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How do I derive this equation? HELP 
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#1
Sep1308, 07:05 PM

P: 9

I started with the equation:
y=yinitial + (vinitialsintheta)t 1/2gt^2 I plugged in t=x/(vinitialcostheta) to get: y=yinitial + x (tantheta)  [(gx^2)/(2(vinitial^2)(cos^2theta))] Using derivatives, how do I get from this equation: y=yinitial + x (tantheta)  [(gx^2)/(2(vinitial^2)(cos^2theta))] to this equation: (dy)/(dtheta)= x(sec^2theta)[(gx^2(tantheta)(sec^2theta))/(vinitial^2)] I am completely and totally and utterly confused, bewildered, perplexed, and a list of other things. Plese help!!!!! 


#2
Sep1308, 07:31 PM

P: 10

To get from y=yinitial + x (tantheta)  [(gx^2)/(2(vinitial^2)(cos^2theta))]
to this equation: (dy)/(dtheta)= x(sec^2theta)[(gx^2(tantheta)(sec^2theta))/(vinitial^2)]  Taking y and x not to vary with theta.. then they are taken as constants. differentiating yinitial wrt theta gets 0, differentiating x tantheta wrt theta gets x sec^2 theta (differentiate tangent to get sec^2) You could use quotient rule for the last term to get [gx^2 . (2 cos theta sin theta)] / [2 vinitial (cos^4 theta)] before simplifying to get the eventual equation.. 


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