- #1

albertrichardf

- 165

- 11

I know that the speed of an object can be calculated the following way:

[tex] v = \frac {ds}{dt} = \sqrt{(v_x)^2 + (v_y)^2} [/tex]

where ##s## is the position and ##v_x = \frac {dx}{dt}## and ditto for ##v_y##

Based on its definition, one should be able to obtain the same result by calculating ##s = \sqrt{x^2 + y^2}## and then differentiating that. I squared s for simplicity and obtained the following:

[tex] v = \frac {xv_x + yv_y}{\sqrt {x^2 + y^2}} [/tex]

Or squaring again:

[tex] v^2 = \frac {x^2v_x^2 + y^2v_y^2 + 2xyv_xv_y} {x^2 + y^2} [/tex]

This expression should reduce to:

[tex] v_x^2 + v_y^2 [/tex]

But no matter what I try I can't simplify it to that.

Furthermore, I tried this with an actual function: ##x = t^2## and ##y = t##. Their derivatives are ##2t## and ##1## respectively. Squaring those and adding them gives me ##\sqrt {4t^2 + 1} = v##. Trying the other way, ##s = t \sqrt {t^2 + 1}##, the derivative of which is ## \frac {2t^2 + 1}{\sqrt {t^2 + 1}}##. Squaring this gives me

[tex] \frac {4t^4 + 4t^2 + 1}{t^2 + 1} [/tex]

Now I can factor out ##4t^2## from the first two terms of the numerator, which leaves me with:

[tex] \frac {4t^2(t^2 + 1) + 1}{t^2 + 1} [/tex]

And I cannot reduce this to ##4t^2 + 1##

Any help with this?

Thank you