3U Maths Question Thread

[RuiAce]

Hey.
With regards to "with respect to" when integrating or differentiating, is it okay to simplify it to "wrt"?

Thanks!

Fairly sure this is ok. Personally I use dots though: "w.r.t."

[hobocop]

Hi, could I get some help on how the answer to part ii comes about?
This is from the 2001 past paper.
Thanks.

[Opengangs]

Hi, could I get some help on how the answer to part ii comes about?
This is from the 2001 past paper.
Thanks.

Let the coordinates of Q be its parametric form: (2aq, aq^2)
Now, the tangent at Q  is q, so the normal has gradient -1/q
Since the normal at q is perpendicular to the normal at p, then -1/q * -1/t = -1
1/tq = -1
q = -1/t
Thus, Q has coordinates (-2a/t, a/t^2)

[CyberScopes]

Could I please get some help with 2015 q14 ii and iii? Can someone please explain why you need to take dy/dx and how the whole tan thing works in this scenario?

Thanks

[Opengangs]

Could I please get some help with 2015 q14 ii and iii? Can someone please explain why you need to take dy/dx and how the whole tan thing works in this scenario?

Thanks

The angle theta is linked with the parametric velocities of the projectile. So, you will need to find the horizontal and vertical velocity. tan theta = velocity of y / velocity of x.

[RuiAce]

Could I please get some help with 2015 q14 ii and iii? Can someone please explain why you need to take dy/dx and how the whole tan thing works in this scenario?

Thanks

Previously addressed

[winstondarmawan]

Would appreciate help with the following:
https://i.imgur.com/qtnZYZ5.png
TIA!

[Opengangs]

Would appreciate help with the following:
https://i.imgur.com/qtnZYZ5.png
TIA!

Greatest coefficient occurs at t_7:

[CyberScopes]

Greatest coefficient occurs at t_7:

Correct me if im wrong, but if k=6 then shouldnt the term

T[6+1] = 25C6 * 7^19 * 3^6 ?

[Opengangs]

Correct me if im wrong, but if k=6 then shouldnt the term

T[6+1] = 25C6 * 7^19 * 3^6 ?

From our condition, we stated that T_k+1 > T_k. This implies that:
T_0 < T_1 < T_2 < T_3 < T_4 < T_5 < T_6 < T_7 > T_8 > T_9 > ...

As you can see, the signs switch when we hit the eighth term (T_7) since our opening condition justified it as such.
To find T_7, we treat k as 7 and use the term we defined for T_k

[CyberScopes]

From our condition, we stated that T_k+1 > T_k. This implies that:
T_0 < T_1 < T_2 < T_3 < T_4 < T_5 < T_6 < T_7 > T_8 > T_9 > ...

As you can see, the signs switch when we hit the eighth term (T_7) since our opening condition justified it as such.
To find T_7, we treat k as 7 and use the term we defined for T_k

Ahhh yes ok, my mistake. Im slowly coming to the realisation that im missing so many small important detail, its gona screw me for the exam RIP

[bsdfjnlkasn]

Hey there!

I need some help understanding the inequality for the last part of this question. We've established that the time taken for the object to fall is 4.9 seconds. I've attached the solutions but was wondering if there was an alternate method that someone could propose and explain? I understand the at most 50 business means -50 on the LHS of the inequality and +50 on the RHS, but I don't understand what it's being subtracted/added to and whether we put S, the position of the sailor in the middle of the inequality or S+distance travelled in 4.9 seconds....

Would really appreciate having this explained!

[winstondarmawan]

Would appreciate help with the following:
https://i.imgur.com/a47QdTC.png
https://i.imgur.com/kDAAEtL.png
For (ii), I'm not sure how to make n the subject.
https://i.imgur.com/3eNIKVl.png
TIA!

[hobocop]

Hi, could I please get some help with this?
Thanks.

[Opengangs]

Hi, could I please get some help with this?
Thanks.

Convert to auxiliary form: 5sin4t + 12cos4t = 13sin(4t + alpha)
Now, just differentiate w.r.t t to find velocity.

velocity = 52cos(4t + alpha)
Notice that the maximum value for cosine and sine is equal to 1, as both wave functions oscillate between the two points.
This implies that the maximum velocity is 52 (C)