Derivative of a Tough Equation: Simplified Explanation and Steps

[PMP]

Would you tell me what is:

$$\frac{d\sqrt{(v_0cos \theta t)^2 + (v_0 \sin \theta t - 1/2gt^2)^2}}{dt}$$

Please. Thanks in advance.

 

[neutrino]

It's a simple application of the chain rule. What have you done so far?

 

[PMP]

I have just basic knowledge of derivatives, this exercise is one difficult for my level, so I would just like to know if I derivated this correctly. Would you help me, neutrino?

 

[arildno]

Hint:
Simplify the radicand first!

 

[neutrino]

so I would just like to know if I derivated this correctly.

Post what you have done so that we can check if it is done correctly.

Treat $$v_0cos \theta t)^2 + (v_0 \sin \theta t - 1/2gt^2)^2$$ as a single function of t, say g, under the square root. If the whole function is denoted by f, then $$f = \sqrt{g}$$. Now use the chain rule $$\frac{df}{dt} = \frac{df}{dg}\frac{dg}{dt}$$

 

[PMP]

Thanks arildno, I have already done that, and have my result. I really just want to see if it is correct. Is it:

$$2v_0^2t +g^2t^3 - 3v_0\sin \theta gt^2$$ ?

The denominator doesen't matter because I will equal that derivative to zero.

 

[neutrino]

That is right.

 

[PMP]

Ok, thank you all of you. My basic derivative technique works. :)

How would I find the maximum value of theta, such that the this derivative is positive for t belonging to the interval [0, 2v₀sin(theta)/g]?

 

[PMP]

No ideas? :)

 

[PMP]

Knock, knock!

 

[neutrino]

Who's there? :P

If you want to maximise the above derivative wrt to theta, then differentiate it wrt to theta and set it zero.

 

[PMP]

Who's there? :P

That person already ran way.

I should differentiate wrt to theta $$2v_0^2t +g^2t^3 - 3v_0\sin \theta gt^2$$ or the initial $$\sqrt{(v_0cos \theta t)^2 + (v_0 \sin \theta t - 1/2gt^2)^2}$$ ?

Because if it is to differentiate the above I would get $$\cos \theta = 0$$, right? And it is not the correct result.
Thanks.

 

[neutrino]

Depends on which function you want to maximise.

How would I find the maximum value of theta, such that the this derivative is positive for t belonging to the interval [0, 2v0sin(theta)/g]?

 

[PMP]

This is a projectile position vector and I need to find theta such that the magnitude of the position vector during the projectile motion is always increasing. So I did dr/dt>=0. Got the expression above, but now if I differentite this wrt to theta I will not get the correct result. The deravite is $-3v_0\cos\theta gt^2$. Right?

 

[PMP]

neutrino, knock, knock...

 

[Hootenanny]

neutrino, knock, knock...

My turn, my turn... who's there...?

This is a projectile position vector

Is $\theta$ your launch angle?

 

[PMP]

Yes, it is. Thanks for being prepared to help.

 

[Hootenanny]

Yes, it is. Thanks for being prepared to help.

No problem, then why may I ask you to state the complete problem as given in your text.

 

[PMP]

This not in a text. But it is like this: What is the maximum lauch angle for a projectile be always going away from me? That is, for the magnitude of the position vector be always increasing in time. I don't want you to solve the problem, just to say if I am in the right way. Thanks in advance!

 

[PMP]

I guess. The initial conditions are:

x_0=0
y_0=0

But that is the magnitude of the postion vector. Is it right?

 

[Hootenanny]

You do not need to find the second derivative, simply find the interval of theta such that

$$2v_0^2t +g^2t^3 - 3v_0\cdot gt^2\sin \theta > 0$$

Do you understand why?