a conveyor belt is dropping sand in a conical pile.users.xecu.net/jacobs/calc1/relrates.pdf · a...
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A conveyor belt is dropping sand in a conical pile.
V =1
3πx2h
Let us suppose that the height is always twice as big as thebase radius
h = 2x
V =1
3πx2h =
1
3π
(h
2
)2
y =π
12h3
Suppose the volume of sand is increasing at 8 cubic meters perminute. How fast is the height of the sandpile increasing at thepoint in time when the height is 4 meters.
Suppose the volume of sand is increasing at 8 cubic meters perminute. How fast is the height of the sandpile increasing at thepoint in time when the height is 4 meters.
dV
dt= 8 Find
dh
dt
V =π
12h3
dV
dt=
d
dt
( π
12h3
)
V =π
12h3
dV
dt=
d
dt
( π
12h3
)dV
dt=
π
12
d
dt
(h3
)
V =π
12h3
dV
dt=
d
dt
( π
12h3
)dV
dt=
π
12
d
dt
(h3
)dV
dt=
π
12· 3h2 dh
dt=
π
4h2 dh
dt
V =π
12h3
dV
dt=
d
dt
( π
12h3
)dV
dt=
π
12
d
dt
(h3
)dV
dt=
π
12· 3h2 dh
dt=
π
4h2 dh
dt
dh
dt=
4
πh2
dV
dt
If dVdt = 8 m3/min and h = 4 meters then,
dh
dt=
4
πh2
dV
dt=
4
π · 16 m2· 8 m3
min=
2
π
m
min
Hurricane
Hurricane
Suppose that the closest point of approach to Daytona Beachis 300 miles and that the storm is moving along the coast at 10miles per hour.
At the point in time when the storm is 500 miles from DaytonaBeach, calculate the rate at which the distance between thestorm and Daytona Beach is changing.
Piston.
Let P be the pressure (in atmospheres)Let V be the volume (in liters)
PV = nRT
If n, R and T are constant then
PV = (constant)
Let us suppose that the constant is 4 liter-atmospheres
PV = 4
PV = 4
If the piston is being compressed and the pressure is increasingat 1
8 atmospheres per minute, find the rate at which the volumeis changing at the point in time when P = 1 atmosphere andV = 4 liters.
PV = 4
If the piston is being compressed and the pressure is increasingat 1
8 atmospheres per minute, find the rate at which the volumeis changing at the point in time when P = 1 atmosphere andV = 4 liters.
dP
dt=
1
8Find
dV
dt
PV = 4
d
dt(PV ) =
d
dt(4)
PV = 4
d
dt(PV ) =
d
dt(4)
PdV
dt+ V
dP
dt= 0
PV = 4
d
dt(PV ) =
d
dt(4)
PdV
dt+ V
dP
dt= 0
dV
dt= −V
P
dP
dt
PV = 4
d
dt(PV ) =
d
dt(4)
PdV
dt+ V
dP
dt= 0
dV
dt= −V
P
dP
dt
At V = 4 liters and P = 1 atmosphere, dVdt will be:
dV
dt= −4 liters
1 atm· 18
atm
min= −1
2
liters
min
An observer is tracking a plane flying at 5,000 feet. The planeflies directly over the observer on a horizontal path at a fixedrate of speed of 520 feet per second. As time goes by, thedistance from the plane to the observer changes.
What is the rate of change in the distance between the observerand the plane at the point in time when the plane has traveled12,000 feet from the point above the observer.
Let x(t) be the horizontal distance traveled by the plane.Let y(t) be the distance from the observer to the plane.Given dx
dt = 520 feet/sec. Find dydt
(5, 000)2 + x2 = y2
(5, 000)2 + x2 = y2
d
dt
((5, 000)2 + x2
)=
d
dt
(y2)
(5, 000)2 + x2 = y2
d
dt
((5, 000)2 + x2
)=
d
dt
(y2)
0 + 2xdx
dt= 2y
dy
dt
(5, 000)2 + x2 = y2
d
dt
((5, 000)2 + x2
)=
d
dt
(y2)
0 + 2xdx
dt= 2y
dy
dt
dy
dt=
x
y
dx
dt
(5, 000)2 + x2 = y2
d
dt
((5, 000)2 + x2
)=
d
dt
(y2)
0 + 2xdx
dt= 2y
dy
dt
dy
dt=
x
y
dx
dt
We are given that dxdt = 520. Find dy
dt when x = 12, 000.
dy
dt=
12, 000
13, 000
feet
feet· 520 ft
sec= 480
ft
sec
A ladder that is 10 feet long is leaning against a wall. Thetop of the ladder is slipping down the wall and the base of theladder is sliding away from the wall at 1
3 ft/min. Find how fastthe top of the ladder is slipping down the wall at the point intime when the bottom of the ladder is 6 feet away from thewall.
Given that dxdt = 1
3 , finddydt
x2 + y2 = 102
x2 + y2 = 102
d
dt
(x2 + y2
)=
d
dt
(102
)
x2 + y2 = 102
d
dt
(x2 + y2
)=
d
dt
(102
)2x
dx
dt+ 2y
dy
dt= 0
x2 + y2 = 102
d
dt
(x2 + y2
)=
d
dt
(102
)2x
dx
dt+ 2y
dy
dt= 0
dy
dt= −x
y
dx
dt= −6
8· 13= −1
4
ft
min
A man 6 ft. tall is directly under a light which is 16 feet high.He walks to the right at a steady speed of 2 feet per second.As time goes by the length of his shadow increases. How fastis the length of the shadow increasing?
Let x be the distance from the base of the light to the man.We are given that dx
dt = 2.
Let s be the length of the shadow. Find dsdt
Use similar triangles
16
x+ s=
6
s
16
x+ s=
6
s
16s = 6(x+ s)
16s = 6x+ 6s
16
x+ s=
6
s
16s = 6(x+ s)
16s = 6x+ 6s
10s = 6x
s =6
10x =
3
5x
s =3
5x
ds
dt=
3
5
dx
dt=
3
5· 2 ft
sec= 1.2
ft
sec
A man in a hot-air balloon is ascending at the rate of .002 milesper second. How fast is the distance from the balloon to thehorizon increasing when the balloon is 1
5 of a mile high.
Assume that the earth is a ball of radius 4,000 miles.
Offshore lighthouse.
The lighthouse is 1 mile offshore. It is revolving at ω radiansper minute. The beam of light from the lighthouse is movingalong the shoreline as the light rotates. When x = 1
2 , the lightis moving at 5 miles/minute. Calculate ω
A small ferris wheel has a radius of 10 meters and rotates atthe rate of 1 revolution every 2 minutes. Let (x, y) be thecoordinates of a point on the wheel. Find dy
dt at the point intime when the point is 16 meters above the ground.