Calc Problem

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  • #1
playboy

Homework Statement




Snow began to fall during the morning of February 2 and continued steadily into the afternoon.
At noon a snowplow began removing snow from a road at a constant rate. The plow traveled 6 km from noon to 1 P.M. but only 3 km from 1 P.M. to 2 P.M. When did the snow begin to fall?



Homework Equations




t = time

x(t) = distance

dx\dt = speed of the plow

b - the number of hours before noon that it began to snow



The Attempt at a Solution




I know that the speed of the plow is (6 + 3) /2 hours = 4.5km\hour


I am trying to figure out an expression for the height of the snow at time t and then use the given ingormation that the rate of removal R is constant, and thus, find the time it began to snow

Can somebody help me out please?
 

Answers and Replies

  • #2
D H
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The average speed of the plow isn't going to help much here, as the plow's speed is changing (and doing so constantly). This is a calculus problem.

The answer to the two questions below should help you solve the problem at hand:

How does the plow's instantaneous speed relate to the constant rate at which the plow removes snow from the road?

How would you interpret the phrase "Snow began to fall during the morning of February 2 and continued steadily into the afternoon" mathematically?
 
  • #3
playboy
I am more lost than ever.

How does the plow's instantaneous speed relate to the constant rate at which the plow removes snow from the road?

I dont think the instantaneous speed has an anything to do with the problem correct?
The plow could have been going really fast at one point and perphaps at rest at another point?


How would you interpret the phrase "Snow began to fall during the morning of February 2 and continued steadily into the afternoon" mathematically?

This simply means that the snow fell at a constant rate from morning into afternoon,
we can call this rate D.
 
  • #4
D H
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The instantaneous speed has everything to do with the problem. You are given two distances, both of which are definite integrals of the instantaneous speed.

Formulate these integrals and you will be close to having solved the problem. You will need to have an expression for the instantaneous speed to formulate these integrals.
 
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  • #5
playboy
Distance 1, 6km, is on the interval [0,1]

Distance 2, 3 km, is on the interval [1,2]

int(0,1) 6 dt + int(1,2) 3 dt = 9

How does that sound?
 
  • #6
D H
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Not very good. You are integrating distance. You need to integrate velocity,

[tex]\int_{12:00}^{13:00} v(t) dt = 6\text{\, km}[/tex]

[tex]\int_{13:00}^{14:00} v(t) dt = 3\text{\, km}[/tex]

Now you just have to come up with an expression for [itex]v(t)[/itex]. Hint: You know that the removal rate is a constant. How does this relate to velocity and the amount of snow on the ground?
 
  • #7
playboy
Hint: You know that the removal rate is a constant. How does this relate to velocity and the amount of snow on the ground?

The removal rate is Constant.

Therefore, the velocity of the plow is the same as the velocity of snowfall?
 
  • #8
D H
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The removal rate is constant, but the velocity is not. Snow is falling all afternoon long, and it had a head start on the snowplow.
 
  • #9
D H
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How does the height of the accumulated snow on the road affect the velocity at which the plow can move?
 
  • #10
D H
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Another hint: What units would you use to reflect the constant rate at which snow is falling on the road and the constant rate at which the plow is removing snow from the road?
 
  • #11
playboy
The removal rate is constant, but the velocity is not. Snow is falling all afternoon long, and it had a head start on the snowplow.

I'm confused between "removal rate" and "velocity" ... I thought they were equivalent.


How does the height of the accumulated snow on the road affect the velocity at which the plow can move?

The higher the hight, the smaller the velocity.


Another hint: What units would you use to reflect the constant rate at which snow is falling on the road and the constant rate at which the plow is removing snow from the road?

Units for snow falling on road --> [tex]m^2/hour[/tex] or perhaps [tex]km^2/hour[/tex]
Units for snow removal --> [tex]m^3/hour[/tex] or perhaps [tex]km^3/hour[/tex]
 
  • #12
D H
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The higher the hight, the smaller the velocity.

Good. Now you're getting somewhere.

