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koblitz mid2 win18 sol.pdf
koblitz_mid2_win18_sol.pdf
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MATH | 124 | Calculus with Analytic Geometry I | e...
MATH | 124 | Calculus with Analytic Geometry I | exam |Answers to Second Midterm 1. a ( t ) 2 +5 2 = c ( t ) 2 . Taking d/dt of both sides and then substituting a ( t ) = 12 and c ( t ) = 13, we get 2 · 12 ˙ a =2 · 13 ˙ c , and so the desired ratio ˙ c/ ˙ a is equal to 12 / 13. 2. The curve has equation √ x 2 +( y - 10) 2 =2 √ ( x - 5) 2 + y 2 . Next take d/dx of both sides, then substitute x = 8, y = 4, and solve for y ′ . It saves a little work to square both sides befo
MATH | 124 | Calculus with Analytic...
MATH | 124 | Calculus with Analytic Geometry I | exam |Answers to Second Midterm 1. a ( t ) 2 +5 2 = c ( t ) 2 . Taking d/dt of both sides and then substituting a ( t ) = 12 and c ( t ) = 13, we get 2 · 12 ˙ a =2 · 13 ˙ c , and so the desired ratio ˙ c/ ˙ a is equal to 12 / 13. 2. The curve has equation √ x 2 +( y - 10) 2 =2 √ ( x - 5) 2 + y 2 . Next take d/dx of both sides, then substitute x = 8, y = 4, and solve for y ′ . It saves a little work to square both sides befo
Page 1
Answers to Second Midterm
1.
a
(
t
)
2
+5
2
=
c
(
t
)
2
. Taking
d/dt
of both sides and then substituting
a
(
t
) = 12
and
c
(
t
) = 13, we get 2
·
12 ˙
a
=2
·
13 ˙
c
, and so the desired ratio ˙
c/
˙
a
is equal to
12
/
13.
2. The curve has equation
√
x
2
+(
y
-
10)
2
=2
√
(
x
-
5)
2
+
y
2
. Next take
d/dx
of both sides, then substitute
x
= 8,
y
= 4, and solve for
y
′
. It saves a little work
to square both sides before taking the derivative, in which case after taking the
derivative you have 2
x
+2(
y
-
10)
y
′
= 4(2(
x
-
5)+2
yy
′
). Substituting
x
= 8,
y
=4
and solving for
y
′
, you get 16
-
12
y
′
= 24 + 32
y
′
so that
y
′
=
-
8
/
44 =
-
2
/
11.
(If you don’t first square both sides, then when you substitute
x
= 8,
y
= 4 into
x
+(
y
-
10)
y
′
√
x
2
+(
y
-
10)
2
=2
x
-
5+
yy
′
√
(
x
-
5)
2
+
y
2
,
and multiply through by 20, you get the same thing.) The equation of the tangent
line (which passes through (8
,
4) with slope
-
2
/
11) is then
y
=
-
2
11
x
+
60
11
.
3. (a)
L
is the derivative of
x
x/
2
at the point (4
,
16).
(b) Set
y
=
x
x/
2
, so that ln(
y
)=
1
2
x
ln(
x
), and, taking
d/dx
of both sides,
y
′
y
=
1
2
(1 + ln(
x
)). At (4
,
16) you get
y
′
=
y
2
(1 + ln(
x
)) = 8(1 + ln(4)).
4. Let
f
(
x
) =Arcsin(
x
), so that
f
′
(
√
2
2
)=1
/
√
1
-
(
√
2
/
2)
2
=1
/
√
1
/
2=
√
2.
Then
f
(
√
2
2
+
h
)
≈
Arcsin(
√
2
/
2) +
h
√
2=
π
4
+
h
√
2.
5.
(a)
dy
dx
=
√
3 sin(
θ
)
2
-
√
3 cos(
θ
)
;
(b)
(2
-
√
3 cos(
θ
)(
√
3 cos(
θ
))
-
√
3 sin(
θ
)(
√
3 sin(
θ
))
(2
-
√
3 cos(
θ
))
2
.
(c) Set the numerator in part (b) equal to zero and use the identity cos
2
+ sin
2
= 1.
You get 2 cos(
θ
)
-
√
3 = 0, that is, cos(
θ
)=
√
3
/
2, so that
θ
=
π/
6 radians (or
30
◦
, that is, 1/12 revolution).
(d) angular velocity is
π/
2 rad/sec, so at
t
=1
/
3 sec; (e)
t
= 2 sec; (f)
t
= 1 sec.
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