Homework 5 - Math 1451-008 (Howle)
Due Friday 4/20/2012 in class
Name:
R Number:
This cover sheet must be attached as the top page of your homework.
See homework requirements in the syllabus.
1. A Norman window has the shape of a semicircle atop a rectangle so that the diameter of the
semicircle is equal to the width of the rectangle.
What is the area of the largest possible
Norman window with a perimeter of 50 feet?
v(0) = 96 f t/s.
a(t) = −32 f t/s2 , deter-
2. A ball it thrown directly upward from ground level with an initial velocity
Assuming that the ball's only acceleration is that due to gravity, i.e,
mine the maximum height reached by the ball and determine the time it takes to return to
ground level.
3. Given the following integral:
ˆ
3
4x2 + 2 dx
1
(a) Estimate the value of the integral using a Riemann sum
n=4
Sn =
Pn
k=1 f (a + k∆x)∆x with
and using right endpoints.
(b) Evaluate the denite integral
using the denition.
Use equal width subintervals and
evaluate the function at right endpoints. So in this case, you are evaluating
lim
n→∞
where
∆x =
n
X
k=1
b−a
n .
1
f (a + k∆x) ∆x,
1.
A Norman window has the shape of a semicircle atop a rectangle so that the
diameter of the semicircle is equal to the width of the rectangle. What is the area
of the largest possible Norman window with a perimeter of 50 feet?
Solution: Let x be the width of the rectangle (then the radius of the semicircle is x2 ) and y
be the height of the rectangle. The perimeter of the window is then the permeter of half of a
circle plus
x
plus
2y :
P
=
1
πx + x + 2y = 50.
2
The area of the window is the area of the semicircle plus the area of the rectangle:
2
x
1
A= π
2
2
We need to eliminate one variable from
constraint for that. Solving for
y,
A
+ xy.
so we can dierentiate. We can use our perimeter
we have:
1
2y = 50 − πx − x
2
1
1
y = 25 − πx − x.
4
2
Therefore, we can write the area formula as a function of a single variable,
1
x 2
1
1
π
+ x 25 − πx − x
2
2
4
2
1 2 1 2
1 2
πx + 25x − πx − x
=
8
4
2
1 2 1 2
= − πx − x + 25x
8
2
x.
A =
Dierentiating, we can nd critical numbers:
dA
dx
1
x 1+ π
4
1
= − πx − x + 25
4
1
0 = − πx − x + 25
4
= 25
25
≈ 14.0025
1 + 0.25π
x =
Using the perimeter constraint again, we can solve for the corresponding
y ≈ 7.0012.
Therefore, the area of the window is
y
value, giving
A = 18 πx2 + xy ≈ 175.031.
We can check that we have found a maximum with the 2nd derivative test. Note that
− π4 − 1,
which is always negative. Therefore, the graph of
the critical point we found must be a relative maximum.
2
A(x)
d2 A
dx2
=
is always concave down, and
2.
A ball it thrown directly upward from ground level with an initial velocity v(0) =
96 f t/s. Assuming that the ball's only acceleration is that due to gravity, i.e,
a(t) = −32 f t/s2 , determine the maximum height reached by the ball and determine
the time it takes to return to ground level.
Solution: The velocity v(t) is the integral of a(t) and the position s(t) is the integral of v(t).
We will use the given information about
v(0)
and starting position to solve for the constants
from the indenite integrals. Since the ball starts from ground level,
s(0) = 0.
a(t) = −32
ˆ
v(t) =
a(t) dt
ˆ
= − 32 dt
= −32t + C1
v(0) = 96
96 = −32(0) + C1 ⇒ C1 = 96
v(t) = −32t + 96
ˆ
s(t) =
v(t) dt
ˆ
=
−32t + 96 dt
= −16t2 + 96t + C2
s(0) = 0 ⇒ C2 = 0
s(t) = −16t2 + 96t
Now we have formulas for
v(t)
and
s(t).
The ball is at maximum height when
v(t) = 0,
i.e.,
0 = −32t + 96
32t = 96
t = 3 sec
It has returned to ground level when
s(t) = 0:
0 = −16t2 + 96t
= t (−t + 6)
t = 0, 6 sec
The ball started from ground level at
and returned to ground level at
−16(3)2 + 96(3) = 144 f t
t = 6
t = 0,
reached it's max height at
seconds.
t = 3
seconds,
So it's max height (at 3 sec) was
s(3) =
and it took 3 more seconds to return to ground level (total travel
time of 6 seconds - 3 up and 3 down).
3
3.
Given the following integral:
ˆ
3
4x2 + 2 dx
1
(a)
Estimate the value of the integral using a Riemann sum Sn
k∆x)∆x
with
n
=
4
and
using
right
=
Pn
k=1 f (a
+
endpoints.
With 4 equal width rectangles and right endpoints for function evaluation, we have
2
1
∆x = b−a
n = 4 = 2 . Then:
xk = a + k∆x
f (xk )
1
3
1 + 1( 2 ) = 2
11
1 + 2( 21 ) = 2
18
1 + 3( 12 ) = 25
27
1
1 + 4( 2 ) = 3 = b
38
So the estimate using the sum of 4 rectangles is
ˆ
4
X
3
4x2 + 2 dx ≈
1
f (1 + k∆x) ∆x
k=1
1
2
= (11 + 18 + 27 + 38)
= 47
(b)
Evaluate the denite integral using the denition . Use equal width subintervals and evaluate the function at right endpoints. So in this case, you are
evaluating
n
X
lim
n→∞
where
∆x =
b−a
n
f (a + k∆x) ∆x,
k=1
.
Using the denition of the denite integral (with equal width subintervals and function
evaluations at right endpoints, we have that
ˆ
3
2
4x + 2 dx =
1
lim
n→∞
n
X
k=1
n
X
2
n
2k
= lim
f 1+
n→∞
n
k=1
2
n
=
=
=
lim
n→∞
lim
n→∞
lim
n→∞
lim
n→∞
k=1
n
X
n
X
k=1
n
X
k=1
n
X
k=1
=
2
n and
f (a + k∆x) ∆x
2
n
=
b−a
n
∆x =
f 1+k
2k
4 1+
n
!
2
4k 4k 2
4 1+
+ 2
n
n
16k 16k 2
6+
+ 2
n
n
4
2
n
+2
!
!
!
+2
2
n
2
n
=
=
=
lim
n→∞
lim
n
X
k=1
n
X
12 32k 32k 2
+ 2 + 3
n
n
n
!
n
n
32k X
32k 2
12 X
+
+
n
n2
n3
k=1
k=1
k=1
n→∞
!
n
n
n
12 X
32 X
32 X
1+ 2
k+ 3
k2
n k=1
n k=1
n k=1
lim
n→∞
!
using summation formulas for the three sums, we get:
32 n(n + 1)(2n + 1)
12
32 n(n + 1)
+ 3
(n) + 2
n→∞
n
n
2
n
6
!
3
2
2
16n + 16n 32n + 48n + 16n
+
lim 12 +
n→∞
n2
3n3
16 32 16
16
lim 12 + 16 +
+
+
+ 2
n→∞
n
3
n
3n
32
12 + 16 +
3
116
3
=
=
=
=
=
lim
If you do the denite integral the easy way, i.e., with the 1st Fundamental Theorem of
Calculus, you will see that this is the correct value.
5