3U Maths Question Thread

[bluecookie]

Prove: d/dx tan^-1(cosecx+cotx)=-1/2

[RuiAce]

Prove: d/dx tan^-1(cosecx+cotx)=-1/2

Just note that internationally csc is used to denote cosec. And it's what LaTeX supports. So if you see csc, reinterpret it as cosec.

[yasmineturner]

Hello,

I was wondering if someone could please assist me with the following question:

"An object is projected vertically upwards with a speed of 49m/s. Two seconds later another object is projected vertically upwards from the same spot at the same speed. Find when and where the objects meet."

Thankyou so much 

[Syndicate]

Hello,

I was wondering if someone could please assist me with the following question:

"An object is projected vertically upwards with a speed of 49m/s. Two seconds later another object is projected vertically upwards from the same spot at the same speed. Find when and where the objects meet."

Thankyou so much 

x1 = x2
t1 = t2+2

x1 = 49 x t1 - 4.9 x t1^2
x2 = 49 x (t1+2) -4.9 x (t1+2)^2

equate the equations.

49t - 4.9t^2 = 49t + 98 - 4.9t^2 -19.6t -19.6
19.6t = 117.6
t1 = 6 seconds
t2 = 4 seconds

sub t =6 into the first equation.
x = 117.6 m

So at t = 6 seconds, both the projectiles have the same displacement of 117.6 metres.

EDIT: just to add: I have assumed that you are doing physics by using a constant acceleration formula ( x = ut + 1/2at^2). I don't really know if this can be applied to 3U maths (hopefully one of the mods can confirm if this working out is valid in HSC maths).

[RuiAce]

x1 = x2
t1 = t2+2

x1 = 49 x t1 - 4.9 x t1^2
x2 = 49 x (t1+2) -4.9 x (t1+2)^2

equate the equations.

49t - 4.9t^2 = 49t + 98 - 4.9t^2 -19.6t -19.6
19.6t = 117.6
t1 = 6 seconds
t2 = 4 seconds

sub t =6 into the first equation.
x = 117.6 m

So at t = 6 seconds, both the projectiles have the same displacement of 117.6 metres.

EDIT: just to add: I have assumed that you are doing physics by using a constant acceleration formula ( x = ut + 1/2at^2). I don't really know if this can be applied to 3U maths (hopefully one of the mods can confirm if this working out is valid in HSC maths).

Not at home right now to do the full answer. (I'll probably attend to it when I come home)

But no. For any equations of projectile motion, you must start with the fundamental assumptions of x"=0 and y"=-g, and integrate your way back up to the displacement.

[Syndicate]

Not at home right now to do the full answer. (I'll probably attend to it when I come home)

But no. For any equations of projectile motion, you must start with the fundamental assumptions of x"=0 and y"=-g, and integrate your way back up to the displacement.

I basically applied the stuff from VCE physics (as this kind of question in more sort of related to VCE physics, than any of the VCE maths imo) 

Even if you integrate your way back up, wouldn't that still yield the same answer? (Sorry, I haven't really done such a question for a while now)

[legorgo18]

Hello friends its me again

Dont understand this inequality integration thing at all would love some guidance through these 2 examples

1) Without evaluating the integrals, explain whether int pi (up) 0 (low) sin^4 x dx > int pi(up) 0 (low) sin4x dx is true or false

2) Use lnt = int t(up) 1 (low) dx/x for t>1 to deduce that 1- 1/t < or equal to lnt < or equal to 1/2 (t- 1/t) for t> or equal to 1.

tyyyy

[RuiAce]

I basically applied the stuff from VCE physics (as this kind of question in more sort of related to VCE physics, than any of the VCE maths imo) 

Even if you integrate your way back up, wouldn't that still yield the same answer? (Sorry, I haven't really done such a question for a while now)

I'll check later. But yes, I don't believe your answer will be wrong

[jamonwindeyer]

I basically applied the stuff from VCE physics (as this kind of question in more sort of related to VCE physics, than any of the VCE maths imo) 

Even if you integrate your way back up, wouldn't that still yield the same answer? (Sorry, I haven't really done such a question for a while now)

