Question

a particle moves along the x axis so that at any time t>0, its velocity is v(t)=4 - 6t^2. if the particle is at position x=7 at time t=1, what is the position of the particle at time t=2?

Answer 1

to find position, s(t), i integrated v(t) and got this\[s(t)=4t-2t ^{3}+c\]

Answer 2

how do i find the value of c?

Answer 3

when t=1,s=7

Answer 4

so c=7?

Answer 5

7=4-2+c
c=5

Answer 6

find s when t=2

Answer 7

thank you

Answer 8

\[r=2\cos(\theta/2) => dr/d \theta=-2\sin(\theta/2)(1/2)=-\sin(\theta/2)\]

Answer 9

yw.

Answer 10

\[x=rcos(\theta) => dx/d(\theta)=-rsin(\theta)+\cos(\theta)dr/d(\theta)=-rsin(\theta)-\cos(\theta)\sin(\theta/2)\]

Answer 11

you can ignore these numbers; i'm trying to solve something

Answer 12

\[y=rsin(\theta) =>dy/d(\theta)=rcos(\theta)+\sin(\theta)dr/d(\theta)=rcos(\theta)+\sin(\theta)\sin(\theta/2)\]

Answer 13

\[dy/dx= [ rcos(\theta) + \sin(\theta)\sin(\theta/2) ]\div [rcos(\theta)+\sin(\theta)\sin(\theta/2)]\]

Answer 14

"Substituting θ=pi into this formula gives us 0, which is the slope of the tangent line to the curve at this point. therefore the line normal to this curve will have undefined slope." the book says the answer is "undefined."

Answer 15

as you have written \[\frac{ dx }{ d \theta }=-r \sin \theta-\cos \theta \sin \frac{ \theta }{ 2 }\]
\[\frac{ dy }{ d \theta }=r \cos \theta+\sin \theta \sin \frac{ \theta }{ 2 }\]
\[\frac{ dy }{ dx }=\frac{ r \cos \theta+\sin \theta \sin \frac{ \theta }{2} }{ -r \sin \theta-\sin \theta \sin \frac{ \theta }{ 2 } }\]
when we put \[\theta=\pi ,\sin \pi =0\]
so denominator is 0 ,hence slope undefined.