Question

derive (dx/dϴ) the given equation :
x = Vcosϴ[(Vsinϴ-(V^(2)(sinϴ)^(2)+2gy)^(1/2))/g]

Answer 1

Hey Snitch!
Just so I understand your equation properly,
is this how it should be formatted?\[\large\rm x=V \cos \theta\frac{Vsin \theta- \sqrt{V^2 \sin^2 \theta+2gy}{}}{g}\]

Answer 2

\[x = Vcosϴ [\frac{ Vsinϴ+\sqrt{V^2(\sinϴ)^2+2gy} }{ g }]\]

Answer 3

sorry it's supposed to be Vsinϴ+- in the numerator inside the bracket

Answer 4

V, g constant,
is y constant? or some function of x and theta or something?

Answer 5

constant

Answer 6

i made a mistake i'm so sorry it should be -2gy

Answer 7

\[\large\rm x=V \cos \theta\left[\frac{Vsin \theta\pm\sqrt{V^2 \sin^2 \theta-2gy}}{g}\right]\]
I suppose we could apply product rule, ya?\[\rm \color{royalblue}{(V \cos \theta)'}\left[\frac{Vsin \theta\pm\sqrt{V^2 \sin^2 \theta-2gy}}{g}\right]+(V \cos \theta)\color{royalblue}{\left[\frac{Vsin \theta\pm\sqrt{V^2 \sin^2 \theta-2gy}}{g}\right]'}\]

Answer 8

Ahh darn, I thought that would fit on the page lol

Answer 9

Hopefully you get the idea though,
blue is the stuff we need to differentiate.

Answer 10

The first term should be pretty straight forward.
The other one will require a little work.
What do you think? :d

Answer 11

\[\frac{ dx }{ dϴ } = (\frac{ Vsinϴ-\sqrt{V^2\sin^2ϴ-2gy} }{ g })(-Vsinϴ) + (Vcosϴ)(\frac{ Vcosϴ-\sqrt{V^22\cosϴ\sinϴ-2gy} }{ g}\]

Answer 12

shoult it be like this?

Answer 13

oop it should be +/-

Answer 14

The last part was cut, V^(2)2cosϴsinϴ-2gy

Answer 15

\[\rm \color{orangered}{(-Vsin \theta )}\left[\frac{Vsin \theta\pm\sqrt{V^2 \sin^2 \theta-2gy}}{g}\right]+(V \cos \theta)\color{royalblue}{\left[\frac{Vsin \theta\pm\sqrt{V^2 \sin^2 \theta-2gy}}{g}\right]'}\]Your first term looks good.
Let's drop it for now so it's easier to fit the other one on the page.

So we have to take care of this,\[\large\rm (V \cos \theta)\color{royalblue}{\left[\frac{Vsin \theta\pm\sqrt{V^2 \sin^2 \theta-2gy}}{g}\right]'}\]

Answer 16

Let's pull the 1/g to the front so it's even easier to read,\[\large\rm \left(\frac{V}{g} \cos \theta\right)\color{royalblue}{\left[Vsin \theta\pm\sqrt{V^2 \sin^2 \theta-2gy}\right]'}\]

Answer 17

Your derivative of sine looks ok.
Square root takes a little bit more effort though.
You can convert to 1/2 power and apply power rule if you feel more comfortable doing that.

But honestly, square root derivative shows up so often that it's worth remember its derivative,\[\large\rm \frac{d}{dz}\sqrt{z}=\frac{1}{2\sqrt z}\]The derivative of square root is one over two square roots.

Answer 18

oh right i forgot about the square root

Answer 19

\[\large\rm \left(\frac{V}{g} \cos \theta\right)\color{orangered}{\left[Vcos\theta\pm\frac{1}{2\sqrt{V^2 \sin^2 \theta-2gy}}\right]}\]So this would be our first step in dealing with the square root.
Derivative of square root is one over two square roots.
But we need chain rule, ya?

Answer 20

Chain rule,\[\large\rm \left(\frac{V}{g} \cos \theta\right)\color{orangered}{\left[Vcos\theta\pm\frac{\color{royalblue}{\left(V^2\sin^2\theta-2gy\right)'}}{2\sqrt{V^2 \sin^2 \theta-2gy}}\right]}\]We have to multiply by the derivative of the inner function.

Answer 21

I hope the color scheme is making sense.
I'm using blue when we need to take a derivative,
and orange when we've finished taking a derivative.

