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dctp05
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Related Rates Verification, Help Please!
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?
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.
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.
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.
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