Parametric equation finding the gradient of the normal.. :s Watch

hamstein
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Hi again, I've found dy/dx to be 4/4t how do I sub the co ordinates in :confused:

Thanks.
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nuodai
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This question has come up 3 times in 2 days now :p: See here
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hamstein
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(Original post by nuodai)
This question has come up 3 times in 2 days now :p: See here
"If you follow that through you will end up with 1/t. Now, you should know that if you sub in an x value into this, you will get the gradient of the tangent, but we want the gradient of the normal, so find its negative reciprocal (n = -1/m).

That should leave you with -p as the gradient of the normal, now you know that this is equal to PQ, so equate their gradients:

and that gets you the answer."


I managed to get -t how comes t's have been turned into p's? :confused:
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Unfinished Business
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(Original post by hamstein)
"If you follow that through you will end up with 1/t. Now, you should know that if you sub in an x value into this, you will get the gradient of the tangent, but we want the gradient of the normal, so find its negative reciprocal (n = -1/m).

That should leave you with -p as the gradient of the normal, now you know that this is equal to PQ, so equate their gradients:

and that gets you the answer."


I managed to get -t how comes t's have been turned into p's? :confused:
The co-ordinates of P are just ts replaced with ps...
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hamstein
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'Now we need the equation of the normal at R. So at R, x=8 y=8, therefore t=2.'

Why do we have to make x and y = 8?
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hamstein
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How did she get dy/dx = 1/2 :s
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hamstein
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Okay dw I've got it lol
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