How to Find the Maximum and Minimum Speed of a Particle?

hellboy324
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Homework Statement


The position vector of a particle at time t is given by r(t)= 2sin(2t)i + cos(2t)j + 2tk where t >=0.

Homework Equations


v(t) = 4cos(2t)i - 2sin(2t)j + 2k
speed = | v(t) | =√(16cos^2(2t)+4sin^2(2t)+4) = √(12cos^2(2t)+8)

The Attempt at a Solution


I found the velocity and speed, but I have no idea what to do to find out the maximum and minimum speed for this question, someone can teach me?
 
Physics news on Phys.org
How do you usually find the extrema of a function?
 
Things are easier in that the question is asking about scalar speed. What does it mean for the derivative when speed hits a peak/trough?
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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