Derivative of a Tough Equation: Simplified Explanation and Steps

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The discussion centers on finding the derivative of a complex equation related to projectile motion, specifically the expression involving the square root of a combination of trigonometric and gravitational terms. Participants emphasize using the chain rule for differentiation and suggest simplifying the radicand first. There is a focus on determining the maximum launch angle, theta, such that the derivative remains positive within a specified time interval. The conversation highlights the importance of analyzing the expression derived from the initial equation to ensure the magnitude of the position vector is always increasing. Overall, the participants are collaboratively working through the mathematical challenges presented by the problem.
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.
 
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It's a simple application of the chain rule. What have you done so far?
 
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?
 
Hint:
Simplify the radicand first!
 
PMP said:
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}
 
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.
 
That is right.
 
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]?
 
No ideas? :)
 
  • #10
Knock, knock!
 
  • #11
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.
 
  • #12
neutrino said:
Who's there? :P
That person already ran way. :smile:

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.
 
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  • #13
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]?
 
  • #14
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?
 
  • #15
neutrino, knock, knock...
 
  • #16
PMP said:
neutrino, knock, knock...
My turn, my turn... who's there...? :wink:
PMP said:
This is a projectile position vector
Is \theta your launch angle?
 
  • #17
Yes, it is. Thanks for being prepared to help.
 
  • #18
PMP said:
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.
 
  • #19
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!
 
  • #20
I guess. The initial conditions are:

x_0=0
y_0=0

But that is the magnitude of the postion vector. Is it right?
 
  • #21
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?
 
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