Calculus 2
MATH 2423 Section 010, Fall 2013
Review for Exam 2 Monday, October 14, 2013
This is only a sample. Try all the home work problems.
1. A spherical tank of radius 2 ft is half full of water. Find the work required to pump the
water out of the outlet which is 3 ft above the top of the tank. Water weighs 62.5 lb/ft3 .
2. Find a formula for the inverse of the function y =
x2 −2
5 ,
x≤0
3. Write the definitions for the following functions. (There are some more that you should
know. E.g. ax , exp(x), ln(x)...).
(1) y = loga x
(2) y = cos−1 x
(3) y = sinh x
(4) y = sinh−1 x
4. Write the definition of y = cos−1 x. (The domain for x must be clearly written). Then
d
1
[Explain the choice of sign (why it is
derive the derivative formula: dx
[cos−1 x] = − √1−x
2
“−”) clearly].
5. Find
dy
dx .
(x2 + 1)5
= ln √
1√− x
= (5x + 1) 2
= e5−7x
2
2
= cosh
s (3x − 2) − sinh (3x − 2)
(1) y
(2) y
(3) y
(4) y
(x + 1)5
(x + 2)20
−1
= sin (7x)
2
= e−x
= (x + 1)x√
= cosh−1 ( x)
= cosh3 (3x − 2)
(5) y = ln
(6)
(7)
(8)
(9)
(10)
y
y
y
y
y
6. Evaluate
the integrals.
Z
(1)
coth x dx
Z ln x
2
(2)
dx
Z x
dx
(3)
2+4
9x
Z
dx
√
(4)
9 − 4x2
Z
dx
√
(5)
4x2 − 9
Z
sinh x
(6)
dx
Z 1 + cosh x
dx
√
(7)
16x2 + 1
7. Memorize the derivative formulas for sin−1 x, cos−1 x, tan−1 x, sinh x, cosh x, tanh x,
sinh−1 x, cosh−1 x, tanh−1 x. [Necessary for #5 and #6]. See the last page
92
3
· 62.5π (ft-lb).
√
2. y = − 5x + 2.
1.
3.
(1)
(2)
(3)
(4)
y
y
y
y
= loga x ⇐⇒ x = ay
= cos−1 x ⇐⇒ x = cos y and 0 ≤ y ≤ π
x
−x
= sinh x = e −e
2
= sinh−1 x ⇐⇒ x = sinh y
4. Let y = cos−1 x. Then
(1)
x = cos y,
Take
0 ≤ y ≤ π.
d
dx .
1 = − sin y ·
dy
dx
so that
dy
1
=−
.
dx
sin y
Now we need to write sin y in terms of x. We have
p
p
sin y = ± 1 − cos2 y = ± 1 − x2
(from (1)).
However,
the angle condition 0 ≤ y ≤ π as shown in (1) forces sin y ≥ 0 so that sin y =
√
+ 1 − x2 . Thus,
d
1
1
[sin−1 x] = −
= −√
.
dx
sin y
1 − x2
5.
(1)
(2)
(3)
(4)
(5)
(6)
10x
+ 1+x
2
√
√
5 2(5x + 1) 2−1
−7e5−7x
0
5(2+3x)
− 2(2+3x+x
2)
1
2−2x
7
1−49x2
2
−2xe−x
√
(7)
x
(8) (x + 1)x x+1
+ ln(x + 1)
1
(9) √√ √
√
√
2
x−1
x+1 x
(10) 9 cosh2 (3x − 2) · sinh(3x − 2).
6.
(1) ln | sinh(x)| + C
ln(x)
(2) 2ln(2) + C
(3) 16 tan−1 3x
2 + C
(4) 12 sin−1 2x
+C
3 (5) 12 cosh−1 2x
3 +C
(6) ln(1 + cosh x) + C
(7) 14 sinh−1 (4x) + C
1. We introduce a coordinate system so that 3 ft above the top of the tank as 0. Therefore,
the top of the tank is 3, and the bottom is 7. Water is in between 5 ≤ x ≤ 7. Consider a
thin slice at stage x with thickness ∆x.
radius =
p
22 − (x − 5)2
volume = π · 22 − (x − 5)2 · ∆x
force = π · 22 − (x − 5)2 · ∆x · 62.5
work = π · 22 − (x − 5)2 · ∆x · 62.5 · x
Therefore,
Work = lim
X
n→∞
π · 22 − (x − 5)2 · ∆x · 62.5 · x
7
Z
22 − (x − 5)2 x dx
= 62.5π
5
92
· 62.5π(ft-lb).
