#Multivariable Calc Help Needed
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rare pier
For this I got (-2,0)
For this I got:
a = (-10/(1+4x^2+4y^2)) (-2x, -2y)
Now I'm not sure what to do for this:
@thin token
thin token
will you have
$a = \frac{-10}{1+4x^2+4y^2}*(-2x,-2y)$
rough magnetBOT
Dragon_42
rare pier
the thing is tho... for a, will I use the accleration at the point (1,0) ?
thin token
yes use 1,0
rare pier
Why do u we use (1,0), that's what I don't understand sir
thin token
a stays constant is the note you are given
and it is 1 sec so very same change in time and distance
It is also an approximation too
So there be some error in your answer but that is why you have the approximation symbol being used
rare pier
But look, in the question is says the inital velocity of the puck = (0, 1) in m/s
but then it also says: at time t_0, the puck is at postion (1,0)
and for the acceleration, we need a position vector right?
to fill in for x and y?
thin token
use so you have
$ (u_1,v_1) = (0,1) + (1-0)(\frac{-10(-2x,-2y)}{1+4x^2+4y^2})$
rare pier
it didn't pop up
thin token
ya sad lol
rare pier
U know how to find my acceleration, I need to sub in a point(x,y)
why do we sub in (1,0)
thin token
but yes plugin x = 1 and y =0
Cause this is all a rough approximation so you truly only know what point
rare pier
yeah, that's what I don't get why. Why and where did we choose x = 1 and y = 0?
Is it because it is the point of origin?
thin token
Yes
rare pier
because at time 0, the puck is in that position of (1,0)
thin token
At start point we know the time and position
thin token
if you are given new x.y plug them in
rare pier
yeah I plug in x = 1 and y = 0 right?
rare pier
+close