Problem #1
Think of a cone (45degrees) is cut with a horizontal plane in a
Standard x,y, and z coordinate plane. The trace of the plane on a hollow cone is a circle
with radius OD  OA  r .
If the plane rotates with angle of  away from xy plane, then the trace will be an ellipse
with longer side OC  L and shorter side OB  S . The sum of these two is major axis
where minor axis does not change.
Look at it in 2-D: there are two triangles OCD , and OAB . We know the following:
DOC  BOA   and ODC  135 , BA0  45 and OCD  45   , OBA  135  
Use law of Sine:In OCD we have
Sin (135) Sin (45   )
and in OAB

L
r
Sin (45) Sin (135   )
Expand the sine function and simplify we have two equations

S
r
 1 Cos  Sin 
r
 L 
 L  Cos  Sin
r
or 
Solve for Cosine and Sine in terms of L, S, and r.

 1  Cos  Sin 
 r  Cos  Sin 
 S
 S
r
Cos 
r(L  S )
r(L  S )
and Sin 
Solve for tangent of the angle which is the slope of
2 LS
2 LS
the line in 2D. Tan 
(L  S )
(L  S )
The longer side is L  r  f and the shorter side is S  r  f where f is the deviation of
new center from old center (Focal distance).
Tan 
(r  f )  (r  f ) 2 f
f
f

