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can anyone help me with this: x = 2sin t; y = 3cos t; z(x,y) = 3x^2 + 2y find: d(dz)/d^2(t)
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you mean \[\frac{d^2z}{dt^2}\] ?
@hosiduy
yes
just put x(t) and y(t) in the equation of z and take derivative twice
then what will u get?
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i still don't understand clear your way, can you make that more clear?
hiii z(x,y)=3 x^2+2y first substitute the values of x and y in z(x,y), you'll get z(t)=6sin^2t+6cost dz/dt=6(2sint cost -sint) =6(sin2t-sint) {2sint cost =sin2t} d^2z/dt^2 = 6(2cost2t-cost) :) cheers
\[x=2 \ sint \\ y=3 \ \cos t \\ z(t)=3x^2+2y=3 (2 \ \sin t)^2+2( 3 \ \cos t)=12 \sin^2t+6 \cos t\]
@hosiduy now take derivative of z(t) twice am i clear now?
oh, i understand now, thank all guys :D
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yw
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