HSC 2015 MX1 Marathon (archive) (1 Viewer)

[Drsoccerball]

Re: HSC 2015 3U Marathon

Which expansions are you comparing the coefficients of?

You expand the LHS ceebs writing out the expansion

 

[davidgoes4wce]

Re: HSC 2015 3U Marathon

From the Normanhurst trials 2015.

Q4. Find the acute angle between the tangents to the graphs y=x and y=x^3 at the point (1,1).

A) 27 degrees B) 30 degrees C) 45 degrees D) 63 degrees


This is their solution:



I had a value of 29.52 degrees, which is closer to 30 degrees.

 

[rand_althor]

Re: HSC 2015 3U Marathon

I had a value of 29.52 degrees, which is closer to 30 degrees.

Did you end up with though?

 

[davidgoes4wce]

Re: HSC 2015 3U Marathon

Did you end up with though?

Yes i did

 

[rand_althor]

Re: HSC 2015 3U Marathon

Well when I use my calculator to find , I get . Are you sure you entered it into your calculator correctly?

 

[davidgoes4wce]

Re: HSC 2015 3U Marathon

Well when I use my calculator to find , I get . Are you sure you entered it into your calculator correctly?

OK there is my problem I was using 'GRAD'.

 

[rand_althor]

Re: HSC 2015 3U Marathon

Okay I think I've figured out your problem. Your calculator isn't in degrees, it is in gradians.

EDIT: Nevermind didn't see your post.

 

[davidgoes4wce]

Re: HSC 2015 3U Marathon



I understood part (i) and part (ii)

The answer for (i) was ln 2 and the answer for (ii) was 25/36

I have no idea on where to start on part (iii) any suggestions?

 

[Carrotsticks]

Re: HSC 2015 3U Marathon

I understood part (i) and part (ii)

The answer for (i) was ln 2 and the answer for (ii) was 25/36

I have no idea on where to start on part (iii) any suggestions?

The actual value, via integration, is ln(2).

So ln(2) ~ 25/36 (assuming your calculation is correct)

e^25/36 ~ 2

e ~ ....

 

[davidgoes4wce]

Re: HSC 2015 3U Marathon

OK thanks I understand that now

 

[davidgoes4wce]

Re: HSC 2015 3U Marathon

Any ideas on how often Simpsons Rule pops up in the 2 Unit or 3 Unit exams? I must admit its just a formula that has come back to me today and tutoring my student last night, I guess thats why exam papers are good because it brushes off the cobwebs after not learning it for a period of time.

 

[Drsoccerball]

Re: HSC 2015 3U Marathon

Any ideas on how often Simpsons Rule pops up in the 2 Unit or 3 Unit exams? I must admit its just a formula that has come back to me today and tutoring my student last night, I guess thats why exam papers are good because it brushes off the cobwebs after not learning it for a period of time.

vat

 

[Carrotsticks]

Re: HSC 2015 3U Marathon

Any ideas on how often Simpsons Rule pops up in the 2 Unit or 3 Unit exams? I must admit its just a formula that has come back to me today and tutoring my student last night, I guess thats why exam papers are good because it brushes off the cobwebs after not learning it for a period of time.

Appears often in 2U. Rarely in 3U.

 

[davidgoes4wce]

Re: HSC 2015 3U Marathon



had a bit of a problem with the last part (iv)

 

[davidgoes4wce]

Re: HSC 2015 3U Marathon

^^^^^^^This question was the last question in the Normanhurst Trials 2015



My thinking was to do the derivative and then plot it on a graph but the solution has a totally different solution to me.

 

[davidgoes4wce]

Re: HSC 2015 3U Marathon



Here is my graph of the velocity function , my thinking was the highest velocity occurs at t=3pi/4

 

[braintic]

Re: HSC 2015 3U Marathon

Here is my graph of the velocity function , my thinking was the highest velocity occurs at t=3pi/4

You'd better read the question again - it's not asking for the highest velocity.

 

[davidgoes4wce]

Re: HSC 2015 3U Marathon

Acceleration is the greatest at the ends of a function. I should do dv/dt=0, is where acceleration is greatest. Hence, t=pi/2 is the time when velocity is increasing most rapidly.

 

[rand_althor]

Re: HSC 2015 3U Marathon

Acceleration is the greatest at the ends of a function. I should do dv/dt=0, is where acceleration is greatest. Hence, t=pi/2 is the time when velocity is increasing most rapidly.

What do you mean by ends of a function? Also dv/dt is the rate of change of velocity, aka acceleration, so dv/dt=0 would give you where acceleration is zero, not where it is greatest. You do have the right answer though.

 

[davidgoes4wce]

Re: HSC 2015 3U Marathon

What do you mean by ends of a function? Also dv/dt is the rate of change of velocity, aka acceleration, so dv/dt=0 would give you where acceleration is zero, not where it is greatest. You do have the right answer though.

My bad I should have said the end points in Simple harmonic Motion (i.e where the particle turns)

At the end points the restorative force and acceleration are at a maximum.