Question

Help me please

Answer 1

Particle's positon as a function of time is given by x=-t^2 + 4t +4 find the maximum value of position co-ordinate of particle

Answer 2

@zepdrix @Miracrown

Answer 3

Hey blue bird.
So to maximize position we need to take a derivative, yes?

Answer 4

yes

Answer 5

So what'd you get for a derivative? :3

Answer 6

x'= -2t + 4

Answer 7

\[\Large\rm x(t)=-t^2+4t+4\]\[\Large\rm x'(t)=-2t+4\]
Mmm ok great.

Answer 8

This derivative function represents the rate of change of position.
It tells us when we're moving up, or moving down.

We want to know when we're stationary.
That will represent a point where we're at the top of a hill, or the bottom of a valley.

So we need to find critical points by letting our derivative function equation zero.
From there we can determine if one of those points is a maximum ( top of a hill ).

Answer 9

\[\Large\rm 0=-2t+4\]

Answer 10

We should equal it to 0 then because the slope would be zero i guess

Answer 11

Yess you just typed it :)

Answer 12

So what is the value of your critical point?
What t value?

Answer 13

2

Answer 14

Ok good. t=2 is a critical point of this function.
How can we determine whether it's a max or min? :)
Any ideas?

Answer 15

No i really don't know !

Answer 16

We have this point t=2, it's a stationary point.
So it's either at the top of a hill ( maximum ),
or it's at the bottom of a valley ( minimum ).