Proofs

• Feb 2nd 2008, 04:32 PM
shilz222
Proofs
1. Prove that $\frac{d}{dt} (\bold{r} + \bold{s}) = \frac{d \bold{r}}{dt} + \frac{d \bold{s}}{dt}$

So $\frac{d}{dt} (\bold{r}+\bold{s}) = \lim_{h \to 0} \frac{\left[ \bold{r}(t+h) - \bold{r}(t) + \bold{s}(t+h) - \bold{s}(t) \right]}{h}$

$= \lim_{h \to 0} \frac{\bold{r}(t+h) - \bold{r}(t)}{h} + \lim_{h \to 0} \frac{\bold{s}(t+h) - \bold{s}(t)}{h} = \frac{d \bold{r}}{dt} + \frac{d \bold{s}}{dt}$

2. Prove that $\frac{d}{dt}(\bold{r} \cdot \bold{s}) = \frac{d \bold{r}}{dt} \cdot \bold{s} + \bold{r} \cdot \frac{d \bold{s}}{dt}$

So $\lim_{h \to 0} \frac{\bold{r}(t+h) \cdot \bold{s}(t+h) - \bold{r}(t) \cdot \bold{s}(t)}{h}$ $= \lim_{h \to 0} \frac{\bold{r}(t+h)\cdot \bold{s}(t+h) - \bold{r}(t) \cdot \bold{s}(t+h) + \bold{r}(t) \cdot \bold{s}(t+h) - \bold{r}(t) \cdot \bold{s}(t)}{h}$

$= \lim_{h \to 0} \left(\frac{\bold{r}(t+h) - \bold{r}(t)}{h} \cdot \bold{s}(t+h) + \bold{r}(t) \cdot \frac{\bold{s}(t+h) - \bold{s}(t)}{h} \right)$

$= \left( \lim_{h \to 0} \frac{\bold{r}(t+h) - \bold{r}(t)}{h} \right) \cdot \left( \lim_{h \to 0} \bold{s}(t+h) \right) + \bold{r}(t) \cdot \left(\lim_{h \to 0} \frac{\bold{s}(t+h) - \bold{s}(t)}{h} \right)$ $= \frac{d \bold{r}}{dt} \cdot \bold{s} + \bold{r} \cdot \frac{d \bold{s}}{dt}$

3. Prove that $\frac{d}{dt} (\bold{r} \times \bold{s}) = \frac{d \bold{r}}{dt} \times \bold{s} + \bold{r} \times \frac{d \bold{s}}{dt}$.

Thiis is basically the same method as two? Are the above correct?

Thanks
• Feb 3rd 2008, 04:23 AM
pellikkan
the proof for 1 seems convincing except for placement of terms, but the one for 2 doesn't, as it stands.
To me, it would be more convincing if you substituted the expansion
r(x+h)=r(x)+hr'(x)+..., and likewise for s, into the second line, and then
threw out terms second order and above in h.