Question

I am struggling to figure out the 1F-7B/C questions in problem set 1, the solution given is not helping me and I can't even begin to see the steps. If somebody could give a better solution that'd be helpful
Thanks

Answer 1

Sorry I should have linked it
http://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/part-a-definition-and-basic-rules/problem-set-1/MIT18_01SC_pset1sol.pdf

Answer 2

Ok first one
\[m=\frac{m_0}{\sqrt{1-v^2/c^2}}\]
Here we have the quotient rule, that is
\[\frac{d}{dv}\frac{f(v)}{g(v)}=\frac{f'(v)g(v)-g'(v)f(v)}{g(v)^2}\]
Now set the numerator and the denominator as functions like
\[f(x)=m_0\]
and
\[g(x)= \sqrt{1-v^2/c^2}\]
Then take the derivative of each one to fill in the formula.
First the derivative of the numerator f(x), in this case f(x) is a CONTSTANT because the question states 'Assume all letters represent constants, except for the dependent and independent variables'. So that means 'm' is the dependent variable and 'v' is the independent variable.
So again
\[f(x) = m_0\]
seeing as this is not 'm' or 'v' (its actually 'm0') we know it's a CONSTANT,
and the derivative of a constant is 0. Thus
\[f'(v) = 0\]
Now we have the numerator done, lets do the denominator, namely
\[g(v)=\sqrt{1-v^2/c^2}\]
First rewrite this so we have an exponent instead of the square root symbol
\[g(v)=(1-v^2/c^2)^\frac{1}{2}\]

Remember to take a derivative of an exponent its
\[x^n = nx^{n-1}\]
So first take the derivative of the 'outside' function, that is the square root, and then the 'inside' of the function
\[\frac{d}{dv}(1-v^2/c^2)^{\frac{1}{2}}= \frac{1}{2}(1-v^2/c^2)^{\frac{1}{2}-1} \cdot \frac{d}{dv}1-v^2/c^2\]
Now to actually take the derivative of
\[1-v^2/c^2\]
we would need the quotient rule again, but be careful remember they are all contstants except for 'v' and 'm', so when taking the derivative of 'c' it will be a constant and so ZERO
\[\frac{d}{dv}(1-v^2/c^2)= \frac{c^2 \frac{d}{dv} (-v^2) - v^2 \frac{d}{dv}c^2}{(c^2)^2} = \frac{-2vc^2}{c^4}= \frac{-2v}{c^2}\]
So the derivative of g(v) is
\[\frac{d}{dv}g(v)= \frac{1}{2}(1 - v^2/c^2)^{-\frac{1}{2}} \cdot \frac{-2v}{c^2}=\frac{-v(1-v^2/c^2)^{-\frac{1}{2}}}{c^2}=\frac{-v}{c^2\sqrt{1-v^2/c^2}}\]
Look carefully above, the derivative of c squared is zero so when multiplied by v squared the whole thing becomes zero, so the only thing above that matters is the first part, the derivative of negative v squared is negative 2v multiplied by csquared.
Now we have the derivatives of both numerator and denominator we can complete the initial derivative
Remember it was
\[\frac{dm}{dv}\frac{f(v)}{g(v)}= \frac{f'(v)g(v)-g'(v)f(v)}{g(v)^2}\]
So lets fill it in ( remember the derivative of f(v) is zero so
\[\frac{dm}{dv}=\frac{d}{dv}\frac{f(v)}{g(v)}=- \frac{-vm_0}{c^2\sqrt{1-v^2/c^2)}} \cdot \frac{1}{(1-v^2/c^2)} = \frac{vm_0}{c^2(1-v^2/c^2)^{\frac{3}{2}}}\]

Answer 3

Oops, above I put f(x) and g(x), that should read f(v) and g(v)