VCE Methods Question Thread!

[MB_]

Pretty tricky question...    You can ultilize the fact the turning and infection point happened to be at the x-axis, so you can write the equation for graph as:


Expand out the whole thing, and you can equation each coefficients to the original equation and find h/a.

I'm a bit confused with what I have to do after expanding, could you explain that again?

[MandhreeE]

Hey guys, I was hoping someone could explain step by step how to answer the following two questions because the answers are very brief and I can't seem to follow them. Thank you so much!
Hope everyone's exams are going well
PS I have attached a picture of it but I don't if I have done it correctly. :/

[rnagar12]

Hey guys, I was hoping someone could explain step by step how to answer the following two questions because the answers are very brief and I can't seem to follow them. Thank you so much!
Hope everyone's exams are going well
PS I have attached a picture of it but I don't if I have done it correctly. :/

I think this should be correct. Hope it helps.

[MandhreeE]

I think this should be correct. Hope it helps.

Thank you so much! just wondering though how did you get U^2 in the first question?

[MB_]

Thank you so much! just wondering though how did you get U^2 in the first question?

Because \(u=\sqrt{x}, x=u^2\)

[qwertyu1234567]

I think this should be correct. Hope it helps.

How did you get ln(1/2)?

[esthertham58]

Hi everyone,
new to this thread!
do you guys get " True" on your calculator when you try  to solve the derivative of something? And what should I do? Thanks!

[MightyBeh]

How did you get ln(1/2)?

-ln(2) and ln(1/2) are the same thing. It comes from the power law, because (1/2) can also be written 2^(-1), so the negative power can come out and you have a cleaner logarithm overall.

Hi everyone,
new to this thread!
do you guys get " True" on your calculator when you try  to solve the derivative of something? And what should I do? Thanks!

Could you post a picture? I've never had that happen.

[MB_]

How did you get ln(1/2)?

\(-log_a(x)=log_a(\frac{1}{x})\)

\(-log_e(2)=log_e(\frac{1}{2})\)

[wyzard]

I'm a bit confused with what I have to do after expanding, could you explain that again?

Oops, I meant equate the coefficients. After the expansion, the coefficients for each term must equal each other. For instance, the constant term at the back equals to 108a, so h must be equal to 108a by comparison.

[esthertham58]

-ln(2) and ln(1/2) are the same thing. It comes from the power law, because (1/2) can also be written 2^(-1), so the negative power can come out and you have a cleaner logarithm overall.
Could you post a picture? I've never had that happen.

Picture attached- thanks!

[esthertham58]

Sorry sent the same picture twice!

[P.GUAN]

is 5B correct?

[MightyBeh]

Picture attached- thanks!

Looks like you're trying to multiply the function v by ln(9). You need to write it as v(t)  or whatever your main variable is because in your current expression there's not actually any ts for your derivative to be a function so it interprets the expression to be a constant and is 0. That's why it's coming up as 'true' and not giving you a number.

Try this:

Code: [Select]

solve(d/dt(ln(9)*v(t)) = 0, d)

is 5B correct?

Pretty sure but couldn't you have also just done Pr(X=2) + Pr(X=3)? Seems like it'd be a bit faster. Not sure if it'd be worth the whole two marks though.

[P.GUAN]

For 5B it says 2 or more, does that effect the answer? I think I only did it to exactly 2.