Answer 22

yes thank you for helping me :)

Answer 23

oh the derivative of the blue should be V^(2)2sinϴcosϴ?

Answer 24

\[V^22\sinϴ\cosϴ\]?

Answer 25

\[\large\rm \left(\frac{V}{g} \cos \theta\right)\color{orangered}{\left[Vcos\theta\pm\frac{\left(2V^2\sin\theta\cos\theta\right)}{2\sqrt{V^2 \sin^2 \theta-2gy}}\right]}\]Mmm yes very good.
The constant on the end differentiates to 0.

Answer 26

From this point,
it depends how much simplifying your teacher expects you to do.
It could end up being rather burdensome if you try to hard :P

Answer 27

He says we have to simplify it :(

Answer 28

Let's ummm..... thinking.... where to start..

Answer 29

oh we have to equate dx/dϴ = 0 then we must find ϴ

Answer 30

Oh oh oh, Ok that makes things much easier then.

Answer 31

Lemme write down what we have so far.\[\rm -\frac{V}{g}\sin \theta\left(Vsin \theta\pm\sqrt{V^2\sin^2\theta-gy}\right)+\frac{V}{g}\cos \theta\left(V \cos \theta\pm\frac{2V^2\sin \theta \cos \theta}{2\sqrt{V^2\sin^2\theta-gy}}\right)\]Yes?

Answer 32

yup

Answer 33

Multiply both sides by g,
divide by V,\[\rm -\sin \theta\left(Vsin \theta\pm\sqrt{V^2\sin^2\theta-gy}\right)+\cos \theta\left(V \cos \theta\pm\frac{2V^2\sin \theta \cos \theta}{2\sqrt{V^2\sin^2\theta-gy}}\right)\]

Answer 34

Add the sine term to the other side,\[\rm \sin \theta\left(Vsin \theta\pm\sqrt{V^2\sin^2\theta-gy}\right)=\cos \theta\left(V \cos \theta\pm\frac{2V^2\sin \theta \cos \theta}{2\sqrt{V^2\sin^2\theta-gy}}\right)\]

Answer 35

Grr this problem is a little annoying :) lol

Answer 36

Oh let's get rid of the 2 in the numerator and denominator on our last term,\[\rm \sin \theta\left(Vsin \theta\pm\sqrt{V^2\sin^2\theta-gy}\right)=\cos \theta\left(V \cos \theta\pm\frac{V^2\sin \theta \cos \theta}{\sqrt{V^2\sin^2\theta-gy}}\right)\]I see a bunch of different paths we could take from here.
Maybe it's just best to distribute the sine and cosine.

Answer 37

\[\rm V \sin^2\theta\pm\sin \theta\sqrt{V^2\sin^2\theta-gy}=V \cos^2\theta\pm\frac{V^2\sin \theta \cos^2\theta}{\sqrt{V^2\sin^2\theta-gy}}\]

Answer 38

To be honest, I'm not sure if this is going to be the best way to do this,
but I'm going down a road and we'll see where it leads us :)

Answer 39

Subtracting cos squared to the left side,
adding the other term over to the right,\[\rm V \sin^2\theta-V \cos^2\theta=\pm\frac{V^2\sin \theta \cos^2\theta}{\sqrt{V^2\sin^2\theta-gy}}\mp\sin \theta\sqrt{V^2\sin^2\theta-gy}\]Applying our Cosine Double Angle Formula,\[\rm V \cos2\theta=\pm\frac{V^2\sin \theta \cos^2\theta}{\sqrt{V^2\sin^2\theta-gy}}\mp\sin \theta\sqrt{V^2\sin^2\theta-gy}\]

Answer 40

Common denominator gives us,\[\rm V \cos2\theta=\pm\frac{V^2\sin \theta \cos^2\theta\mp\sin \theta(V^2\sin^2\theta-gy)}{\sqrt{V^2\sin^2\theta-gy}}\]

Answer 41

Hmm not sure what to do from here :(
What is the purpose of all of this?
We're trying to find critical/stationary points for our function x?

Answer 42

Ftom a engineering mechanics problem, our prof says we have to find theta in order to find Rmax in our problem. I'm just having a hard time in deriving this equation.

Answer 43

Thank you for helpin me :)

Answer 44

I think I need a math break, brain will esplode >.<
I can't figure how to find theta from this mess.
Maybe looking at it later will help.

Ya hopefully that at least heads you in the proper direction D:

Answer 45

Thank you :)