 The letter e represent this ratio or e   Tan
r
( r  f )  ( r  f ) 2r
r
Now if 0    45, or 0  e  1Ellipse
  45, or e  1 Parabola
  45, or e  1 Hyperbola
Problem #2
We begin with the drawing of the ellipse. In an ellipse, the lines going to the point (x,y)
form a triangle from the focal points of the ellipse. The sum of the distances of these two
lines is a constant, so we begin with the equation:
√(𝑥 − (−𝑐))2 + (𝑦 − 0)2 + √(𝑥 − 𝑐)2 + (𝑦 − 0)2 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
We know that the sum of the distances must be 2a, because the lines at the x axis are the
equivalent of 2a. so we get the equation:
√(𝑥 − (−𝑐))2 + (𝑦 − 0)2 + √(𝑥 − 𝑐)2 + (𝑦 − 0)2 = 2𝑎
Next we use algebra to derive the equation of the ellipse
√(𝑥 + 𝑐)2 + (𝑦)2 + √(𝑥 − 𝑐)2 + (𝑦)2 = 2𝑎
√(𝑥 − 𝑐)2 + (𝑦)2 = 2𝑎 − √(𝑥 + 𝑐)2 + (𝑦)2
(𝑥 + 𝑐)2 + 𝑦 2 = (2𝑎 − √(𝑥 + 𝑐)2 + (𝑦)2 )2
𝑥 2 + 2𝑐𝑥 + 𝑐 2 + 𝑦 2 = 4𝑎2 − 4𝑎√(𝑥 − 𝑐)2 + 𝑦 2 + (𝑥 − 𝑐)2 + 𝑦 2
𝑥 2 + 2𝑐𝑥 + 𝑐 2 + 𝑦 2 = 4𝑎2 − 4𝑎√(𝑥 − 𝑐)2 + 𝑦 2 + 𝑥 2 − 2𝑐𝑥 + 𝑐 2 + 𝑦 2
2𝑐𝑥 = 4𝑎2 − 4𝑎√(𝑥 − 𝑐)2 + 𝑦 2 − 2𝑐𝑥
𝑎√(𝑥 − 𝑐)2 + 𝑦 2
4𝑐𝑥 − 4𝑎2 = 4𝑎√(𝑥 − 𝑐)2 + 𝑦 2 𝑐𝑥 − 𝑎2 =
(𝑐𝑥 − 𝑎2 )2 = 𝑎2 ((𝑥 − 𝑐)2 + 𝑦 2 )
𝑐 2 𝑥 2 − 2𝑎2 𝑐 2 𝑥 2 + 𝑎4 = 𝑎2 𝑥 2 − 2𝑎2 𝑐 2 𝑥 2 + 𝑎2 𝑐 2 + 𝑎2 𝑦 2
𝑎4 − 𝑎2 𝑐 2 = 𝑎2 𝑥 2 + 𝑐 2 𝑐 2 + 𝑎2 𝑦 2
𝑎2 (𝑎2 −𝑐 2 ) = 𝑥 2 (𝑎2 − 𝑐 2 ) + 𝑎2 𝑦 2
We know that 𝑎2 = 𝑏 2 + 𝑐 2 so 𝑎2 − 𝑐 2 = 𝑏 2
𝑎2 𝑏2
𝑎2 𝑏2
𝑏2 𝑥 2
𝑎2 𝑦2
= 𝑎2 𝑏2 + 𝑎2 𝑏2
𝑐 2 𝑥 2 + 𝑎4 = 𝑎2 𝑥 2 + 𝑎2 𝑐 2 + 𝑎2 𝑦 2
𝑥2
𝑦2
1 = 𝑎2 + 𝑏2
𝑎2 𝑏 2 = 𝑏 2 𝑥 2 + 𝑎2 𝑦 2
Hyperbola
Parabola
The difference from (x,y) to the foci is √(𝑥 − (−𝑐))2 + (𝑦 − 0)2 + √(𝑥 − 𝑐)2 + (𝑦 − 0)2 = 2𝑎
Algebraic derivation: √(𝑥 + 𝑐)2 + (𝑦)2 + √(𝑥 − 𝑐)2 + (𝑦)2 = 2𝑎
(𝑥 + 𝑐)2 + 𝑦 2 = (2𝑎 − √(𝑥 + 𝑐)2 + (𝑦)2 )2
√(𝑥 − 𝑐)2 + (𝑦)2 = 2𝑎 − √(𝑥 + 𝑐)2 + (𝑦)2
𝑥 2 + 2𝑐𝑥 + 𝑐 2 + 𝑦 2 = 4𝑎2 − 4𝑎√(𝑥 − 𝑐)2 + 𝑦 2 + (𝑥 − 𝑐)2 + 𝑦 2
𝑥 2 + 2𝑐𝑥 + 𝑐 2 + 𝑦 2 = 4𝑎2 − 4𝑎√(𝑥 − 𝑐)2 + 𝑦 2 + 𝑥 2 − 2𝑐𝑥 + 𝑐 2 + 𝑦 2
2𝑐𝑥 = 4𝑎2 − 4𝑎√(𝑥 − 𝑐)2 + 𝑦 2 − 2𝑐𝑥
4𝑐𝑥 − 4𝑎2 = 4𝑎√(𝑥 − 𝑐)2 + 𝑦 2
𝑐𝑥 − 𝑎2 = 𝑎√(𝑥 − 𝑐)2 + 𝑦 2
(𝑐𝑥 − 𝑎2 )2 = 𝑎2 ((𝑥 − 𝑐)2 + 𝑦 2 )
𝑐 2 𝑥 2 − 2𝑎2 𝑐 2 𝑥 2 + 𝑎4 = 𝑎2 𝑥 2 − 2𝑎2 𝑐 2 𝑥 2 + 𝑎2 𝑐 2 + 𝑎2 𝑦 2
𝑐 2 𝑥 2 + 𝑎4 = 𝑎2 𝑥 2 + 𝑎2 𝑐 2 + 𝑎2 𝑦 2
𝑎4 − 𝑎2 𝑐 2 = 𝑎2 𝑥 2 + 𝑐 2 𝑐 2 + 𝑎2 𝑦 2
Now we multiply by -1 to make both sides positive 𝑎2 (𝑎2 −𝑐 2 ) = 𝑥 2 (𝑎2 − 𝑐 2 ) − 𝑎2 𝑦 2
Then substitute 𝑏 2 = 𝑐 2 − 𝑎2
𝑎2 𝑏2
𝑎2 𝑏2
𝑏2 𝑥 2
𝑎2 𝑦2
𝑥2
= 𝑎 2 𝑏 2 − 𝑎 2 𝑏2
𝑦2
1 = 𝑎2 − 𝑏2
Parabola(The figure on top of the page)
Let the vertex be at 𝑣 = (ℎ, 𝑘)
The directrix at 𝑦 = (𝑘 − 𝑚, 𝑚)
The focus at 𝑓 = (ℎ, 𝑘 + 𝑚)
And the point 𝑝 = (𝑥, 𝑦)
2
2
𝑉 ((𝑥 − ℎ)2 + (𝑦 − (𝑘 + 𝑚)) ) = 𝑉((𝑥 − 𝑥)2 + (𝑦 − (𝑘𝑚)) )
𝑥 2 − 2ℎ𝑥 + ℎ2 + 𝑦 2 − 2𝑦(𝑘 + 𝑚) + 𝑘 2 + 2𝑚𝑘 + 𝑚2 = 𝑦 2 − 2𝑦(𝑘 − 𝑚) + (𝑘 − 𝑚)2
𝑥 2 − 2ℎ𝑥 + ℎ2 + 𝑦 2 − 2𝑦𝑘 − 2𝑦𝑚 + 𝑘 2 + 2𝑚𝑘 + 𝑚2 = 𝑦 2 − 2𝑦𝑘 − 2𝑦𝑚 + 𝑘 2 − 2𝑚𝑘 + 𝑚2
𝑥 2 − 2ℎ𝑥 + ℎ2 − 2𝑚𝑦 = 2𝑦𝑚 − 2𝑚𝑘
4𝑦𝑚 = 𝑥 2 − 2ℎ𝑥 + ℎ2 + 4𝑚𝑘
−4𝑦𝑚 = −𝑥 2 + 2ℎ𝑥 − ℎ2 − 4𝑚𝑘
4𝑚(𝑦 − 𝑘) = (𝑥 − ℎ)2
(𝑦 − 𝑘) =
1
(𝑥
4𝑚
− ℎ)2
Problem #3 We have x(t) = (vcosθ)t + X 0 and y(t) = −4.9t 2 + v(sinθ)t + Y0
Since it is launched at the origin, X 0 and Y0 equals 0. Then Solve for t: t =
x
x
Plug in t into y(t): y(t) = −4.9(vcosθ)2 + v(sinθ)(vcosθ) or
Now set y = 0 to find the maximum range (x-intercept)
4.9x2
2
v cos2 θ
xtanθv 2 cos2 θ = 4.9x 2 solve for 𝑥 =
= xtanθ
Differentiate y to find maximum height (when y’ = 0)
9.8x
0 = −(v2 cos2 θ) + tanθ
y max 
v 2 cos 2 θtanθ = 9.8x
−4.9x2
y(t) = v2 cos2 θ + xtanθ
−4.9x2
0 = v2 cos2θ + xtanθ
v2 sinθcosθ
4.9
=
v 2 Sin (2 )
g
9.8x
)+
v2 cos2 θ
y′ = −(
x of Maximum height =
tanθ
v2 Sin(2θ)
2(9.8)
v 2 Sin 2
2g
Michael Tran
CT and CP are radiuses
I did not understand the proof
Fix it
4) Show that the parametric equation for all three different cycloid can
 x(t )  a  bSin 
. Then find the slope in terms of angle  .
 y (t )  a  bCos
be written as 
x
vcosθ
Find point P as reference for the parametric equation.
̅̅̅̅
̆ , so length of line ̅̅̅̅
𝑂𝑇 = 𝑃𝑄
𝑂𝑇 = 𝑎𝜃
To find the x variable of point P, we will subtract line ̅̅̅̅
𝑂𝑇 from line ̅̅̅̅
𝑃𝑄 which turns out to
be:
X(t) = 𝑎𝜃 – b sin 𝜃
̅̅̅̅.
To find the y variable of point P, we will find 𝑄𝑇
Y(t) = a - b cos 𝜃

When b/a > 1, this equation will yield an epicycloid.

When b/a = 1, this equation will yield a standard cycloid.