Looks perfect to me Syndicate! Cheers! To yasmine, IF you weren't given the formula to use in the question, you'll need to derive them as Rui suggested - Let us know if you need help with that! But Syndicate's answer is perfect once you have those equations

[jamonwindeyer]

Hello friends its me again

Dont understand this inequality integration thing at all would love some guidance through these 2 examples

1) Without evaluating the integrals, explain whether int pi (up) 0 (low) sin^4 x dx > int pi(up) 0 (low) sin4x dx is true or false

2) Use lnt = int t(up) 1 (low) dx/x for t>1 to deduce that 1- 1/t < or equal to lnt < or equal to 1/2 (t- 1/t) for t> or equal to 1.

tyyyy

Hey! This relies on a little bit of intuition

For the first one, remember that definite integrals correspond to areas! The integral of a function from \(\int^\pi_0f(x)dx\) for any function corresponds to the area between the function and the x-axis. So, the question is, is the area between \(\sin^4{x}\) and the x-axis larger than the area between \(\sin{4x}\) and the x-axis (both from 0 to \(\pi\))? The answer is yes, that is immediately clear if we examine them graphically (the \(\sin{4x}\) is in light blue):



For \(\sin{4x}\), the value is zero! The areas below cancel the areas above! So, the statement MUST be true!

Sorry, I'm having a little trouble interpreting Q2 right now; I'll try again in the morning if no one else has had a crack

[RuiAce]

Looks perfect to me Syndicate! Cheers! To yasmine, IF you weren't given the formula to use in the question, you'll need to derive them as Rui suggested - Let us know if you need help with that! But Syndicate's answer is perfect once you have those equations

In that case, I will trust Jamon's instinct

Hello friends its me again

Dont understand this inequality integration thing at all would love some guidance through these 2 examples

1) Without evaluating the integrals, explain whether int pi (up) 0 (low) sin^4 x dx > int pi(up) 0 (low) sin4x dx is true or false

2) Use lnt = int t(up) 1 (low) dx/x for t>1 to deduce that 1- 1/t < or equal to lnt < or equal to 1/2 (t- 1/t) for t> or equal to 1.

tyyyy

I feel like this one is done in two separate parts.
GeoGebra simulation attached

Then, for the int < 1/2(t-1/t) part, treat it like one of the earlier differentiation questions. Note that the integral is just equal to ln(t).

But I will look for a solution again later. Because that's not likely the solution they wanted. I just can't simulate the appropriate graph at 12:40AM

[yasmineturner]

x1 = x2
t1 = t2+2

x1 = 49 x t1 - 4.9 x t1^2
x2 = 49 x (t1+2) -4.9 x (t1+2)^2

equate the equations.

49t - 4.9t^2 = 49t + 98 - 4.9t^2 -19.6t -19.6
19.6t = 117.6
t1 = 6 seconds
t2 = 4 seconds

sub t =6 into the first equation.
x = 117.6 m

So at t = 6 seconds, both the projectiles have the same displacement of 117.6 metres.

EDIT: just to add: I have assumed that you are doing physics by using a constant acceleration formula ( x = ut + 1/2at^2). I don't really know if this can be applied to 3U maths (hopefully one of the mods can confirm if this working out is valid in HSC maths).

Thankyou 

[RuiAce]

In that case, I will trust Jamon's instinctI feel like this one is done in two separate parts.
GeoGebra simulation attached

Then, for the int < 1/2(t-1/t) part, treat it like one of the earlier differentiation questions. Note that the integral is just equal to ln(t).

But I will look for a solution again later. Because that's not likely the solution they wanted. I just can't simulate the appropriate graph at 12:40AM

Coming back to this, I believe that you should just stick to the differentiation approach.

[shreya_ajoshi]

How do you do this:
Find the area enclosed between the curve y=3x(x^2 - 1), the x-axis and the lines x = 0, x = 3

[RuiAce]

How do you do this:
Find the area enclosed between the curve y=3x(x^2 - 1), the x-axis and the lines x = 0, x = 3