Sample examinations Calculus I (201-NYA-05) Autumn 2010

Sample examinations
Calculus I (201-NYA-05)
1. Use the graph of the function f below to determine the following where possible.
y
a. lim f (x)
x→−∞
y = f (x)
b. lim f (x)
x→∞
c.
d.
lim f (x)
x→−3
10. Given
f (x) =
x→2
g.
9. Given the curve with equation x2 − xy + y 2 = 9.
dy
.
a. Find
dx
b. Find all point on the curve where the tangent line is horizontal;
c. A point P (x, y) moves along the curve so that dx
= −2 when x = 0 and y = 3.
dt
At what rate (and in which direction) is the line tangent to the curve at P rotating at
this instant?
lim f (x)
x→2−
e. lim f (x)
f.
Autumn 2010
lim f (x)
x→−1
lim f (x)
x
x→0+
x3 − 8
,
x3 + 8
f 0 (x) =
48x2
+ 8)2
(x3
f 00 (x) =
and
−192x(x3 − 4)
,
(x3 + 8)3
sketch the graph of f . Make sure your solution includes all (if any) intercepts,
asymptotes, intervals of monotonicity, extrema, intervals of concavity and points
of inflection.
h. f (−3)
11. Find the absolute extrema of f (x) = x1/3 − x2/3 on [ −1, 1 ].
i. All values of x at which f is continuous but not differentiable.
12. Find the intervals of monotonicity and concavity of f , where f (x) = x2 ex .
2. Evaluate each of the following limits.
13. Let ϑ be the radian measure of an angle in a right-angled triangle and let x and
y be, respectively, the lengths of the sides adjacent to and opposite to that angle.
Suppose that x and y vary with time. At the instant when x is 2 and is increasing at
4 units per second, y is 2 and is decreasing at 1 unit per second. At what rate is ϑ
changing at this instant?
x2 + 2x − 8
a. lim
x→2
x4 − 16
r
2 + 9x
d. lim
x→−∞
5 + 4x
b.
lim
t→
1 −
π
3
sin t
1 − 2 cos t
e. lim
x→2
4−x
c. lim √
x→4
x+5−3
x log x − log(x2 )
x−2
3. Discuss fully the continuity of f , where


|x − 2| if x 6 1,


√
x − 1 if 1 < x < 5 and
f (x) =


2x


if x > 5 and x 6= 10.
10 − x
4. The function f is defined by
(
xp sin(1/x) if x > 0, and
f (x) =
0
if x 6 0,
where p is a real number.
a. For which values of p is f continuous but not differentiable at 0?
b. For which values of p is f 0 continuous at 0?
5. Use the definition of derivative to find
6. Find
f 0 (x),
2x
where f (x) =
.
7−x
dy
for each of the following.
dx
a. y = (3x2 − 9) sec4 (5x + 3)
b. y = log
d. y = sin(2x − 1)3 + cos3 (2x − 1)
√
(x2 − 9)4
2
√
c. y = e 3+tan x
3
x x+7
e. y = (sin x)log x
14. Michael has 340 m of fencing to enclose two separate fields, one of which will
be a rectangle twice as long as it is wide, and the other will be a square. Due to
zoning regulations, the width of the rectangular field has to be at least 20 m and at
most 50 m. Find the maximum and minimum total area.
15. Find an expression (or the exact value of) each of the following limits.
sin(x + h) − sin x
h
n
X
2
c. lim
e2i/n
n→∞
n
i=1
a. lim
h→0
b.
lim
x→0−
d. lim
ϑ→0
|x|
x
sin ϑ
ϑ
16. Show that the function f , defined by
p
f (x) = 100 − x2 ,
satisfies the hypotheses of the Mean Value Theorem on [ −6, 8 ]. Then find all numbers c that satisfy the conclusion of the Mean Value Theorem.
17. Evaluate each of the following integrals.
Z 1π
Z
Z 2
3
√
(2y + 1)2
sec ϑ + tan ϑ
dy b.
dϑ c.
