Maths Methods 3/4 Help Thread 2011

[BlueSky_3]

The height of the tide, h metres, at the entrance to a point is given by h= 3 sin( t(pi x a)/4 ) + 8.
where t is the number of hours after midnight and a is a positive integer. The number of times the height is 10m within the first 12 hour day is:

Thanks in advance.

[brightsky]

solve 10 = 3 sin (t(pi*a))/4) + 8 within the domain 0 =< t =< 12 then count the number of solutions.

[BlueSky_3]

Oh its multiple choice and the answer is 6a? Do you happen to know how?

[brightsky]

in that case:
y = 3 sin (t(pi*a))/4) + 8
5 =< y =< 11
so we know it passes 10 twice at every period
period = 8/a
the part of the graph we're looking at is t E [0,12]
so how many periods go into 12?
12a/8
so how many times does it pass 10, 24a/8 = 3athis part is actually wrong, my bad. looked good at the time lol
i think the question should be a 24 hour day if the answer is 6a..but my solution is a bit dodgy

EDIT: if the question is in exam 2, just sketch the graph and find how many times it crosses y = 10.

[luken93]

in that case:
y = 3 sin (t(pi*a))/4) + 8
5 =< y =< 11
so we know it passes 10 twice at every period
period = 8/a
the part of the graph we're looking at is t E [0,12]
so how many periods go into 12?
12a/8
so how many times does it pass 10, 24a/8 = 3athis part is actually wrong, my bad. looked good at the time lol
i think the question should be a 24 hour day if the answer is 6a..but my solution is a bit dodgy

EDIT: if the question is in exam 2, just sketch the graph and find how many times it crosses y = 10.

Wouldn't it BE at 10m twice every period (in first and second quads)? If so, multiply your 3a x 2 = 6a?

[brightsky]

in that case:
y = 3 sin (t(pi*a))/4) + 8
5 =< y =< 11
so we know it passes 10 twice at every period
period = 8/a
the part of the graph we're looking at is t E [0,12]
so how many periods go into 12?
12a/8
so how many times does it pass 10, 24a/8 = 3athis part is actually wrong, my bad. looked good at the time lol
i think the question should be a 24 hour day if the answer is 6a..but my solution is a bit dodgy

EDIT: if the question is in exam 2, just sketch the graph and find how many times it crosses y = 10.

Wouldn't it BE at 10m twice every period (in first and second quads)? If so, multiply your 3a x 2 = 6a?

yeah that's what i meant, probably didn't express it clearly enough. either way, it doesn't change the answer. (i multiplied by 2 in the last step already)

[BlueSky_3]

Yeah I had a similar approach to yours Brightsky but the latter part of your solution confused me, how did you get 24a/8? And yeah I assume its in exam 2 since its multi-choice but a quick sketch only reveals that the no. of solutions per period is 2 and nothing about the relationship between the period and no.of solutions?

[Pixon]

Err...I'm pretty sure Brightsky is correct to say this question should be a "24 hour day". (I mean...a "12 hour day" is not a thing )

The question becomes too difficult for a MCQ if you deal with incomplete periods where you might have just one intersection for a period. Thus it makes sense for the question to ask for a 24 hour day where you can be sure at t=24, the period starts again. (The period is 8/a and 24 is a multiple of 8 ). With a 24 hour day, you get 6a.

edit: I should clarify that the period I'm referring to is from [0,8/a] considering that we're starting from t=0. (any other starting point for a period would run into a similar issue to what I stated above)

[brightsky]

Yeah I had a similar approach to yours Brightsky but the latter part of your solution confused me, how did you get 24a/8? And yeah I assume its in exam 2 since its multi-choice but a quick sketch only reveals that the no. of solutions per period is 2 and nothing about the relationship between the period and no.of solutions?

yeah as i said, my solution isn't really correct. it's too hard to find by hand how many solutions there are if we have 'incomplete periods' so to speak as pixon put it. if the domain was t E [0,12], then i think there would be 4a solutions instead of 3a. but meh, if you have a calculator, draw the sine graph then draw the graph y = 0, and look within the given domain to find how many times it crosses the line. otherwise just solve the equation for the given domain and count the number of solutions that pops up.

[paulsterio]

The height of the tide, h metres, at the entrance to a point is given by h= 3 sin( t(pi x a)/4 ) + 8.
where t is the number of hours after midnight and a is a positive integer. The number of times the height is 10m within the first 12 hour day is:

Thanks in advance.

Period = 2pi/n
n = (pi*a)/4

so period = 8/a

there will be 2 intersections per period, hence in 12 hours, there will be 12/(8/a) x 2 = 3a

i think the answer of 6a is wrong

[Romperait]

Considering pixon and brightsky have both stated the question to be incorrect...why would you go against them? Furthermore...consider a=1. You end up with 4 intersections, 3a=3 which is not 4.

[paulsterio]

Nah, I see my mistake now, I didn't consider the partial periods, but yeah I think the question itself is incorrect now, in regards to the 12 hour day, should be 24

[nacho]

Pr(X=x) = (x-4)(1+2x)-1/30
where X may take values of 0, 1, 2 and 3

Find the probability of X exceeding its mean.
I drew a Probability distribution up,
did that thing were you go from left to right and add the probabilities of Pr(X=x) until its 0.5 or greater.
which was Pr(X<2)
so then i found Pr(X>2) = 7/30
any idea as to what i did wrong? The answer is 17/30

[Greatness]

Multiply the function by x then do the integral from 0 to 3 to find the mean, then find the probabilty greater than that.

hmm..actually i dont think thats right :/

[b^3]

Multiply the function by x then do the integral from 0 to 3 to find the mean, then find the probabilty greater than that.

Isn't this a discrete distribution since X can only take 0,1,2,3?
And are you sure the equation is right? I'm getting negative probabilities here...