Good.morning sir, With regards to the first exam question Q8 page 253: why you only took t to be Pi/3 and you did not take into account the value of t to be 2Pi/3 which is also a solution in the 2nd Quadrant. We know that sine is positive between 0 to Pi. Thank you for your videos... much appreciated.
ur teaching is very good it helps me think on how i should approach questions logically
hi sir how did you solve and find dy/dx using the calculator at 4:06?
You know cosec =1/sinx cot= 1/tanx cos=1/sinx so just sub that in calculator
That’s how I got the answer
sir how did you find the value of dy/dx at 4:06
sub t=pi/3 into dy/dx function =4/3
@@ibymalik3658 Thank you for your reply but i can't get the answer as the dy/dx contains cosecX and cotX
shouldn't it be - Cosec 2(π/3) × Cot 2(π/3) ÷ cos(π/3)
@@x1.083 in your calculator for cosecx you use 1/sinx and cot you use 1/tanx
@@ibymalik3658 ohh, thank you brother
Good.morning sir, With regards to the first exam question Q8 page 253: why you only took t to be Pi/3 and you did not take into account the value of t to be 2Pi/3 which is also a solution in the 2nd Quadrant. We know that sine is positive between 0 to Pi.
Thank you for your videos... much appreciated.
pmed2612 it is a valid solution but you’ll have to check if it gives us the coordinates for A by substituting it into the equation for x and y.
Will you be doing any mechanics or stats?
Once I’m done with year 2 pure, I’ll be working on year 2 stats followed by year 2 mechanics.
@@zeeshanzamurred9280 Okay, thanks
At 4:16 isn't the gradient -4/3 not 4/3
The minus sign is with -cosec2t not the whole fraction so i think it’s +4/3. Correct me if I’m wrong.
implicit differentiation can you do like 3^x = y-2xy please as a question
No problem, I’ll do that for you 👍🏻
@@zeeshanzamurred9280 ok boss
4:40 how do you end up with that answer 🗿