# Thread: Help with particle motion, derivatives, etc.

1. ## Help with particle motion, derivatives, etc.

I have never seen any examples of these problems, and they showed up on a review sheet for the final exam in May. If you can help me through these or explain to me what I should do for each step it would help me a lot. Thanks.

1. A particle moves along a line so that at any time (t) its position is given by x(t)=πt+cos(πt). btw, π is pi.
a) Find the velocity at time t.
b) Find the acceleration at time t.
c) What are all the values of t, 0≤t≤3, for which the particle is at rest?
d) What is the maximum velocity over the interval 0≤t≤3?

2. Let f be the function defined by f(x)=(x^2 +1)e^-x for all x such that -4≤x≤4.
a) For what value of x does f reach its absolute maximum? Justify.
b) Find the x-coordinates of all points of inflection of f. Justify.

3. Let p and q be real numbers and let f be the function defined by:

f(x)={1+3p(x-2)+(x-2)^2 for x≤2
f(x)={qx+p for x>2

a) Find the value of q, in terms of p, for which f is continuous at x=2.
b) Find the values of p and q for which f is differentiable at x=2.
c) If p and q have the values determined in part b, is f" a continuous function? Justify.

2. The position, velocity and acceleration functions are all related in that:

Velocity is the derivative of the position function, and acceleration is the derivative of the velocity function, and the second derivative of the position function.

Therefore if the position function is defined as $x(t)=\pi t+cos(\pi t)$ (taking $\pi$ as a constant):

A. $x'(t)=v(t)=\pi-\pi sin(\pi t)$

B. $x''(t)=v'(t)=a(t)=-\pi^2 cos(\pi t)$

C. For this problem, understand that for an object to be at rest, its velocity must be equal to 0. Therefore solve the equation given in A for 0:

$v(t)=\pi-\pi sin(\pi t) \Longrightarrow 0=\pi-\pi sin(\pi t) \Longrightarrow \pi sin(\pi t)=\pi \Longrightarrow sin (\pi t)=1$

For what value(s) of t, between 0 and 3 will make the above equation 1?

D. You have your velocity function - now you are being asked for what values of t will you have maximum velocity. In others words, optimize the velocity function.

3. For your second problem, you are doing much of the same, so I will not go into detail in that, simply note that to find the maximum (or minimum of a fucntion) you need to find the derivatve and set it to zero. If you function has multiply zeros, then test (in the original function) for the critical point that has the largest value. For the inflection point, the method is the same, except you are take the second derivative, and finding the critical points. Once your critical points have been established, test points around them to see if your values of f(x) are changing sign. If so, then the critical point is an inflection point. Review your notes on this if you are having trouble but it is a simple exercise.

For your third problem, remember that a function is continuous at a point if and only if the limit as x approaches a from either direction (the left or the right) is the same, and f(a) actually exists. Assuming it does exist:

First equation: $f(2)=1+3p(2-)+(2-2)^2 \Longrightarrow f(2)=1$
Second equation: $f(2)=2q+p$

Equating the two equations:
$1=2q+p \Longrightarrow \frac{1-p}{2}=q$

I will leave parts B and C to you. If you want a check on your answer post work here.

4. We aren't supposed to use a calculator on any of these. So for par c of number 1, what work am I supposed to show to justify my answers of t=1/2 and 2 1/2? Thanks.

5. $v(t) = 0$

$0= \pi-\pi\sin(\pi t)$

$\pi\sin(\pi t)= \pi$

$\sin(\pi t)= 1$

Form the grpah of the sine function

$\pi t= \frac{\pi}{2}$

$t= \frac{1}{2}$

As the period of the fucntion is 2 the next solution will be at $t= \frac{5}{2} = 2.5$

6. And on the second one, I get stuck at f'(x)=(-e^-x)((x-1)^2) and I'm not sure exactly how to progress from there. Any help?

7. The great thing about optimization problems, is that when it comes time to set our differentiated function equal to zero, ANYTHING in the denominator dies a swift death at the hands of Mr. Multiplication-By-Zero:

$f'(x)=\frac{(x-1)^2}{-e^x} \Longrightarrow 0=\frac{(x-1)^2}{-e^x} \Longrightarrow 0=(x-1)^2$

From here it should be simple to see what your solution is. Perhaps you weren't seeing the next step because you were writing $-e^-x$ instead of $\frac{1}{-e^x}$