[arvin_infinity]

So I know how to do everything here except the limits (part b) ! I understand that we need to convert the limits for x to limits for t! but just dunno how we work out the limits of x in the first place!!



+rep

[snow leopard]

Hint: ask yourself where the limits are... notice anything?

Spoiler:

Show

The limit values lie on the x axis which means y=0 at those points. Sub in y=0 into y=9sin2t to obtain 0=9sin2t, solve for t, plug t into x=3cost, and you'll find the limits.

Btw, where is this question taken from?

[arvin_infinity]

(Original post by snow leopard)
Hint: ask yourself where the limits are... notice anything?

The limit values lie on the x axis which means y=0 at those points. Sub in y=0 into y=9sin2t to obtain 0=9sin2t, solve for t, plug t into x=3cost, and you'll find the limits.

Btw, where is this question taken from?

Oh yeh that was the problem..I didn't know what the limits are

so can you not just by looking at the figure just tell what the t values for the shaded area are? it's because there is no t in the figure and its just x and y!?

also I understand why you equate y to zero but then you will get 3 answers for that
t=0, π/2, 3π/2
how do you know which one is which? cause I tried π/2, 3π/2 and it doesn't work

[arvin_infinity]

(Original post by arvin_infinity)
Oh yeh that was the problem..I didn't know what the limits are

so can you not just by looking at the figure just tell what the t values for the shaded area are? it's because there is no t in the figure and its just x and y!?

also I now understand why you equate y to zero but then you will get 3 answers for that
t=0, π/2, 3π/2
how do you know which one is which? cause I tried π/2, 3π/2 and it doesn't work

---got the question off a practice paper-I think its like old questions

[snow leopard]

(Original post by arvin_infinity)
Oh yeh that was the problem..I didn't know what the limits are

so can you not just by looking at the figure just tell what the t values for the shaded area are? it's because there is no t in the figure and its just x and y!?

also I understand why you equate y to zero but then you will get 3 answers for that
t=0, π/2, 3π/2
how do you know which one is which? cause I tried π/2, 3π/2 and it doesn't work

Actually t=0, π/2, π, 3π/2

if t=0, x=1
if t=π/2, x=0
if t=π, x=-3
if t=3π/2, x=0

See diagram now: we need a t that gives us a positive x and a t that gives us 0. t=0 and t=π/2 do the job. These are your limits for integration with respect to t.

(Original post by arvin_infinity)
---got the question off a practice paper-I think its like old questions

Which board...

[arvin_infinity]

(Original post by snow leopard)
Actually t=0, π/2, π, 3π/2

if t=0, x=1
if t=π/2, x=0
if t=π, x=-3
if t=3π/2, x=0

See diagram now: we need a t that gives us a positive x and a t that gives us 0. t=0 and t=π/2 do the job. These are your limits for integration with respect to t.



Which board...

Oh its edexcel!
Ok then maybe that is where Ive gone wrong ..how do you work out t then?
cause I did once using sintcost and another method both methods gave me 3 answers


if t=0, x=3 right?


if t=π/2, x=0
if t=π, x=-3
if t=3π/2, x=0


I understand the whole substitution and finding the x now..but when you say we want a t which gives us x=0 you ignoring the t=3π/2 ??

I am also stuck on part C btw

[snow leopard]

(Original post by arvin_infinity)
Oh its edexcel!
Ok then maybe that is where Ive gone wrong ..how do you work out t then?
cause I did once using sintcost and another method both methods gave me 3 answers


if t=0, x=3 right?


if t=π/2, x=0
if t=π, x=-3
if t=3π/2, x=0


I understand the whole substitution and finding the x now..but when you say we want a t which gives us x=0 you ignoring the t=3π/2 ??

I am also stuck on part C btw

My mistake, if t=0, x=3. If you're unsure on how to solve sin2t=0 I suggest you revisit C2. t=3π/2 isn't being ignored, we could have used that instead, but the question specifically uses π/2 so we go with that.

For part (c) you use integration by substitution.

[arvin_infinity]

t=3π/2 isn't being ignored, we could have used that instead, but the question specifically uses π/2 so we go with that.

Ohh right..get it now

(Original post by snow leopard)
x=3. If you're unsure on how to solve sin2t=0 I suggest you revisit C2.
.

Hmm..done it ages ago just looked at again and realised where I've gone wrong!

For part (c) you use integration by substitution

1)Wondered how you would recognise it and think yeh substitution!!
2) MS done it differently which I couldn't actually figure out what method it is..
3)here is how I've done it notice a mistake somewhere? cause I got a different answer to the MS
sint=u
cost=du/dt
∫u²=...

6[u³] limits are between 0 and 1
=6

the actual answer is 18

[Angry cucumber]

Sorry to but in guys, but what year is this question from? I've done the questions and got answers but I'd like to know whether I'm correct!

[arvin_infinity]

(Original post by Angry cucumber)
Sorry to but in guys, but what year is this question from? I've done the questions and got answers but I'd like to know whether I'm correct!

Ahah no worries.. the final answer is 36cm²
its like 108-4(18)

18 being the shaded area!

I could do part d but in the mark scheme below was mentioned which I didn't quite understand
"At x = 0 there is clearly a point of inflection"

[snow leopard]

(Original post by arvin_infinity)
1)Wondered how you would recognise it and think yeh substitution!!
2) MS done it differently which I couldn't actually figure out what method it is..
3)here is how I've done it notice a mistake somewhere? cause I got a different answer to the MS
sint=u
cost=du/dt
∫u²=...

6[u³] limits are between 0 and 1
=6

the actual answer is 18

I'm not sure, I suppose you just have to 'see' substitution's suitable. There are other techniques to solve it; I suggested substitution but you could have converted sin2tsint into -0.5(cos3t-cost) using the trigonometric identities and integrate that. It just depends on what you prefer.

Your working is correct (assuming A=18). If someone else could double check...

[arvin_infinity]

(Original post by snow leopard)
I'm not sure, I suppose you just have to 'see' substitution's suitable. There are other techniques to solve it; I suggested substitution but you could have converted sin2tsint into -0.5(cos3t-cost) using the trigonometric identities and integrate that. It just depends on what you prefer.

Your working is correct (assuming A=18). If someone else could double check...

Of coursssse..just realised why -0.5(cos3t-cost) and I get 18 which is the right answer..

although couldn't quite figure out how to do the substitution! perhaps you give me some hint on how to do the substitution