precalc

@justjm

I am soo stupid. Sorry I didn't post the problem and accidently posted the other thing twice

just need help on a) I don't know what to do so I guess once I know how to do a I can do the rest

Try 4 ×34 and then divide by 3 and see what it gets you

This question serves as an introduction to polar coordinates. So for (a) the '4' minutes is just meant to trick you. Since it's going 3 RPM, after 4 minutes, it's done 12 revolutions but you're still at the same location where you were initially. Hence θ=34° and r=24. Now you can convert to rectangular coordinates using \((x,y)=(rsinθ, rcosθ)\)

so just radius times sin and cos of 34?

\(\color{#0cbb34}{\text{Originally Posted by}}\) @darkknight so just radius times sin and cos of 34? \(\color{#0cbb34}{\text{End of Quote}}\) yes

btw its (rcosθ,rsinθ)

thx

What about for a problem like c where it doesn't end up in the same spot?

\(\color{#0cbb34}{\text{Originally Posted by}}\) @darkknight btw its (rcosθ,rsinθ) \(\color{#0cbb34}{\text{End of Quote}}\) OH shoot you're right sorry my bad. xD

let me take a look..

I believe you need to use angular kinematics \(\frac{d\theta}{dt}=\omega\) \(d\theta=\omega dt \) \( \int d\theta= \int \omega dt \) \(\theta = \omega t+\theta_o\) [you don't need to do the above, just know the formula] \(\theta = \omega t+\theta_o\) \(\omega = 3\ \frac{rev}{\min}\), \(t=6 s\) \(∴ \theta =\left(3\ \frac{rev}{\min}\cdot\frac{1\ \min}{60\ \sec}\right)6\ \sec\ =\ 0.3 ~rev \) \(0.3\ rev\ \cdot\frac{360°}{1\ rev}=108°\) \(108°+(-14°)=94°\)

Similarly in (d) you'll need to use the same equation(s) \(\theta=\omega t\) \(\theta_f = \omega t + \theta_i\)

and once i get the angle (for example 94 degrees for c) I can just plug it into r cos theta and r sin theta and find the coordinates right?

\(\color{#0cbb34}{\text{Originally Posted by}}\) @darkknight and once i get the angle (for example 94 degrees for c) I can just plug it into r cos theta and r sin theta and find the coordinates right? \(\color{#0cbb34}{\text{End of Quote}}\) yeah and r is the same ofc

wow! Thank you

\(\color{#0cbb34}{\text{Originally Posted by}}\) @darkknight wow! Thank you \(\color{#0cbb34}{\text{End of Quote}}\) np