=
3
2. Inverse function is
x=
Solving this for y, we get
y2 − 2
, y ≤ 0.
5
√
y = ± 5x + 2
But, since y ≤ 0, we take the negative one. So the inverse function is,
√
y = − 5x + 2.
5. Find
(1)
dy
dx .
y = ln
(x2 + 1)5
√
1−x
= 5 ln(x2 + 1) −
1
ln(1 − x).
2
Therefore,
dy
1
1
1
=5· 2
· 2x − ·
· (−1)
dx
x +1
2 1−x
1
10x
= 2
+
x + 1 2(1 − x)
(4) y = cosh2 (3x − 2) − sinh2 (3x − 2) = 1. Therefore
dy
dx
= 0.
(Another way) One can really differentiate to get
y 0 = 2 cosh(3x − 2) · sinh(3x − 2) · 3 − 2 sinh(3x − 2) cosh(3x − 2) · 3
= 6 cosh(3x − 2) · sinh(3x − 2) − 6 sinh(3x − 2) cosh(3x − 2)
= 0.
(5)
s
y = ln
=
(x + 1)5
(x + 2)20
5
20
ln(x + 1) −
ln(x + 2)
2
2
Therefore,
5
1
1
dy
= ·
− 10 ·
dx
2 x+1
x+2
5
1
4
=
−
2 x+1 x+2
5(2 + 3x)
=−
.
2(2 + 3x + x2 )
d
1
du
sin−1 u = √
·
.
2
dx
1 − u dx
d
(7) Use the general fact: dx
(eu ) = eu · du
dx .
x
(8) y = (x + 1)
d
(9) Use the general fact: dx
cosh−1 u = √u12 −1 · du
dx .
(6) Use the general fact:
(10) y = cosh3 (3x − 2)
h
i0
dy
= 3 cosh2 (3x − 2) · cosh(3x − 2)
dx
= 3 cosh2 (3x − 2) · sinh(3x − 2) · 3
= 9 cosh2 (3x − 2) · sinh(3x − 2).
6. Evaluate
the integrals.
Z
(1) coth x dx
h
Let u = sinh x. Then du = cosh xdx
Z
i
Z
cosh x
dx
sinh x
Z
1
=
du
u
= ln |u| + C
coth x dx =
= ln | sinh x| + C
2ln x
(2)
dx
x
h
i
Let u = ln x. Then du = x1 dx
Z
Z
2ln x
dx =
x
Z
2u du
2u
+C
ln 2
2ln x
=
+ C.
ln 2
=
Z
dx
(3)
9x2 + 4
i
h
3
.
Then
du
=
dx
Let u = 3x
2
2
Z
1
dx =
2
9x + 4
Z
1
2
· du
+4 3
Z
1
1
=
· du
2
6
u +1
Z
1
1
· du
=
2
6
u +1
= tan−1 u + C
−1 3x
= tan
+ C.
2
4u2
Z
(4)
√
h
dx
Let u =
9 − 4x2
2x
3 .
Z
Z
(5)
√
h
dx
Let u =
4x2 − 9
Z
(6)
1
√
dx =
9 − 4x2
i
Z
1
3
√
· du
2
9 − 9u 2
Z
1
1
√
=
du
2
1 − u2
1
= sin−1 u + C
2
2x
1
= sin−1
+C
2
3
2x
3 .
Z
Then du = 23 dx
Then du = 23 dx
dx
√
=
4x2 − 9
i
Z
1
3
√
· du
9u2 − 9 2
Z
1
1
√
du
=
2
2
u −1
1
= cosh−1 u + C
2
1
−1 2x
+C
= cosh
2
3
h
i
sinh x
dx Let u = 1 + cosh x. Then du = sinh xdx
1 + cosh x
Z
Z
sinh x
1
dx =
du
1 + cosh x
1+u
= ln |1 + u| + C
= ln (1 + cosh x) + C.
Z
(7)
√
dx
16x2 + 1
h
Let u = 4x. Then du = 4dx
Z
i
dx
√
=
16x2 + 1
Z
√
1
1
du
u2 + 1 4
·
1
sinh−1 u + C
4
1
= sinh−1 (4x) + C
4
=
Formulas to memorize:
d
1
[sin−1 x] = √
dx
1 − x2
1
d
[cos−1 x] = − √
dx
1 − x2
1
d
[tan−1 x] =
dx
1 + x2
d
1
[sinh−1 x] = √
dx
1 + x2
d
1
[cosh−1 x] = √
2
dx
x −1
d
1
1
[tanh−1 x] =
is wrong
2
2
dx
1−x
x −1