Deadly M2 vectors question... please help me understand :(

Thanks in advance, if it's easier to read my actual written working out, then I can upload that too.

I differentiated the sqrt(sint) first, and got (cost/2sqrtsint). I then differentiated the (sqrt2)/2(sqrtcos(2t)) using the quotient rule and got
-(sqrt2)sintcost/(sqrtcos(2t)).

Hello!

I've been sitting here all evening trying to complete this question. I feel as if I understand what the question is asking, but I don't seem to find myself getting any closer to an answer...

A particle has position vector given by (sqrt sint)i and (sqrt2)/2(sqrtcos(2t)). Find the exact values of the minimum and maximum distances from the origin of the particle.

The second I saw 'maximum and minimum' I knew it was an optimisation question with differentiation.

I differentiated the sqrt(sint) first, and got (cost/2sqrtsint). I then differentiated the (sqrt2)/2(sqrtcos(2t)) using the quotient rule and got
-(sqrt2)sintcost/(sqrtcos(2t)).

Using the trigonometric identity cos(2t) = cos^(2)t - sin^(2)t, would it make sense that the square root of cos(2t) is equal to cost + sint? Something tells me that this logic is incorrect.

I added these results together and made it equal to 0, because max and min points occur when dy/dx=0. However, it's become a bit of a sticky sum because I thought the square roots would cancel further into the question in order to make the exact results a bit 'nicer'.

I know that I use the second order derivative to prove the maximum or minimum point, but I'm a bit stuck with the part above.

Thanks in advance, if it's easier to read my actual written working out, then I can upload that too.

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You appear to be optimising the wrong thing. As I understand from your working, you've differentiated each of the pairwise components and then optimised their sum, but this isn't optimising their distance. (Well, at least not in the Euclidean sense usually used in mechanics problems!)

What's the formula for distance between vectors and ?

We can see what you did via the working on its own e.g.

"My working was:

$\frac{d \sqrt{\sin t}}{dt} = \frac{\cos t}{2 \sqrt {\sin t}}, \frac{d}{dt} \frac{ \sqrt{2} }{ 2 \sqrt{\cos 2t}} = \cdots$"

The latex for that was:

[noparse] $\frac{d \sqrt{\sin t}}{dt} = \frac{\cos t}{2 \sqrt {\sin t}}, \frac{d}{dt} \frac{ \sqrt{2} }{ 2 \sqrt{\cos 2t}} = \cdots$
[/noparse]

I'm fairly new to student room so I didn't know that you could do all that latex stuff! Here is a photo of the question. I rubbed out most of my working out though to restart.

As Jarred pointed out, you're minimising the wrong thing.

What's the formula for the distance between two points?

In this case one point is the origin, and the other is expressed in terms of a parameter, t.

Giving you a distance of .... (a function of t)

And that's the formula you want to find the max/min of.

I'm fairly new to student room so I didn't know that you could do all that latex stuff!