1
151 WebCalc Fall 2002-copyright Joe Kahlig
In Class Questions
MATH 151-Fall 02
October 15
1. At noon Ship A leaves a port traveling north at 35km/hr. Ship B leaves the same port
traveling east at 1pm at 25 km/hr. At what rate is the distance between them changing at
3pm?
Ship A
dy
we know that dx
dt = 35km/hr and dt = 25km/hr. At 3pm
x = 70 and y = 50. Using the formula z 2 = (x + 35)2 + y 2
we find that z = 116.297km. Now take a derivative.
x
z
dy
dx
2z dz
dt = 2(x + 35) dt + 2y dt
35
plug in what we know and solve for
Answer:
dz
dt
Ship B
dz
dt .
y
= 42.3485km/hr
2. A water tank has the shape of an inverted right circular cone with an altitude of 12ft and
a base radius of 6ft. If water is being pumped into the tank at a rate of 10 gal/min(1.337
f t3 /min, approximate the rate at which the water level is rising when the water is 3 ft deep.
6
dv
dt
We know
= 1.337, h = 3. We also
have the relationship between h and r by using similar triangles.
r
6
=
or that r = 12 h
h
12
2
1
V = 13 πr 2 h = 13 π 12 h h = 12
πh3
dV
dh
1
= πh2 ∗
now plug in the info and solve for dh
dt
dt
4
dt
dh
Answer: dt = .1891f t/min
r
12
h
3. A swimming pool is 20 ft wide and 40ft long, 3ft deep at the shallow end and 9 feet deep at
its deepest point. If the pool is being filled at a rate of 1.2 cu. ft/min, how fast is the water
level rising when the depth at the deepest point is 5ft?
We know that dV
dt = 1.2 and the area of a
trapezoid is .5 ∗ (b1 + b2 )h, where b1 and b2 are
the bases of the trapezoid.
3
x1
x2
6
h
V = 20 ∗ 12 (b1 + b2 )h = 10(b1 + b2 )h
Now b1 = 12 and b2 = x1 + 12 + x2 .
Using similar triangles(see bottom pictures),
we see that xh1 = 66 or x1 = h and
8
that xh2 = 16
6 or x2 = 3 h. This gives
V = 10(12 + h + 12 +
8
3 h)h
= 240h +
dh
=
+ 220
3 h dt now plug
and solve for dh
dt .
Answer: dh
dt = .001978f t/min
dV
dt
240 dh
dt
110 2
3 h
in the info
6
12
16
6
16
6
x1
h
x2
h
6
2
151 WebCalc Fall 2002-copyright Joe Kahlig
4. A revolving beacon in a light house makes one revolution every 15 sec. The beacon is 200ft
from the nearest point P on a straight shoreline. Find the rate at which a ray from the light
moves along the shore at a point 400ft from P.
y
x
theta
light house
200
dθ
rev
rad
rad
=4
∗ 2π
= 8π
and x = 400.
dt
min
rev
min
x
The equation that we use is tan(θ) =
200
Taking a derivative gives:
dθ
1 dx
sec2 (θ)
=
dt
200 dt
y
Since sec(θ) =
and 4002 + 2002 = y 2 (since x = 400) we get
200
y2
4002 + 2002
=
2002
2002
plugging back into the derivative gives
sec2 (θ) =
4002 + 2002
1 dx
8π =
2
200
200 dt
dx
Answer: dt = 25132.7412f t/min