ATAR Notes: Forum
VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: abhi223 on April 10, 2018, 06:05:48 pm
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Can someone plz help me with the following question, thanks!
1. A curve is specified by the parametric equations
x=2cot(t)
y=3cosec(t) where t is (0/pi)
Find the Cartesian equation of the curve and determine its axes intercepts if any and the equations of any asymptotes.
My biggest issue is knowing how to account for the domain of t given.
Without considering the domain I tried it to see how far I can go and here is my working out: (sorry for the poor formatting, im bad with doing mathy stuff computers)
(x/2)=cot(t) y/3=cosec(t)
sin(t)=2cos(t)/x cos(t)=3/y
bcs sin(t)^2 + cos(t)^2 = 1, i did:
(2cos(t)/x)^2 + (3/y)^2 = 1
and i realised that i had cos(t) in that equation so i subbed in (3/y) there because i solved for that already above in the y part of the equation and i ended up with:
(6/xy)^2 + (3/y)^2 =1
and i solved for y in my cas and i got two y='s
y= (-3(x^2 +4)^(1/2))/(x)
y= (+3(x^2 +4)^(1/2))/(x)
and... i m stuck
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Can someone plz help me with the following question, thanks!
1. A curve is specified by the parametric equations
x=2cot(t)
y=3cosec(t) where t is (0/pi)
Find the Cartesian equation of the curve and determine its axes intercepts if any and the equations of any asymptotes.
My biggest issue is knowing how to account for the domain of t given.
Without considering the domain I tried it to see how far I can go and here is my working out: (sorry for the poor formatting, im bad with doing mathy stuff computers)
(x/2)=cot(t) y/3=cosec(t)
sin(t)=2cos(t)/x cos(t)=3/y
bcs sin(t)^2 + cos(t)^2 = 1, i did:
(2cos(t)/x)^2 + (3/y)^2 = 1
and i realised that i had cos(t) in that equation so i subbed in (3/y) there because i solved for that already above in the y part of the equation and i ended up with:
(6/xy)^2 + (3/y)^2 =1
and i solved for y in my cas and i got two y='s
y= (-3(x^2 +4)^(1/2))/(x)
y= (+3(x^2 +4)^(1/2))/(x)
and... i m stuck
There's a different trigonometric you can use. Hint: check the formula sheet.
Hope that helps. Post if you still get stuck. :)
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cot^2 (x) + 1 = cosec^2 (x) this one? Thanks, never thought about referring to it. But still is my original approach appropriate for getting the correct answer? using the new identity i got a hyperbola:
-(x^2)/(2^2) + (y^2)/(3^2) = 1
Also, what should i be doing with the domain for t given?
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cot^2 (x) + 1 = cosec^2 (x) this one? Thanks, never thought about referring to it. But still is my original approach appropriate for getting the correct answer? using the new identity i got a hyperbola:
-(x^2)/(2^2) + (y^2)/(3^2) = 1
Also, what should i be doing with the domain for t given?
That's the identity I was looking at :)
Preferably you'd want to use the identity that requires the least work, however you might be able to observe that the only difference between the two identities, is that the new one that's you've used is simply the first on you tried, except everything is divided by sin2(x) :)
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Thanks for ur help! :) but what i dont get is where did i go wrong in my original working out :P bcs i got an incorrect answer that way. But ur advie has been amazing so thanks again for that! :)
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Thanks for ur help! :) but what i dont get is where did i go wrong in my original working out :P bcs i got an incorrect answer that way. But ur advie has been amazing so thanks again for that! :)
It should work out to be the same if you divide everything by sin^2. You also don't really need y as the subject - maybe leave it square to avoid the square root and see how you go? Regardless, if you understand how to use different identities, then you should be fine.
Also apologies, I didn't read part of your last post properly. Since the domain is simply all of the possibly x-values, you simply need to find the "range" of the function with x as the subject (using the given domain of t).
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Also, what should i be doing with the domain for t given?
It should work out to be the same if you divide everything by sin^2. You also don't really need y as the subject - maybe leave it square to avoid the square root and see how you go? Regardless, if you understand how to use different identities, then you should be fine.
Also apologies, I didn't read part of your last post properly. Since the domain is simply all of the possibly x-values, you simply need to find the "range" of the function with x as the subject (using the given domain of t).
In general, to find the domain and range, sketch both \(x\left(t\right)\) and \(y\left(t\right)\) over the given domain for \(t\). The range of \(x\left(t\right)\) will give you the domain of the graph of the Cartesian relation and the range of \(y\left(t\right)\) will give you the range of the graph of the Cartesian relation
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so for x(t) where t is (0,pi), x will be all R numbers so the domain is all x values in my final equation.
and for y(t) where t is (0,pi), y will be from (3,inf) and so the range for the final equation will be that as well?
and i assume i have to find a section of the graph where both the range and domain figured out above intersect, and so for the hyperbola opening up and down and with y intercepts at 3 and -3 respectively, and so i only draw the line on the original graph which is opening upwards?
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so for x(t) where t is (0,pi), x will be all R numbers so the domain is all x values in my final equation.
and for y(t) where t is (0,pi), y will be from (3,inf) and so the range for the final equation will be that as well?
and i assume i have to find a section of the graph where both the range and domain figured out above intersect, and so for the hyperbola opening up and down and with y intercepts at 3 and -3 respectively, and so i only draw the line on the original graph which is opening upwards?
Yes, that's right. This will give you the top half of the hyperbola only