(1 − x + x2 ) x dx
a.
1π
y
cos
ϑ
1
4
Z
4
x
18. Verify the integral formula
dx = √
+ C.
3(4 − 3x2 )3/2
3 4 − 3x2
19. Find z given that
7. Find an equation of the tangent line to the graph of
sin x
y=
1 + cos x
at the point where x =
8. Given
2
π.
3
x2 + 1
f (x) =
.
(x + 1)2
a. Find f 0 (x) and simplify your answer as much as possible.
b. Determine whether f has a local maximum or minimum at x = 1. You may use
the fact that
−4(x − 2)
f 00 (x) =
.
(x + 1)4
d2 z
= 2 sin t − 4 cos t + et ,
dt2
dz =3
dt t=0
and
20. Let S be the region bounded by the graph of f (x) =
z(0) = −5.
3+x
and the x-axis
x
between x = 1 and x = 9.
a. Approximate the area of S by finding the Riemann sum with four equal subintervals and taking midpoints as sample points.
b. What is the exact area of S?
Z 5
21. Evaluate
|x − 2| dx by interpreting it in terms of area.
0
22. Use the Fundamental Theorem of Calculus to find a function f and a number a
such that
Z x
f (t)
8+
√ dt = 4 log x.
a t t
Sample examinations (solution outlines)
1. Inspecting the graph of f reveals that a.
Calculus I (201-NYA-05)
lim f (x) = ∞, b. lim f (x) = 2,
b. By the solution to the first part of this problem, f is differentiable at 0 if, and only
if, p > 1, in which case f 0 (0) = 0. Notice also that f 0 (x) = 0 if x < 0, so f 0 is
continuous from the left at 0. On the other hand, if x > 0 then by the Product Rule
and the Chain Rule,
x→∞
x→−∞
c. lim f (x) = 1, d. lim f (x) = 2, e. lim f (x) is undefined (since the left
x→2−
x→−3
x→2
limit is 2 and the right limit is infinite), f. lim f (x) = −1, g. lim f (x) = 0,
x→−1
x→0+
f 0 (x) = pxp−1 sin(1/x) + xp cos(1/x) · (−1/x2 )
h. f (−3) = 3 and i. f is continuous but not differentiable at −1.
= pxp−1 sin(1/x) − xp−2 cos(1/x).
2. a. Since the numerator and denominator both vanish at 2, they share the factor
x − 2. Independence and direct substitution then give
By the argument from part a of this question,
(x − 2)(x + 4)
x2 + 2x − 8
lim
= lim
x→2 (x − 2)(x + 2)(x2 + 4)
x→2
x4 − 16
x+4
= lim
x→2 (x + 2)(x2 + 4)
=
b. Since sin t →
1√
3
2
lim
t→
1 −
π
3
lim pxp−1 sin(1/x)
x→0+
is zero if p > 1 or p = 0, and undefined if p 6 1 and p 6= 0, and
lim xp−2 cos(1/x)
x→0+
3
.
16
and 1 − 2 cos t → 0− as t →
1 −
π ,
3
Autumn 2010
is zero if p > 2 and is undefined if p 6 2. This implies, since |sin(1/x)| = 1 when
cos(1/x) = 0 and |cos(1/x)| = 1 when sin(1/x) = 0, that f 0 is continuous at 0
if, and only if, p > 2.
it follows that
sin t
= −∞.
1 − 2 cos t
5. By the definition of the derivative of a function,
f (t) − f (x)
t−x
2t
2x
−
7−t
7−x
= lim
t→x
t−x
t(7 − x) − x(7 − t)
= 2 lim
t→x (t − x)(7 − t)(7 − x)
t−x
= 14 lim
t→x (t − x)(7 − t)(7 − x)
1
= 14 lim
t→x (7 − t)(7 − x)
14
=
.
