anonymous
  • anonymous
PLEASE HELP a particle moves so that x, its distance from the origin at time t, t is greater then or equal to 0 is given by x(t)=cos^3(t). the first interval in which the particle is moving to the right it 1.)0< t
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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mathmale
  • mathmale
I assume you're in calculus. Is that correct? x(t) = (cos t)^3 is the horiz. distance from the origin of the particle. Given a function of t such as this one, how do you derive a function for the VELOCITY of the particle as a function of t?
anonymous
  • anonymous
yes
anonymous
  • anonymous
i am

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anonymous
  • anonymous
i am truly unsure
anonymous
  • anonymous
i thought it was by finding the derivative
mathmale
  • mathmale
Yes, the derivative represents "rate of change," "speed" or "magnitude of velocity"
mathmale
  • mathmale
So, if x(t) = (cos x)^3, what is the derivative, x '(t)?
anonymous
  • anonymous
x(t)=cos^3(t) so x'(t)=3cos^2(t).(-sin(t))
mathmale
  • mathmale
Hint: Use the Power Rule first, then the Chain Rule, including the derivative of cos t.
mathmale
  • mathmale
that looks good. I do prefer x'(t)=3(cos t)^2(-sin t).
mathmale
  • mathmale
when is the particle moving to the right? Would the velocity be + or -? It may help if you determine when the velocity is zero (i. e., find the roots of x'(t)=0.
anonymous
  • anonymous
just a question: so i can move the exponent where ever i like? for example cos^2(t) can become (cos(t))^2
mathmale
  • mathmale
Strictly speaking your cos^2(t) is OK; It's just that I personally prefer (cos t)^2 for its slightly greater clarity.
mathmale
  • mathmale
\[\cos ^{2}t\]
mathmale
  • mathmale
is another alternative, very clear.
anonymous
  • anonymous
ok, however i am not sure how to find the roots of our equation
mathmale
  • mathmale
x '(t) = 3(cos t)^2 * (-sin t) = 0 is true when t=0, right? It's also true when t= ?? , thanks to the cos t factor.
mathmale
  • mathmale
Here's the result of an Internet search I did for "cos t." It shows the graph of this function. Remember that we're concerned ONLY with t=0 or greater.
mathmale
  • mathmale
https://www.google.com/search?q=cos+t&oq=cos+t&aqs=chrome..69i57j0l5.2319j0j7&sourceid=chrome&es_sm=0&ie=UTF-8
anonymous
  • anonymous
when t =6 i believe
mathmale
  • mathmale
Is cos 6 = 0?
mathmale
  • mathmale
Are you working with a calculator with built-in trig functions?
anonymous
  • anonymous
yes
anonymous
  • anonymous
@mathmale
mathmale
  • mathmale
Set your calculator to use DEGREES (not RADIANS). Find cos 6. Is it zero?
mathmale
  • mathmale
Review: Your goal is to identify the first INTERVAL on which the cosine function is positive, as this corresponds to the first INTERVAL on which the particle is moving to the right.
mathmale
  • mathmale
Caden, I've found a graph of the derivative of x(t) = (cos t)^3. See the following: http://m.wolframalpha.com/input/?i=d%2Fdx+%28cos+x%29%5E3&x=0&y=0 This derivative is positive wherever the graph is ABOVE the t-axis. Can you approximate the t value at which this derivative is first positive? the t value at which it appraoches or touches x=0?
mathmale
  • mathmale
The derivative of x(t) = (cos t)^3 first becomes positive as t approaches t = ??????? The same derivative becomes 0 as t approaches t = ??????? These values are the beginning and end of your time interval for which the particle is moving in the positive x-direction.
mathmale
  • mathmale
@Caden: I get distracted sometimes when working on OpenStudy, and people then wonder where I am. But that happens the other way around, too. Your last response was 11 minutes ago, according to OpenStudy. Are you still interested in working with me here and now?
mathmale
  • mathmale
I'm leaving my computer for a while. Perhaps we can re-connect later.

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