Easy integration q (2 Viewers)

[lyounamu]

Quick check on this question. I think solution is screwed.

Q 2U97-9b iii)

Is it just me who gets 14cm all the time? (which yields 0.7mg)





http://www.boredofstudies.org/courses/maths/2u/1202631531_1990_Mathematics_HSC.pdf

Page 20/31 sorry!

 

[Js^-1]

1997 HSC paper?

 

[-tal-]

Umm.. problem: where's the question?

 

[lyounamu]

Sorry. I posted up above.

Sorry again!

Page 20.

 

[-tal-]

I see lots of integration questions, but I can't seem to see the one you're talking about...

edit: right.. now i do

 

[minijumbuk]

I got 0.7 mg as well.

 

[lyounamu]

I got 0.7 mg as well.

Thanks. I am assured now.

 

[minijumbuk]

Who set the solutions?

 

[Dota55]

0.7 is correct methinks...

 

[Js^-1]

Yeh just got 0.7...

 

[lyounamu]

Who set the solutions?

I am guessing that another BoSer did this. But everyone is prone to some sort of mistakes, I guess.

 

[-tal-]

eh, too slow. Just got 0.7

 

[lyounamu]

Can everyone look at page 24 Q 2U93-5b? ii)

I got x = 2, 4 and 6 but they only said 2 and 4.

I am wondering why x = 6 is not the answer .Thanks.

 

[Dota55]

Can anyone here solve part (ii) of the question top of page 25?

 

[Dota55]

Can everyone look at page 24 Q 2U93-5b? ii)

I got x = 2, 4 and 6 but they only said 2 and 4.

I am wondering why x = 6 is not the answer .Thanks.

Its part of the curve from x=4 to x=8

 

[lyounamu]

Can anyone here solve part (ii) of the question top of page 25?

That's easy.

AB = square root of ((2+1)^2 + (4-1)^2)) = square root of 18.

Perpendicular distance between AB and P = absolute value of (-t+t^2-2)/square root of 2

Therefore, A = 3 . absolute value of (-t+t^2-2) = -3t + 3t^2 - 6 or 3t - 3t^2 + 6
dA/dt = -3 + 6t or 3-6t
So t = 1/2 or t = -1/2 when the maximum area occurs.

Put each value in and you will get: A = -6.75 or 3.75

Since the A > 0, maximum area is 3.75 which is 3/4 of 4.5

 

[lyounamu]

Its part of the curve from x=4 to x=8

You sure? It sure doensn't look like a curve that can be described by one function.

 

[Dota55]

You sure? It sure doensn't look like a curve that can be described by one function.

No, you're right. Should've had a harder look.

Umm....tbh i don't know why the points x=2,4 wouldn't be differentiable in the first place. Sorry.

 

[lyounamu]

No, you're right. Should've had a harder look.

Umm....tbh i don't know why the points x=2,4 wouldn't be differentiable in the first place. Sorry.

They cannot be differentiated because it is possible to find the gradient at those points.

 

[Dota55]

That's easy.

AB = square root of ((2+1)^2 + (4-1)^2)) = square root of 18.

Perpendicular distance between AB and P = absolute value of (-t+t^2-2)/square root of 2

Therefore, A = 3 . absolute value of (-t+t^2-2) = -3t + 3t^2 - 6 or 3t - 3t^2 + 6
dA/dt = -3 + 6t or 3-6t
So t = 1/2 or t = -1/2 when the maximum area occurs.

Put each value in and you will get: A = -6.75 or 3.75

Since the A > 0, maximum area is 3.75 which is 3/4 of 4.5

don't get....why A=3....