Download Solutions to Mid Term 2

MID TERM II
CALCULUS I (25504 SEC 3)
Problem 1 (20 points). Compute the following
d (a)
(sin(x))cos(x) . Let y = sin(x)cos(x) . Then ln(y) = cos(x). ln(sin(x)) and
dx
d
1 dy
=
(cos(x). ln(sin(x)))
y dx
dx
cos(x)
= − sin(x) ln(sin(x)) + cos(x)
sin(x)
2
dy
cos (x)
cos(x)
Therefore
− sin(x) ln(sin(x)) +
.
= sin(x)
dx
sin(x)
(b) lim x2 ln(x). The limit is an indeterminate form. We write x2 ln(x) =
x→0+
ln x
x−2
in order to be
able to apply l’Hospital rule.
lim x2 ln(x) = lim
x→0+
= lim
x→0+
x→0+
ln(x)
x−1
=
lim
x−2
x→0+ −2x−3
x2
=0
−2
d
(c) Use logarithmic differentiation to compute
dx
1
1
1
(1 − 3x) 5 (1 − 2x) 3
(x + 1)2 (x − 1)4
!
.
1
(1 − 3x) 5 (1 − 2x) 3
. Using the properties of logarithm, we get
(x + 1)2 (x − 1)4
1
1
ln(y) = ln(1 − 3x) + ln(1 − 2x) − 2 ln(x + 1) − 4 ln(x − 1)
5
3
f 0 (x)
d
(ln(f (x))) =
we get
Since
dx
f (x)
y0
1
−3
1
−2
1
1
=
+
−2
−4
y
5 1 − 3x
3 1 − 2x
x+1
x−1
3
2
2
4
y0 = y −
−
−
−
5(1 − 3x) 3(1 − 2x) x + 1 x − 1
Again let y =
(d) lim
x→2π
1 − cos(x)
. The limit is an indeterminate form, for which we can apply l’Hospital rule.
sin(x)
lim
x→2π
1 − cos(x)
sin(x)
= lim
=0
x→2π cos(x)
sin(x)
1
2
MID TERM II CALCULUS I (25504 SEC 3)
Problem 2 (15 points). A cylindrical container is to be made without a lid to contain 27πcm3
of liquid. Find the radius and height of the container that requires the least amount of material.
Let r and h be the radius and height of the cylinder. Then the volume V = πr2 h = 27π implies
27
that h = 2 .
r
Now the area of the cylindrical container (without the lid) is given by:
27
2
+ πr2 = 54πr−1 + πr2
A = 2πrh + πr = 2πr
r2
We need to minimize A(r) where r > 0. We begin by computing the derivative:
d
(A(r)) = −54πr−2 + 2πr = 2πr−2 −27 + r3
dr
0
Therefore, A (r) = 0 for r3 = 27, i.e, r = 3. Moreover we have
• A0 (r) = 2πr−2 (r3 − 27) < 0 for 0 < r < 3.
• A0 (r) = 2πr−2 (r3 − 27) > 0 for r > 3.
Hence A(r) is absolutely minimum for r = 3cm. This implies
27
h = 2 cm = 3cm
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MID TERM II
CALCULUS I (25504 SEC 3)
3
Problem 3 (20 points). Consider the following function
f (x) = x3 − 9x2 + 24x − 5
(a) Find the critical points of f (x) and classify them as local maxima or local minima.
f 0 (x) = 3x2 − 18x + 24 = 3(x2 − 6x + 8). This quadratic polynomial can be factorized as:
f 0 (x) = 3(x − 2)(x − 4)
Thus the critical points of f are x = 2 and x = 4. Moreover, we have
– f 0 (x) > 0 for x < 2.
– f 0 (x) < 0 for 2 < x < 4.
– f 0 (x) > 0 for x > 4.
By the first derivative test, we conclude that x = 2 is a local max and x = 4 is a local min.
(b) On which intervals is the function increasing/decreasing?
From the previous part, f (x) is increasing on (−∞, 2) and (4, ∞). f (x) is decreasing on
(2, 4).
(c) On which intervals is the graph of f (x) concave up/down? Find the inflection points of f (x).
f 00 (x) = 6x − 18 = 6(x − 3). Therefore we have
– f 00 (x) < 0 for x < 3.
– f 00 (x) > 0 for x > 3.
