Mathematics 116 Section 6.1 p452 - Wellesley College

Mathematics 116 HWK 5c Solutions

Section 6.1 p452

Problem 25, ?6.1, p. 452. Sketch the region that lies between the curves y = cos x and

y = sin 2x and between x = 0 and x =

2

.

Note that the region consists of two separate parts.

Find the area of this region.

Solution.

First,

the

sketch.

Note

that

for

x

ranging

from

0

to

2

,

the

curve

y

=

cos x

will

begin

at y = 1 and fall to y = 0, while the curve y = sin 2x will begin at 0, rise to 1 and then fall to

0. y

So initially y = sin 2x rises

= cos x lies above y =

above cos x,

y= and

sin 2x, somewhere both curves have

in between x fallen to 0 by

=0 the

and

x

=

2

time x =

the curve

2

.

Here's

a sketch, with the two separate regions labeled R and S and a couple of typical approximating

rectangles drawn in.

For the region R, the height of a typical approximating rectangle will be yT - yB = cos x - sin 2x, while for region S, the height will be yT - yB = sin 2x - cos x.

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A. Sontag September 23, 2001

Math 116 HWK 5c Solns continued ?6.1 p452

We will need to identify the point of intersection that occurs a little to the right of x = .5. We can equate the two y-values and solve for x, but in order to do this we'll have to use a trig identity.

cos x = sin 2x

cos x = 2 sin x cos x

cos x - 2 sin x cos x = 0

cos x(1 - 2 sin x) = 0

cos x = 0,

sin x

=

1 2

x = 2, x = 6

Thus there is a point of intersection,

of

intersection

occurs

when

x

=

6

.

as

we

anticipated,

at

the

endpoint

x

=

2

,

and

the

other

point

Now put all this information together:

2

1

2

1

11 1

13 1

Area of S = (sin 2x - cos x) dx = - cos 2x - sin x = [ - 1] - [- ? - ] = - + =

2

2

22 2

24 4

6

6

6

1

6 1 11

11

Area of R = (cos x - sin 2x) dx = sin x + cos 2x = [ + ? ] - [0 + ] =

0

2

0 2 22

24

Total Area = Area of R + Area of S = 1 + 1 = 1 44 2

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A. Sontag September 23, 2001

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