# Homework Help: How do I derive this equation? HELP

1. Sep 13, 2008

### psychfan29

How do I derive this equation?!? HELP!!!

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. Sep 13, 2008

### eftalia

Re: How do I derive this equation?!? HELP!!!

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..