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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