1091模模模組組組06-12班班班 微微微積積積分分分1 期期期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. (17 pts) Evaluate the following limits or show that they do not exist.
(a) lim
x→−∞(
√
x2+x + 1 + x).
(b) lim
x→0(1 − cos x) ⋅ s 1
x2 {
.(Here JxK denotes the ’greatest integer function’ and satisfies x − 1 < JxK ≤ x.) (c) lim
x→0
sin x
√
1 − cos(2x). (d) lim
x→0[
(1 + x)1x
e ]
1 x
.
Solution:
(a)
Marking scheme :
1M - Attempt to rationalise 1M - Rationalise correctly
1M - Divide both the numerator & denominator by x (correctly) 1M - Correct numerical answer
Remark : If a candidate attempts to use L’Hospital’s rule after rationalisation, the last 2M will only be awarded if the last displayed numerical answer is correct.
(Sample solution)
x→−∞lim (
√
x2+x + 1 + x) = lim
x→−∞
x + 1
√
x2+x + 1 − x
´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶
(2M )
= lim
x→−∞
1 +x1
−
√
1 +x1 +1
x −1
´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶
(1M )
= − 1 2
°
(1M )
.
(b)
Marking scheme :
2M - Correct bounds of the Gaussian/floor function 1M - Correct evaluation of the limit lim
x→0 1−cos x
x2
1M - Use of squeeze theorem (Sample solution)
1 x2 −1 <
s 1 x2
{
≤ 1 x2
´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶
(2M )
Ô⇒
1 − cos x
x2 − (1 − cos x) < (1 − cos x) ⋅ s 1
x2 {
≤
1 − cos x x2 Since lim
x→0
1 − cos x x2 =lim
x→0
1 − cos x
x2 − (1 − cos x) = 1 2
®
(1M )
,
the Squeeze Theorem implies that lim
x→0(1 − cos x) ⋅ s 1
x2 {
= 1
2. (1M )
(c)
Marking scheme :
1M - use of double angle formula or the identity 1 − cos2(2x) = sin2(2x) (appropriately) 1M - consider the left and right limits
1M - correct evaluation of one of these limits 1M - correct conclusion
Remark.
1) At most 1M will be given (very generously) to candidates who attempt with L’Hopital’s rule.
2) At most 2M will be awarded to candidates who have only evaluated one-side of the limit.
(Sample solution 1)
Since sin x
√
1 − cos(2x)
=
sin x
√
1 − (1 − 2 sin2x)
´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶
(1M )
=
sin x
√
2 ⋅ ∣ sin x∣ and
⎧⎪
⎪⎪
⎪⎪
⎪⎪
⎪
⎨
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎩
x→0lim+
sin x
√
2 ⋅ ∣ sin x∣ = 1
√ 2
°
(1M ) x→0lim−
sin x
√
2 ⋅ ∣ sin x∣ = − 1
√ 2
´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶
(1M )
,
we conclude that the limit does not exist (1M).
(Sample solution 2)
Since sin x
√
1 − cos(2x)
=sin x ⋅√
1 + cos 2x
√
sin2(2x)
´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶
(1M )
=sin x ⋅√
1 + cos 2x
∣sin 2x∣ and
⎧⎪
⎪⎪
⎪⎪
⎪⎪
⎪
⎨
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎩
x→0lim+
sin x ⋅√
1 + cos 2x
∣sin 2x∣ = lim
x→0+
sin x x ⋅
2x sin 2x⋅
√
1 + cos 2x
2 =
√ 2 2
°
(1M ) x→0lim−
sin x ⋅√
1 + cos 2x
∣sin 2x∣ = −
√ 2 2
´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶
(1M )
,
we conclude that the limit does not exist (1M).
Marking scheme :
1M - for taking logarithms of the given expression appropriately 1M - for tidying up the expression after taking log (correctly)
(Sample solution 1) Let y = [(1 + x)1x
e ]
1 x
. Then ln y = 1
x⋅ln ((1 + x)1x
e )
´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶
(1M )
=
ln(1+x) x −ln e
x =
ln(1 + x) − x x2
´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶
(1M )
. There-
fore,
ln (lim
x→0y) = lim
x→0ln y = lim
x→0
ln(1 + x) − x x2
0
=0 lim
x→0 1 1+x−1
2x
´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶
(2M )
=lim
x→0
−1
2(1 + x) = − 1 2.
