Question

Question #babaa

Answer 1

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`