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Zela101
 one year ago
Vector Valued Functions: Describe the graph of the equation
Zela101
 one year ago
Vector Valued Functions: Describe the graph of the equation

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dan815
 one year ago
Best ResponseYou've already chosen the best response.1you can use t as an angle and think about it liek a polar graph

dan815
 one year ago
Best ResponseYou've already chosen the best response.1or you can use t as a 3rd dimension and graph it as a 3d plot whatever u like

Zela101
 one year ago
Best ResponseYou've already chosen the best response.0Was this a correct approach, dan?

Zela101
 one year ago
Best ResponseYou've already chosen the best response.0Alright, thanks! I was making sure of i'm getting the concept before going ahead and begin with the other problems.

dan815
 one year ago
Best ResponseYou've already chosen the best response.1if u want to think in polar also one more thing to note is that

dan815
 one year ago
Best ResponseYou've already chosen the best response.1you are starting with the tangent vectors along this circle then u have the perpendiculars along the circle

dan815
 one year ago
Best ResponseYou've already chosen the best response.1<x,y> is perpendicular to <y,x>

dan815
 one year ago
Best ResponseYou've already chosen the best response.1u just have a different sclaring factor on these vectors

Zela101
 one year ago
Best ResponseYou've already chosen the best response.0I see. Does it also apply in 3d graphs?

dan815
 one year ago
Best ResponseYou've already chosen the best response.1it would be a slightly different interprataion if t wasnt seen as the angle

dan815
 one year ago
Best ResponseYou've already chosen the best response.1but r'(t) should still be perpendicular to r(t)

dan815
 one year ago
Best ResponseYou've already chosen the best response.1sint r(t) dot r'(t) = 0 no matter how u plot this

dan815
 one year ago
Best ResponseYou've already chosen the best response.1not all r(t) dot r'(t) = 0 ofcourse though

dan815
 one year ago
Best ResponseYou've already chosen the best response.1so this would be a plane with that normal direction vector

dan815
 one year ago
Best ResponseYou've already chosen the best response.1its a line or a curve i mean not a plane

dan815
 one year ago
Best ResponseYou've already chosen the best response.1its just that line along the plane y=3

Zela101
 one year ago
Best ResponseYou've already chosen the best response.0Lol that was a silly mistake I made

dan815
 one year ago
Best ResponseYou've already chosen the best response.1like that on that y=3 plane we have this line existing

Zela101
 one year ago
Best ResponseYou've already chosen the best response.0And this line would be z=1+(3/2)x

dan815
 one year ago
Best ResponseYou've already chosen the best response.1it looks flipped because x is positive when u go left on the planed noe
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