Hence, lim
x→0y = e−12
°
(1M )
. (Sample solution 2)
limx→0[
(1 + x)1x
e ]
1 x
=lim
x→0e
ln
⎡
⎢
⎢⎢
⎢
⎢
⎣ (1+x)1
x e
⎤
⎥
⎥⎥
⎥
⎥
⎦ x
´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶
(1M )
=ex→0lim
ln(1+x)−x x2
´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶
(1M )
0
=0 ex→0lim
1 1+x−1
2x
´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶
(2M )
=ex→0lim
−1 2(1+x)
= e−12
°
(1M )
.
2. (12 pts)
(a) f (x) = ex2⋅sec x. Find f′(x).
(b) f (x) = x ⋅ cos−1x −√
1 − x2. Find f′(x).
(c) f (x) = (1 + sin x)cot x. Find f′(x).
Solution:
(a)
f′(x) = ex2⋅sec x⋅ d
dx(x2⋅sec x)(2 pts)=ex2⋅sec x⋅ (2x ⋅ sec x + x2⋅sec x ⋅ ⋅ tan x). (2 pts) (b)
f′(x) = cos−1x + x ⋅ −1
√ 1 − x2
− d dx
√
1 − x2 (2 pts)
=cos−1x + x ⋅ −1
√ 1 − x2
−
−1
√ 1 − x2
=cos−1x.(2 pts)
(c) (Method 1)
f (x) = (1 + sin x)cot x=ecot x⋅ln(1+sin x); (1 pt) f′(x) = ecot x⋅ln(1+sin x)
⋅ d
dx[cot x ⋅ ln(1 + sin x)] (2 pts)
= (1 + sin x)cot x⋅ [−csc2x ⋅ ln(1 + sin x) + cot x ⋅ cos x
1 + sin x]. (1 pt) (Method 2)
f (x) = (1 + sin x)cot x
ln f (x) = ln(1 + sin x)cot x=cot x ⋅ ln(1 + sin x) (1 pt)
⇒ f′(x)
f (x) = −csc2x ⋅ ln(1 + sin x) + cot x ⋅ cos x
1 + sin x (2 pts)
⇒f′(x) = f (x) ⋅ [− csc2x ⋅ ln(1 + sin x) + cot x ⋅ cos x 1 + sin x]
= (1 + sin x)cot x⋅ [−csc2x ⋅ ln(1 + sin x) + cot x ⋅ cos x
1 + sin x] (1 pt)
3. (15 pts) Consider the function f (x) =
⎧⎪
⎪
⎨
⎪⎪
⎩
sin(3x) + cos(2x) + A ,if x ≤ 0
x2⋅ln x ,if x > 0
where A is an auxiliary constant. It is given that f is continuous everywhere.
(a) (4 pts) Find the value of A.
(b) (5 pts) Is f differentiable at x = 0 ? Explain.
(c) (6 pts) Prove that for x > 1, the equation f (x) = e has a unique solution.
Solution:
(a)
Marking scheme :
1M - Correct definition of continuity
2M - Correct derivation and evaluation of lim
x→0+x2ln x 1M - Correct value of A
Remark : 1M is taken off if a candidate writes lim
x→0+x2ln x = 0 without any explanations.
Sample Solution of (a).
By continuity, we have lim
x→0+f (x) = f (0)
´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶
(1M )
. Since
x→0lim+f (x) = lim
x→0+x2ln x = lim
x→0+
ln x 1/x2 = lim
x→0+
1/x
−2/x3
´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶
(1M )
= 0
®
(1M )
,
we have A = −1
´¹¹¹¹¹¹¸¹¹¹¹¹¹¶
(1M )
.
(b)
Marking scheme :
1M - Correct definition of differentiability 1M - Correct evaluation of lim
h→0+
f (h)−f (0) h
2M - Correct derivation and evaluation of lim
h→0−
f (h)−f (0) h
1M - Correct conclusion Remark :
1. At most 1M will be (generously) awarded to candidates who proved the irrelevant fact that lim
x→0+f′(x) ≠ lim
x→0−f′(x).
