Isaac Physics &#x27;Algebraic Manipulation 5.3&#x27;

OP

Original post by domm1

https://isaacphysics.org/questions/manipulation_5_3?board=25dcbcd5-47cd-4369-8379-f34a43ad7910

(Images attached below).

I'm confused on where my error is. Can anyone help spot it for me please? I feel like I got decently close as it seemed all fine however I'm still left with a (cosTheta)^2 term at the end.

DISCLAIMER: I know my workings drag out for way longer than they need to, however I'm trying not to rush through and make a stupid mistake somewhere within my workings because of it (but it seems even when I work through it slower I still make mistakes). :s-smilie:

Reply 2

Err yes. That does go on ...
You should eliminate t at the start (sub the t<->x expression directly into y) to get
y = Ax + Bx^2
For suitable expressions A and B. Then use a trig identity for 1/cos^2 = sec^2 <-> tan^2.
It should be ~5 or 6 lines.

Partially checking your answer now ...
There shouldn't be a 1/2 in the first expression in the answer.
The second term is 1/2 not 1/4.

(edited 3 years ago)

Reply 3

OP

Original post by mqb2766

Err yes. That does go on ...
You should eliminate t at the start (sub the t<->x expression directly into y) to get
y = Ax + Bx^2
For suitable expressions A and B. Then use a trig identity for 1/cos^2 = sec^2 <-> tan^2.
It should be ~5 or 6 lines.

Partially checking your answer now ...
There shouldn't be a 1/2 in the first expression in the answer.

Well, at least I know why it does now. We haven't learnt reciprocal trig functions (I think that's what they're called) yet, so no wonder it kept going on and on... :frown:

I had no idea you had to use them lmao.

Reply 4

Original post by domm1

Well, at least I know why it does now. We haven't learnt reciprocal trig functions (I think that's what they're called) yet, so no wonder it kept going on and on... :frown:

I had no idea you had to use them lmao.

I think it went on and on because you equated t at the start, rather than subbing the linear x-t relationship into y(t) to give you y(x) instead.

Pythagoras gives
cos^2 + sin^2 = 1
Divide by cos^2
1 + tan^2 = 1/cos^2
You don't need to know about sec. Just use this identity at the end, as the normal parabolic trajectory is given in terms of sec^2, so I guess they didn't want you copying the usual answer.

(edited 3 years ago)

Reply 5

ghostwalker

17

Think that's one of the most epic pieces of algebra I've seen at A-level.

Reply 6

OP

Original post by mqb2766

I think it went on and on because you equated t at the start, rather than subbing the linear x-t relationship into y(t) to give you y(x) instead.

Pythagoras gives
cos^2 + sin^2 = 1
Divide by cos^2
1 + tan^2 = 1/cos^2
You don't need to know about sec. Just use this identity at the end, as the normal parabolic trajectory is given in terms of sec^2, so I guess they didn't want you copying the usual answer.

Now I think about it, that was definitely the smarter thing to do. :redface:

So for my final part are you saying I can take a factor of (gx^2)/(4v^2) out of the last fraction to give 1/(cosTheta)^2 within the brackets and then substitute in 1+tan^2 for that to get the correct answer? Or was there an algebraic mistake earlier on that has given me the incorrect expression?

(edited 3 years ago)

Reply 7

OP

Original post by ghostwalker

Think that's one of the most epic pieces of algebra I've seen at A-level.

Thanks!

I was hoping it'd come to an end soon about a quarter of the way through but it just kept getting worse and worse!

Reply 8

Original post by domm1

So for my final part are you saying I can take a factor of (gx^2)/(4v^2) out of the last fraction to give 1/(cosTheta)^2 within the brackets and then substitute in 1+tan^2 for that to get the correct answer? Or was there an algebraic mistake earlier on that has given me the incorrect expression?

The 4 divisor is 2 ( previous post). You have
gx^2/(2v^2cos^2) = (gx^2/(2v^2))*(1/cos^2) = (gx^2/(2v^2))*...

While I'd agree with the epic / ghostwalkers assessment, Id redo it the elegant way in 5-6 lines. You'd sort out the 1/2 factor problem and learn how to make life easier in future.

(edited 3 years ago)

Reply 9

OP

Original post by mqb2766

The 4 divisor is 2 ( previous post). You have
gx^2/(2v^2cos^2) = (gx^2/(2v^2))*(1/cos^2) = (gx^2/(2v^2))*...

While I'd agree with the epic / ghostwalkers assessment, Id redo it the elegant way in 5-6 lines. You'd sort out the 1/2 factor problem and learn how to make life easier in future.

I am going to attempt the more efficient method afterwards for your reason mentioned above, however I still can't seem to get the correct answer. I changed the '4v^2(cosTheta)^2' to '2v^2(cosTheta)^2', and took the factor outside, changed the term inside the bracket into 1+(tanTheta)^2, however this is still not correct?

y = (xtan)/2 - (gx^2)/(2v^2)*(1+tan^2), where have I gone wrong?

Reply 10

Original post by domm1

I am going to attempt the more efficient method afterwards for your reason mentioned above, however I still can't seem to get the correct answer. I changed the '4v^2(cosTheta)^2' to '2v^2(cosTheta)^2', and took the factor outside, changed the term inside the bracket into 1+(tanTheta)^2, however this is still not correct?

y = (xtan)/2 - (gx^2)/(2v^2)*(1+tan^2) where have I gone wrong?

Get rid of the /2 in the first term (#3).
It also helps reduce errors to not split the numerator so
gx^2(1+tan^2)/(2v^2)

(edited 3 years ago)

Reply 11

OP

Original post by mqb2766

Get rid of the /2 in the first term (#3).
It also helps reduce errors to not split the numerator so
gx^2(1+tan^2)/(2v^2)

Thank you I got it now, I'll have a go at the more efficient route now.

Reply 12