Question #babaa

Question

1174 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2014-09-24T21:49:57

Let $f(t)=vec{r}(t)cdot(vec{r}'(t)\times vec{r}''(t))$ .

$f(t)=(-4t^{-1}-2tlnt+2lnt+t+1)e^t$

$f'(t)=(4t^{-2}-2t^{-1}+t-2tlnt)e^t$

Let us look at some details.

By differentiating component by component with respect to $t$ ,

$vec{r}(t)=(e^t,-lnt,t^2)$
$vec{r}'(t)=(e^t,-t^{-1},2t)$
$vec{r}''(t)=(e^t,t^{-2},2)$

$vec{r}'(x)\times r''(t) =|{:(vec{i}, vec{j},vec{k}),(e^t,-t^{-1},2t),(e^t,t^{-2},2):}|$

$=vec{i}|{:(-t^{-1},2t),(t^{-2},2):}|-vec{j}|{:(e^t,2t),(e^t,2):}|+\vec{k}|{:(e^t,-t^{-1}),(e^t,t^{-2}):}|$

$=(-4t^{-1},2te^t-2e^t,(t+1)t^{-2} e^t)$

Let $f(t)=vec{r}(t)cdot(vec{r}'(t)\times vec{r}''(t))$

$=(e^t,-lnt,t^2)cdot(-4t^{-1},2te^t-2e^t,(t+1)t^{-2} e^t)$

$=-4t^{-1} e^t-2te^tlnt+2e^t lnt+(t+1)e^t$

$=(-4t^{-1}-2tlnt+2lnt+t+1)e^t$

By Product Rule,

$f'(t)=(4t^{-2}-2lnt-2+2t^{-1}+1)e^t+(-4t^{-1}-2tlnt+2lnt+t+1)e^t$

$=(4t^{-2}-2t^{-1}+t-2tlnt)e^t$