Question

If r(t)= 3t^2 i + 1t j + (t^2-10t) k gives the position of a particle at time , find the time at which the speed of the particle is minimized.

Answer 1

looking for t=?

Answer 2

I get t=1/2

Answer 3

i need answer for the same question ..... looking for t= ?

Answer 4

t =1/2 is not correct :(

Answer 5

\[v(t)=[6t,1,2t-10]\]

Answer 6

i need t=? not v(t)

Answer 7

\[s(t)=\|v(t)\|=\sqrt{(6t)^2+1^2+(2t-10^2}\]

\[=\sqrt{40t^2-40t+101}\]

Answer 8

now you want to find the time for which the speed is a minimum.

Answer 9

ok

Answer 10

take \[\frac{d}{dt}s(t)\]

set equal to zero and solve for t

Answer 11

still im not getting the right answer what u get for t ?

Answer 12

I get 1/2

Answer 13

this question is from web work so it telling me is not corect

Answer 14

is the problem typed correctly...is r(t) correct above?

Answer 15

If r(t)= -(t^2)i+(8t)j+((t^2)-3t)k gives the position of a particle at time , find the time at which the speed

Answer 16

thats my questions

Answer 17

well then you should get a different answer

Answer 18

ok can u solve for this one

Answer 19

coz i did same way
but still i m getting wrong

Answer 20

3/4

Answer 21

can u help me plz

Answer 22

ya now is corect

Answer 23

can u help me with diffrent question tooo

Answer 24

if you do the procedure I did above with your problem you will get the 3/4

Answer 25

ya now i get it i just miss _2

Answer 26

i need help with this question

Answer 27

Find the length of the curve given by r(t)=((2)^(1/2))*ti/2+e^(t/2))j+(e^(-t/2))k where -2<=t<=1

Answer 28

just compute

\[\int\limits_{-2}^{1}\sqrt{\left[\frac{dx(t)}{dt}\right]^2+\left[\frac{dy(t)}{dt}\right]^2+\left[\frac{dz(t)}{dt}\right]^2}dt\]

Answer 29

ok then ?

Answer 30

first part is 1/2

Answer 31

then ?

Answer 32

plz r u helping me coz i dont no how \[\int\limits_{?}^{?}\] this eq