When b/a < 1, this equation will yield a trochoid.
Slope at any point on this graph will be denoted by:
M=
𝑑𝑦
𝑑𝜃
𝑑𝑥
𝑑𝜃
𝑏𝑠𝑖𝑛𝜃
= 𝑎−𝑏𝑐𝑜𝑠𝜃
Problem #5
We assume that the position of is what we want to solve, is the radian from the
tangential point to the moving point , and is the radian from the starting point to the
tangential point.
Since there is no sliding between the two cycles, then we have that
By the definition of radian (which is the rate arc over radius), then we have that
From these two conditions, we get the identity
By calculating, we get the relation between
and , which is
From the figure, we see the position of the point
clearly.
Problem #6. Show that the parametric equation of all hypocycloid can be written as
ì
a-b
q)
ïï x(t) = (a - b)Cosq + bCos(
b
í
ï y(t) = (a - b)Sinq - bSin( a - b q )
ïî
b
A hypocycloid is a curve traced out by a fixed point P on a circle C of radius b as C rolls on
the inside of a circle with center O and radius a.
If the point, P, starts at (a,0) then Arc PS = Arc As. Thus, ∠ PQS = (a/b)
(a/b)
q
–
q.
q
and ∠ PQT =
The center of the smaller circle, Q, has the coordinates
((a-b)Cos q , (a-b)Sin q ) So, P has the coordinates
X = (a-b)cos q + bcos(∠ PQT) = (a-b)cos q + bcos((a/b)
Y = (a-b)sin q + bsin(∠ PQT) = (a-b)sin q + bsin((a/b)
q
q
–
–
q)
q)
Which simplifies to
ì
a-b
q)
ïï x(t) = (a - b)Cosq + bCos(
b
í
ï y(t) = (a - b)Sinq - bSin( a - b q )
ïî
b
Problem #9- Find arc length and the area under the curve of a cycloid after one rotation
of generating disc.
Solutions: Parametric equations of cycloid are 𝑥 = 𝑟𝜃 − 𝑟𝑠𝑖𝑛𝜃, 𝑦 = 𝑟 − 𝑟𝑐𝑜𝑠𝜃
𝑑𝑥
𝑑𝜃
𝑑𝑦
= 𝑟 − 𝑟𝑐𝑜𝑠𝜃 = 𝑟(1 − 𝑐𝑜𝑠𝜃), 𝑑𝜃 = 𝑟𝑠𝑖𝑛𝜃
2𝜋
L=∫
0
2𝜋
𝑑𝑥 2
𝑑𝑦 2
√( ) + ( ) 𝑑𝜃 = ∫ √𝑟 2 (1 − 𝑐𝑜𝑠𝜃)2 + 𝑟 2 𝑠𝑖𝑛2 𝜃 𝑑𝜃
𝑑𝜃
𝑑𝜃
0
2𝜋
2𝜋
2𝜋
𝑟 ∫ √1 − 2𝑐𝑜𝑠𝜃 + (𝑐𝑜𝑠 2 𝜃 + 𝑠𝑖𝑛2 𝜃) 𝑑𝜃 = 𝑟 ∫ √2 − 2𝑐𝑜𝑠𝜃 𝑑𝜃 = 𝑟 ∫ √2(1 − 𝑐𝑜𝑠𝜃) 𝑑𝜃
0
0
Since 𝑠𝑖𝑛2 𝜃 =
1−𝑐𝑜𝑠2𝜃
,
2
2𝜋
𝜃
𝜃
𝜃
𝑐𝑜𝑠𝜃)(𝑟 − 𝑟𝑐𝑜𝑠𝜃)𝑑𝜃 = 𝑟
2𝜋 𝑐𝑜𝑠2𝜃
2
1) 𝑑𝜃 = 𝑟 2 ∫0 (
𝜃
√2(1 − 𝑐𝑜𝑠𝜃) = √4𝑠𝑖𝑛2 = 2sin ( )
2
2
So L = 2r ∫0 𝑠𝑖𝑛 (2 ) 𝑑𝜃 = 2𝑟 [−2𝑐𝑜𝑠 (2 )]
2𝜋
∫0 𝑟(1 −
0
2𝜋
0
2𝜋
∫0 (𝑐𝑜𝑠 2 𝜃
2
3
− 2𝑐𝑜𝑠𝜃 + 2) = 𝑟 2 [
2𝜋
= 8r
𝑠𝑖𝑛2𝜃
4
𝐴 = ∫0 𝑦𝑑𝑥 =
2𝜋 1+𝑐𝑜𝑠2𝜃
2
− 2𝑐𝑜𝑠𝜃 + 1)𝑑𝜃 = 𝑟 2 ∫0 (
3
− 2𝑠𝑖𝑛𝜃 + 2 𝜃]
2𝜋
0
− 2𝑐𝑜𝑠𝜃 +
3
= 𝑟 2 (2𝜋 ∙ 2) = 3𝜋𝑟 2
Problem #10- Find Surface area of rotating a cycloid after one rotation of generating disc
around x axis.
Solution- General cycloid equation is
ì x(t) = a(q - Sinq )
í
î y(t) = a(1- Cosq )
(a is the radius of the circle)
First, think about∆𝑠, which is the surface of ring with radius = y, width =√(∆𝑥)2 + (∆𝑦)2 .
∆𝑆 ≈ 2𝜋𝑦√(∆𝑥)2 + (∆𝑦)2 Let domain of definition be 0 ≦ 𝜃 < 2𝜋, then Surface area of
rotating a cycloid is going to be
dx 2
2𝜋
𝑑𝑦 2
2𝜋
𝑆 = ∫0 2𝜋𝑦√(𝑑𝜃) + (𝑑𝜃) 𝑑𝜃 = ∫0 2𝜋{𝑎(1 − 𝑐𝑜𝑠𝜃)}√{𝑎(1 − 𝑐𝑜𝑠𝜃)}2 + {𝑎(𝑠𝑖𝑛𝜃)}2 𝑑𝜃
2𝜋
2𝜋
= 2𝜋𝑎2 ∫0 (1 − 𝑐𝑜𝑠𝜃)√(1 − 𝑐𝑜𝑠𝜃)2 + (𝑠𝑖𝑛𝜃)2 𝑑𝜃 = 2𝜋𝑎2 ∫0 (1 − 𝑐𝑜𝑠𝜃)√2(1 − 𝑐𝑜𝑠𝜃) 𝑑𝜃
2
2
2
2𝜋
2𝜋
𝜃
𝜃
𝜃
𝜃
= 2𝜋𝑎2 ∫0 2 (𝑠𝑖𝑛 (2 )) √4 {𝑠𝑖𝑛 (2 )} 𝑑𝜃 (half angle formula)= 8𝜋𝑎2 ∫0 (𝑠𝑖𝑛 (2 )) |𝑠𝑖𝑛 (2 )| 𝑑𝜃
𝜋
𝜃
Let t = 2 , then 𝑑𝜃 = 2𝑑𝑡, and θ: 0 to2π, t = 0 to π. 𝑆 = 8𝜋𝑎2 ∫0 {sin(𝑡)}2 (𝑠𝑖𝑛(𝑡)) (2𝑑𝑡)
𝜋
2
3
𝜃
2
2
2
𝜃
2
= 16𝜋𝑎 ∫ (𝑠𝑖𝑛(𝑡)) 𝑑𝑡 = 32𝜋𝑎 ∫ (𝑠𝑖𝑛(𝑡)) (𝑠𝑖𝑛(𝑡)) 𝑑𝑡 = 32𝜋𝑎 ∫ (1 − 𝑐𝑜𝑠 2 (𝑡))(𝑠𝑖𝑛(𝑡)) 𝑑𝑡
0
2
0
0
𝜃
By substation rule, cos(t)=u, then du= -sin(t)dt. And t: 0 to 2 , u: 1 to 0.
0
𝑑𝑢
1
1
1
= 32𝜋𝑎2 ∫1 (𝑠𝑖𝑛(𝑡)) (1 − 𝑢2 ) (− sin(𝑡)) = −32𝜋𝑎2 ∫0 (𝑢2 − 1) 𝑑𝑢 = −32𝜋𝑎2 [3 𝑢3 − 𝑢] =
0
64
𝜋𝑎2
3
Problem #11- If you double the radius of generating disc, by what factors the Arc length,
the Area and the Surface area changes?
𝛽
𝑑𝑥
𝑑𝑦
When the disc is at radius 1, Arc length is: 𝐿 = ∫𝛼 √(𝑑𝜃)2 + (𝑑𝜃)2 𝑑𝑡
Cycloid equation in parametric form is: 𝑥 = 𝑟(𝜃 − sin(𝜃)) and 𝑦 = 𝑟(1 − cos(𝜃))
