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Calculus1 16 Online
OpenStudy (anonymous):

Need help about linear density how is this p(4) in the given pictures below...

OpenStudy (anonymous):

OpenStudy (anonymous):

OpenStudy (anonymous):

please help.

OpenStudy (anonymous):

why is it p(4) in the 2nd picture...can someone explain?

OpenStudy (phi):

The english is not very clear. when they say "the linear density... varies directly ... with distance from an external point 2 meters from the end" they are (trying?) to say d = k (6-x) we get 6 from: rod is 4 meters long , so set up the problem so the rod goes from x =0 to x=4. two beyond "the end" puts us at 6. the "distance from a point on the rod" to the point x=6 is 6-x

OpenStudy (phi):

they mumble 2 meters from an end, where the density is 5 kg/m they are saying the density at the x=4 end is 5 kg/m I would have written it as "The density at the right-side end is 5 kg/m"

OpenStudy (anonymous):

but why is it p(4) in the 2nd picture, how come?

OpenStudy (phi):

we use that info to find the "constant of variation" d= k (6-x) 5 = k (6-4) 2.5= k so the density varies as d= 2.5 (6-x) from x=0 to x=4

OpenStudy (anonymous):

huh? I do not get it why 4 is on the p(4) in the 2nd picture... can you direct to the point...

OpenStudy (phi):

they are saying rho (which apparently stands for density here) at x=4 is 5 \[ \rho(4) = 5\]

OpenStudy (phi):

and we know the density at x=4 is 5 because they tell us it is (though the wording is not very clear)

OpenStudy (anonymous):

what is that x=4 ? and how come that p(4)=5.

OpenStudy (anonymous):

ok ok thanks .

OpenStudy (phi):

the density is "a function of x" in other words, you need a "formula" to figure out what the density is at each x location Based on the problem they decided to put the bar on the x-axis so its left side is at x=0, and its right side is at x=4 (and the units are meters) the formula for the density , as a function of x is (after working it out) \[ \rho(x) = 2.5(6-x) \] that is typical "math" way to say, some quantity called rho changes with another quantity called x. And we have set up the problem so that x means position along the bar (as long as we keep within the interval x=0 to x=4)

OpenStudy (phi):

There is a lot that is not stated in this problem, and so a lot of ways to get confused.

OpenStudy (anonymous):

but why is p(4) not p(0) is on the 2nd picture...

OpenStudy (phi):

Maybe if we walk through the question slowly? "The linear density at any point of a rod 4 m long varies directly as..." that tells us 1) density will change. Let's call the density rho. 2) "varies directly" means expect an equation rho= k * some variable where k is the constant of variation 3) the bar is 4 meters long. At this point I would automatically think: "put the bar on the x-axis" |dw:1429195891662:dw|

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