Question

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 10 years ago

Answer 1

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?

Answer 2

yes

Answer 3

i am

Answer 4

i am truly unsure

Answer 5

i thought it was by finding the derivative

Answer 6

Yes, the derivative represents "rate of change," "speed" or "magnitude of velocity"

Answer 7

So, if x(t) = (cos x)^3, what is the derivative, x '(t)?

Answer 8

x(t)=cos^3(t)
so
x'(t)=3cos^2(t).(-sin(t))

Answer 9

Hint: Use the Power Rule first, then the Chain Rule, including the derivative of cos t.

Answer 10

that looks good. I do prefer x'(t)=3(cos t)^2(-sin t).

Answer 11

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.

Answer 12

just a question: so i can move the exponent where ever i like? for example cos^2(t) can become (cos(t))^2

Answer 13

Strictly speaking your cos^2(t) is OK; It's just that I personally prefer (cos t)^2 for its slightly greater clarity.

Answer 14

\[\cos ^{2}t\]

Answer 15

is another alternative, very clear.

Answer 16

ok, however i am not sure how to find the roots of our equation

Answer 17

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.

Answer 18

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.

Answer 19

https://www.google.com/search?q=cos+t&oq=cos+t&aqs=chrome..69i57j0l5.2319j0j7&sourceid=chrome&es_sm=0&ie=UTF-8

Answer 20

when t =6 i believe

Answer 21

Is cos 6 = 0?

Answer 22

Are you working with a calculator with built-in trig functions?

Answer 23

yes

Answer 24

@mathmale

Answer 25

Set your calculator to use DEGREES (not RADIANS). Find cos 6. Is it zero?

Answer 26

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.

Answer 27

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?

Answer 28

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.

Answer 29

@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?

Answer 30

I'm leaving my computer for a while. Perhaps we can re-connect later.