Multivariable calculus help anyone please? :(

- anonymous

Multivariable calculus help anyone please? :(

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- anonymous

##### 1 Attachment

- anonymous

How did the get l2(t) = (9-9t,9t) ?

- anonymous

It's not the equation of the line for the triangle plate either.

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## More answers

- anonymous

@oldrin.bataku

- UnkleRhaukus

|dw:1378015226650:dw|

- UnkleRhaukus

|dw:1378015316818:dw|

- anonymous

Yep. I got that.

- UnkleRhaukus

|dw:1378015416051:dw|

- anonymous

Yep. I got that.

- UnkleRhaukus

|dw:1378015498731:dw|

- anonymous

Just want to know about the hypotenuse though.

- anonymous

Yeah. What's the relation there? I got y=-x+9 but that's wrong.

- UnkleRhaukus

yeah thats kinda right
for the second interval , (the hypotneuse)
\[\ell_2(t)=(9-t,t)\]
but that would be for \[\text{if \(t\) is}\qquad0\leq t\leq 9\]

- UnkleRhaukus

notice how for some reason in the question thay have
0≤t≤1

- anonymous

How do the inequalities change it?

- UnkleRhaukus

well the inequality is defining the domain of t ,
this has the same effect as scaling the terms in the the expression

- anonymous

What is t exactly? The "region" ?

- UnkleRhaukus

its kinda like the position on one of those boundry lines

- anonymous

Right. Yeah.

- anonymous

Some arbitrary position. Got it.

- UnkleRhaukus

|dw:1378016110635:dw|

- anonymous

Still confused as to how the inequalities affect the scaling...

- anonymous

|dw:1378016191403:dw|

- anonymous

Kinda like that.

- UnkleRhaukus

(position/position time)
I guess that if the \(t\) is the same for all the lines, then
one interpretation might be that something is moving around the boundry and it moves quickest on that hypotenuse line

- anonymous

Makes sense because the gradient changes the most rapidly.

- anonymous

But how does 9-t become 9-9t? That's bugging me.

- anonymous

Actually I can kinda see it.

- anonymous

Because as the domain gets smaller and smaller, the function changes more rapidly as the interval becomes shorter.

- anonymous

So since the equation is linear it changes 9x as fast.

- anonymous

At least that's my reasoning.

- UnkleRhaukus

line two goes form (9,0) to (0,9)
\[ℓ_2(t)=(9−9t,9t)\qquad0≤t≤1\]
lets see what happens if we set
\[s=9t\]
\[ℓ_2(s)=(9−s,s)\qquad 0≤s≤9\]

- UnkleRhaukus

from*

- UnkleRhaukus

- anonymous

What was the point of setting s to that value? :/ .

- anonymous

Sorry if I am being stubborn. Something is just not clicking.

- UnkleRhaukus

it just makes that particular interval simpler, to see whats happening ,

- anonymous

Is my reasoning correct for reason the slope/gradient changed?

- UnkleRhaukus

can you try and re-word it

- anonymous

Like if the gradient is 9 for 0

- UnkleRhaukus

the gradient on the hypotenues is -1

- anonymous

Right sorry. But is the rest of it correct?

- anonymous

Because the region was scaled down by 1/9 so the rate of change increases by 9 times as fast.

- UnkleRhaukus

its not the region that has been scaled, it it the parameter t that has been scaled up on ℓ_2

- anonymous

Because it's over a much smaller interval, so the rate of change is much greater.

- anonymous

Yeah okay, that makes sense. And since this is a linear relationship it would change 9x faster.

- UnkleRhaukus

yeah that is kinda right

- anonymous

Kinda?

- anonymous

But okay, that makes tons of sense now.

- anonymous

Thanks :) .

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