Finding Maximum of v^x_s: Solve for \theta_{max}

[Xkaliber]

Hi all,

I am having a problem understanding how to find a certain maximum of this equation and am not sure if I am going about this the proper way.

$$v^x_s = \frac{v \sin{\theta}}{1-v \cos{\theta}}$$

Find an expression for the angle $$\theta_{max}$$ at which $$v^x_s$$ has its maximum value for a given speed v. Show that this angle satisfies the equation $$\cos{\theta_{max}}=v$$.

Answer: Using my knowledge of calculus, I believe I should take the derivative of the above equation with respect to $$\theta$$

Using the quotient rule, this gives me $$\frac{v \cos{\theta}-v^2 (\sin{\theta})^2}{1-v \cos{\theta}}$$

I should now find values of theta that make the equation 0 or undefined. However, I do not know what v is, which is throwing me off on what value theta should be. Any help would be greatly appreciated.

 

[xman]

You're on your way set the numerator to zero, and solve for $$\theta$$ You'll see the answer makes sense.

 

[HallsofIvy]

However, I do not get ]$$\frac{v \cos{\theta}-v^2 (\sin{\theta})^2}{1-v \cos{\theta}}$$ as the derivative.
For one thing, the denominator must be $(1- vcos \theta)^2$. For another, the numerator will involve both $sin^2 \theta$ and $cos^2 \theta$ which will then combine.

 

[Xkaliber]

Opps! I found my error and problem solved. Thanks