Math 161 - Solutions To Sample Exam 2 Problems Worksheet Page 5
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8Solutions To Sample Exam 2 Problems – Math 161
5
Solution:
dy
Given: When y = 3000,
= 300
Shuttle
dt
dx
dθ
Find:
and
when y = 3000
dt
dt
x
Camera
y
θ
2000
(a) By Pythagorean’s Theorem, we have
dy
dx
2
2
2
2000
+ y
= x
=⇒
2y
= 2x
dt
dt
dx
y · (dy/dt)
=⇒
=
.
dt
x
2
2
2
Also note that 2000
+ y
= x
, so when y = 3000, we have x ≈ 3605.6. Therefore,
dx
3000 · 300
=
≈ 249.6,
dt
3605.6
so we see that the distance between the shuttle and the camera is increasing at a rate of about 249.6 feet
per second.
(b) Using right triangle trigonometry, we have
y
dθ
1
dy
2
tan θ =
=⇒
sec
θ ·
=
·
2000
dt
2000
dt
2
dθ
cos
θ
dy
2
=⇒
=
(after multiplying both sides by cos
θ)
·
dt
2000
dt
2
dθ
(2000/x)
dy
2000
=⇒
=
.
(since cos θ =
)
·
dt
2000
dt
x
Now, since we know from part (a) that x ≈ 3605.6, we have
2
dθ
(2000/3605.6)
=
· 300 ≈ 0.046,
dt
2000
so the angle is increasing at a rate of about 0.046 radians per second.
7. A gardener wants to design a two-piece garden in the shape of a rectangle
Expensive Fence
with a semicircular region extending from one side (see diagram to the
right). He will plant his prized rutabagas in the semicircular part of the
garden, so he will need a special kind of fence costing $5 per foot along the
semicircular border of the garden. He can use a cheaper fence costing $2 per
foot along the other two sides of the yard to keep out those rascally rabbits.
(No fence is needed along the dotted lines). Given that the gardener has
$200 total to spend on fencing material, what should the dimensions be so
that he gets the largest possible combined area? What is this maximum
area?
Solution:
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