When plugging it in and finding the integral, you get L=8r. With a radius of 1, 8(1)=8 and
with the radius of 2, 8(2)=16. In terms of arc length, doubling the radius of the generating
disc will also double the arc length. In traveling a distance of 2π, doubling a radius will
result in doubling the distance traveled for 2π. Therefore, the factor that arc length
changes is 2 times. Another way to see it is that the arc created by the generating disc is
a result of the circle completing 1 rotation. Therefore the distance travelled in 2π is equal
to the circumference of the generating disc. C=2𝜋𝑟 which shows that if you plug in the
value of r=1, then r=2, you can see that the factor for doubling the radius is 2 times.
𝛽
For area: A=∫𝛼 𝑔(𝜃)𝑓 1 (𝜃)𝑑𝑡 using the substitution rule, y= g(t) and x=𝑓 1 (𝑡). Plugging in the
parametric equations for cycloid into the equation for area, you get A=3𝜋𝑟 2 .
With a radius of 1, the area of the cycloid is 3π. If you double the radius to 2 results in
the area to be 12π. In this case, the factor in the area of the cycloid by doubling the
radius is 4 times the difference.
𝛽
𝑑𝑥 2
𝑑𝑦 2
For surface area: S=∫𝛼 2𝜋𝑦√(𝑑𝜃) + (𝑑𝜃) 𝑑𝑡. Using x=rcos(𝜃) and y=rsin(𝜃) and plugging it
in, you get an equation of S=𝟒𝝅𝒓𝟐 (I have other answer
𝟔𝟒
𝝅𝒓𝟐
𝟑
)
With a radius of 1, the surface area of a cycloid is 4π. The surface area when doubling the
radius to 2 is 16𝝅. (I have
𝟐𝟓𝟔
𝝅
𝟑
)Therefore, doubling the radius, the area changes by 4
times as well.
Are you steel solving for cycloid or a circle??????
Fix it
Do the calculus work
Problem #12- Find the area inside both functions. (Graph them)
r1  Sin (2 )
r2  Cos
𝑟1 = sin(2𝜃)
𝑟2 = 𝑐𝑜𝑠(𝜃)
When graphing the equations, we obtain the graph above.
To find the area between the curves, we must first calculate the points of intersection. To
do this we set the equations of the curves equal to each other and solve for the angle:
sin(2𝜃) = cos(𝜃)
cos(𝜃) (2 sin(𝜃) − 1) = 0
sin(2𝜃) − cos(𝜃) = 0
cos(𝜃) = 0
2 sin(𝜃) cos(𝜃) − cos(𝜃) = 0
2 sin(𝜃) − 1 = 0
∴ 𝜃 = 0,
𝜋 7𝜋
, 6
6
Cosx=0 then x =90 , or 270
Sinx=1/2 then x= 30, or 150
All the angles are wrong and you left the integral undone. The integrant in area formula is
incorrect. Fix it
Now we know the angles at which the graphs intersect. Based on these values, we obtain
our limits of integration.
We know that to find the area of a polar function we must use the equation:
𝑏
𝐴 = 1/2 ∫ 𝑟 2 𝑑𝜃
𝑎
We also know that since the graphs are symmetrical across the polar axis we can take the
integral from 0 radians to 𝜋 radians and multiply the integral by 2 to find the total area
between the two curves.
Therefore our equation would look like this:
1 𝜋/6
𝐴 = 2 [ ∫ [cos(𝜃) − sin(2𝜃)]2 𝑑𝜃 ]
2 0
To find the area between two curves you must subtract the innermost function from the
outermost function. This is why we subtract the sin(2𝜃) from the cos(𝜃).
Then we simplify:
1 𝜋/6
𝐴 = 2 [ ∫ [cos(𝜃) − sin(2𝜃)]2 𝑑𝜃 ]
2 0
1 𝜋/6
= 2 [ ∫ [𝑐𝑜𝑠 2 (𝜃) − 2 sin(2𝜃) cos(𝜃) + 𝑠𝑖𝑛2 (2𝜃)]𝑑𝜃 ]
2 0
.
.
.
*This integral is unbelievably long and would take an hour for me just to type it all up so I
will leave that you your mathematical imagination! 
But basically, the area between the two curves would be represented by the integral above.
Problem #13- Find the tangent of the angle between the tangent lines of both graphs at
the point of intersection.
r1  Sin (2 )
r2  Cos
 3
You are missing
2
,
2 Fix it
Find points of intersection (see graph from #12):
  /6
  5 / 6
Find derivative of r1 with respect to theta:
dr1
d

sin( 2 )
d d 
u  2
du / d  2
dr1
 2 cos(u )  2 cos( 2 )
d
Use that to find dy1/dx1:
y1  r1 sin(  )
x1  r1 cos( )
dy1 dr1

sin(  )  r1 cos( )  2 cos( 2 ) sin(  )  sin( 2 ) cos( )
d d
dx1 dr1

cos( )  r1 sin(  )  2 cos( 2 ) cos( )  sin( 2 ) sin(  )
d d
dy1 dx1 / d 2 cos( 2 ) sin(  )  sin( 2 ) cos( )


dx1 dy1 / d 2 cos( 2 ) cos( )  sin( 2 ) sin(  )
Plug in pi/6 to find the slope of r1 at intersection point.
dy1 2 cos( / 3) sin(  / 6)  sin(  / 3) cos( / 6)
5


dx1 2 cos( / 3) cos( / 6)  sin(  / 3) sin(  / 6)
3
Find derivative of r2 with respect to theta:
dr2
d

cos( )   sin(  )
d d
Use that to find dy2/dx2:
y2  r2 sin(  )
x2  r2 cos( )
dy2 dr2

sin(  )  r2 cos( )   sin 2 ( )  cos 2 ( )  cos( 2 )
d d
dx2 dr2

cos( )  r2 sin(  )   sin(  ) cos( )  cos( ) sin(  )  2 sin   cos( )   sin( 2 )
d d
dy2 dx2 / d
cos( 2 )


  cot( 2 )
dx2 dy2 / d  sin( 2 )
Plug in pi/6 to find the slope of r2 at intersection point.
dy2
1
  cot( 2 )   cot( / 3)  
dx2
3
Now we have (at theta = pi/6):
r1 
'
r2 
'
5
3
1
3
dy
dr
m
 m Fix it
dx
d

These are
or
Using alpha = angle of r1's intersection, beta = angle of r2's intersection:
5
)
3
1

  tan 1 ( )  
6
3
  tan 1 (
The tangent of the angle between the tangent lines of both graphs at the point of
intersection is:
tan(tan 1 (
5

)  ) Work this out and simplify it. Fix it
6
3
16) What is the question??
If the graph never overlaps at any point, that would mean the points (0, π/4), and (0, 5π/4)
both would have to have the same slope.
General equation to find slope of parametric curves:
𝑑𝑦
𝑑𝑟
sin 𝜃 + 𝑟 cos 𝜃
𝑑𝑦
𝑑𝜃
𝑑𝜃
=
=
𝑑𝑥
𝑑𝑟
𝑑𝑥
cos 𝜃 − 𝑟 sin 𝜃
𝑑𝜃
𝑑𝜃
But since r= 0 at points (0, π/4) and (0, 5π/4), this equation simplifies to
𝑑𝑦
= tan 𝜃
𝑑𝑥
Plugging in the two points we find that
𝑑𝑦
= tan 𝜃 = tan 𝜋/4 = 1
𝑑𝑥
𝑑𝑦
= tan 𝜃 = tan 5𝜋/4 = 1
𝑑𝑥
The slope at both points is equal to 1, thus the curve never overlaps.
Find the Area bound by inside 𝑟 2 = 2cos(2𝜃) and outside of r=1
Before even thinking about how to solve for the area, we must consider the angles of
intersection of the functions in order to determine an interval for which to integrate over.
This can be done by equating the magnitudes of functions and solving for the angles.
r1= √2cos(2𝜃)
r2= 1
Equating the functions results in
√2 cos(2𝜃) = 1
Squaring both sides of the equation gives us
2 cos(2𝜃) = 1
Dividing both sides of the equation results in
cos(2𝜃) =
If 𝜃 =
𝜋
3
1
2
1
for the cos(𝜃) = 2, then 𝜃 =
𝜋
6
1
for the cos(2𝜃) = 2
Due to the periodic nature of the cos function, we can find the all of the angles where cos
is ½ using the equations
𝜃=
𝜋
+ 𝑛𝜋
6
𝜋
𝜃 = − − 𝑛𝜋
6
𝜋 𝜋 5𝜋
7𝜋
, 𝑎𝑛𝑑 6
6
The angles of intersection are as − 6 , 6 ,
Always check to see the angles give you same r value for both graphs
Now that we know where the two functions intersect each other we can calculate the area
1
by utilizing the integral of ∫ (𝑟)2 𝑑𝜃
2
The Area can then be written as
𝜋
𝜋
7𝜋
7𝜋
6
6
6
6
1 6
1 6
1 6
1 6
𝐴 = ∫ 2 cos(2𝜃) 𝑑𝜃 − ∫ 𝑑𝜃 + ∫ 2 cos(2𝜃) 𝑑𝜃 − ∫ 𝑑𝜃
𝜋
𝜋
5𝜋
2 −
2 −
2
2 5𝜋
These are simple integrals whose integrations are trivial
7𝜋
7𝜋
𝜋
𝜋
1
6 − [𝜃] 6 + [[1 sin 2𝜃] 6 − [𝜃] 6
𝐴 = [ sin 2𝜃] −𝜋
−𝜋
5𝜋
5𝜋
2
2
6
6
6
6
√3 𝜋
𝐴 = 2( + )
2
6
Alternatively, if one is keen on the symmetry present within the functions, the Area can
be expressed as
𝜋
6
𝜋
6
𝐴 = ∫ 2 cos(2𝜃) 𝑑𝜃 − ∫ 𝑑𝜃
−
𝜋
6
−
Which will also yield
√3 𝜋
𝐴 = 2( + )
2
6
𝜋
6
Find the area of the region inside r= 1+sin(ϴ) and outside of r= 3sin(ϴ)
1) First Find the points of intersection
1 + 𝑠𝑖𝑛𝜃 = 3𝑠𝑖𝑛𝜃
1 2𝑠𝑖𝑛𝜃
1
=
∴ 𝑠𝑖𝑛𝜃 =
2
2
2
∴ 𝑡ℎ𝑒 𝑡𝑤𝑜 𝑐𝑢𝑟𝑣𝑒𝑠 𝑖𝑛𝑡𝑒𝑟𝑠𝑒𝑐𝑡 𝑎𝑡
𝜋
5𝜋
𝑎𝑛𝑑 𝑎𝑡
6
6
2) The area of the desired region will be the (Area of the Cardioid 𝑟 = 1 + 𝑠𝑖𝑛𝜃) –
(Area of the Circle 𝑟 = 3 ∗ 𝑠𝑖𝑛𝜃) with limits from
𝜋
6
𝜋
6
6
5𝜋
6
𝜋
𝑡𝑜 6 .
1
6 1
𝐴=∫
(1 + 𝑠𝑖𝑛𝜃)2 𝑑𝜃 − ∫
(3𝑠𝑖𝑛𝜃)2 𝑑𝜃
5𝜋 2
5𝜋 2
𝜋
(Note: because both curves are symmetric of the vertical line ( 2 ), the area curve
can be written as follows:)
3𝜋
3𝜋
6
6
1 2
1 2
𝐴 = 2 [ ∫ (1 + 𝑠𝑖𝑛𝜃)2 𝑑𝜃 − ∫ (3𝑠𝑖𝑛𝜃)2 𝑑𝜃 ]
5𝜋
2
2 5𝜋
3𝜋
2
𝐴=∫
5𝜋
6
3𝜋
2
1 + 2𝑠𝑖𝑛𝜃 + 𝑠𝑖𝑛2 𝜃 𝑑𝜃 − ∫
5𝜋
6
9𝑠𝑖𝑛2 𝜃𝑑𝜃
3𝜋
2
𝐴=∫
5𝜋
6
1
{ 𝑛𝑜𝑡𝑒: 𝑠𝑖𝑛2 𝜃 = (1 − 𝑐𝑜𝑠2𝜃)}
2
1 + 2𝑠𝑖𝑛𝜃 − 8𝑠𝑖𝑛2 𝜃𝑑𝜃
3𝜋
2
𝑦𝑖𝑒𝑙𝑑𝑠
𝐴 = ∫ 4𝑐𝑜𝑠2𝜃 + 2𝑠𝑖𝑛𝜃 − 3 𝑑𝜃 →
5𝜋
6
3𝜋
2
𝐴 = [2𝑠𝑖𝑛2𝜃 − 2𝑐𝑜𝑠𝜃 − 3𝜃] 5𝜋
6
𝜋
𝜋
𝐴 = [0 − 0 − ] − [−√3 − 2 − ]
2
2
𝑦𝑖𝑒𝑙𝑑𝑠
𝐴 = 2 + √3 →
𝐴 = 3.732 𝑢𝑛𝑖𝑡𝑠 2
You have found the area inside of circle and outside of cardioid
𝜋
6
𝜋
6
6
1
6 1
𝐴=∫
(1 + 𝑠𝑖𝑛𝜃)2 𝑑𝜃 − ∫
(3𝑠𝑖𝑛𝜃)2 𝑑𝜃
5𝜋 2
5𝜋 2
Can you do the area outside of circle and inside of the cardioid?
01/24/2014
Chapter 10
Question 19: Find the total area inside of the loop of 𝑟 = 1 + 2𝑠𝑖𝑛𝜃
-Graph:
𝑟 = 1 + 2𝑠𝑖𝑛𝜃
- Loop starts at the origin and ends at the origin so we need points
where 1 +
2𝑠𝑖𝑛𝜃 = 0.
- Solving this equation on the interval [0, 2π]
1 + 2𝑠𝑖𝑛𝜃 = 0
𝑠𝑖𝑛𝜃 =
𝜃=
1
−1
2
7𝜋 11𝜋
, 6
6
11𝜋
So 𝐴 = ∫7𝜋6 ( 1 + 2𝑠𝑖𝑛𝜃)2
2
6
1
11𝜋
𝐴 = 2 ∫7𝜋6 ( 1 + 4𝑠𝑖𝑛𝜃 + 2(1 − 𝑐𝑜𝑠(2𝜃))𝑑𝜃
6
3
2
𝐴 = 𝜋 − √3
21.
2
Find all HORIZONTAL and VERTICAL slopes to the curve r  sin 
Rectangular
Arc
Length
Polar
b
b
L   x'2  y'2 dt
a
b
S  2  f ( x) 1  f ' ( x) dx
2
a
b
S  2  r sin  r  r ' d
2
a
a
b
b
S  2  f ( y) 1  f ' ( y) 2 dy
a
b
L   r 2  r '2 d
L   1  f ' ( x) 2 dx
a
Surface
Area
Parametric
2
S  2  r cos r 2  r '2 d
a
Graph
b
S  2  y(t ) x'2  y'2 dt
a
b
S  2  x(t ) x'2  y'2 dt
a
a. Convert to rectangular
r 2  sin 
Given
r 3  r sin 
Multiply r to both sides
r y
Substitution
r  x2  y2
Pythagorean Theorem
3
3
2 2
y  (x  y )
2
More substitution
2
y 3  x2  y2
Algebra
2
y 3  y2  x2
Algebra
 2  13
 dy
 y  2y
3
 dx  2 x


Derive
dy
2x

dx 2  13
y  2y
3
Somewhat Simplify
b. Find conditions for
dy
 0 (HORIZONTAL)
dx
all values where x=0;
2
3
Substitution
y  y2  0
2
Algebra
y 3  y2
y=-1,y=0,y=1
c. Find conditions for
dy
  (VERTICAL)
dx
1
2 
all values where y 3  2 y =0
3
1
2 3
y  2 y
3
Algebra
1
2
 2 y ( y 3 )
3
4
1
  y3
3
4
 1 3
In other words, for all y    = 0.2311, -0.2311
 3
2
x   y 3  y 2 =0.5685, - 0.5685
d. ANSWER:
Horizontal slopes at
(0,-1)
(0,0)
(0,1)
Vertical slopes at
(0.2311, 0.5685)
(0.2311,- 0.5685)
(-0.2311, 0.5685)
(-0.2311, -0.5685)
Gordon Wang
22)- Find arc length for each and arrange them from highest to lowest
a)
2
2
3
𝑓(𝑥) = 𝑥 3
𝑓 ′ (𝑥) =
4
𝑥
9
1≤𝑥≤8
1
−
3
1
8
4
𝐿 = ∫1 √1 + (9 𝑥 −3 )2 𝑑𝑥
8
𝐿 = ∫1 √1 +
16
2
𝑑𝑥
81𝑥 3
2
𝐿=
8 81𝑥 3 +16
∫1 √
2 𝑑𝑥
81𝑥 3
8 1
𝐿 = ∫1
1
9𝑥 3
8 1
𝐿 = ∫1
1
9𝑥 3
2
√81𝑥 3 + 16 𝑑𝑥
2
√81 𝑥 3 + 1 𝑑𝑥
16
let u =
81
𝐿 = ∫814
81 2
𝑥 3,
16
𝐿=∫
1
𝑑𝑥
8𝑥 3
8
√𝑢
243
16
85
4 8
97
243
16
27
du =
+ 1 𝑑𝑢
√𝑧 𝑑𝑧
85
𝐿=
8 2 3 4
[ 𝑧 2 ]|97
243 3
16
𝐿=
b)
3
852
16
[
729 4
3
−
972
]
16
𝑟 = 𝑒 3𝜃
𝑟′ = 3𝑒
0 ≤ 𝜃 ≤ 2𝜋
3𝜃
2
𝐿 = ∫0 √(𝑒 3𝜃 )2 + (3𝑒 3𝜃 )2 𝑑𝜃
2
𝐿 = ∫0 √10𝑒 6Ө 𝑑𝜃
2
𝐿 = √10 ∫0 𝑒 3𝜃 𝑑𝜃
1
𝐿 = √10[3 𝑒 3𝜃 ]|20
𝐿=
c)
√10 6
𝑒
3
−
√10
)
3
𝑟 = 2𝑎 sin 𝜃
0 ≤ 𝜃 ≤ 2𝜋
1
𝑟 ′ = sin 𝜃 𝑑𝜃
2
2𝜋
𝐿 = ∫0 √(2𝑎 sin 𝜃)2 + (2𝑎 cos 𝜃)2 𝑑𝜃
2𝜋
𝐿 = ∫0 √4𝑎2 sin2 𝜃 + 4𝑎2 cos2 𝜃 𝑑𝜃
2𝜋
𝐿 = ∫0 2𝑎 𝑑𝜃
𝐿 = 2𝑎𝜃|2𝜋
0
𝐿 = 4𝜋𝑎
d)
𝜃
𝑟 = sin2(2 )
0≤𝜃≤𝜋
1
1
therefore 𝑟 = 2 − 2 cos 𝜃
1
𝑟 ′ = 2 sin 𝜃
2
2
𝜋
1
1
1
𝐿 = ∫0 √(2 − 2 cos 𝜃) + (2 sin 𝜃) 𝑑𝜃
𝜋
1
1
𝜋
1
1
1
1
𝐿 = ∫0 √4 − 2 cos 𝜃 + 4 cos2 𝜃 + 4 sin2 𝜃 𝑑𝜃
𝐿 = ∫0 √2 − 2 cos 𝜃 𝑑𝜃
𝜋
𝜃
𝐿 = ∫0 √sin2 (2 ) 𝑑𝜃
𝜋
𝜃
𝐿 = ∫0 sin(2 ) 𝑑𝜃
𝜃
𝐿 = 2 [− cos (2 )] |𝜋0
𝐿=2
e)
𝑥(𝑡) = 2𝑡
−1 ≤ 𝑡 ≤ 2
{
𝑦(𝑡) = 𝑡 2 − 1
𝑥′(𝑡) = 2
{
𝑦′(𝑡) = 2𝑡
2
𝐿 = ∫−1 √22 + (2𝑡)2 𝑑𝑡
2
𝐿 = ∫−1 √4 + 4𝑡 2 𝑑𝑡
2
𝐿 = 2 ∫−1 √1 + 𝑡 2 𝑑𝑡
let t = tanx, dt = sec2xdx and solve as indefinite integral
𝐿 = 2 ∫ √sec 2 𝑥 sec 2 𝑥 𝑑𝑥
𝐿 = 2 ∫ sec 3 𝑥 𝑑𝑥
by parts:
u = secx
v = tanx
du = secxtanxdx
dv = sec2xdx
2
3
2
3
𝐿 = sec 𝜃 tan 𝜃 + ln | sec 𝜃 + tan 𝜃 |
return back to "t" notation and evaluate over limits
2
2
𝐿 = [𝑡√𝑡 2 + 1]|2−1 + ln |√𝑡 2 + 1 + 𝑡| |2−1
3
3
2
2
𝐿 = 3 [2√5 + √2] + 3 ln |
f)
√5+2
|
√2−1
𝑥(𝑡) = 𝑡 sin 𝑡
0≤𝑡≤𝜋
{
𝑦(𝑡) = 𝑡 cos 𝑡
𝑥 ′(𝑡) = 𝑡 cos 𝑡 + sin 𝑡
{ ′(𝑡)
𝑦
= −𝑡 sin 𝑡 + cos 𝑡
𝜋
𝐿 = ∫0 √(𝑡 cos 𝑡 + sin 𝑡)2 + (−𝑡 sin 𝑡 + cos 𝑡)2 𝑑𝑡
expand and simplify to:
𝜋
𝐿 = ∫0 √𝑡 2 + 1𝑑𝑡
let t = tanx, dt = sec2xdx and solve as indefinite integral
𝐿 = ∫ √sec 2 𝑥 sec 2 𝑥 𝑑𝑥
by parts:
u = secx
v = tanx
du = secxtanxdx
dv = sec2xdx
1
1
𝐿 = 2 sec 𝜃 tan 𝜃 + 2 ln | sec 𝜃 + tan 𝜃 |
return back to "t" notation and evaluate over limits
1
𝑡
1
1
] |𝜋 + 2 ln | 2 + 𝑡| |𝜋0
√𝑡 2 +1 0
√𝑡 +1
1
𝜋
1
1
[
] + 2 ln | 2 + 𝜋|
2 √𝜋2 +1
√𝜋 +1
𝐿 = 2[
𝐿=
Highest to lowest (let a = 1 for part c):
b>a>c>e>d>f
HW # 1
YIZHI FANG [LARA]
23.
A）Write out the question and do not use pictures
Question:
Formula:
𝟏
𝟏
− 𝟐 √𝒙
𝟐√𝒙
Thus,we get:
S=
𝟏
𝟏
− 𝟐 √𝒙)𝟐
√𝒙
√𝟏 + (𝟐
𝑨𝒏𝒔𝒘𝒆𝒓 𝟏. 𝟕𝟖𝝅
B
Question:
Formula:
r’=3r=3
S=
𝒄𝒐𝒔(𝜽)𝒊𝒏𝒔𝒕𝒆𝒂𝒅 𝒐𝒇 𝒔𝒊𝒏(𝜽), 𝒔𝒐 𝒕𝒉𝒆 𝒂𝒏𝒔𝒘𝒆𝒓 𝒘𝒊𝒍𝒍 𝒃𝒆
𝟐√𝟏𝟎𝝅(𝒆𝟑𝝅 −𝟔)
𝟑𝟕
WOW!!!!!
e
3
sin  (e 3 ) 2  (3e 3 ) 2 d