Hence, f (x) is concave down on (−∞, 3) and concave up on (3, ∞). x = 3 is the inflection
point of f (x).
(d) Use the information of the previous parts to sketch the graph of f (x).
4
MID TERM II CALCULUS I (25504 SEC 3)
Problem 4 (15 points). A man walks along a straight path at a speed of 5ft/sec. A searchlight
is located on the ground 25ft away from the path and is kept focused on the man. At what rate is the
searchlight rotating when the man is 15ft away from the point on the path closest to the searchlight?
O
•
x
P
•
25ft
θ
•
S
Let S be the searchlight, O the point closest to the searchlight, and P the position of the person
dx
(see figure above). We are given that
= 5ft/sec, and OS is of 25ft. Therefore,
dt
x = 25 tan(θ)
Taking derivative with respect to t gives
dθ
dx
= 25 sec2 (θ)
dt
dt
15
3
2
When x = 15ft, we get tan(θ) = 25 = 5 . Using sec (θ) = 1 + tan2 (θ), we get
9
34
sec2 (θ) = 1 +
=
25
25
dx
dx
dθ
2
Substituting sec2 (θ) = 34
25 and dt = 5ft/sec in the equation dt = 25 sec (θ) dt we get
34 dθ
5 = 25
25 dt
dθ
5
and hence
= rad/sec.
dt
34
MID TERM II
CALCULUS I (25504 SEC 3)
5
Problem 5 (20 points). True/False. Give proofs for the true statements and counterexamples
for the false ones.
(a) If f (x) and g(x) are two functions which are both increasing on an interval, then so is
f (x)g(x).
False. For example, let f (x) = g(x) = −1/x. Then the functions are both increasing on
(0, ∞), since f 0 (x) = 1/x2 > 0. But the product f (x)g(x) = 1/x2 is decreasing on (0, ∞),
since (f g)0 (x) = −2/x3 < 0 on (0, ∞).
(b) The equation x2 − cos(x) = 0 has exactly one solution in [0, 1].
True. Let f (x) = x2 − cos(x). Then f (0) = −1 and f (1) = 1 − cos(1) > 0. By the
Intermediate Value Theorem, f (x) = 0 has at least one solution in [0, 1].
Moreover, f 0 (x) = 2x + sin(x) > 0 on (0, 1). By the Mean Value Theorem, there is at most
one solution of f (x) = 0 on [0, 1].
(c) Let f (x) be a function, defined and continuous everywhere, which satisfies f (x + 2) = f (x).
Then f (x) takes its absolute maximum value at infinitely many points.
True. Since f (x) is continuous, it has an absolute maximum value in the closed interval
[0, 2]. If c ∈ [0, 2] is the point so that f (c) is absolutely maximum, then so are the point
c + 2, c + 4, c + 6, · · · .
(d) For every x > 0, we have ex > 1 + x.
True. Let f (x) = ex − 1 − x. Then f (0) = 0 and f 0 (x) = ex − 1 > 0 for x > 0. Hence f (x)
is increasing on (0, ∞) which gives ex − 1 − x > 0 for all x > 0.
(Bonus) If f (x) is a one to one function (i.e, a 6= b imples f (a) 6= f (b)) then f 0 (x) is always non–zero.
False. Let f (x) = x3 . Then f (x) is a one to one function, however f 0 (0) = 0.
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MID TERM II CALCULUS I (25504 SEC 3)
Problem 6 (10 points). Find
dy
if
dx
ex/y = 3x + y
Use it to find the equation of the tangent line at (0, 1).
Taking derivative of the equation ex/y = 3x + y with respect to x gives:
dy
d x/y e
=3+
= 3 + y0
dx
dx
Using the chain rule, we can compute
d x/y d
e
= ex/y . (x/y)
dx
dx
!
d
d
y
(x)
−
x
(y)
dx
dx
= ex/y
y2
0
x/y y − xy
=e
y2
Thus we get
y − xy 0
e
= 3 + y0
y2
1
x
⇐⇒ ex/y − ex/y 2 y 0 = 3 + y 0
y
y
x/y
ex/y y1 − 3
0
y =
ex/y yx2 + 1
Substituting x = 0 and y = 1, we obtain the slope of the tangent line at (0, 1):
e0 − 3
= −2
0+1
Thus the equation of the tangent line at (0, 1) is
slope =
y − 1 = −2x