How far below the x-axis is the beam when x=0? (1 Viewer)

[Faytle]

What did you guys get for this Q? It was right at the end of Q9.

If I remember correctly, my answer was that the beam is k5b^4/12 units below the x-axis. Most likely wrong, but I'm curious to know how everyone else approached it!

 

[rodders1]

Hey

that is what i got. What did you think about the rest of the exam. I thought it was good

 

[Foreshadowed]

I got the same thing.

I integrated it and found c, and the constant defines whether the graph shifts up or down so yeh.

 

[Faytle]

Oh, wow. Think we're on the right track then!

I thought it was a relatively easy paper. Just starting to realise a few silly mistakes that I made though .

 

[chook 05]

naw i got -1/6kb^4


maybe my f(x) is wrong?

f(x) = 1/2 k (b^2 x^2 - 1/3 x^4 - 1/3 b^4)

 

[2S1D3]

yea thats what i got

 

[ke2004]

hey im crap as at probablity- so can anyone tell me the answers to 7c??
thanks!

 

[zxcvbee]

got the same buddy!

 

[dolbinau]

was it a fraction you found the cd and simplified? I think I got something like that but didn't simplify

 

[munchiecrunchie]

OP yep thats correct.

 

[tybermon j]

i think i got that as well..
That has made me rethink my decision to jump off a cliff

 

[henry08]

I had no idea. I guessed 5 lol.

 

[ipod21]

i left my answer as y= - k5b^4/12

im wrong casue i left it like that...fuck!

 

[sasha279]

How did you find C?

I let f(b) = f(-b) but it didn't work... Got C = C hehe ^_^;

 

[farkmedead]

How did you find C?

I let f(b) = f(-b) but it didn't work... Got C = C hehe ^_^

did u do it properly? it took me a while to get it right too, but you find that c=o.

 

[YoungToby182]

Dont worry about it, in the past now

Theres nothing you can do now.

 

[Lanser7]

i got 10kb^4/24 because c=-10kb^4/24 and because it said how far BELOW the x axis that made it positve.

ps i only just realsied that simplifies to 5kb^4/12

 

[Wingom]

i'm kinda pissed.. coz i got the method right.. just a silly silly mistake...
which altered my answer, but I agree with it being 5kb^4/12..

 

[Zephyrio]

Looks good to me.

 

[r.uss]

-5kb^4/12 is correct. It would have been your constant when you integrated f'(x).

I'm seriously really wondering how on earth they're going to differentiate between everyone...