Question

The position function of a particle is given by r(t)=⟨4t^2,−4t,t^2+1t⟩. At what time is the speed minimum?

Answer 1

i have v(t) = <8t, -4, 2t+1>

Answer 2

no idea what to do now

Answer 3

This might help.

You are given the position vector r(t) = <t^2, 5t, t^2-16t>. Note that r(0)=<0,0,0>, which means the particle starts out at the origin.

The velocity vector is v(t)=<2t, 5, 2t-16>. Note that v(0)=<0, 5, -16>, so in fact its initial velocity is nonzero.

The speed of the particle is s(t) = sqrt( 4t^2 + 25 + (2t-16)^2 ) = sqrt( 8t^2 -64t + 281)

One can check that the speed is always nonzero (either using the formula for s(t), or more simply by examination of v(t) and noting that v(t) is never the zero vector).

Minimizing the speed involves finding the critical point for s(t) = sqrt( 8t^2 -64t + 281), which as you note occurs when t=4, although I get s(4) = sqrt(153), not 156.

Answer 4

ty

Answer 5

I have s(t) = sqrt(68t^2 + 4t + 17)

Answer 6

what do I do now? @TruWubs

Answer 7

how did you find that the critical point is when t = 4?

Answer 8

you can take the derivative of the speed s(t)

Answer 9

then set to 0 and solve for t?

Answer 10

@perl

Answer 11

yes

Answer 12

ah, thank you :)

Answer 13

Sorry for not replying Open study is being glitchy he explained fine :)

Answer 14

np, ty both, yeah I love this site but lots of connection issues with it

Answer 15

@TruWubs I'm getting s(t) = sqrt(68t^2 + 4t + 17), when i take derivative of that set it to 0 and solve for t i get -1/34, which is incorrect, ive also tried entering 1/34

Answer 16

I'm sorry I tried again and -1/34 is correct, i was typing it wrong, but how is it possible that its negative time?

Answer 17

ok let me check

Answer 18

i know for sure the answer -1/34 is correct, at least according to the auto grading system, but that makes no sense to me

Answer 19

okay well

Answer 20

Is this r(t)
r(t)=<4t^2,−4t,t^2+1t>

Answer 21

@perl yes that is correct

Answer 22

lets think about it if u take r'(t), this is giving u the velocity vectors then |r'(t)| is telling u the speed over time

Answer 23

I get -1/34

Answer 24

so did you take the derivative of s(t) ?

Answer 25

@perl yes I got -1/34 and its correct, im wondering why a negative time is correct