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R(t) = acos^3(t) i + asin^3(t) j + k: Find derivative of vectors?

0 votes

I'm a bit stuck. The answer to this problem is:
-3asin(t)cos^2(t) i + 3asin^2(t)cost j

what I have is:
cos^3(t) - 3asin(t)cos^2(t) i + sin^3(t) + 3asin^2(t)cos(t) j

Please help. Much appreciated! :)

asked Mar 8, 2013 in CALCULUS by andrew Scholar

1 Answer

+1 vote

Differentiate  each side with respective t

R1 ( t ) = d/dt (a cos3(t) i + a sin3(t) j +k)

R1 ( t ) = d/dt (a cos3(t) i )+ d/dt ( a sin3(t) j ) + d/dt (k)

R1 ( t ) = a [ d/dt(cos3(t) i ] + a [ d/dt (sin3(t) j ] + 0

Recall : d/dt [cos3(A) ] = 3cos2(A) (-sin(A) ) and d/dt [sin3(A)] = 3Sin2(A) cos(A)

R1 ( t ) =  a [  3cos2(t) i (-sin ( t ) ) ( 1 ) ] + a [ 3(sin2(t) j (cos ( t ) (1)

R1 ( t ) = - 3asin(t) cos2(t) i  +3a sin2(t)cos(t) j

 

answered Mar 8, 2013 by diane Scholar

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