2. In this part, if candidates write lim
x→0+x ln x = 0 without any explanations, no deductions will be made.
3. Candidates will lose 1M for the evaluation of lim
h→0−
f (h)−f (0)
h if their answers to (a) is incorrect. (So they can earn at most 4M in this part)
4. Candidates may get 2M or 3M if they have only considered one of the one-sided limits of the difference quotient, depending on which side has been computed.
Sample Solution of (b).
lim
h→0+
f (h) − f (0) h
´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶
(1M )
= lim
h→0+h ln h = lim
h→0+
ln h 1/h = lim
h→0+
1/h
−1/h2 = 0
®
(1M )
h→0lim−
f (h) − f (0)
h = lim
h→0−
sin(3h) + cos(3h) − 1
h = lim
h→0−3 ⋅ sin(3h)
3h +
cos(3h) − 1 h
´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶
(1M )
= 3
®
(1M )
Therefore, the limit lim
h→0
f (h) − f (0)
h does not exist and hence f is not differentiable at x = 0 (1M).
(c)
Marking scheme : Existence
1M - Mentioning f (or certain piece of of f ) is continuous 1M - Evaluating f at two (appropriate) values
1M - Conclusion (it doesn’t matter if ‘IVT’ is explicitly mentioned or not) Uniqueness
1M - computing the derivative of f (correctly) 1M - mentioning that f′≠0 or f′>0
1M - referencing MVT/Rolle/‘f is (strictly) increasing’ to complete the proof Remark : 1M will be taken off if a student only writes f′≥0.
Sample Solution of (c).
Existence
Since f is continuous (1M)
and f (1) = 0 and f (e) = e2, (1M)
the Intermediate Value Theorem implies that ∃c ∈ (1, e) such that f (c) = e. (1M) Uniqueness
Moreover, f′(x) = x(2 ln x + 1) (1M)
so f′(x) > 0 for x ≥ 1 (1M)
and hence this solution is unique as a consequence of the Rolle’s Theorem/M.V.T. (1M) OR this solution is unique because f is strictly increasing for x ≥ 1. (1M) Alternative proof for uniqueness
Suppose α and β are two distinct solutions to the equation f (x) = e.
Then Rolle’s Theorem implies that f′(c) = 0 for some c between α and β. (1M)
However, f′(x) = x(2 ln x + 1) (1M)
can never be zero/ is strictly positive for every x ≥ 1. (1M)
4. (10 pts) The graph given below is defined by the equation
ln(x2+1) + y2− (cos(πxy))2−ln 2 = 0.
(a) (6 pts) Find dy dx ∣
(x,y)=(1,1)
.
(b) (4 pts) Let f (x) be the function such that the above equation can be implicitly written as y = f (x) near (1, 1). Use the linear approximation of f (x) at x = 1 to approximate f (0.98).
Solution:
(a) Differentiation of F (x, y) with respect to x gives us 2x
x2+1+2yy′=2 cos(πxy)(− sin(πxy))(π(1 + xy′)).
Plugging (x, y) = (1, 1) into the equation, we obtain 1 + 2y′ =0, so y′(1) = −12. Thus, the equation of the tangent line is
y = L1(x) = 1 −1
2(x − 1).
Grading Schemes. : 1pt for derivative of ln x 1pt for derivative of cos x
3pt for chain rule (loss 3 points if fail three times)
1pt for the slope, provided the computation above is correct line.
(b) The linear approximation, being the tangent line at x = 1, of f (0.98) is L1(0.98) = 1.01.
Grading Schemes. :
2 pt for knowing the tangent line is the linear approximation Ò1 pt (concept) Ó1 pt (tangent line)
2 pt for computation correctness Ò1 pt (substitution) Ó1 pt (computation)
P.S. If the slope in (a) is wrong, deduct 1 pt from (computation) and (tangent line) in (b) only.
5. (8 pts) Assume that f (x) is continuous on [0,1] and differentiable on (0,1) with f (0) = 1, f (1) = 0.
