Braintic Solutions (1 Viewer)

[braintic]

Time to join the bandwagon and add my (hopefully) finished solutions.

 

[Kurosaki]

Thanks braintic!

 

[Carrotsticks]

Great solutions, braintic.

I liked how you provided more than one method when possible.

 

[dan964]

Fairly decent

Comments:
*** I also liked (sarcastic) how you used implicit differentiationn which is not even in 3U course for Q13(b)(ii)
*** Q13 (b)(iii) - you assume the reflective property without proving it. I did it a similar way without making that assumption.
I let t=0, t=1 to get the points O and S both of which are part of the locus. I used the distance formula to prove that SQ is a.

My other method is to find the centre S by again letting t=0, t=1. And then finding
x^2 and (y-a)^2 using the substitution t=tan theta/2.

***Also your method for Q12(c) is very ugly/shifty but still gets the same answer. I tend to write
as 1/2 v^2 = 2x+2e^-x/2 + c then evaluate the constant by plugging in v=4, x=0. I know Terry Lee uses that same method sometimes as well.
***Liked your visual methods for Q10

***There was an easier way to Q14(a)(ii)
Note it is a right angle isosceles triangle
i.e. x=-y. Therefore, Vt cos(th) = -1/2gt^2 +Vt sin(th) (t cannot be equal to 0 at D)
Divide by t, make t the subject
then subsitute back in to find x.
Then use trig ratios in right angle triangle. A lot easier.

I like your second method, it is a lot nicer, for Q14(b)(ii)

 

[Carrotsticks]

Fairly decent

Comments:
*** I also liked (sarcastic) how you used implicit differentiationn which is not even in 3U course for Q13(b)(ii)
*** Q13 (b)(iii) - you assume the reflective property without proving it. I did it a similar way without making that assumption.
I let t=0, t=1 to get the points O and S both of which are part of the locus. I used the distance formula to prove that SQ is a.

My other method is to find the centre S by again letting t=0, t=1. And then finding
x^2 and (y-a)^2 using the substitution t=tan theta/2.

***Also your method for Q12(c) is very ugly/shifty but still gets the same answer. I tend to write
as 1/2 v^2 = 2x+2e^-x/2 + c then evaluate the constant by plugging in v=4, x=0. I know Terry Lee uses that same method sometimes as well.
***Liked your visual methods for Q10

***There was an easier way to Q14(a)(ii)
Note it is a right angle isosceles triangle
i.e. x=-y. Therefore, Vt cos(th) = -1/2gt^2 +Vt sin(th) (t cannot be equal to 0 at D)
Divide by t, make t the subject
then subsitute back in to find x.
Then use trig ratios in right angle triangle. A lot easier.

I like your second method, it is a lot nicer, for Q14(b)(ii)

- In his defence, you could just as easily argue that he is using the Chain Rule.

- S is not a part of the locus, it is the centre of it.

- If you obtained Q by substituting t=1 and then used the distance formula from that to the focus S, then you would have lost at least one mark because you are then fixing Q, which defeats the purpose of a locus equation.

- What is so ugly/shifty about that method? Your method is LITERALLY the exact same thing, if you think about what happens to the constant of integration when computing definite integrals.

- Your method for the projectile motion problem works, but I do not feel that it is any more or less advantageous than using the Cartesian equation. The mechanics are pretty much the same from there (you eliminated t=0, he eliminated x=0 etc...)