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Gradient scalar field

thanks
(edited 12 years ago)
Reply 1
Avatar for nuodai
nuodai
17
Nope... how have you worked out your partial derivatives? Post your working and I'll show you where you've gone wrong.
Reply 2
Avatar for mathslover786
mathslover786
OP
Original post by nuodai
Nope... how have you worked out your partial derivatives? Post your working and I'll show you where you've gone wrong.


think ive got it....is the answer 2xz(y-1) i + x^2 z j + x^2(y-1) k ?
Reply 3
Avatar for nuodai
nuodai
17
Original post by mathslover786
think ive got it....is the answer 2xz(y-1) i + x^2 z j + x^2(y-1) k ?


Yup :smile:
Reply 4
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mathslover786
OP
Original post by nuodai
Yup :smile:


Thank You! :biggrin:
Reply 5
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mathslover786
OP
Original post by nuodai
Yup :smile:


another quick question:


Show that the following points form a right angled triangle :

A( 1 , 2 , 3 ) , B( 3 , 8 , 13 ) and C( –2 , 8 , 14 ) .

Also, find the area of the triangle, and the equation of the plane on which the triangle sits .

cos(theta) between two of the letters has to be 1 i know but it doesnt seem to work...HELP!
Reply 6
Avatar for nuodai
nuodai
17
Original post by mathslover786
another quick question:


Show that the following points form a right angled triangle :

A( 1 , 2 , 3 ) , B( 3 , 8 , 13 ) and C( –2 , 8 , 14 ) .

Also, find the area of the triangle, and the equation of the plane on which the triangle sits .

cos(theta) between two of the letters has to be 1 i know but it doesnt seem to work...HELP!


There's your error. The letters are the position vectors of the three vertices of the triangle, not the sides. The sides are AB→\overrightarrow{AB}, AC→\overrightarrow{AC} and BC→\overrightarrow{BC}, and it's two of these that have to be perpendicular for it to be a right-angled triangle.
Reply 7
Avatar for mathslover786
mathslover786
OP
Original post by nuodai
There's your error. The letters are the position vectors of the three vertices of the triangle, not the sides. The sides are AB→\overrightarrow{AB}, AC→\overrightarrow{AC} and BC→\overrightarrow{BC}, and it's two of these that have to be perpendicular for it to be a right-angled triangle.


ahhhhh yeah i get it now lol...i was using them as the length of the sides do you have your own youtube channel...it would be verry useful :biggrin:
Reply 8
Avatar for mathslover786
mathslover786
OP
Original post by mathslover786
think ive got it....is the answer 2xz(y-1) i + x^2 z j + x^2(y-1) k ?


for the Curl of this i got 0 is this correct?

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