MATH 1080
Test 2 -Version A-SOLUTIONS
Fall 2014
π
. 6
a. (8 pts) Find the exact length of the curve on the given interval. dy sec x tan x
=
= tan x dx
sec x
1. Consider the curve defined by y = ln ( sec x ) , 0 ≤ x ≤
π
L = ∫ 6 1+ tan 2 x dx
0
π
= ∫ 6 sec 2 x dx
0
π
= ∫ 6 sec x dx
0
= ln sec(x) + tan(x)
= ln(
2
3
1
+
3
π
6
0
) − ln(1)
L = ln( 3)
b. (3 pts) Set up an integral (Do not simplify or evaluate!) that represents the area of the surface obtained by rotating the curve about the x-­‐axis on the given interval. R = y = ln ( sec x )
π
SA = ∫ 6 2π ln ( sec x ) 1+ tan 2 x dx
0
Or ( )
y = ln ( sec x ) ⇔ x = arcsec e y
( )
ln
SA =
2
3
∫
0
⎛
2πy 1+ ⎜⎜
⎜⎝ e y
and
dx
=
dy
ey
ey
(e )
y
2
−1
2
⎞
ey
⎟ dy
⎟
2
e y − 1 ⎟⎠
( )
c. (3 pts) Set up an integral (Do not simplify or evaluate!) that represents the area of the surface obtained by rotating the curve about the y-­‐axis on the given interval. R=x
( )
ln
π
6
SA = ∫ 2πx 1+ tan x dx
0
2
or SA =
2
3
∫
0
( )
2πarcsec e y
⎛
1+ ⎜⎜
⎜⎝ e y
2
⎞
ey
⎟ dy
⎟
2
e y − 1 ⎟⎠
( )
1
MATH 1080
Test 2 -Version A-SOLUTIONS
Fall 2014
3
2. Consider the integral ∫ f (x) dx . 1
a. (4 pts) Using the Trapezoidal Rule with n = 4 , set up the expression that gives the 3
approximation of ∫ f (x) dx . 1
Δx =
3
∫
1
3− 1 1
=
4
2
f ( x )dx ≈
3
2
⎡ f (1) + 2 f (1.5) + 2 f ( 2 ) + 2 f ( 2.5) + f ( 3) ⎤⎦ 2⎣
1
1
∫ f ( x )dx ≈ 4 ⎡⎣ f (1) + 2 f (1.5) + 2 f ( 2) + 2 f ( 2.5) + f (3)⎤⎦
1
b. (4 pts) Using Simpson’s Rule with n = 4 , set up the expression that gives the 3
approximation of ∫ f (x) dx . 1
Δx =
3
3− 1 1
=
4
2
∫ f ( x )dx ≈
1
3
⎡ f (1) + 4 f (1.5) + 2 f ( 2 ) + 4 f ( 2.5) + f ( 3) ⎤⎦ 3⎣
1
2
1
∫ f ( x )dx ≈ 6 ⎡⎣ f (1) + 4 f (1.5) + 2 f ( 2) + 4 f ( 2.5) + f (3)⎤⎦
1
b
3. (2 pts) Give an example of an integral, ∫ f (x) dx , that is impossible to evaluate exactly. a
1
It is impossible to evaluate the integrals ∫ e
0
− x2
1
dx or
∫ sin ( x ) dx because we can’t find an 2
0
antiderivative. Or It is also impossible to evaluate the integral when the function is determined from collected data and there is no formula for the function. 2
MATH 1080
Test 2 -Version A-SOLUTIONS
4. Let R be the region bounded by the function f ( x ) =
[0,1]. Fall 2014
1
1 − x2
and x -­‐axis on the interval a. (8 pts) Evaluate the area of R . This is an improper integral and must be evaluated using limits. 1
t
1
1
A= ∫
dx = lim− ∫
dx 0
2
2
t →1 0
1−
x
1
−
x
t
= lim− sin −1 x (
)
= lim ( sin t − sin 0 ) 0
t →1
−1
−1
−
t →1
A =
π
2 The improper integral converges to π
. Thus the area of the region is finite. 2
b. (8 pts) Use the disk/washer method to evaluate the volume of the solid generated when R is revolved about the x -­‐axis. 2
t
⎛ 1 ⎞
1
π∫
dx ∫0 π ⎜⎝ 1 − x2 ⎟⎠ dx = lim
−
0 1 − x2
t →1
t⎛ 1
1
1 1 ⎞
= π lim− ∫ ⎜ ⋅
+ ⋅
⎟ dx t →1 0 ⎝ 2 1 − x
2 1+ x ⎠
1
=
π
lim− ( − ln 1 − x + ln 1 + x )0 t
2 t →1
π
lim ( − ln 1 − t + ln 1 + t ) 2 t →1−
= ∞ The improper integral is divergent. Therefore, the volume is not finite. =
3
MATH 1080
Test 2 -Version A-SOLUTIONS
Fall 2014
5. Consider the curve defined parametrically by x = t + sint, y = t − cost for 0 ≤ t ≤ 2π . a. (5pts) Find a Cartesian equation of the tangent(s) to the curve at the given point when π
t = . 4
dy
dx
= 1+ sint and
= 1+ cost
dt
dt
dy dy dt 1+ sint
=
=
dx dx dt 1+ cost
mtan =
dy
=1
dx t=π
When t =
4
⎛π
π
2 π
2⎞
, ( x, y ) = ⎜ +
, −
⎟.