Units for snow falling on road --> [tex]m^2/hour[/tex] or perhaps [tex]km^2/hour[/tex]

Try again. Think of how the weatherman expresses snowfall rates.

Units for snow removal --> [tex]m^3/hour[/tex] or perhaps [tex]km^3/hour[/tex]

Good. Obviously, the snow removal rate cannot equal the snowfall rate since they have different units.

Now factor in the time-varying height of the snow and the fixed width of the plow to determine the velocity of the plow.


Edited to add:
Kudos on attempting to learn LaTeX. Nothing beats it (and certainly not ASCII math) for making math look nice.
 
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  • #13
playboy
Units for snow falling on road ---> [tex]m/hour[/tex]

Yeah, Latex is awesome, Something I really have to master
 
  • #14
D H
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Excellent. Now for the velocity of the plow ...
 
  • #15
D H
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Have to take a break. Home duties are calling. You're almost there. You should get to a point where you have two expressions involving a lot of unknowns, including the two rates, the width of the plow, and time at which it started snow. A simple algebraic manipulation will eliminate all but one (the time). Solve for the time, and voila, you have the answer.
 
  • #16
playboy
Excellent. Now for the velocity of the plow ...

:)

Have to take a break. Home duties are calling. You're almost there. You should get to a point where you have two expressions involving a lot of unknowns, including the two rates, the width of the plow, and time at which it started snow. A simple algebraic manipulation will eliminate all but one (the time). Solve for the time, and voila, you have the answer.

Thank you so much for your time! I really appreciate it,

I am still working on this problem..
 
  • #17
playboy
1. Homework Statement


Snow began to fall during the morning of February 2 and continued steadily into the afternoon.
At noon a snowplow began removing snow from a road at a constant rate. The plow traveled 6 km from noon to 1 P.M. but only 3 km from 1 P.M. to 2 P.M. When did the snow begin to fall?



So here is what I have so far:


[tex]\int_{12:00}^{13:00} v(t) dt = 6\text{\, km}[/tex]


[tex]\int_{13:00}^{14:00} v(t) dt = 3\text{\, km}[/tex]


RATE OF SNOW REMOVAL is [tex]km^3/hour[/tex]/hour


RATE OF SNOW FALL is [tex]km/hour[/tex]/hour


Time of snowfall is [tex]b[/tex] hours BEFORE noon.


INSTANTANEOUS veolocity of the plow, [tex]v(t)[/tex] ... i still cannot find.


Now factor in the time-varying height of the snow and the fixed width of the plow to determine the velocity of the plow.


Excellent. Now for the velocity of the plow ...


Is the "time-varying height of the snow" the "rate of snowfall" [tex]km\hour[/tex] ? The snow did have a head start to fall, so should something be added?

I suppose the width of the plow could just be a constant, which I will call [tex]w[/tex]
 
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  • #18
D H
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Good. In a short span of time [itex]\Delta t[/itex], what is the volume of snow [itex]\Delta v_{\text{snow}}[/itex] removed by the plow? (Assume a constant velocity [itex]v(t)[/itex] over this short span of time.) How does this relate to the known rate at which the plow removes snow, and what happens in the limit [itex]\Delta t \to 0[/itex]?
 
  • #19
playboy
Good. In a short span of time [itex]\Delta t[/itex], what is the volume of snow [itex]\Delta v_{\text{snow}}[/itex] removed by the plow? (Assume a constant velocity [itex]v(t)[/itex] over this short span of time.) How does this relate to the known rate at which the plow removes snow, and what happens in the limit [itex]\Delta t \to 0[/itex]?

Im very unsure of this...I thought it...and all I can come up with is that the plow can remove only [tex]x*km^3[/tex] of snow...

I don't really know how to fit time into it unless i turn it into a rate.
 
  • #20
D H
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Between time [itex]t[/itex] and [itex]t+\Delta t[/itex], the plow will have moved about [itex]v(t)\Delta t[/itex] meters down the road. Now suppose the snow height at this time is [itex]h(t)[/itex]. Since the plow has a width [itex]w[/itex], it will have removed a volume of snow [itex]\Delta V_{\text{snow}} = w\; h(t)v(t)\Delta t[/itex].

