# An object of mass 10 moves with position function r(t)=<1+3cos(t),4cos(t),−1+5sin(t)> Find the force vector acting on this object at time t. Show that it points from the object towards the point (1,0,−1).

An object of mass 10 moves with position function $r\left(t\right)=<1+3\mathrm{cos}\left(t\right),4\mathrm{cos}\left(t\right),-1+5\mathrm{sin}\left(t\right)>$ Find the force vector acting on this object at time t. Show that it points from the object towards the point $\left(1,0,-1\right)$
I believe that I have found the correct force vector: $F\left(t\right)=10<-3\mathrm{cos}t,-4\mathrm{cos}t,-5\mathrm{sin}t>$
However, I am not sure how to figure out how to show that it points from the object towards the point $\left(1,0,-1\right)$
Any help would be greatly appreciated.
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

na1p1a2pafr
What you need to do is describe the line along $\stackrel{\to }{F}$ at time t. You know that it is going through $\stackrel{\to }{r}\left(t\right)$. So the equation of this line is

Plugging in what you calculated so far,
${\stackrel{\to }{l}}_{\alpha }\left(t\right)=<1+3\mathrm{cos}t-30\alpha \mathrm{cos}t,4\mathrm{cos}t-40\alpha \mathrm{cos}t,-1+5\mathrm{sin}t-50\alpha \mathrm{sin}t>$
Now let's calculate the intersection by saying that ${\stackrel{\to }{l}}_{\alpha }\left({t}_{1}\right)={\stackrel{\to }{l}}_{\beta }\left({t}_{2}\right)$. Then
$1+3\left(1-10\alpha \right)\mathrm{cos}{t}_{1}=1+3\left(1-10\beta \right)\mathrm{cos}{t}_{2}\phantom{\rule{0ex}{0ex}}4\left(1-10\alpha \right)\mathrm{cos}{t}_{1}=4\left(1-10\beta \right)\mathrm{cos}{t}_{2}\phantom{\rule{0ex}{0ex}}-1+5\left(1-10\alpha \right)\mathrm{sin}{t}_{1}=-1+5\left(1-10\beta \right)\mathrm{sin}{t}_{2}$
If we want this intersection to be time independent, we see that $\alpha =\beta =0.1$ Plugging in this value of $\alpha$, you get a single point $<1,0,-1>$