Find min. speed of particle with trajectory....

**ZeroVector** · October 17th 2010, 10:42 PM

Find the minimum speed of a particle with trajectory $c(t) = (t^3, t^{-2}), t \geq \frac{1}{2}$

So here's what I told my professor and he confirmed it as right (though I am still unsure if it was really right):
1. Write the speed formula:
$speed = \sqrt{ 9t^4 + 4t^{-6} }$

2. Write the derivative of the speed and find where it's equal to 0:
$\frac{d}{dt}( \sqrt{ 9t^4 + 4t^{-6} } )$

3. Write the second derivative of the speed and find where it's > 0:
$\frac{d}{dt}{(\frac{d}{dt}( \sqrt{ 9t^4 + 4t^{-6} } ))}$

So if this is correct, is there a way to simplify this so that I won't need to expand the equation that much? Or am I doing something wrong?

**NOX Andrew** · October 18th 2010, 09:32 AM

Your reasoning is correct. I assume you are finding the second derivative of speed to apply the second derivative test to the critical numbers. If so, you could use the first derivative test because finding the second derivative seems cumbersome without a graphing calculator.

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