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parametric eq question

Hello, I am currently trying to learn parametric equations and I'm confused about the method for this question. I did 0=1/4sint to get pi but the MS states to equal tcost=0. Please could some one explain the process? thank you
(edited 1 year ago)
Reply 1
Original post by lavely
Hello, I am currently trying to learn parametric equations and I'm confused about the method for this question. I did 0=1/4sint to get pi but the MS states to equal tcost=0. Please could some one explain the process? thank you

What was your logic? You should be trying to find the intercept with the x-axis, so ...
Reply 2
Oh Okay thank you, I'm not sure what my logic was since I didn't know how to approach it :frown: This may be a silly question but how do you know when to find the the intercepts with the x axis or the y axis? I think the involvement of the extra variable t is confusing me... I hope this makes sense


(when I saw the MS answer was 1/2pi I thought it was something to do with using the smaller range )
Original post by davros
What was your logic? You should be trying to find the intercept with the x-axis, so ...
(edited 1 year ago)
Reply 3
Original post by lavely
Oh Okay thank you, I'm not sure what my logic was since I didn't know how to approach it :frown: This may be a silly question but how do you know when to find the the intercepts with the x axis or the y axis? I think the involvement of the extra variable t is confusing me... I hope this makes sense


(when I saw the MS answer was 1/2pi I thought it was something to do with using the smaller range )

You're trying to use the information given in the diagram to help you. The point O is where x = 0 and y = 0 so doesn't help you find k, because it just corresponds to t = 0. You can see that the curve hits the x-axis again later on, and at this point y = 0 (but x > 0). So you want to start by setting y = tcos(t) = 0. You also know that t can't be 0 because otherwise x would have to be 0 too. So your equation becomes much simpler to solve :smile:
(edited 1 year ago)
Reply 4
Original post by davros
You're trying to use the information given in the diagram to help you. The point O is where x = 0 and y = 0 so doesn't help you find k, because it just corresponds to t = 0. You can see that the curve hits the x-axis again later on, and at this point y = 0 (but x > 0). So you want to start by setting y = tcos(t) = 0. You also know that t can't be 0 because otherwise x would have to be 0 too. So your equation becomes much simpler to solve :smile:


thank you so much for your help, it makes sense now!

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