Author Topic: Confused with Position/Velocity vector (Read 693 times) Tweet Share

Cuntryboner · « **on:** June 01, 2010, 11:25:03 pm »

A particle moves so that it's position vector at a time t is given by

$r(t)=8cos^{3}(2t)i+8sin^{3}(2t)j when t\ge0$ .

if the speed can be expressed in the form $csin(bt)$ then find the values of $b and c$ .
I am fine up to...

$\dot{r}(t)=-48sin(2t)cos^{2}(2t)i+48cos(2t)sin^{2}(2t)j$

$|\dot{r}(t)|=\sqrt{-48^{2}sin^{2}(2t)cos^{4}(2t)+48^{2}cos^{2}(2t)sin^{4}(2t)}$

$=\sqrt{48^{2}sin^{2}(2t)cos^{2}(2t)(cos^{2}(2t)+sin^{2}(2t))}$

$=48sin(2t)cos(2t)$

but I can't manipulate to get the answer, which is

$|\dot{r}(t)|=24sin(4t)$

Any help, would be gr8

Yitzi_K · « **Reply #1 on:** June 01, 2010, 11:36:45 pm »

It's the double angle formula: $sin(2x)=2sin(x)cos(x)$

so $24sin(4t)=24sin(2(2t))$ , which when expanded is $48sin(2t)cos(2t)$

Cuntryboner · « **Reply #2 on:** June 01, 2010, 11:40:12 pm »

Thanks for Factoring $24sin(2(2t))$ , I am more the wiser!!

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Author Topic: Confused with Position/Velocity vector (Read 693 times) Tweet Share

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Confused with Position/Velocity vector

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