Author Topic: Dekoyl's Questions (Read 24844 times) Tweet Share

dekoyl · « **Reply #135 on:** October 28, 2009, 11:06:28 am »

Could someone help me understand what each component represents and the subsequent question?

A ball has position vector $r(t)=(8t)i+(26t)j+\left(\frac{5\sqrt{2}}{2}sin\left(\frac{\pi t}{2}\right)\right)k$ , where i and j are unit vectors horizontally and vertically forward respectively, and k is a unit vector in the vertical direction.

Since I couldn't 'visualise' the above components, I had trouble finding (and understanding the solution) of what angle the ball leaves the ground at.

Thanks!

humph · « **Reply #136 on:** October 28, 2009, 11:52:05 am »

Visualise it by keeping the vertical direction fixed at first, so you just imagine it moving it in the plane. It's clear then that it's a straight line pointing north-by-north-east. Then factor in the vertical movement, which is sinusoidal, so it's moving up and down with wave motion as it travels north-by-north-east.

dekoyl · « **Reply #137 on:** October 28, 2009, 03:43:56 pm »

An easy question :/

Is $sin(46^{o}) = cos(44^{o})$ ? Is there a symmetrical identity that shows this?

Thanks

TrueTears · « **Reply #138 on:** October 28, 2009, 04:05:25 pm »

Mao · « **Reply #139 on:** October 29, 2009, 11:36:58 am »

the velocity vector is constant in the i and j direction, and it oscillates in the k direction

This implies it is a wave-like motion travelling in the 8i + 26j direction.

dekoyl · « **Reply #140 on:** October 31, 2009, 07:48:15 pm »

~~What am I doing wrong here?~~

Find the exact area of the region $\{z:\frac{\pi}{6}\leq Arg(z) \leq\frac{\pi}{3}\}$ INTERSECT $\{z: Re(z) + \sqrt{3}Im(z) \leq 4\}$

I then found the equations of the lines given by $\{z:\frac{\pi}{6}\leq Arg(z) \leq\frac{\pi}{3}\}$ and did:
Area = $\int_0^1\left(\sqrt{3}x-\frac{1}{\sqrt{3}}x\right)dx + \int_1^2\left(\frac{1}{\sqrt{3}}(4-x) - \frac{1}{\sqrt{3}}x\right)dx$

The solutions found the area by doing some things then finding the area of the triangle.

Nevermind. Method not wrong but rather I typed it in the calc wrong. Ha.

TonyHem · « **Reply #141 on:** October 31, 2009, 07:55:00 pm »

I dunno if this is right but is this how you interpreted the question?
I can kind of see a triangle there

dekoyl · « **Reply #142 on:** October 31, 2009, 07:58:07 pm »

Yep that's how I saw it. I found the intersections (which was x=1, x=2) and separated them accordingly.

There is a triangle but they had to do stuff like finding the distance of the points of the triangle from the origin and yeah. I'm just not sure what I did wrong, though.

dekoyl · « **Reply #143 on:** October 31, 2009, 09:10:08 pm »

Hmm.

To convert metres cubed into litres, do I times it by 1000? Is there any other conversions I should be aware of?

Thanks!

dejan91 · « **Reply #144 on:** October 31, 2009, 09:13:04 pm »

What about L to mm cubed?? Came across one today.

TrueTears · « **Reply #145 on:** October 31, 2009, 09:15:57 pm »

I always work everything out from $1 cm^3 = 1 mL$

Everything else is derived from that so for example:

$1 m^3 = ? L$

$100 cm = 1 m$

$100^3 cm^3 = 1 m^3$

Thus $100^3 cm^3 = 100^3 mL$

$1 L = 1000 mL$

So $1000 L = 100^3 mL = 1 m^3$

So yeah just derive it quickly for yourself.

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