way)

ln y = 5 ln(2x + 1) + 3 ln(x²+1) − 6 ln(3x − 2) + 4 ln(x³+1) hence,

y^′ y =

10 2x + 1+

6x x²+1−

18 3x − 2+

12x²

x³+1 [3 points]

y^′(0) = y(0)(10 + 0 + 9 + 0) =19

64 [1 points]

[Grading Criterion]

(a)(b)

Simply the functions. [1 points]

Differentiate the equation correctly. [3 points]

(c)

Differentiate the equation correctly. [3 points]

Get the correct value of f^′(0).[1 points]

Page 9 of 9