3 Unit Maths HSC Exam Revision (1 Viewer)

[randomnessss]

Not sure for the above question but i think you use logs and u'll get a ans with d.p?

The gradient of a curve is given by y'= . the curbe passes through points (0,2)

Find the question of the curve. How would u answ this?

 

[Rezen]

 

[random-1006]

It is given that current=I= dq/dt, where q is charge , measured in coulombs.

Given that current ( in amps) varies with time t ( in seconds) by I ( t) = 5/ ( 3 +9t^2), find the total amount of charge that passes a fixed point in the 9th second, correct to 2 sig fig.

 

[hscishard]

Not sure for the above question but i think you use logs and u'll get a ans with d.p?

The gradient of a curve is given by y'= . the curbe passes through points (0,2)

Find the question of the curve. How would u answ this?

e itself is a number(constant). e^x is a function.

 

[hscishard]

it is given that current=i= dq/dt, where q is charge , measured in coulombs.

Given that current ( in amps) varies with time t ( in seconds) by i ( t) = 5/ ( 3 +9t^2), find the total amount of charge that passes a fixed point in the 9th second, correct to 2 sig fig.

0.0068

 

[random-1006]

0.00683?

lol, give me 2-3 minutes, i just made that up, have to do it myself lol

 

[random-1006]

0.0068

pretty sure thats wrong,

 

[hscishard]

pretty sure thats wrong,

Well I did get confused.

I did:

I = dQ/dt
I(9) = 5/ 3 + 729
=5/732
=dQ/dt = amount travelling at 9 seconds.

 

[random-1006]

Well I did get confused.

I did:

I = dQ/dt
I(9) = 5/ 3 + 729
=5/732
=dQ/dt = amount travelling at 9 seconds.

i knew it would be a good question

if you do uni physics this technique will become more common ( however i reckon it would still be a reasonable harder 2 unit question to ask

mmmm " total" , this question should look familar to another question you have seen before, lol im not going to give answer yet, just think about it a bit, you will prob get it.

Hint: if i asked you to find the total of : 1, 2 and 3, how would you do it??

 

[hscishard]

i knew it would be a good question

if you do uni physics this technique will become more common ( however i reckon it would still be a reasonable harder 2 unit question to ask

mmmm " total" , this question should look familar to another question you have seen before, lol im not going to give answer yet, just think about it a bit, you will prob get it.

Hint: if i asked you to find the total of : 1, 2 and 3, how would you do it??

I think I know what to do..
You integrate it and find the area from 0 to 9.
But isn't the question asking for the amount at 9 seconds?

 

[random-1006]

I think I know what to do..
You integrate it and find the area from 0 to 9.
But isn't the question asking for the amount at 9 seconds?

in the 9th second, there is a few more tricks than just saying "integrate", its not all that standard, i like questions to test everything lol

 

[hscishard]

Omg. I really suck at reading. Is it asking the amount that's travelling in the 9th second?
If not..
???

 

[random-1006]

Omg. I really suck at reading. Is it asking the amount that's travelling in the 9th second?
If not..
???

ok, in the 9th second.

It said ( im pretty sure it did when i made it up), the total amount of charge that goes past a given ( FIXED) point

a common mistake for this question would be to say that the 9th second is from t=9 to t=10, this is WRONG.

the ninth second is from t=8 to t=9, so your limits would be 9 to 8, but even then i can still see a fair few people making two potential mistakes on that integral, i would make sure you can actually do it, instead of just saying ahh thats a "...." function, and then walk away

 

[hscishard]

ok, in the 9th second.

It said ( " im pretty sure it did when i made it up"), the total amount of charge that goes past a given ( FIXED) point

a common mistake for this question would be to say that the 9th second is from t=9 to t=10, this is WRONG.

the ninth second is from t=8 to t=9, so your limits would be 9 to 8, but even then i can still see a fair few people making two potential mistakes on that integral, i would make sure you can actually do it, instead of just saying ahh thats a "...." function, and then walk away

I have no idea on how to integrate it. Is it inverse trig? Logs don't work..

Man. I was thinking the amount travelling in the 9th second, like 9.00000 exactly. Oh..I was thinking on how much it was increasing. f...

 

[random-1006]

I have no idea on how to integrate it. Is it inverse trig? Logs don't work..

Man. I was thinking the amount travelling in the 9th second, like 9.00000 exactly.

no, if i wanted that i would say what is value of q at t=9, or what is q(9)

remember, in an exam you have an integral sheet lol. Whereever you see an integral, just take 5 seconds and glance at the integral sheet.

 

[random-1006]

so answer = 5 integral 1 /( [ 9 ( [1/3] + t^2) dt
= 5/9 integral 1/ [ ( 1/sqrt3)^2 +t^2 ] dt
= [5/9] / [1/sqrt3] tan^ (-1) [ sqrt3 * t ] between limits of 8 and 9
= [ 5 sqrt3]/ 9 [ tan ^ (-1) [ 9sqrt3] - tan ^ (-1) [ 8sqrt3] ]

common mistakes would be to forget to divide by the highlighted bit, and also, the calculator must be in radian mode when evaluating, reckon would make a good question to sort people out in Q1, worth 3 marks i reckon

 

[random-1006]

 

[hscishard]

3^(1/3) x 2^(1/2)

 

[random-1006]

3^(1/3) x 2^(1/2)

lol wasnt expecting you to do it that way, to be honest i wasnt thinking for someone to do it that way when i wrote the question, i was thinking someone would do

ln ( x^6) = ln9 + ln8
x^6= 8 x 9
x= +- 6th root of ( 72)

but since domain of log > 0 , we discard negative soln

your answer should be the same numerically

 

[hscishard]

lol wasnt expecting you to do it that way, to be honest i wasnt thinking for someone to do it that way when i wrote the question, i was thinking someone would do

ln ( x^6) = ln9 + ln8
x^6= 8 x 9
x= +- 6th root of ( 72)

but since log > 0 , we discard negative soln

your answer should be the same numerically

Totally didn't see that way.

HSC is in one month XD.$0.05 that everyone's super speedily studying.