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Related Rates Verification, Help Please

  • Thread starter dctp05
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Related Rates Verification, Help Please!!!

Homework Statement


The capacity of Bob's truck is 72 lbs. He needs to carry as much tar paper as possible and 1 square yard weighs 5 lbs. If the paper comes off the roll at 2 square feet per second, how long will it take to fill his truck?



Homework Equations


72 lbs= 5 lbs * x yd2, so Bob's truck can fit 14.4 yd2 of tar paper.
I assume that I'm trying to find how many seconds it takes for 14.4 yd2 of tar paper to get in the truck.

I'll replace 72 with y,
so, the equation would be y=5x.

[itex]\frac{d}{dt}[/itex](y=5x) is [itex]\frac{dy}{dt}[/itex]=5[itex]\frac{dx}{dt}[/itex]

Anyway I know that [itex]\frac{dx}{dt}[/itex]=2ft2/sec, which is .2222yd2/sec when converted.


The Attempt at a Solution


I would just substitute .2222yd2/sec into the derived equation to get 1.1111 lbs/sec. This means that the tar paper goes into the truck at 1.1111 lbs/sec. However, I don't know how to work this into the original equation, do I just divide 72 by 1.1111? I think I'm just being a bit stupid with this. Somebody please help this shouldn't take to long.
 
Last edited:

Answers and Replies

  • #2
307
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Hey dctp,

Your equation, [itex]\frac{dy}{dt} = 5 \frac{dx}{dt}[/itex], tells you the rate of weight being added to the truck, denoted by [itex]\frac{dy}{dt}[/itex], as a function of the speed at which the tar paper is being added, [itex]\frac{dx}{dt}[/itex].

Since you want to know how much weight is in the truck as a function of time, you would integrate your equation to obtain:
[tex]y(t) = (5\textrm{lbs/yd}^2)\cdot (0.2222\textrm{yd}^2\textrm{/s})\cdot\Delta t[/tex]
Can you figure it out from here?
 
  • #3
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Hey dctp,

Your equation, [itex]\frac{dy}{dt} = 5 \frac{dx}{dt}[/itex], tells you the rate of weight being added to the truck, denoted by [itex]\frac{dy}{dt}[/itex], as a function of the speed at which the tar paper is being added, [itex]\frac{dx}{dt}[/itex].

Since you want to know how much weight is in the truck as a function of time, you would integrate your equation to obtain:
[tex]y(t) = (5\textrm{lbs/yd}^2)\cdot (0.2222\textrm{yd}^2\textrm{/s})\cdot\Delta t[/tex]
Can you figure it out from here?
Sorry, the problem is, I haven't learned integration yet, so I can't use it. But I think I might have it using another method.

If [itex]\frac{dx}{dt}[/itex]=.2222yd2/sec, I could substitute into the derivative, [itex]\frac{dy}{dt}[/itex]=5[itex]\frac{dx}{dt}[/itex] to get [itex]\frac{dy}{dt}[/itex]=1.1111 lbs/sec. So its saying that tar paper is being loaded onto the truck at 1.1111 pounds per second. Since I want to know when lbs equals 72, I could just divide 72 by 1.1111, which would give me 64.8 sec to load the truck to its maximum.

Do you get the same answer through integration? Please verify, if that is allowed.
 
  • #4
307
3


No matter how you end up doing the question, you do end up integrating :), it's just that the integration in this question is so trivial that it can be "swept under the rug" without anyone noticing they're doing it.

In this case, you've done it by considering the physical implications of your equation. The solution you obtained is correct.

But to show you the similarity between the two equations:
Mine:
[tex]y(t) = (5\textrm{lbs/yd}^2)\cdot (0.2222\textrm{yd}^2\textrm{/s})\cdot\Delta t[/tex][tex]= 1.1111\textrm{lbs/s} \cdot \Delta t[/tex]
Yours:
[tex]\frac{dy}{dt}=1.1111\textrm{lbs/s}[/tex]
To get from yours to mine, recall that dy is a small changes in y, and dt is a small change in t, and therefore,
[tex]\Delta y = 1.1111\textrm{lbs/s} \cdot \Delta t[/tex].
 
  • #5
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Okay I see the connection. I'm a bit fuzzy concerning the details, but I'll learn it within a few weeks at school, so it's all right. I'll revisit it then.

Thanks :)
 

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