Hi :) just wanted to let you and others know that you've got an error in your answers for the first parametric equation question, part i). It should simplify to 2/t + 1/(2t^2 -t)
Samuel
2024-09-05 07:05:41 +0000 UTC
He found the gradient before and after the stationary point. Before the stationary point the gradient is negative, after the stationary point it is positive. That's how it is for a minimum point, it helps if you can imagine the graph.
Jaafar
2023-10-09 15:51:51 +0000 UTC
aren't we just suppose to get the 2nd derivative and then sub the x value we found and determine if we get a ans greater or less than zero?
Abdur rahman
2023-07-26 02:18:50 +0000 UTC
hey jacob, in q5 part 2 how did you get a minimum value without doing 2nd derivative? and why did you use two values 0.51 a min and 0.53 a max?
Abdur rahman
2023-07-26 02:14:15 +0000 UTC
I will make an A* with this videos
Kenneth Shonhiwa
2022-08-19 02:27:20 +0000 UTC
Pliz we need p6
Kenneth Shonhiwa
2022-08-17 10:50:46 +0000 UTC
If I did that then it's a mistake, I probably accidentally wrote 1 instead of t.
Intuitive
2022-07-26 18:05:21 +0000 UTC
Hey thanks for the awesome video, in part two question 1 at the end of your working you go from 2/2t(2t +1) to equaling 1/2t^2 +1 I would have thought it would equal 1/2t^2 +t. Could you please explain why 1 is used instead of t
Zara Ford
2022-07-26 03:04:23 +0000 UTC
I am so sorry . I just realised that the other y=( -2x ). Not just( -2 ) . So sorry.
Rramb
2022-04-16 08:14:44 +0000 UTC
Hello . Wanted to ask about the last question . Last part . When u got y=0 and y= -2 ... U did it correct when y= 0 but when y=-2 there is a mistake . If u switch y with -2 inside the brackets . U will get( x-6 ) not (-5x) correct me if I am wrong pls. Thanks alot and I hope u reply
Rramb
2022-04-16 08:11:30 +0000 UTC
I'm not quite sure what your asking. Or what you mean by quadrants? It asks for solutions for theta between 0 and pi, so we simply give the solutions that occur within that range. Because something like cos(theta) = 1/2 has infinite number of solutions (as y = cos(theta) oscillates forever), thus we just select the solutions that occur between 0 and pi. Maybe, watch my Trig Identity video for more clarity on this, I go through plenty of examples in there.
Intuitive
2021-10-23 04:25:32 +0000 UTC
Super helpful video, thanks so much, for the parametric equations question part ii, just wondering, since cos(theta) = 1/2, and 1/2 is positive so its positive in the first and fourth quadrant, why do we omit pi - 1/3pi for the answer in the fourth quadrant like we would in a trig question, am I way off or do we just take the first answer for these types of qns ?
Ry Lee Lee
2021-10-22 15:46:14 +0000 UTC
Yep, thanks. Will be needing alot more help for later chaps. Thanks alot for your hardwork!
Ally A
2021-09-21 13:05:25 +0000 UTC
Well the numerator : -1*(-x^2 - 2xy) gives x^2 + 2xy and the Denominator: -1*(x^2 - y^2) gives -x^2 + y^2 or simply written as y^2 - x^2. Does that make sense?
Intuitive
2021-09-21 11:02:49 +0000 UTC
Hey, in the last question that you did in the video, I didn't quite get how you removed the minus sign. Could you explain that pls?
Ally A
2021-09-20 17:20:13 +0000 UTC
o sorry, I didn't see they specify the range, x> 0 and x=a, so a>0 , and ln(2-3^1/2) < 0 so they reject it.
yoki
2021-07-03 22:32:55 +0000 UTC
Hello, at question 6, I look at the mark scheme, they only accept one value, and question it self says "value" instead of "values", I don't understand why they don't accept a = ln(2-3^1/2)