4
⎝4 2 4 2 ⎠
⎛π
⎛π
2⎞
2⎞
y−⎜ −
⎟ = x−⎜ +
⎟
⎝4 2 ⎠
⎝ 4 2 ⎠ y= x− 2
b. (3 pts) Find all points on the curve where the tangent line is horizontal. Provide both t and (x, y) . You must justify your answer! dy
3π
= 0 when 1+ sint = 0, t =
dt
2
dy
dx
= 0 and
≠ 0 . dx
This occurs when dt
dt
=1≠ 0
dt t= 3π
2
The point on the curve where the tangent line is horizontal is when ⎛ 3π
3π
3π ⎞
t = , ( x, y ) = ⎜ − 1, ⎟ . 2
2⎠
⎝ 2
c. (3 pts) Find all points on the curve where the tangent line is vertical. Provide both t and (x, y) . You must justify your answer! dx
= 0 when 1+ cost = 0, t = π
dt
dx
dy
= 0 and
≠ 0 . This occurs when dy
dt
dt
=1≠ 0
dt t=π
The point on the curve where the tangent line is vertical is when t = π, ( x, y ) = ( π,π + 1) . 4
MATH 1080
Test 2 -Version A-SOLUTIONS
Fall 2014
6. Consider the curve defined parametrically by x = et , y = te−t . a. (6 pts) Set up an integral (Do not simplify or evaluate!) that represents the length of the curve on the interval e ≤ t ≤ e2 . dy
dx
= e−t − te−t = e−t (1− t )
and
= et
dt
dt
−t
−t
dy
dy
e − te
1− t
dt
=
=
= 2t
dx dx dt
et
e
e2
L = ∫
e
(e )
t
2
+ ⎡⎣ e−t (1− t ) ⎤⎦ dt 2
b. (4 pts) Set up an integral (Do not simplify or evaluate!) that represents the area of the surface obtained by rotating the curve about the x-­‐axis on the interval e ≤ t ≤ e2 . e2
SA = ∫ 2πte−t
e
(e )
t
2
+ ⎡⎣ e−t (1− t ) ⎤⎦ dt 2
c. (4 pts) Set up an integral (Do not simplify or evaluate!) that represents the area of the surface obtained by rotating the curve about the y-­‐axis on the interval e ≤ t ≤ e2 . e2
SA = ∫ 2πet
e
(e )
t
2
+ ⎡⎣ e−t (1− t ) ⎤⎦ dt 2
5
MATH 1080
Test 2 -Version A-SOLUTIONS
7. Consider the curve defined in polar coordinates by r = 1− cosθ . a. (3 pts) Sketch the graph of the curve. π 3π
2π
θ 0 π
2
2
r 0 1
2 1
0
b. (8 pts) Find the exact area of the region enclosed by the curve. The curve is traced out for θ = 0 to θ = 2π . Therefore, we have 2π 1
2
A = ∫0 (1 − cos θ ) dθ 2
1 2π
= ∫ 1 − 2 cos θ + cos 2 θ dθ 2 0
1 2π ⎛
1 + cos 2θ ⎞
= ∫ ⎜1 − 2 cos θ +
⎟ dθ 2 0 ⎝
2
⎠
(
)
2π
1⎡
1
1
⎤
θ − 2sin θ + θ + sin 2θ ⎥ ⎢
2⎣
2
4
⎦0
3
A = π 2
=
Fall 2014
6
MATH 1080
Test 2 -Version A-SOLUTIONS
Fall 2014
7
8. Consider the two curves defined in polar coordinates by r1 = 1 and r2 = 2sin 2θ for 0 ≤ θ ≤ 2π . a. (5 pts) Set up an integral(s) (Do not simplify or evaluate!) that represents the area of the region that lies inside both curves. Shade this region in the graph below. 1
π 5π
π 5π
