Related Rates of Change Per Unit Time

[tmlrlz]

Homework Statement

In the special theory of relativity the mass of a particle moving at speed v is given by the expression
m/(1-(v²/c²))
where m is the mass at rest and c is the speed of light. At what rate is the mass of the particle changing when the speed of the particle is (1/2)c and is increasing at the rate of 0.01c per second?

Homework Equations

The Attempt at a Solution

For this problem i tried using related rates. The question is asking for dm/dt when v = (1/2)c and they give dv/dt = 0.01c
dm/dt = (dm/dv)*(dv/dt)
Suppose that y = m/√(1-(v²/c²)) i took the derivative of y
the one assumption i made which is what I'm confused about is that the derivative of y = 0
[(dm/dv)*(√(1-(v²/c²))) - 1/2*((1-(v²/c²))^-1/2*(-2v/c²)*(m)]/(1-(v²/c²) = 0

dm/dv(√(1-(v²/c²))) = -m(1-(v²/c²))^-1/2(v/c²)

dm/dv = [-m(v/c²)]/(1-(v²/c²))

dm/dt = [-m(v/c²)]/(1-(v²/c²)) * (0.01c)
= -0.01mv/(c-v²/c)
At v = 1/2c
dm/dt = -mc/200 * 4/3c
= -m/150

I would just like to know if my approach is correct and specifically, if i was correct in assuming that the derivative of y = 0. Thank You.

 

[mighty2000]

Your approach is mostly correct, but there are a few errors in your calculations. Let's go through them step by step.

First, your assumption that the derivative of y is 0 is not correct. In fact, the derivative of y with respect to v is not 0, but rather -m/(1-(v^2/c^2))^(3/2). This can be found by using the quotient rule to take the derivative of y.

Next, in your calculation of dm/dv, you forgot to include the factor of 1/2 in front of the v/c^2 term. The correct expression should be dm/dv = [-m(v/c^2)]/(1-(v^2/c^2))^(3/2).

Finally, when you substitute in v = 1/2c, your final expression for dm/dt should be -m/600, not -m/150. This is because you are substituting in v = 1/2c, not v = 1/2.

So, to summarize, your approach is correct but there are a few mistakes in your calculations. The correct expression for dm/dt is -m/600. I hope this helps clarify things for you. Keep up the good work!