How can you relate this to the plow's snow removal rate, call it [itex]\dot V_{\text{plow}}[/itex]?
 
  • #21
playboy
Between time [itex]t[/itex] and [itex]t+\Delta t[/itex], the plow will have moved about [itex]v(t)\Delta t[/itex] meters down the road. Now suppose the snow height at this time is [itex]h(t)[/itex]. Since the plow has a width [itex]w[/itex], it will have removed a volume of snow [itex]\Delta V_{\text{snow}} = w\; h(t)v(t)\Delta t[/itex].

How can you relate this to the plow's snow removal rate, call it [itex]\dot V_{\text{plow}}[/itex]?

I find this problem more like Physics than Calculus..:frown:

THE VOLUME OF SNOW REMOVAL IS [itex]\Delta V_{\text{snow}} = w\; h(t)v(t)\Delta t[/itex] which makes sense since (Km) * (Km) * (Km/time) * time = km^3 which is volume!

THE RATE OF SNOW REMOVAL IS (Km^3)/hour

Is [itex]\dot V_{\text{plow}}[/itex] Supposed to be the removal rate or speed of the plow?
 
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  • #22
D H
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THE RATE OF SNOW REMOVAL IS [tex]Km^2[/tex]/hour

That is like saying "the speed of my car is km/hr". Those are the units in which the removal rate is expressed (and they should be in volume/time, not area/time).

Is [itex]\dot V_{\text{plow}}[/itex]

This is the volume removal rate. To avoid some confusion, I used capital V for volume, lower case v for velocity.
 
  • #23
playboy
That is like saying "the speed of my car is km/hr". Those are the units in which the removal rate is expressed (and they should be in volume/time, not area/time).



This is the volume removal rate. To avoid some confusion, I used capital V for volume, lower case v for velocity.


I dont know why Latex was spiting out those numbers..I had it correct in the code :yuck:


THE RATE OF SNOW REMOVAL IS (Km^3)/hour
 
  • #24
playboy
Sorry, I am really messing this thread up as I am having problems with Latex...

So isn't the VOLUME REMOVAL RATE, [itex]\dot V_{\text{plow}}[/itex], = [tex]Km^3[/tex]/hour?
 
  • #25
D H
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That is not the rate, those are the units in which the rate is expressed.

In post #20, I gave the clues needed to relate the fixed removal rate [itex]\dot V_{\text{plow}}[/itex] and the velocity of the plow [itex]v(t)[/itex]. Do you need me to spell out the answer?
 
  • #26
playboy
That is not the rate, those are the units in which the rate is expressed.

In post #20, I gave the clues needed to relate the fixed removal rate [itex]\dot V_{\text{plow}}[/itex] and the velocity of the plow [itex]v(t)[/itex]. Do you need me to spell out the answer?

Please let me try it on my own for a bit, if you don't mind :cool:
 
  • #27
playboy
Between time [itex]t[/itex] and [itex]t+\Delta t[/itex], the plow will have moved about [itex]v(t)\Delta t[/itex] meters down the road. Now suppose the snow height at this time is [itex]h(t)[/itex]. Since the plow has a width [itex]w[/itex], it will have removed a volume of snow [itex]\Delta V_{\text{snow}} = w\; h(t)v(t)\Delta t[/itex].

How can you relate this to the plow's snow removal rate, call it [itex]\dot V_{\text{plow}}[/itex]?

THE AMOUNT OF SNOW VOLUME AT TIME t IS: [itex]\Delta V_{\text{snow}} = w\; h(t)v(t)\Delta t[/itex]

That is not the rate, those are the units in which the rate is expressed.

In post #20, I gave the clues needed to relate the fixed removal rate [itex]\dot V_{\text{plow}}[/itex] and the velocity of the plow [itex]v(t)[/itex]. Do you need me to spell out the answer?