Where do the curves intersect? 2sin 2θ = 1 ⇒ sin 2θ = ⇒ 2θ = , ⇒ θ = , 2
6 6
12 12
5π
12
⎡ π12 1
1 2 ⎤
2
A = 4 ⎢ 2 ∫ ( 2sin 2θ ) d θ + ∫ (1) d θ ⎥
2
π
⎢⎣ 0 2
⎥⎦
12
π
4
⎡ π12 1
1 2 ⎤
2
or A = 4 ⎢ 2 ∫ ( 2sin 2θ ) d θ + 2 ∫ (1) d θ ⎥
2
π
⎢⎣ 0 2
⎥⎦
12
5π
π
12
2
⎡ π12 1
⎤
1 2
1
2
2
or A = 4 ⎢ ∫ ( 2sin 2θ ) d θ + ∫ (1) d θ + ∫ ( 2sin 2θ ) d θ ⎥
2
2
π
5π
⎢⎣ 0 2
⎥⎦
12
12
π
π
π
4
2
⎡ 12 1
⎤
1 2
1
2
2
or A = 4 ⎢ ∫ ( 2sin 2θ ) d θ + 2 ∫ (1) d θ + ∫ ( 2sin 2θ ) d θ ⎥
2
2
π
5π
⎢⎣ 0 2
⎥⎦
12
12
5π
12
⎡π2 1
⎤
1
2
2
2
or A = 4 ⎢ ∫ ( 2sin 2θ ) d θ − ∫ ⎡( 2sin 2θ ) − (1) ⎤ d θ ⎥
⎣
⎦
2
π
⎢⎣ 0 2
⎥⎦
12
π
π
4
⎡ 21
⎤
1
2
2
2
or A = 4 ⎢ ∫ ( 2sin 2θ ) d θ − 2 ∫ ⎡( 2sin 2θ ) − (1) ⎤ d θ ⎥
⎣
⎦
2
π
⎢⎣ 0 2
⎥⎦
12
π
π
12
⎡ 21 2
⎤
1
2
2
or A = 4 ⎢ ∫ (1) d θ − ∫ ⎡(1) − ( 2sin 2θ ) ⎤ d θ ⎥
⎣
⎦
2
⎢⎣ 0 2
⎥⎦
− π12
12
⎡ 21 2
⎤
1
2
2
or A = 4 ⎢ ∫ (1) d θ − 2 ∫ ⎡(1) − ( 2sin 2θ ) ⎤ d θ ⎥
⎣
⎦
2
⎢⎣ 0 2
⎥⎦
0
π
π
b. (5 pts) Set up an integral(s) (Do not simplify or evaluate!) that represents the area of the region that lies inside of the rose but outside the circle. Shade this region in the graph below. 5π
12
⎡ 5π 12 1
2
1 2 ⎤
A = 4 ⎢ ∫ ( 2sin 2θ ) dθ − ∫ (1) dθ ⎥
2
π
⎢⎣ π 12 2
⎥⎦
12
5π
⎡ 12 1
⎤
2
2
or A = 4 ⎢ ∫ ⎡⎢( 2sin 2θ ) − (1) ⎤⎥ dθ ⎥
⎦ ⎥
⎢⎣ π 12 2 ⎣
⎦
π
⎡ 41
⎤
2
2
or A = 4 ⎢ 2 ∫ ⎡⎢( 2sin 2θ ) − (1) ⎤⎥ dθ ⎥
⎦ ⎥
⎢⎣ π 12 2 ⎣
⎦
⎡
⎤
π
5π
⎢π2
⎥
12
12
⎛
2
2
1
1
1 2 ⎞
or A = 4 ⎢ ∫ ( 2sin 2θ ) dθ − ⎜ 2 ∫ ( 2sin 2θ ) dθ + ∫ (1) dθ⎟ ⎥
⎢0 2
2
π
⎝ 0 2
⎠⎥
12
⎢

⎥
⎢⎣
⎥⎦
or any equivalent area from part (a)
MATH 1080
Test 2 -Version A-SOLUTIONS
Fall 2014
9. Consider the conic described by the equation y 2 − 4x 2 + 2 y − 16x = 19 . a. (8 pts) Rewrite the equation in standard form and state the type of conic equation it represents. y 2 + 2 y − 4 x 2 + 4x = 19 + 1− 16
(
(y
2
) (
)
+ 2 y + 1) − 4 ( x + 4x + 4 ) = 33
2
y + 1 2 − 4 x + 2 2 = 4
( ) ( )
( y + 1) − ( x + 2)
4
2
2
=1
This is a hyperbola. b. (3 pts) Graph the conic from part a. Find and label the vertices and center of the curve. The center is the point (-­‐2,-­‐1). The vertices are the points (-­‐2,1) and (-­‐2,-­‐3). c. (3 pts) Using the standard form from part a., provide a parameterization using a π
π
trigonometric identity for the top half of the curve with − < θ < . 2
2
2
2
2
2
Using the identity tan θ + 1 = sec θ ⇒ sec θ − tan θ = 1 , one parameterization that gives π
π
the the top half of the curve with − < θ < is 2
2
y +1
= secθ and x + 2 = tanθ
2
Then
y = 2secθ − 1
x = tanθ − 2
−
π
π
<θ< .
2
2
8