Question

Take the Derivative of R with Respect to θ and set it to equal 0.
Equation is attached below

Answer 1

Can someone pleas help？

Answer 2

I have the answer but i dont know how to get there

Answer 3

\[\Large\bf\sf R=\color{royalblue}{\frac{v^2}{g}}\cos \theta\left(\sin \theta+\sqrt{\sin^2 \theta+\frac{2gh}{v^2}}\right)\]If we move the denominator it becomes a little easier to read.
So that blue part in front is just a constant, we won't worry about it too much.

So to differentiate, we'll need to apply the product rule. Ok? :o

Answer 4

product rule and chain rule and use wolframalpha.com

Answer 5

And do i then put it to 0?

Answer 6

\[\bf\sf R=\frac{v^2}{g}\color{orangered}{\left(\cos \theta\right)'}\left(\sin \theta+\sqrt{\sin^2 \theta+\frac{2gh}{v^2}}\right)+\frac{v^2}{g}\cos \theta\color{orangered}{\left(\sin \theta+\sqrt{\sin^2 \theta+\frac{2gh}{v^2}}\right)'}\]Here is your product rule setup. You'll need to differentiate the orange terms.
Hmm that second one is going to take a bit of work.

Ya after you get through all of that, setting it equal to zero will allow you to solve for theta and find .... critical points or equilibrium points.. whatever the application may be here :O

Answer 7

That should be R' =

Answer 8

So the answer should be\[\sin ^{2}\theta=2+(2gh/v^2)^(-1) \]

Answer 9

I didnt end up with the above answer