rates of change

[mlane]

PROBLEM: A gas station stand at the intersection of a north-south road and an east-west rd. A police car is traveling toward the gas station from the east, chasing a stolen truck which is traveling north away from the gas station. The speed of the police car is 100mph at the moment it is 3 miles from the gas station. At the same time, the truck is 4 miles from the gas station going 80mph.
At this moment:
a) Is the distance increasing or decreasing? How fast?
b) How does your answer change if the truck is going 70mph?

Problem 2
Sand falls from a hopper at a rate of .1 cubic meters per hour and forms a conical pile beneath. If the side of the cone makes an angle of pi/6 radians with the vertical, find the rate at which the height of the con increases. At what rate does the radius of the base increas? Give both answeres in terms of h, the geight of the pile in meters.

I am suppose to solve these using liebnitz notation for figuring rates of change but cannot get the correct answer.
I have several other problems using similar concepts but somewhere I am missing something I could use some step by step instruction.

 

[soroban]

Hello, mlane!

Here's the first one . . .

A gas station stand at the intersection of a north-south road and an east-west road.
A police car is traveling toward the gas station from the east, chasing a stolen truck
which is traveling north away from the gas station.
The speed of the police car is 100mph at the moment it is 3 miles from the gas station.
At the same time, the truck is 4 miles from the gas station going 80mph.

At this moment:
a) Is the distance increasing or decreasing? How fast?
b) How does your answer change if the truck is going 70mph?

. . . . .T
. . . . . *                  The police car is at P,
. . . . . | \             x miles from the intersection.
. . . . . | . \
. . . . y | . . \R           The truck is at T,
. . . . . | . . . \       y miles from the intersection.
. . . . . | . . . . \
. . . - - +-----------* -    Let R = distance PT.
. . . . . | . .x . . .P

From Pythagorus, we have: .R² .= .x² + y²

Differentiate with respect to time: .2R(dR/dt) .= .2x(dx/dt) + 2y(dy/dt)

. . and we have: .dR/dt .= .(x/R)(dx/dt) + (y/R)(dy/dt)

We are told that: .dx/dt = -100 mph . (The police car's distance is decreasing.)
. . . . . . and that: .dy/dt .= .80 mph . . (The truck's distance is increasing.)

At a particular moment: .x = 3, y = 4 .---> . R = 5

Plug in those values:

. . dR/dt . = . (3/5)(-100) + (4/5)(80) . = . -60 + 64 . = . +4

(a) At that moment, the distance is <u>increasing</u> at 4 mph.

Note: Since the police car is going faster, it will overtake the truck eventually.
. . . But at that moment, the straight-line distance is increasing.
. . . It is as if the police is sighting the truck with a radar gun,
. . . aiming behind the gas station . . . or maybe with a sniperscope?

 

[mlane]

thanks for the help

Thanks for your reply
I figured out my mistake earlier today I was not using a neg for the police car. This problem had me especially confused because I thought I knew what I was doing until the neg messed me up. Now that I know my mistake it seems silly to be having so much trouble with this one.

 

[Gene]

2.) The Radius of the base is h/sqrt(3)
The Volume is (1/3)*PI*R²*h =
PI*h²/9
dV/dh=2*PI*h/9
dV/dt=.1
dh/dt=(dV/dt)/(dV/dh)=
.9/(2*PI*h)

dR/dh=1/sqrt(3)
dR/dt=(dR/dh)*(dh/dt) =
(1/sqrt(3))*(.9/(2*PI*h)

Not a clue as to what liebnitz notation is.