Thread: Position of particle when it changes direction?

1. Position of particle when it changes direction?

I have these two identical question with different answer options: I picked the one I thought was right for T, and it turned out being in both question options. Is this right or am I dumbing it down too much? I think it might be too lucky that they both happen to be in the options, maybe I'm calculating the equation for T wrong. This is an online practice quiz only for participation points, but it doesn't show which ones we missed and I think these two might have been two of the ones I missed but I'm not sure (we can take it multiple times). Thanks for your help or any input!

2. Re: Position of particle when it changes direction?

Did you preform any computation, or just guess? Your answer is wrong for both of them. I don't see a correct answer for #8, but I do for #7. You should show your work. That is the only way we can really help you.

In my experience, learning calculus requires a deeper understanding than being able to guess a correct answer from a list of possible answers.

3. Re: Position of particle when it changes direction?

Originally Posted by SlipEternal
Did you preform any computation, or just guess? Your answer is wrong for both of them. I don't see a correct answer for #8, but I do for #7. You should show your work. That is the only way we can really help you.

In my experience, learning calculus requires a deeper understanding than being able to guess a correct answer from a list of possible answers.
Most of the time I do really try to work out each problem, but this one I'm not sure what it's asking or where to start. How would I go about finding what equation T must satisfy? :/

4. Re: Position of particle when it changes direction?

When a particle moving on a straight line changes direction, its velocity goes from either positive to negative or negative to positive. In other words, at the moment that its direction changes, its velocity must be zero. You are given an equation for the particle's position. You should have been taught the relationship between position and velocity. Velocity is the derivative of the position function.

5. Re: Position of particle when it changes direction?

Originally Posted by SlipEternal
When a particle moving on a straight line changes direction, its velocity goes from either positive to negative or negative to positive. In other words, at the moment that its direction changes, its velocity must be zero. You are given an equation for the particle's position. You should have been taught the relationship between position and velocity. Velocity is the derivative of the position function.
Oh wow! I completely forgot. I can be really bad about just cramming for tests and then forgetting things. The derivative of the position function is s'(t)= -e^(-t)(5sin(5t)+cos(5t)) How do we incorporate T into this though? And simplify it? Most of the questions we've done have been a lot more simple and straight forward, at least for position and velocity, this one isn't so simple. If it were short answer I probably would have just left it with that equation but obviously that's not right.

6. Re: Position of particle when it changes direction?

Look at the equation. What do the different symbols mean? What does $t$ mean? What does $T$ mean? Go back to the question. The question tells you how to plug everything in. You gave $s'(t)$. The problem tells you $s'(T) = 0$.

7. Re: Position of particle when it changes direction?

Originally Posted by SlipEternal
Look at the equation. What do the different symbols mean? What does $t$ mean? What does $T$ mean? Go back to the question. The question tells you how to plug everything in. You gave $s'(t)$. The problem tells you $s'(T) = 0$.
t means time, T is the specific time that the velocity is at 0 and the particle turns around, I gave the derivative of the position function and the derivative of the position at time T=0??

8. Re: Position of particle when it changes direction?

Correct, so $s'(T) = 0$ means if you replace every $t$ with $T$ in the derivative function, you get $0$.

$0 = -e^{-T}(5\sin(5T) + \cos(5T))$

Play with that equation to see if you can get one of the options.

9. Re: Position of particle when it changes direction?

Originally Posted by SlipEternal
Correct, so $s'(T) = 0$ means if you replace every $t$ with $T$ in the derivative function, you get $0$.

$0 = -e^{-T}(5\sin(5T) + \cos(5T))$

Play with that equation to see if you can get one of the options.
Hmm, I'm having trouble matching that with an option. Obviously sin/cos=tan and cos/sin=cot but I still can't get it to be exactly like any of the options

10. Re: Position of particle when it changes direction?

Originally Posted by canyouhelp
Hmm, I'm having trouble matching that with an option. Obviously sin/cos=tan and cos/sin=cot but I still can't get it to be exactly like any of the options
Hint, multiply both sides by $e^T$. Remember, $e^T\cdot e^{-T} = e^{T-T} = e^0 = 1$

11. Re: Position of particle when it changes direction?

Originally Posted by SlipEternal
Hint, multiply both sides by $e^T$. Remember, $e^T\cdot e^{-T} = e^{T-T} = e^0 = 1$
Okay, if I multiply both sides by e^T I get 0= -5sin(5T)-cos(5T)?

12. Re: Position of particle when it changes direction?

Correct. Now play with that equation. Try getting one or more of the equations that are options. Play with it.

13. Re: Position of particle when it changes direction?

Originally Posted by SlipEternal
Correct. Now play with that equation. Try getting one or more of the equations that are options. Play with it.
Okay sin(5T)= -1/5cos(5T), divide both sides by cos(5T) and that's tan(5T) = -1/5?

14. Re: Position of particle when it changes direction?

Originally Posted by canyouhelp
Okay sin(5T)= -1/5cos(5T), divide both sides by cos(5T) and that's tan(5T) = -1/5?
Correct. Note that if you take the arctan of both sides, that is one of the options (for #7), but that answer is incorrect. Do you know why?

15. Re: Position of particle when it changes direction?

Originally Posted by SlipEternal
Correct. Note that if you take the arctan of both sides, that is one of the options (for #7), but that answer is incorrect. Do you know why?
No, I'm not sure why that's wrong. :/

Also for #8 it would have to be one of the last two choices right?...How would I get e^-T on one side?

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