Homework 14 Solutions
1
Comment
It is my hope that the graphs accompanying some of the implicit differentiation exercises made
it easier to see if your solutions made sense.
You should be using the pointslope formula for lines, since you have the point and the
slope: (
y

b
) =
m
(
x

a
). There is no need (and it is a waste of time) to reduce the equation
to the slope intercept form.
2
Implicit Differentiation Exercises
Page 125, Number 17:
Find an equation of the tangent line to the ellipse
x
2
+
xy
+
y
2
= 3
at (1
,
1). We have (using the power rule, product rule and chain rule)
2
x
+ (
y
+
xy
0
) + 2
yy
0
=
0
2 + (1 +
y
0
) + 2
y
0
=
0
3 + 3
y
0
=
0
so
y
0
=

1. Using the pointslope formula we have
y

1 =

(
x

1) as an equation of
the tangent line to the ellipse at (1
,
1).
Page 125, Number 18:
Find an equation of the tangent line to the hyperbola
x
2
+ 2
xy

y
2
+
x
= 2 at (1
,
2).
We have (using the power rule, product rule and chain rule):
2
x
+ 2(
y
+
xy
0
)

2
yy
0
+ 1
=
0
2 + 2(2 +
y
0
)

4
y
0
+ 1
=
0
7

2
y
0
=
0
so
y
0
= 7
/
2. Using the pointslope formula we have (
y

2) = 7(
x

1)
/
2 or, if you hate
fractions, 2(
y

2) = 7(
x

1) as equations of the tangent line to the hyperbola at (1
,
2).
Page 126, Number 19:
Find an equation of the tangent line to the cardioid
x
2
+
y
2
=
(2
x
2
+ 2
y
2

x
)
2
at (0
,
1
/
2).
Judging from the graph this line should have a positive slope that is slightly more than
1.
We have (using the power rule, product rule and chain rule):
2
x
+ 2
yy
0
=
2(2
x
2
+ 2
y
2

x
)(4
x
+ 4
yy
0

1)
0 +
y
0
=
2(0 + (1
/
2)

0)(0 + 2
y
0

1)
y
0
=
2
y
0

1
so
y
0
= 1. Using the pointslope formula we have (
y

(1
/
2)) =
x
or, if you hate fractions,
2
y

1 = 2
x
.
1
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Page 126, Number 20:
Find an equation of the tangent line to the astroid
x
2
/
3
+
y
2
/
3
= 4
at (

3
√
3
,
1). Note that 3
√
3 = 3
3
/
2
.
Judging from the graph this line should have a positive slope that is slightly less than 1.
We have (using the power rule, product rule and chain rule):
2
3
x

1
/
3
+
2
3
y

1
/
3
y
0
=
0
2
3
(

3
√
3)

1
/
3
+
2
3
y
0
=
0

(3
3
/
2
)

1
/
3
+
y
0
=
0

1
√
3
+
y
0
=
0
so
y
0
= 1
/
√
3
<
1 and an equation of the tangent line is
y

1 = (
x
+ 3
√
3)
/
√
3 or
√
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