Parametric Equations : Tangents and Normals : ExamSolutions
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- Опубліковано 15 гру 2024
- Tutorial on tangents and normals to parametric equations.
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For anyone without annotations on, Stuart has included in an annotation that t=-2 was a mistake, at 7:08.
🙂 Thanks
RIP UA-cam Annotations. You will be missed. Man fuck youtube for removing it.
i have 2 domestic cats at home
Thanks for that. Annotation added.
at 5:20 , what if in a question, both of the values satisfied the Y equation? which one would you use?
Thanks a lot sir..Very clear explanation
Thanks for watching and pleased to hear it helped.
your explanations are very clear sir. thank you so much❤
Hi Sir, I love your videos and soluitions as they are extremely helpful.question; why did you take t=-2 and not t=+2?
wouldn't (3t^2-1)/2t equation be the equation for the curve of x and y and the derivative of this equation will be the gradient for tangent? Please help. Thanks
+fahad jhljh (3t^2-1)/2t is not an equation, there is no = sign. So no.
Both of them I suppose but I am sure there will be some clause in the question to stop this from being the case.
Hi if I can ask a question, why don’t u use y=mx + c when trying to find your tangent or normals? Why y2-y1=m(x2-x1)? Just wondering I never understood this thanks!
Check out this page, especially the second video. It may help answer your question www.examsolutions.net/tutorials/equation-of-a-line-given-the-gradient-and-point/?level=AS&board=Edexcel&module=Pure%20Maths%20AS&topic=1239
thank you for this video
How did you know that we should be using the x-coordinate 4 and x = t^2 to find t?
You could have used the x or the y part of the parametric equation, but we chose the x =t^2 and the x coordinate = 4 because it is easier to solve for t.
7:05 , did you mean t=-2 at that point? Corrective annotation? :) Thanks for the vids!
Look at Edexcel Jan 2009 Q7 last part. I have a video on the kind of thing yoou are asking. Hope it helps.
thank you so much sir your explanation help me a lot.
+Lungile Ntulie good
Was there any need to work out the y values?
Yes, so that you could see which t value corresponded with the point (4,-6)
-16-66 = - 82. Sorry, but I think you will find that I am correct on this one.
thank you very much sir, the video helped me a lot
michael gich Thanks for using.
THERE IS AN EASIER WAY TO SOLVE THIS (Just sayin')
x=t^2 and y=t^3-t
so y=(t^2)t-t therefore y=xt-t
sub. x & y
-6=4t-t which gives t=-2
eqn. of tangent is [y=-2x-2]
gradient of normal is -1/(gradient) i.e 0.5
eqn. of line is y=mx+c
sub. x,y & new gradient
you get y=0.5x-8 => eqn. of normal
Very simple :) correct me if i'm wrong, coz honestly i dnt know if i'm right
Yes but as you said this would be very long winded.
Thanks for this video .. ^_^
Wasn't the last part, eqn of normal supposed to be 4x+11y+82?????
How many marks would a question like this be?
Hard to say, but I would have said 7 to 10 judging by the stages in the calculation.
@@ExamSolutions_Maths Nice, thanks for the reply, you are amazing!
U hav a mistake on replacing the value of t .... U just kept the value of t as 2 where as it should b -2
kishan shrestha he just wrote t=2 but has used the value t=-2 in the calculation of dy/dx
U kept t =2
How many marks would a question like this be?