calculo jacobiano pag 129 barrientos.docx

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Donde q1=1, q2=3, q3=d3Cos(1+2)=cos1cos2-sen1sen2Para x11= D(I3 cos1cos2-I3sen1sen2+ I2cos1)-I3sen1cos2-I3cos1sen2-I2sen1= -I3sen(1+2)-I2sen1=- (I3S12+I2S1)Para yY=I3sen(1+2)+I2sen1= I3(sen1cos2+cos1sen2)+I2sen1Y21= I3(cos1cos2-sen1sen2)+I2cos1=I3cos(1+2)+I2cos1=I3C12+I2C1Y=I3sen(1+2)+I2sen1= I3(sen1cos2+cos1sen2)+I2sen1