Midterm 2 Solutions, Math 1A, section 1
Thursday, November 6, 2008, 8:109:30 am
1. Let
p
= 0. Show by implicit differentiation that the tangent line to the curve
x
p
+
y
p
= 1
, x >
0
, y >
0
at the point (
x
0
, y
0
) is given by the equation
x
p

1
0
x
+
y
p

1
0
y
= 1. Show that the
x
intercept
a
and
y
intercept
b
of the tangent line satisfy
a
p/
(1

p
)
+
b
p/
(1

p
)
= 1 if
p
= 1.
Solution:
Implicitly differentiate, using the power, sum, and chain rules:
d
dx
(
x
p
+
y
p
) =
px
p

1
+
py
p

1
y
=
d
dx
(1) = 0
.
Solving for
y
, we get
y
=

(
x/y
)
p

1
. We plug in the point (
x
0
, y
0
) such that
x
p
0
+
y
p
0
= 1
into the pointslope formula for the tangent line:
y

y
0
=

(
x
0
/y
0
)
p

1
(
x

x
0
)
.
Multiplying by
y
p

1
0
and putting the constants to one side, we obtain
y
p

1
0
y
+
x
p

1
0
x
=
y
p
0
+
x
p
0
= 1
.
Setting
x
and
y
= 0, we see that the intercepts are (
a, b
) = (
x
1

p
0
, y
1

p
0
). Then we see that
a
p/
(1

p
)
+
b
p/
(1

p
)
= (
x
1

p
0
)
p/
(1

p
)
+ (
y
1

p
0
)
p/
(1

p
)
=
x
p
0
+
y
p
0
= 1
.
2. A ladder 10
ft.
long leans against a vertical wall. If the bottom of the ladder slides away from
the base of the wall at a speed of 2
ft./s.
, how fast is the angle between the ladder and the
wall changing when the bottom of the ladder is 6
ft.
from the base of the wall?
Solution:
Let
α
be the angle between the ladder and the wall, let
x ft.
be the distance from
the base of the ladder and the wall, and let
t s.
be the time. The given information implies
that
dx
dt
= 2. Then sin(
α
) =
x/
10, so
α
= arcsin(
x/
10). Then we differentiate using the chain
and constant multiple rules:
dα
dt
=
d
dt
arcsin(
x/
10) =
1
1

(
x/
10)
2
dx
dt
1
10
=
2
√
10
2

x
2
.
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 Fall '08
 CHRIST
 Derivative, Power Rule, lim cosh, lim sinh

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