THE PLOWS REMOVAL RATE IS: [tex]w*h(t)*v(t)[/tex]

(I am having a tough time with this Latex stuff..it keeps giving the wrong picture when I put in the correct code?)
 
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  • #28
D H
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The volume of snow [itex]\Delta V_{\text{plow}}[/itex] the plow removes from the road during the time interval [itex][ t, t+\Delta t ] [/itex] is

[tex]\Delta V_{\text{plow}} \approx w\; h(t)v(t)\Delta t[/tex]

or

[tex]\frac {\Delta V_{\text{plow}}}{\Delta t} \approx w\; h(t)v(t)[/tex]

In the limit [itex]\Delta t \to 0[/itex], this becomes

[tex]\dot V_{\text{plow}} = w\; h(t)v(t)[/tex]

solving for [itex]v(t)[/itex],

[tex]v(t) = \frac{\dot V_{\text{plow}}}{w\;h(t)}[/tex]

So if you can get an expression for the height of the snow on the road at time t, [itex]h(t)[/itex], you will be just about there.
 
  • #29
playboy
So if you can get an expression for the height of the snow on the road at time t, [itex]h(t)[/itex], you will be just about there.

Snow FALLS AT A RATE OF [tex]Km[/tex]/hour.

Therefore, at time t, we have

h(t) = Rate of snow fall * time?
 
  • #30
D H
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Almost. That is only correct if it started snowing at time [itex]t=0[/itex]. But you don't know when it started snowing. The trick is to invent some variable, call it [itex]t_0[/itex] to represent the time at which it started snowing.
 
  • #31
playboy
Almost. That is only correct if it started snowing at time [itex]t=0[/itex]. But you don't know when it started snowing. The trick is to invent some variable, call it [itex]t_0[/itex] to represent the time at which it started snowing.

I see what your saying.

Would this work:

Let [itex]t_0[/itex] = 12:00 - b
and we must solve for b?
 
  • #32
D H
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You still haven't developed an equation for the snow height, [itex]h(t)[/itex]. You wrote [itex]h(t) = \dot h t[/itex], where [itex]\dot h[/itex] is the constant snowfall rate. This is not correct, because it yields the snow height at the onset of the storm as [itex]h(t_0) = \dot h t_0[/itex]. The height at the onset is presumably zero.
 
  • #33
playboy
You still haven't developed an equation for the snow height, [itex]h(t)[/itex]. You wrote [itex]h(t) = \dot h t[/itex], where [itex]\dot h[/itex] is the constant snowfall rate. This is not correct, because it yields the snow height at the onset of the storm as [itex]h(t_0) = \dot h t_0[/itex]. The height at the onset is presumably zero.

let [itex]t_0[/itex] be the time it started snowing (and this is what we wan to ultimately solve)

let [itex]\dot h[/itex] be the rate of snow fall.

then [itex]h(t) = \dot h (t-t_0)[/itex] where t>[itex]t_0[/itex].
 
  • #34
D H
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Excellent. Now apply this result to the velocity equation, and then integrate velocity (definite integral) wrt time to get the distance traveled over some time interval.
 
  • #35
playboy
So now we have:

[tex]\int_{12:00}^{13:00} v(t) dt = 6\text{\, km}[/tex]


[tex]\int_{13:00}^{14:00} v(t) dt = 3\text{\, km}[/tex]

[tex]v(t) = \frac{\dot V_{\text{plow}}}{w\;h(t)}[/tex]

[itex]h(t) = \dot h (t-t_0)[/itex]

Which gives..

[tex]v(t) = \frac{\dot V_{\text{plow}}}{w\;\dot h (t-t_0))}[/tex]

and ultimately,

[tex]\int_{12:00}^{13:00} \frac{\dot V_{\text{plow}}}{w\;\dot h (t-t_0))} dt = 6\text{\, km}[/tex]


[tex]\int_{13:00}^{14:00} \frac{\dot V_{\text{plow}}}{w\;\dot h (t-t_0))} dt = 3\text{\, km}[/tex]
 

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