(7 − x)2
c. Rationalizing the denominator gives
√
√
4−x
x+5+3
(4 − x)( x + 5 + 3)
lim √
·√
= lim
x→4
x→4
x−4
x+5−3 x+5+3
√
= − lim ( x + 5 + 3)
f 0 (x) = lim
t→x
x→4
= −6,
by independence and direct substitution.
d. Extracting dominant powers of x gives
s
r
2 + 9x
2/x + 9
= lim
=
lim
x→−∞
x→−∞
5 + 4x
5/x + 4
3
,
2
since 2/x → 0 and 5/x → 0 as x → −∞.
e. Writing log(x2 ) = 2 log x and factoring gives
(x − 2) log x
x log x − log(x2 )
= lim
= lim log x = log 2,
x→2
x→2
x−2
x−2
by independence and direct substitution.
6. a. By the Product Rule and the Chain Rule,
dy
= 6x sec4 (5x + 3) + 3(x2 − 3)4 sec4 (5x + 3) tan(5x + 3) · 5
dx
= 6 sec4 (5x + 3) x + 10(x2 − 3) tan(5x + 3) .
lim
x→2
3. Each part of f is defined by a function continuous at every real number in its
domain, so f is continuous at every real number except possibly 1, 5 and 10. Since
f (1) = lim f (x) = 1
x→1−
and
b. If
(x2 − 9)4
√
= 4 log(x2 − 9) − 3 log x −
x3 x + 7
then by the linearity of the derivative and the Chain Rule,
lim f (x) = 0,
y = log
x→1+
it follows that f has a jump discontinuity at 1. Next, f (5) is undefined, but
x→5+
so f has a removable discontinuity at 5. Finally, f (10) is undefined, and
lim f (x) = ∓∞,
x→10±
so f has an infinite discontinuity at 10.
4. a. If p 6 0 then, since |xp sin(1/x)| > |sin(1/x)| for 0 < x < 1, and since
sin(1/x) oscillates between −1 and 1 as x → 0, it follows that f is discontinuous
at 0. On the other hand, |sin(1/x)| 6 1, and hence 0 6 |xp sin(1/x)| 6 |xp |, for
x > 0. If p > 0 then |xp | → 0 as x → 0, and so
lim f (x) = 0 = f (0)
e. If y = (sin x)log x = elog x log(sin x) , then by the Product Rule and the Chain
Rule,
dy
= elog x log(sin x) · (1/x) log(sin x) + (log x)(cos x/ sin x)
dx
= (sin x)log x log(sin x) + x log x cot x /x.
x→0
by the Squeeze Theorem. Hence, f is continuous at 0 if, and only if, p > 0. Next,
lim
x→0−
log(x + 7)
4 · 2x
3
1
9x3 + 70x2 + 63x + 378
dy
= 2
− −
=
.
dx
x −9
x
2(x + 7)
2x(x + 7)(x2 − 9)
c. By the Chain Rule,
√
√
2
2
dy
e 3+tan x sin x
e 3+tan x 2 tan x sec2 x
√
= √
.
=
2
2
dx
2 3 + tan x
3 + tan x cos3 x
d. By the linearity of the derivative and the Chain Rule,
dy
= cos(2x − 1)3 · 3(2x − 1)2 · 2 + 3 cos2 (2x − 1) · − sin(2x − 1) · 2
dx
= 6 (2x − 1)2 cos(2x − 1)3 − cos2 (2x − 1) sin(2x − 1) .
lim f (x) = lim f (x) = 2,
x→5−
1
2
f (x) − f (0)
= 0,
x−0
and by the foregoing argument
f (x) − f (0)
= lim xp−1 sin(1/x)
x−0
x→0+
exists (and is equal to 0) if, and only if, p − 1 > 0, i.e., p > 1, in which case
f 0 (0) = 0. So f is continuous but not differentiable at 0 if, and only if, 0 < p 6 1.
7. If x = 23 π then y = (sin 32 π)/(1 + cos 23 π) =
by the Quotient Rule,
lim
x→0+
1√
3/(1
2
−
1
)
2
=
√
dy
(cos x)(1 + cos x) − (sin x)(− sin x)
1
=
=
,
dx
(1 + cos x)2
1 + cos x
and hence the slope of the tangent line in question is equal to
dy 1
= 2.