Show that there exists c ∈ (0, 1) such that
f′(c) = − f (c) c . (Hint: Consider F (x) = x ⋅ f (x), x ∈ [0, 1].)
Solution:
Consider the function F (x) = x ⋅ f (x) defined on [0, 1]. Then F (x) is continuous on [0,1] and differentiable on (0,1) with F (0) = F (1) = 0. (3 pts) By Rolle’s theorem(or MVT), there exists c ∈ (0, 1) such that F′(c) = f (c) + cf′(c) = 0.(3 pts) That is, f′(c) = − f (c)c for some c ∈ (0, 1).(2 pts)
6. (12 pts) Ryan is flying a kite with an extendible string of length L (see figure). While the kite is rising vertically above the point Q at a rate of 0.4 m/sec, Ryan is running away from Q at a constant rate of 2 m/sec. At the moment when the kite is 3 m above the ground, the length of the string between Ryan and the kite is 5 m. Find the rate of change of the length of the string at this moment.
Solution:
Set x is the distance between the Ryan and the point Q and y is the distance between the point Q and the kite. (2%)
Then the relation between x, y and L is given by
L2 =x2+y2. (2%) (1)
Differentiating (1) on both sides with respect to t, we have 2LdL
dt =2xdx
dt +2ydy
dt. (3%) (2)
When L = 5 and y = 3, we have x =
√
52−32 =4 (1%).
Since Ryan is running away from Q at a constant rate of 2 m/sec and the kite is rising vertically above the point Q at a rate of 0.4 m/sec, we have
dx
dt =2, dy
dt =0.4. (2%) Substituting x = 4, y = 3, L = 5, dx
dt =2 and dy
dt =0.4 into (2), we have 2(5)dL
dt =2(4)(2) + 2(3)(0.4) ⇒ dL
dt =
2(4)(2) + 2(3)(0.4)
10 =1.84
Thus, the rate of change of the length of the string is 1.84 (m/s). (2%)
7. (10 pts) On the xy-plane, a line segment passes through three points P (x, 0), Q(1, a) and R(0, y), with y > a > 0, x > 1. Fix Q, and use the first or second derivative test to show that the length of the segment P R can be minimized by adjusting the locations of P and R.
Solution:
Must show a)the problem formulation, b)differentiation to find the criticals, and c)1st/2nd derivative test
(Method 1)
a) Let s denote the slope of any line segment thus formed. Then s = y−a0−1 = a−0
1−x ⇒y = a+x−1a =
a x
x−1 > a for x > 1.
Let ` denote the segment’s length, then
`(x) =
√
x2+y2= x−1x
√
(x − 1)2+a2 for x > 1 ⋯ (4%) b) ⇒ `′(x) = −
√
(x−1)2+a2
(x−1)2 +√ x
(x−1)2+a2 = (x−1)
3−a2 (x−1)2
√
(x−1)2+a2
=0 at x = x∗=1 + 3
√
a2 ⋯ (3%)
c) Since `′(x) < 0 for x < x∗, and `′(x) > 0 for x > x∗, a minimum length is confirmed by first derivative test. ⋯ (3%)
(Method 2)
a) Let θ denote the angle contained by the segment and the x-axis. Then the segment’s length is given by g(θ) = sin θa +cos θ1 =sec θ + a csc θ. 0 < θ < π/2 ⋯ (4%)
b) Therefore g′(θ) = tan θ sec θ − a cot θ csc θ = tan θ sec θ(1 − a cot3θ) which shows that g′=0 at θ∗=cot−1 √31
a ⋯ (3%)
c.1) Since g′<0 for θ < θ∗, g′>0 for θ > θ∗, a minimum length is confirmed by first derivative test. ⋯ (3%)
c.2) Or, g′′ =sec θ (tan2θ + sec2θ) +a csc θ (cot2θ + csc2θ) > 0, for any 0 < θ < π/2. Hence, a minimum length is confirmed by second derivative test. ⋯ (3%)
(Method 3)
Let m < 0 be the slope passing through Q(1, a). Then we have ←→
P R ∶ y − a = m(x − 1). That is,
←→P R ∶ y = mx + a − m which implies P (m−am , 0) and R(0, a − m). (3 pts) Consider the function f (m) ∶= P R2=
(m − a)2
m2 + (a − m)2= (m − a)2⋅ [1 + 1
m2], m < 0. (3 pts) Then
f′(m) = 2(m − a) ⋅ [1 + 1
m2] + (m − a)2⋅ (
−2
m3) = (m − a) ⋅ [2 + 2
m2 + (m − a) ⋅ −2 m3]
is the absolute minimum value of f (m). (2 pts) That is, Min(P R) =
√
f (−a13) = (1 + a23)
3 2.
