Find the expression for the unit vector tangential to the curve given in spherical coordinates by r = 1, [tex]\phi[/tex] = 2[tex]\theta[/tex], 0 [tex]\leq[/tex] [tex]\theta[/tex] [tex]\leq[/tex] [tex]\pi[/tex](adsbygoogle = window.adsbygoogle || []).push({});

the equation for the differential length vector for spherical coordinates:

dl = drar + rd[tex]\theta[/tex]a[tex]\theta[/tex] + rsin([tex]\theta[/tex])d[tex]\phi[/tex]a[tex]\phi[/tex]

where ar, a[tex]\theta[/tex], and a[tex]\phi[/tex] are unit vectors

i used the equations for the curve to determine that:

dr = 0, d[tex]\phi[/tex] = 2d[tex]\theta[/tex]

by substituting into the differential length vector, i got:

rd[tex]\theta[/tex][a[tex]\theta[/tex] + 2sin([tex]\theta[/tex])a[tex]\phi[/tex]]

i know this is not a unit vector, but i am lost. i think there would need to be either a cosine or a way to get rid of that sine in the differential length vector equation for the expression to be a unit vector.

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# Homework Help: Unit vector tangential to curve

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