=
dx 2
1− 1
x= 3 π
2
2
3. Next,
Sample examinations (solution outlines)
Calculus I (201-NYA-05)
so the line defined by y = 1 is the horizontal asymptote of the graph of f . Where
defined,
48x2
f 0 (x) = 3
(x + 8)2
is positive or zero, and is zero only if x = 0, so f is increasing on ( −∞, −2 ) and
on ( −2, ∞ ). Therefore, f has no local extrema. Finally,
Therefore, an equation of the line
tangent to the given curve at the point where
√
x = 32 π is y = 3 + 2 x − 23 π .
8. a. Writing f (x) = (x2 + 1)(x + 1)−2 , and applying the Product Rule and the
Chain Rule, gives
f 0 (x) = 2x(x + 1)−2 − 2(x2 + 1)(x + 1)−3
= 2 x(x + 1) − (x2 + 1) (x + 1)−3
f 00 (x) =
= 2(x − 1)(x + 1)−3 .
b. Since f 0 (1) = 0 and f 00 (1) = −4(−3)2−4 =
is a local minimum value of f .
3
4
> 0, it follows that f (1) =
Autumn 2010
−192x(x3 − 4)
(x3 + 8)3
√
is zero if, and only if, x is 0 or 3 4, and is defined on the domain of f . Examining
√
the sign of f 00 between −2, 0 and 3 4 reveals that the graph of f is concave up
√
√
3
on ( −∞, −2 ) and on ( 0, 4 ), and concave down
on ( −2, 0 ) and on ( 3 4, ∞ ),
√
with points of inflection at (0, −1) and 3 4, − 31 . Below is a sketch of the graph of
f , with unit lengths marked along the coordinate axes, asymptotes drawn as dashed
lines, and the points of inflection emphasized.
y
1
2
9. a. Where y is an implicit function of x defined by the equation x2 −xy+y 2 = 9,
the Chain Rule gives
dy
dy
dy
y − 2x
2x − y − x
+ 2y
=0
or
=
.
dx
dx
dx
2y − x
dy
b. The tangent line to the graph of the given equation is horizontal where dx
= 0,
i.e., where y = 2x (and 2y 6= x). Replacing y by 2x in the given equation yields
√
√
x2 − x(2x) + (2x)2 = 9, or x2 = 3, which gives x = ± 3 and y = ±2 3.
2
2
Therefore, the tangent line to the graph of x − xy + y = 9 is horizontal at the
√
√
points (± 3, ±2 3).
c. The angle α between the tangent line to the curve at a given point and the positive
dy
, and therefore
x axis satisfies tan α = dx
2 2
d y
d2 y
dα
dy
dα
=
,
or
1
+
=
.
sec2 α
2
dx
dx
dx
dx
dx2
Now
d dy
d2 y
=
dx2
dx dx
dy
dy
− 2 (2y − x) − (y − 2x) 2
−1
dx
dx
=
(2y − x)2
dy
3 x
−y
dx
=
(2y − x)2
3
y − 2x
=
x
·
−
y
(2y − x)2
2y − x
y=
√
3
−6(x2 − xy + y 2 )
(2y − x)3
−54
=
,
(2y − x)3
=
and so
dy =
dx x=0,
1
2
and
y=3
x
4, − 31
11. Since f is continuous on R and
f 0 (x) =
1 −2/3
x
3
− 32 x1/3 =
x3 − 8
x3 + 8
1 −2/3
x
(1
3
√
− 2 3 x),
1
,
8
each of which belongs to [ −1, 1 ]. Since
f (−1) = −2,
f (0) = 0,
f 81 = 14
and
f (1) = 0,
1
1
the largest and smallest values of f on [ −1, 1 ] are f 8 = 4 and f (−1) = −2.