8. (16 pts) Consider the function f (x) = x(x − 8)
√
x2−4, ∣x∣ > 2 with f′′(x) = 4(x2−24x + 8) (x2−4)52 .
Fill in 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.
(a) (6 pts) The asymptotes of f (x) are: x = ±2, y = −x + 8, y = x − 8 . (b) (5 pts) f (x) is increasing on the interval(s): (−4, −2) ∪ (2, ∞) .
f (x) is decreasing on the interval(s): (−∞, −4) .
Local maximum point(s) of f (x) : (x; y) =None ( No local maximum ). . Local minimum point(s) of f (x) : (x; y) =(−4, 8√
3) .
(c) (3 pts) f (x) is concave upward on the interval(s):(−∞, −2) ∪ (12 + 2√
34, ∞) .
f (x) is concave downward on the interval(s):(2, 12 + 2√
34) .
The inflection point(s) would occur at x =12 + 2√
34 .
(d) (2 pts) Sketch the graph of y = f (x). Indicate, if any, asymptotes, intervals of increase or decrease, concavity, local extreme values, and points of inflection.
Solution:
(a) Since lim
x→±∞f (x) = ∞, there is no horizontal or vertical asymptote of f (x). It is easy to see that the vertical asysmptotes are x = ±2. (2 pts) Now, we determine the slant asymptote of the f (x). We firstly compute the limits
x→∞lim f (x)
x = lim
x→∞
x(x − 8) x√
x2−4= lim
x→∞
x − 8
√ x2
√ 1 −x12
= lim
x→∞
x − 8
∣x∣
√ 1 −x12
= lim
x→∞
x − 8 x
√ 1 −x12
=1; (1 pt)
x→−∞lim f (x)
x = lim
x→−∞
x − 8
√ x2
√ 1 −x12
= lim
x→−∞
x − 8
∣x∣
√ 1 −x12
= lim
x→−∞
x − 8
−x
√ 1 −x12
= −1. (1 pt)
Then we compute th limits
x→∞lim[f (x) − x] = lim
x→∞[
x(x − 8)
√ x2−4
−x] = lim
x→∞
x[(x − 8) −√
x2−4]
√ x2−4
= lim
x→∞
x[(x − 8)2− (x2−4)]
√
x2−4 ⋅ [(x − 8) +
√ x2+4]
= lim
x→∞
x(−16x + 68)
√
x2−4 ⋅ [(x − 8) +
√
x2+4]
= −16
2 = −8; (1 pt)
x→−∞lim [f (x) + x] = lim
x→−∞[x(x − 8)
√ x2−4
+x] = lim
x→−∞
x[(x − 8) +
√
x2−4]
√ x2−4
= lim
x→−∞
x[(x − 8)2− (x2−4)]
√
x2−4 ⋅ [(x − 8) −
√
x2+4]
= lim
x→−∞
x(−16x + 68)
√
x2−4 ⋅ [(x − 8) −
√ x2+4]
=
−16
−2 =8.(1 pt) Therefore, the line y = x − 8 and y = −x + 8 are the slant asymptote of f (x).
(c) After solving f′′(x) = 0, we have x = 12 ± 2√
34. Note that 0 < 12 − 2√
34 = 8
12+2√
34 <2. So f (x) is concave upward on (−∞, −2) ∪ (12 + 2
√
34, ∞) (1 pt) and concave downward on (2, 12 + 2√
34). (1 pt) The inflection point occurs at x = 12 + 2√
34. (1 pt) (d) The graph is as following:
