1. Find the length of the curve y =
Z x q
√ t − 1 dt
1
1 ≤ x ≤ 16
Solution
We have: y
0
= q
√ t − 1 and
L =
=
=
Z
16 q
1 + ( y 0 ) 2 dt =
1
Z
16 q
√ t dt =
Z
16 t
Z
16 q
1
1 / 4 dt
1 +
=
1 1
√ t −
4
5 t
5 / 4
¯
16
¯
¯
¯
1
1
= dt
4
5
(32 − 1)
124
5
2. A uniform disk of radius r is to be cut by a line so that the center of mass of the smaller piece lies halfway along the radius. How close to the center of the disk should the cut be made?
Solution
To appear later...
3. Consider a rectangle R of width 6 and height 3 with sides parallel to the coordinate axis whose left bottom corner is at ( − 2 , 0). A rectangular hole of width 3 and height
1 is made in R so that the left bottom corner of the hole is in (0 , 1). The hole sides are also parallel to the coordinate axis. Find the centroid of the remaining part of
R .
Solution
We have the following setup (the shadowed part of the rectangle is taken off):
3
6 y pp pp pp pp pp pp pp
2
1
-2 -1 0 1
1
2
-
3 4 x

It follows from the symmetry that the y -coordinate of the centroid is in the middle of the area, that is y = 1 .
5.
Concerning the x -coordinate, the part of the region to the left of the dotted line is a rectangle, so its centroid is at ( − 1 .
5 , 1 .
5). Due to the symmetry, the centroid of the region which is to the right of the dotted line is in its center of symmetry, that is, in the point (1 .
5 , 1 .
5). Furthermore, the area of the right part is 4 times larger than the area of the left part. The same holds for their masses. To compute x we can think of the area as point masses m and 4 m located at points x = − 1 .
5 and x = 1 .
5, respectively. So, the centroid of this mass system is at x = 0 .
9.
4. A curve is defined by the parametric equations x =
Z t cos u
1 u du y =
Z t sin u
1 u du
Find the length of the arc of the curve from the origin to the nearest point where there is a vertical tangent line.
Solution
A vertical tangent line is specified by the following conditions: dx
= 0 dt dy dt
= 0 t
Since dx/dt = cos t/t and dy/dt = sin t/t , we have: dx/dt = cos t/t = 0 implies
=
π
2 k for k = 1 , 2 , . . .
. The nearest point to t = 1 is t =
π
2
. Therefore,
L =
Z
π/ 2
1 v u à u dx t dt
!
2
+
à dy
!
2 dt dt =
Z
π/ 2 s
µ cos t ¶
2
1 t
+
µ sin t ¶
2 t dt
Z
π/ 2 s cos 2 t + sin
2 t
=
= ln
1
π
2 t 2 dt =
Z
π/ 2
1 dt t
= ln t |
π/ 2
1
= ln
π
2
− ln 1
5. Show that the curve r = sin θ tan θ has the line x = 1 as a vertical asymptote. Show also that the curve lies entirely within the vertical strip 0 ≤ x ≤ 1.
Solution
We have x = r cos θ = sin
2
θ . Since 0 ≤ sin
2
θ ≤ 1, we get 0 ≤ x ≤ 1.
Furthermore, for the vertical asymptote we must have dy/dx → ∞ . Taking into account that y = r sin θ = sin
3
θ/ cos θ dy dx
= dy dθ dx dθ
=
³ sin
3 cos θ
θ
´
0
(sin
2
θ ) 0
=
3 sin
2
θ cos
2 cos 2
θ +sin
4
θ
θ
2 sin θ cos θ
=
3 sin θ cos θ + sin
3
θ cos 3 θ
So, dy/dx → ∞ as θ → π/ 2. In this case x = sin
2
θ → 1, hence x = 1 is a vertical asymptote.
2

6. Derive a formula for the area of the surface generated by rotating a polar curve r = f ( θ ), a ≤ θ ≤ b , (where f
0 is continuous and 0 ≤ a < b ≤ π ), about the line
θ = π/ 2. Use this formula to compute the surface area produced by rotating the curve r
2
= cos 2 θ , 0 ≤ θ ≤ π , about the line θ = π/ 2.
Solution
A general formula for rotating a curve y = f ( x ) about the x -axis is
S =
Z b
2 πx ds a
For parametric curve, in particular a polar one, we have ds = v u à u t dx !
2 dθ
+
à dy !
2 dθ dθ = v u u t r 2 +
à dr !
2 dθ dθ
So, we obtain the following formula:
S = 2 π
Z b a v u r cos θ u t r 2 +
à dr !
2 dθ dθ
In our case the curve is symmetric about the x -axis and y -axis. Moreover, it starts from the polar pole (point (0 , 0)) at θ = 0 and reaches the point (1 , π/ 4) being above the x -axis. After this is goes from (1 , π/ 4) back to (0 , 0) as θ varies between π/ 4 and π/ 2. After this the curve moves symmetrically about the y -axis, so its shape reminds the ∞ symbol. Hence, the surface are consists of 2 equal parts caused by rotating of the part of the curve corresponding to 0 ≤ θ ≤ π/ 4. Taking all this into account,
S = 2 π
Z b a v u r cos θ u t r 2 +
à dr !
2 dθ dθ = 4 π
Z
π/ 4
√ cos θ cos θ s cos 2 θ +
0 sin
2
2 θ cos 2 θ dθ
= 4 π
Z
π/ 4
0
√ cos θ cos θ s cos 2 2 θ + sin
2 cos 2 θ
2 θ
Z
π/ 4
= 4 π
= 2 π
√ 0
2 cos θ dθ = 4 π sin θ |
π/ 4
0 dθ = 4 π
Z
π/ 4
0
√ cos θ cos θ s
1 cos 2 θ dθ
3