Linear Density: find the mass of the rod

S

steve555

New member
Joined
Jun 21, 2006
Messages
5
Hi, I need help with this problem:

The linear density of a rod of length 1 m is given by p(x)=1/sqrt(x), in grams per centimeter, where x is measured in centimeters from one end of the rod. Find the mass of the rod.

I started off with this:

. . .d = m/L

. . .d = p(x) = 1/sqrt(x) g/cm

. . .L = is it 10cm?

The method I'm suppose to be using is antiderivatives/indefinite integrals. And my problem is that I am not sure how to start the problem. When the problem said x is measured from one end of the rod, is that the length too? Overall I feel lost and need some help.
 
linear density = mass/length

\(\displaystyle \frac{dm}{dx} = \frac{1}{sqrt{x}}\)

\(\displaystyle dm = \frac{1}{sqrt{x}} dx\)

x varies from 0 to 1 meter ...

\(\displaystyle M = \L \int_0^1 \frac{1}{sqrt{x}} dx\)

can you finish from here?
 
Is the antiderivative of 1/sqrt(x) = ln(sqrt(x)) + C?

If so..

Then I plugged in 0 and 1 for x. Which turns out to be
x(0)=no solution and x(1)=0. Then I found the difference and the answer is 0?

Or should the 1 m be coverted to 10cm? If it is then, I would plugin

x(10)=1.151 <<< would that be correct?
 
steve555 said:
Is the antiderivative of 1/sqrt(x) = ln(sqrt(x)) + C?
Click to expand...
No absolutely not!
\(\displaystyle \L y' = \frac{1}{{\sqrt x }} = x^{ - \frac{1}{2}} \quad \Rightarrow \quad y = 2x^{\frac{1}{2}}\).
 
The directions stated that \(\displaystyle p(x) = \frac{1}{\sqrt{x}}\) in g/cm.

The integral is:
\(\displaystyle \L \int_0^{100} x^{\frac{-1}{2}} dx\)

BTW, Steve, 1m = 100cm, not 10cm. Centi = one one-hundredth.
 
pka, skeeter, and daon thanks for pointing out my mistakes...

1. y'=1(x^(.5)/.5)= 2x^.5+C
2. x(100)=2(100)^.5=20
x(0)=2(0)^.5=0
3. x=20-0=20cm <<<is that the final answer or do I plug in

x=20cm for density which is p(x)=1/sqrt(x) and times that by Length?

Since density=mass/length >>> mass=density x length
 
\(\displaystyle \L M = \int_0^{100} x^{-1/2} dx\)

\(\displaystyle \L M = [2x^{1/2}]_0^{100}\)

\(\displaystyle \L M = 2(100)^{1/2} - 2(0)^{1/2}\)

\(\displaystyle \L M = 20 grams\)

thanks for pointing out the units
 
You must log in or register to reply here.
Top