Question

Calculus! Help for a medal.

Answer 1

So I am given a function which represents the position of a particle. The problem tells me to find how far the particle has traveled in the first 4 seconds.

Answer 2

Would I just plug 4 into the function and solve?

Answer 3

I don't think I need to take the derivative since it's dealing with position/distance.

Answer 4

depends if the function given is of one variable
as its given as position
just put value of the variable as 4 you may find the answer

Answer 5

Yes, the only available variable in the equation is 't' and t=4, so I just plug in 4? There are no other steps?

Answer 6

yes for sure

Answer 7

Thank you!

Answer 8

in case it was second degree variable then you would have to derivate to make it single degree thenput the value in the function

Answer 9

second degree variable?

Answer 10

like t^2?

Answer 11

My actual function is \[f(t)=\frac{1}{3}t^3-\frac{1}{2}t^2-2t+1\]

Answer 12

it depends what they mean by "how it traveled"
the simple answer is the change in distance from t=0 to t=4

(but if it went "out" and them came back to the starting point, that gives zero as the distance traveled)

the harder answer is the "length of the path" i.e. the arc length

Answer 13

*how far it traveled

Answer 14

The information I have provided is all that I have been given unfortunately, but based off what the question is asking, do you know if I just plug in the variable or are there other steps involved?

Answer 15

I think they want the arc length. Have you studied that ?

Answer 16

It doesn't ring a bell.

Answer 17

see
f(t) isgiven just put values of t you ll have distance at time t
f' (t) (derivative of f(t) ) gives you speed at time t
f'' (t) (double derivative of f(t) gives you acceleration at time t

Answer 18

phi dear you are talking about displacement
they are asking for distance
hence if the as for distance arc or circle or elips all does play a role.

Answer 19

See, that's what I thought from the start. Now with it asking for the distance traveled, should I also plug in 0 into t and subtract f(0) from f(4)?

Answer 20

Or will f(4) just do?

Answer 21

if they would have asked distance traveled between 7 and 12th second then u might have done if f(7) and f(12) --- > f(12) - f(7)
now only give f(0) and f(4)
u ll have answer as f(4)-f(0)

Answer 22

So I got 13/3 for f(4) and 1 or 3/3 for f(0)
Using subtraction, I got a final answer of a distance of 10/3.

Answer 23

Could one of you help me with the second part?

Answer 24

It asks to find the maximum velocity of the particle over the interval of [0,4].
I'm pretty sure I have to take the second derivative for this one.

Answer 25

f(0) = 1
subtract and get the answer

Answer 26

That's what I did. I just converted the 1 to a 3/3 to subtract it from 13/3.

Answer 27

Is that not correct?

Answer 28

yeh sure thats fine to substract

Answer 29

Okay cool. Thank you!
Do you happen to have any idea about the second part? (It's the last part. Promise)

Answer 30

It seems they're working on another question, so what about you, Phi? Have an idea?

Answer 31

what is the second part ?

Answer 32

"Could one of you help me with the second part?

It asks to find the maximum velocity of the particle over the interval of [0,4].
I'm pretty sure I have to take the second derivative for this one."

Answer 33

vel= derivative of position

once you have v(x)
to find a min or max, take the derivative of v(x)
and set it equal to zero

Answer 34

what do formula do you get for the velocity ?

Answer 35

I got f"(t)=2t-1

Answer 36

ok, but what did you get for f'(x) ?

Answer 37

I got f'(t)=t^2 -t -2

Answer 38

Oh yea, duh. second derivative is acceleration and not velocity. Sorry bout that.

Answer 39

yes, and that represents the velocity as a function of time.
notice it is a parabola with a \( \cup\) shape , so when you find the min/max, you will be finding the min (at the vertex)
in other words, when you solve f''=0 using 2t-1= 0 --> t= ½
that will be a min

Answer 40

So I set it equal to 0 and got t=-1 and t=2

Answer 41

but for min/max problems, you also have to check the boundaries, because the max will have to be at the boundaries i.e. at t=0 or t= 4 (we know there is no local max inside the interval, because the only local min/max is at t=½ and that is a min)

Answer 42

back up: you found velocity v(x)= t^2 -t-2
to find a local min/max you take the derivative and get v'= 2t-1
and set that = 0 to get 2t-1=0, t= ½
but as explained up above, that is a *min*

Answer 43

So you use acceleration to find the local max and mins?

Answer 44

once you find there is only 1 min/max between 0 and 4
and you found it was a min
that means the max must be one of the "edge" values i.e. v at t=0 or t=4

Answer 45

yes, you would use acceleration to find the local max or min. But I just think of the problem as
given f(t) find its local min/max
and the strategy is f'(t)= 0
solve for t

in other words, I don't think in terms of the physics.

Answer 46

Okay. I think I understand.

Answer 47

So you would test points around the 0 and 4 to see if one of them are the max?

Answer 48

yes, you would evaluate v(t)= t^2 -t-2 at t=0 or t=4

Answer 49

btw, what level of calculus is the class you are taking ?

Answer 50

I'm in high school and I'm taking AB AP Calc

Answer 51

what did you get for the first part of the question ?

Answer 52

For the first part, I got 10/3.

Also, what function do I evaluate the points in? The f(t), f'(t), or f"(t)?

Answer 53

the question was where is the max velocity in the interval [0,4]
where velocity is f'(t)
if we plot f'(t), we would see the answer immediately

Answer 54

Unfortunately, I am not allowed to use graphs. Is there a way by solving that I can show this?

Answer 55

Yes, I am assuming you can't graph it, but it might help to see what is going on.
notice that at t=½ you see the velocity at its min value
notice that the right edge at t=4 we have the largest velocity.
so you would solve this problem by evaluating the velocity at the 1 local max/min at t=½, and at the edges, then picking the biggest number

Answer 56

Okay, so you would plug the 1/2, 0, and 4 into the velocity equation and see which one is the biggest number? (Obviously it won't be 1/2 since it's a min)

Answer 57

exactly.

for a more complicated example, look at this

in this case, you would find 2 local max/min (one of each) and the boundaries would be the min and max (for this interval)