UCLA prof is killing me with this derivation! help!

  • Thread starter Raywashere
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  • #1

Homework Statement



This is for calc based physics, and the new phs prof wants us to derive this equation for a projectile motion lab.

From this range equation: R(theta) = (v_o^2cos(theta) / g) * [sin(theta) + [tex]\sqrt{sin^2(theta+ (2gh/v_o^2}[/tex]]

he wants us to derive this range maximizing equation: (theta)=cot^-1[tex]\sqrt{}1 + (2gh/v_o^2)[/tex]



The Attempt at a Solution



I was able to differentiate the inside of the sqrrt to 1 + 2gh/v_o^2 , because I made sin^2(theta) = to p and that just differentiated to 1, but how the heck do you get cot^-1???

also, another thing that is throwing me off is the fact that g(gravity) is a constant, so won't that just go to 0? but if that happens then the range eqution can't be differentiated...right?

someone help...please.
 

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  • #2
SammyS
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Homework Statement



This is for calc based physics, and the new phs prof wants us to derive this equation for a projectile motion lab.

From this range equation: R(theta) = (v_o^2cos(theta) / g) * [sin(theta) + [tex]\sqrt{sin^2(theta+ (2gh/v_o^2}[/tex]]

he wants us to derive this range maximizing equation: (theta)=cot^-1[tex]\sqrt{}1 + (2gh/v_o^2)[/tex]

The Attempt at a Solution



I was able to differentiate the inside of the sqrt to 1 + 2gh/v_o^2 , because I made sin^2(theta) = to p and that just differentiated to 1, but how the heck do you get cot^-1???

also, another thing that is throwing me off is the fact that g(gravity) is a constant, so won't that just go to 0? but if that happens then the range equation can't be differentiated...right?

someone help...please.
I don't mean to insult you, but your new UCLA phys. prof. probably figures that his students know how to differentiate a wide variety of functions and can handle the product rule & chain rule in particular.

BTW: if p = sin2θ , then dp/dθ ≠ 0, because p is not a constant. In fact, dp/dθ = 2(sinθ)(cosθ).

ALSO: if [tex]\theta = \cot^{-1}\sqrt{1 + (2gh/{v_0}^2)}[/tex], that means [tex]\cot(\theta) = \sqrt{1 + (2gh/{v_0}^2)}[/tex]

What did you get for dR/dθ ?
 

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