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How do I derive this equation? HELP

by psychfan29
Tags: derive, equation
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psychfan29
#1
Sep13-08, 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!!!!!
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eftalia
#2
Sep13-08, 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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