Homework Help: Few Simple Questions

1. Nov 29, 2008

razored

I have a few questions and I was wondering if someone could check my answers.

1) Given the velocity function http://texify.com/img/%5CLARGE%5C%21v%28t%29%20%3D-%28t%2B1%29%20%5Csin%7B%5Cfrac%7Bt%5E%7B2%7D%7D%7B2%7D%7D%20.gif [Broken],[/URL] find all the times on hte open interval 0<t<3 where the particle changes direction. Justify your answer.

I said, the possible places where this occurs is when v=0. So I solved for v =0, and the only number that matches that domain is approximately 2.5 Are there any other answers I am missing? Is my idea of when the particle changes direction v=0 the entire reasoning to when the particle changes direction? I thought of an instance where v=0 but then particle doesn't change direction but rather keeps going in the same direction. Any help?

2)Suppose a function g is defined by http://texify.com/img/%5CLARGE%5C%21k%5Csqr%7Bx%2B1%7D%20%5C%20%5C%20%20for%20%5C%20%5C%200%5Cleq%20x%20%5Cleq%203%20%5C%5Cmx%2B2%20%5C%20%5C%20%20for%20%5C%20%5C%203%3C%20x%20%5Cleq%205%20.gif [Broken] where k and m are constants. If g is differentiable at x=3, what are the values for k and m?

I was usually able to solve this but now there is one extra variable that I do not know how to get rid of. What do I do?

Last edited by a moderator: May 3, 2017
2. Nov 29, 2008

e(ho0n3

I would also look at the case where v = 0 for I imagine the particle is moving forward when v > 0, moving backward when v < 0 and not moving when v = 0. Clearly v = 0 only when the sine term is 0. Can you give me an exact value for t between 0 and 3 that will make the sine term 0?

Remember that if g'(3) exists, then g is continuous at x = 3. With this additional piece of info., what values of k and m will make g continuous at x = 3?

Last edited by a moderator: May 3, 2017
3. Nov 30, 2008

razored

1) $$\sqrt{2 * \pi }$$
2)When k and m are equal?

Are those correct?

4. Nov 30, 2008

e(ho0n3

The first one is correct. The second one is not: Just try it for m = k = 1. Here's a tip: Find the equation of the tangent of k * sqrt{x + 1} at x = 3 and try to manipulate it so that it takes the form mx + 2.