Question #babaa

1 Answer
Sep 24, 2014

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