Help mechanics

Help!
I don’t get this question why do I square r and how do I know that r squared is 9, I don’t have any value to sub in so isn’t r squared the whole 1st line of ms

(edited 1 year ago)

Reply 1

MindMax2000

Universities Forum Helper

Original post by Alevelhelp.1

Help!
I don’t get this question why do I square r and how do I know that r squared is 9, I don’t have any value to sub in so isn’t r squared the whole 1st line of ms

I presume this is for 11 a).

To find the magnitude of a vector (i.e. the scalar quantity), you use Pythagoras c^2 = a^2 + b^2, then use square root find c
sub a for i coordinates and b for j (or vice versa, we don't care)...
r^2 = (3cos2t)^2 + (3sin2t)^2
r^2= 9cos^2(2t) + 9sin^2(2t)
factorise: r^2 = 9(cos^2(2t) + sin^2(2t))
reflect back to trig functions where you realise sin^2x + cos^2x = 1...r^2=9(1)
square root r^2...r=+ or - 3
As scalar quantities are positive (because negative distances isn't a thing in A Level Mechanics, it might be at university level physics)... r= +3
The locus is constant, and if you map the above r vector function on a graphic calculator or desmos.com (sub i for x and j for y), you will find a circle with a radius of 3, hence the question.

Reply 2

davros

Original post by Alevelhelp.1

Help!
I don’t get this question why do I square r and how do I know that r squared is 9, I don’t have any value to sub in so isn’t r squared the whole 1st line of ms

As MindMax says, this is virtually the definition of |r| - I would suggest going back to your textbook and revising this topic if unsure. If you're given a vector in the form ai + bj and you want its modulus, then the first thing that should enter your head is that you need to add a^2 and b^2 and take the positive square root :smile: