Math 111 Quiz #6 Solutions 1. Consider the equation = 1, which ...
Math 111
Quiz #6
1. Consider the equation x2 + 2y2 = 1, which describes an ellipse in the xy-plane.
(a)
Use
implicit
differentiation
to
calculate
dy dx
for
this
ellipse.
x2 + 2y2 = 1
dy 2x + 4y = 0,
dx
so
dy -2x x
= =- .
dx 4y
2y
Solutions
(b) Find the point(s) where the tangent line to the given ellipse has slope 1.
The points where the tangent line to the ellipse has slope 1 will be pairs (x, y) on the ellipse
where
dy dx
=
-x 2y
= 1.
This equation
together
with
the equation
for
the
ellipse
gives
us a
system
of
two equations in two unknowns that we can hopefully solve using basic algebra.
-x
1=
x = -2y
2y
We substitute x = -2y into the equation for the ellipse, x2 + 2y2 = 1:
(-2y)2 + 2y2 = 1 4y2 + 2y2 = 1 6y2 = 1 y = ? 1/6.
When y = 1/6, we have x = -2( 1/6) = -2/ 6, and the point (-2/ 6, 1/ 6) is a point on
the ellipse where the tangent line has slope 1.
When y = - 1/6, we have x = -2(- 1/6) = 2/ 6, and the point (2/ 6, -1/ 6) is the other
point on the ellipse where the tangent line has slope 1.
2. Suppose y = f (x)g(x)h(x), where f (x), g(x), and h(x) are differentiable functions.
(a)
Apply
the
product
rule
to
find
a
formula
for
dy dx
.
(Hint: You'll need to use the product rule twice.)
Start off by thinking of y as being the following product: y = f (x) g(x)h(x) . So applying the product rule to f (x) times g(x)h(g), we get
dy
d
d
= f (x) [g(x)h(x)] + [f (x)]g(x)h(x).
dx
dx
dx
This equation tells us we need to take the derivative of g(x)h(x), so we'll have to use the product rule again. Taking derivatives as indicated, the above equation becomes
dy = f (x) g(x)h (x) + g (x)h(x) + f (x)g(x)h(x).
dx
(b)
Use
logarithmic
differentiation
to
find
a
formula
for
dy dx
.
y = f (x)g(x)h(x)
ln y = ln f (x)g(x)h(x)
ln y = ln f (x) + ln g(x) + ln h(x)
1 dy 1
1
1
= ? f (x) + ? g (x) + ? h (x)
y dx f (x)
g(x)
h(x)
dy
f (x) g (x) h (x)
=y
+
+
dx
f (x) g(x) h(x)
dy
f (x) g (x) h (x)
= f (x)g(x)h(x)
+
+
dx
f (x) g(x) h(x)
(c)
Show
that
the
expressions
for
dy dx
you
got
in
(a)
and
(b)
are
equivalent.
The result from (a) can be expanded to
dy = f (x) g(x)h (x) + g (x)h(x) + f (x)g(x)h(x)
dx = f (x)g(x)h(x) + f (x)g (x)h(x) + f (x)g(x)h (x).
The result from (b) can be simplified as follows:
dy
f (x) g (x) h (x)
= f (x)g(x)h(x)
+
+
dx
f (x) g(x) h(x)
f (x)
g (x)
h (x)
= f (x)g(x)h(x) + f (x)g(x)h(x) + f (x)g(x)h(x)
f (x)
g(x)
h(x)
= f (x)g(x)h(x) + f (x)g (x)h(x) + f (x)g(x)h (x)
So both techniques of integration gave the same derivative, which is what should happen (a function can have only one derivative).
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