Part II: The Answers
268
Answers
401–500
407.
y
e x e
= −
+
+
2
4
4
40
4
The normal line is perpendicular to the tangent line. You can begin by finding the y
value at x = e
2
, because it isn’t given:
f e
e
2
2
4
2 4 2 2 10
( )
( )
=
+ =
+ =
ln
( )
Next, find the derivative of the function:
( )
x
x
=
=
4 1
4
( )
x
f '
Then substitute in the given x value to find the slope of the tangent line:
( )
e
e
2
2
4
=
f '
To find the slope of the normal line, take the opposite reciprocal of the slope of the
tangent line to get − e
2
4
.
Now use the point-slope formula for a line to get the normal line at x = e
2
:
y
e x e
y
e x e
− = −
−
= −
+
+
10
4
4
40
4
2
2
2
4
(
)
408.
− x
y
Taking the derivative of both sides of x
2
+ y
2
= 9 and solving for
dy
dx
gives you the
following:
2
2
0
2
2
x
y
dy
dx
y
dy
dx
x
dy
dx
x
y
+
=
= −
= −