Question

picture question

Answer 1

@jim_thompson5910 @satelite73 @mathstudent55

Answer 2

@aum will you please help

Answer 3

I got 1.5 for that part but it was saying it's wrong. The professor sent out an email saying the picture wasn't very good so it may be something different, but it will be a .25 interval. I tried 1.75 and that wasn't it so it might try 1.25

Answer 4

1.5 should be correct :U
That's very strange...

Answer 5

I guess he means the picture is wrong.

Answer 6

can't you use \[s=r \theta \] to solve it?

Answer 7

Yeah, he said the picture might be off so it might not be what it looks like (which is kinda lame)

Answer 8

s=r(theta) gives us the arc length of a piece of a circle.
Hmm I don't think that's going to help us with our parabola :d

Answer 9

The only thing I can think of is..... maybe....
If we find an equation for the second graph,
I keep calling it a parabola, but maybe it's the first hump of the sine curve.

Answer 10

It completes the first hump in t=4,
so it has a full period of 8.

\[\Large\rm 8=\frac{2\pi}{b}\qquad\to\qquad b=\frac{\pi}{4}\]
And it goes as high as 4.
So maybe we have...\[\Large\rm \theta(t)=4\sin\left(\frac{\pi}{2}t\right)\]
Evaluating at t=3.5 gives us:\[\Large\rm \theta(3.5)=4\sin\left(\frac{\pi}{2}(3.5)\right)\]Which simplifies to,\[\Large\rm \theta(3.5)=\]

Answer 11

\[\Large\rm \theta(3.5)=1.530734\]

Answer 12

Maybe that's the value they're looking for?

Answer 13

Woops I plugged in pi/2 by mistake in that step* Ignore that.
That shoulda been a pi/4.

I put it in my calculator correctly though.

Answer 14

That would make sense. That's the equation I got for sine.