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1. 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.

)
2. (Original post by jazzay123)
Thanks in advance, if it's easier to read my actual written working out, then I can upload that too.
That would be a good idea. This post is very hard to follow.

You wasted a lot of time by formatting up the working without using latex. If you had used latex, you would have written a pretty similar post but it would be readable at the end. In addition, you don't need to give a commentary on your working like so:

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

"My working was:

"

The latex for that was:

$$\frac{d \sqrt{\sin t}}{dt} = \frac{\cos t}{2 \sqrt {\sin t}}, \frac{d}{dt} \frac{ \sqrt{2} }{ 2 \sqrt{\cos 2t}} = \cdots$$
3. (Original post by jazzay123)
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.

)
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 ?
4. (Original post by Jarred)
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 ?

I rubbed out a lot of my working out to restart, as you can see!
5. (Original post by atsruser)
We can see what you did via the working on its own e.g.

"My working was:

"

The latex for that was:

$$\frac{d \sqrt{\sin t}}{dt} = \frac{\cos t}{2 \sqrt {\sin t}}, \frac{d}{dt} \frac{ \sqrt{2} }{ 2 \sqrt{\cos 2t}} = \cdots$$
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.
6. (Original post by jazzay123)
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.
7. (Original post by ghostwalker)
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.
Oh wow, that's what I've been doing all along??😂😂😂😂 I got a max and a min which are root3/2 and root 2/2 respectively, which looks correct! I'll check it with my teacher tomorrow but thankyou so much for putting me on the right track! 😊
8. (Original post by jazzay123)
I'm fairly new to student room so I didn't know that you could do all that latex stuff!
Given that you'd formatted it all up, I felt duty bound to point out that you could have made it nicely latexed for little more effort.

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