# need help.. asymptote

• November 6th 2006, 11:49 AM
bobby77
need help.. asymptote
According to enstine's special theory of relativity,the mass of the body satisfies the equation m(v) = m0 / sqrt(1-(v^2/c^2))
Where m(v) is mass of body when it moves with speed v in relation to observer,m0 is the mass of body at rest in relation to observer(ie a constant for example 1 kg)and c is speed of light
(1) domain and range of m(v)
(2)when velocity is 0 what is the mass?(this is vertical intercept)
(3)determine equation of vertical and horizontal aymptotes
(where is m(v) increasing and decreasing.
(4) where is m(v) concave up and down
• November 6th 2006, 12:39 PM
topsquark
Quote:

Originally Posted by bobby77
According to enstine's special theory of relativity,the mass of the body satisfies the equation m(v) = m0 / sqrt(1-(v^2/c^2))
Where m(v) is mass of body when it moves with speed v in relation to observer,m0 is the mass of body at rest in relation to observer(ie a constant for example 1 kg)and c is speed of light
(1) domain and range of m(v)
(2)when velocity is 0 what is the mass?(this is vertical intercept)
(3)determine equation of vertical and horizontal aymptotes
(where is m(v) increasing and decreasing.
(4) where is m(v) concave up and down

1) The domain is all real values for v such that the expression m(v) is also a real number. This will be true as long as
a) The number under the square root is either zero or positive
b) The denominator of the fraction is not zero.

Looking at the equation for m(v) we see that $v < c$ to make the number under the square root positive. (It can't be equal to c because that would make the denominator 0.) Thus the domain is [0, c).

(NOTE: I am using the convention that v is a scalar, ie the speed, not the magnitude, so it can't be less than 0.)

As to the range, we can see that m(v) is the smallest for v = 0. We can also see that $\lim_{v \to c} m(v) \to + \infty$. So the range of m(v) is $[m_0, \infty )$.

2) I already used this in the answer to 1), but specifically:
$m(0) = \frac{m_0}{ \sqrt{1 - \frac{0^2}{c^2}} } = m_0$

3)
Vertical asymptotes: These occur when the denominator is 0, which I mentioned in 1) only happens when v = c.

Horizontal asymptotes: These occur when $\lim_{v \to \pm \infty} \to constant$. As v can't go to either limit, there is no horizontal asymptote.

4) We wish to investigate the nature of the function, so we take the first and second derivatives:
$m(v) = \frac{m_0}{ \left ( 1 - \frac{v^2}{c^2}\right )^{1/2} }$

$m'(v) = \frac{m_0v}{ \left ( 1 - \frac{v^2}{c^2}\right )^{1/2} }$

$m''(v) = \frac{m_0}{ \left ( 1 - \frac{v^2}{c^2}\right )^{3/2} } + \frac{3m_0v^2}{ \left ( 1 - \frac{v^2}{c^2}\right )^{5/2} }$

First look at m'(v): The only place it is 0 is at v = 0. From there the slope is always positive. Thus m(v) is monotonically increasing.

Now look at m''(v): Does this ever equal 0? No. So there are no inflection points. And as m''(v) is positive on [tex] [0, c) the curvature is always positive. Thus m(v) is concave up on its domain.

-Dan
• November 7th 2006, 12:35 AM
bobby77
please can you show me how to graph m(v)...
• November 7th 2006, 04:51 AM
topsquark
Quote:

Originally Posted by bobby77
please can you show me how to graph m(v)...

First you need to pick some value for m0, I would set it to 1 for simplicity.

Unless you are using some graphing software you need to select a set of values of v to use. I would recommend at least 10 points. For example, pick v = 3c/10:
$m(v) = \frac{1}{\sqrt{1 - \left (\frac{3}{10} \right )^2}} \approx 1.048284837$

And, say v = 9c/10:
$m(v) = \frac{1}{\sqrt{1 - \left (\frac{9}{10} \right )^2}} \approx 2.294157339$

etc.

I have attached a graph below.

Edit: Whoops! I forgot to mention. In the graph I am using the standard Relativist definition that c = 1. It's just a scaling factor. And I have set m0 = 1 as well. (Probably you should adjust the scale so it doesn't look like it's 0, as it does in my graph.)

-Dan