e
3
sin  ( 10 e 3 d
much easier integral and you can use Tic Tac To method

e
6
sin  ( 10d This is
C)
Qusetion:
Formula：
r’=
Thus,we got:
Steps:
I did not understand the limits. Why are we taking the limits?????
D)
Question:
Formula:
x’=
y’=
Thus,we get
S=
=
=
I did not understand this
e)
Qustions:
Formula:
x’=
y’=
Thus,we get:
Steps:
S=
I did not understand the limits. Why are we taking the limits?????
F)
Question:
Formula:
y=(x^3)/3
x=(3y)^(1/3) 0<y<8/3
f’(y)=
Thus,we get:
𝟏
=𝟑（3y）^(-2/3)
𝟐
√𝟏 + (𝟏 (𝟑𝒚)−𝟑 )𝟐
S=
𝟑
Steps:
𝑨𝒏𝒔𝒘𝒆𝒓 ∶ (𝝅(𝟏𝟐√𝟏𝟒𝟓 + 𝐬𝐢𝐧−𝟏 (𝟏𝟐))/𝟏𝟖 = 𝟐𝟓. 𝟕𝟖
Erin Linder
24-Find the area outside of r 2 = Sin2q and inside of
r  2 (DO NOT USE GEOMETRY. Do
not use Geometry. The first equation is not defined in the second quadrant.
FIX IT
2𝜋
𝜋/8(𝜋)
1/2 ∫0 4𝑑𝜃 – 1/2 ∫0
1
𝑠𝑖𝑛2 (2𝜃) 𝑑𝜃 =
1
1
4𝜋 − [(4) θ − (16) sin(4θ)] (4𝜋 − 2 𝜃 2 cos 2𝜃 ] 𝜋0 ) =
𝜋
16𝜋
𝜋2
4𝜋 − ( ) − (
) (4𝜋 − )
2
3
2
NUMBER 25 WAS A PDF WHICH I COULD NOT COPY IT
Homework Set #1, Problem 26
Write out the question
(1,π/2)
2)
Type equation here.
Focus
Eccentricity is defined as the ratio of half the distance between its two foci, to the
length of the Semi-major axis.
E = c/a therefore E = c/1+c
0.8(1+c) = c
0.8+0.8c = c
0.8 = 0.2c
c = 4, therefore 1+c is a, thus a = 5(Major Axis)
Due to the definition of an ellipse, 𝑎2 = 𝑏 2 + 𝑐 2 , where b is minor axis
42 + 𝑏 2 = 52
b2 = 25 − 16
b = 3
𝑋2
𝑏2
+
𝒙𝟐
𝟗
+
(𝑦−𝑘)2
𝑎2
(𝒚−𝟒)𝟐
𝟐𝟓
=1
= 𝟏 , This is the rectangular Form
(rcosα)2 (rsinα − 4)2
+
=1
9
25
25 − 25𝑟 2 sin2 𝛼 + 9𝑟 2 sin2 𝛼 − 72𝑟𝑠𝑖𝑛𝛼 + 144 = 225
16𝑟 2 + 72𝑟𝑠𝑖𝑛𝛼 + 56 = 0
8𝑟(𝑟𝑠𝑖𝑛2 𝛼 + 9𝑠𝑖𝑛𝛼) = −7
𝒓𝒔𝒊𝒏𝟐 𝜶 + 𝟗𝒔𝒊𝒏𝜶 = −𝟕/𝟖𝒓, This is Polar Form
Homework prompt #29
Math 1C
Hyun Lee
Q. Find the arc length of the graph x(t) = et - t, y(t) = 4et/2 and 0 <= t <= ln 2.
Type the formula. NO PICTURES
solve this problem we use the equation obtaining arc
To
length. which is followed by
in the interval [0, ln 2]
We must obtain the value and
x’(t)
= et - 1
y’(t) = 2et/2
Therefore,
=
=
=
=
=
Type equation here.