At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
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?
i am truly unsure
i thought it was by finding the derivative
Yes, the derivative represents "rate of change," "speed" or "magnitude of velocity"
So, if x(t) = (cos x)^3, what is the derivative, x '(t)?
x(t)=cos^3(t) so x'(t)=3cos^2(t).(-sin(t))
Hint: Use the Power Rule first, then the Chain Rule, including the derivative of cos t.
that looks good. I do prefer x'(t)=3(cos t)^2(-sin t).
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.
just a question: so i can move the exponent where ever i like? for example cos^2(t) can become (cos(t))^2
Strictly speaking your cos^2(t) is OK; It's just that I personally prefer (cos t)^2 for its slightly greater clarity.
is another alternative, very clear.
ok, however i am not sure how to find the roots of our equation
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.
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.
when t =6 i believe
Is cos 6 = 0?
Are you working with a calculator with built-in trig functions?
Set your calculator to use DEGREES (not RADIANS). Find cos 6. Is it zero?
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.
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?
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.
@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?
I'm leaving my computer for a while. Perhaps we can re-connect later.