the critical number of f are 0 and
d2 y = − 14 ,
2
dx x=0,
y=3
and therefore
12. If f (x) = x2 ex then by the Product Rule,
1+
1 2 dα
2
dx
= − 41 ,
or
dα
= − 15 .
dx
f 0 (x) = 2xex + x2 ex = (x2 + 2x)ex = x(x + 2)ex ,
so the critical numbers of f are −2, 0. Since ex is positive for all real values of x,
the sign of f 0 (x) is the same as that of x(x + 2), which is positive if x < −2 or
x > 0 and negative if 0 < x < 2. Therefore, f is increasing on ( −∞, −2 ) and on
( 0, ∞ ), and decreasing on ( −2, 0 ). Next,
Finally, by the Chain Rule,
dα
dα dx
=
= − 51 −2 = 25 ,
dt
dx dt
so the line tangent to the given curve at P is rotating (counterclockwise) at a rate of
2
radians per unit of time when P is (0, 3).
5
f 00 (x) = (2x + 2)ex + (x2 + 2x)ex = (x2 + 4x + 2)ex
and, since ex is positive for all real values of x, f 00 (x) is positive, negative or zero
√
with x2 + 4x + 2 = (x + 2)2 − 2, which is zero if x = −2 ± 2, positive if
√
√
√
√
x < −2 − 2 or x > −2 + 2, and negative if −2 − 2 < x < −2 + 2.
√
√
So the graph of f is concave up on ( −∞, −2 − 2 ) and on ( −2 + 2, ∞ ), and
√
√
concave down on ( −2 − 2, −2 + 2 ).
10. Factorizing the denominator of f (x) gives (x + 2)(x2 − 2x + 4), and since
x2 − 2x + 4 = (x − 1)2 + 3 is always positive, the domain of f is R \ {−2}. The
y intercept of the graph of f is (0, −1). Factorizing the numerator of f (x) gives
(x − 2)(x2 + 2x + 4), and since x2 + 2x + 4 = (x + 1)2 + 3 is always positive,
the x intercept of the graph of f is (2, 0). Next, since
lim
x→−2±
f (x) =
lim
x→−2±
(x − 2)(x2 + 2x + 4)
= ∓∞,
(x + 2)(x2 − 2x + 4)
the line defined by x = −2 is the vertical asymptote of the graph of f . Also,
lim f (x) =
x→±∞
lim
x→±∞
1 − 8/x3
= 1,
1 + 8/x3
3
Sample examinations (solution outlines)
Calculus I (201-NYA-05)
13. Since x tan ϑ = y, differentiating with respect to time yields
dx
dϑ
dy
dx
dϑ
dy
tan ϑ + x sec2 ϑ
=
,
or
tan ϑ + x(1 + tan2 ϑ)
=
.
dt
dt
dt
dt
dt
dt
At the instant when x = 2 and y = 2, tan ϑ = 1,
dx
dt
= 4 and
dy
dt
18. Applying the Product Rule and the Chain Rule gives
x
d
1
x(−6x)
√
= √
−
2
dx 3 4 − 3x
2 · 3(4 − 3x2 )3/2
3 4 − 3x2
(4 − 3x2 ) + 3x2
3(4 − 3x2 )3/2
4
,
=
3(4 − 3x2 )3/2
= −1, and so
=
dϑ
= −1,
or
= − 45 .
dt
dt
Therefore, ϑ is changing at a rate of − 45 radians per second at the instant when
x = 2 and y = 2.
4 + 2 1 + 12
dϑ
and therefore
14. If w denotes the width of the rectangle and s denotes the side of the square
then 6w + 4s = 340, or s = 21 (170 − 3w), so the total area of the fields is
A = 2w2 + s2 = 2w2 + 41 (170 − 3w)2 , and it is required to find the smallest and
largest values of A on the closed interval [ 20, 50 ]. Now
w=30
Z
19. By (the first form of) the Fundamental Theorem of Calculus
Z t
dz
=3+
(2 sin τ − 4 cos δ + eτ ) dτ
dt
0
= 4 − 2 cos t − 4 sin t + et ,
and
Z
w=50
z = −5 +
reveals that the smallest total area is 3400 square metres and the largest total area is
5100 square metres.
d. lim
ϑ→0
0
20. a. If [ 1, 9 ] is partitioned into 4 subintervals of equal length then ∆x = 2 and
the midpoints of the subintervals are 2, 4, 6 and 8. The corresponding Riemann sum
of f (x) = (x + 3)/x = 1 + 3/x is
= 57
.
2 1 + 32 + 1 + 34 + 1 + 21 + 1 + 38
4
b. The exact area of the region below the graph of f and above the interval [ 1, 9 ]
on the x axis is
9
Z 9
(1 + 3/x) dx = (x + 3 log x) = 8 + 6 log 3.
sin ϑ
= 1.
ϑ
1
16. f is continuous on [ −10, 10 ] and differentiable on ( −10, 10 ), so it certainly
satisfies the hypotheses of the Mean Value Theorem on [ −6, 8 ]. Therefore,
there is a
real number ξ in ( −6, 8 ) such that f (8) − f (−6) = f 0 (ξ) 8 − (−6) = 14f 0 (ξ).
Now
√
√
−ξ
f 0 (ξ) = p
,
and
f (8) − f (−6) = 36 − 64 = −2,
100 − ξ 2
p
√
implies that 50ξ 2 = 100, or ξ = ± 2.
so ξ satisfies 7ξ = 100 − ξ 2 , whichp
√
√
However, only 2 is a solution of 7ξ = 100 − ξ 2 , and so 2 is the only number
in ( −6, 8 ) which satisfies the conclusion of the Mean Value Theorem.
y = |x − 2|
2
b. Dividing and integrating term by term gives
Z 1π
Z 1π
3
3
sec ϑ + tan ϑ
dϑ =
(sec2 ϑ + sec ϑ tan ϑ) dϑ
1π
1π
cos
ϑ
4
4
1π
3
= (tan ϑ + sec ϑ)
2 3/2
x
3
x
0
22. The given equation implies (taking x = a) that 8 = 4 log a, i.e., log a = 2 or
a = e2 . Next, if f is continuous then differentiating and applying the Fundamental
Theorem of Calculus gives
1π
4
2 + 1.
f (x)
4
√ = ,
x x
x
c. Expanding and integrating term by term gives
Z
Z
√
(1 − x + x2 ) x dx = (x1/2 − x3/2 + x5/3 ) dx
=
5
The region consists of two triangles. The triangle on the left has base 2, height 2
and area 12 (2)(2) = 2. The triangle on the right has base 3, height 3 and area
1
(3)(3) = 92 . Therefore,
2
Z 2
|x − 2| dx = 2 + 92 = 13
.
2
1
3−
(5, 3)
2
= 10 + log 2.
=
1
21. The region whose area is represented by the integral in question is shaded below.
y
17. a. Expanding, and then dividing and integrating term by term, gives
Z 2
Z 2
(2y + 1)2
dy =
(4y + 4 + y −1 ) dy
y
1
1
2
= (2y 2 + 4y + log y)
√
(4 − 2 cos τ − 4 sin τ + eτ ) dτ
= −10 − 4t − 2 sin t + 4 cos t + et .
0
√
t
0
d
sin(x + h) − sin x
=
(sin x) = cos x.
15. a. lim
h→0
h
dx
|x|
b. lim
= lim (−1) = −1.
x→0− x
x→0−
2
Z 2
n
X
2
ex dx = ex = e2 − 1.
c. lim
e2i/n =
n→∞
n
i=1
4
x
dx = √
+ C,
3(4 − 3x2 )3/2
3 4 − 3x2
as required.
dA
(w − 30),
= 4w − 32 (170 − 3w) = 17
2
dw
so the critical number 30 of A lies in the interval ( 20, 50 ), and comparing
A
= 3825,
A
= 3400
and
A
= 5100,
w=20
Autumn 2010
− 25 x5/2 + 27 x7/2 + C.
4
or
√
f (x) = 4 x.