Thread: Maximum and Minimub values need help!

1. Maximum and Minimub values need help!

The position function of an object that moves in a straight line is $s(t)=1+2t - \frac {8}{t^2+1}, 0. Calculate the maximum and minimum velocities of the object over the given time interval.

The position function of an object that moves in a straight line is $s(t)=1+2t - \frac {8}{t^2+1}, 0. Calculate the maximum and minimum velocities of the object over the given time interval.
Velocity is ds/dt. So differentiate. Max and min values in the time interval will occur either at stationary points (set ds/dt=0, solve for t and sub to find s at these points). Also find s(0) and s(2) ie the end points. Then compare the values and pick the biggest and smallest.

3. I think you first have to derive it, set it to zero, find the x's and plug it into the main function to find the minimum and maximum values

4. Is this the correct differiantion of it?

$s'(t)=-16(t^2+1)^-2$

Is this the correct differiantion of it?

$s'(t)=-16(t^2+1)^-2$
2 + 16t(t^2+1)^-2

6. where did u get the 2 from?

Here is what I did
$=1+8(t^2+1)^-1$
$=1+8(-1)(t^2+1)^-2(2t)$
$=1+8(-2t)(t^2+1)^-2$
$=1-16t(t^2+1)^-2$
$=16t(t^2+1)^-2$

where did u get the 2 from?

Here is what I did
$=1+8(t^2+1)^-1$
$=1+8(-1)(t^2+1)^-2(2t)$
$=1+8(-2t)(t^2+1)^-2$
$=1-16t(t^2+1)^-2$
$=16t(t^2+1)^-2$

8. ok, but I took the (2t) and multiplied with the (-1) which gave me (-2t). That is this step here

$=1+8(-1)(t^2+1)^-2(2t)$
$=1+8(-2t)(t^2+1)^-2$

ok, but I took the (2t) and multiplied with the (-1) which gave me (-2t). That is this step here

$=1+8(-1)(t^2+1)^-2(2t)$
$=1+8(-2t)(t^2+1)^-2$
Go back to your original question. It has 3 terms.
Differentiate the 1 to get 0.
Differentiate the 2t to get 2. (This is where the 2 comes from)
Differentiate the last term to get.....see previous post

10. ohh ok thanks. ill finish rest of question and see how i do >.<

ohh ok thanks. ill finish rest of question and see how i do >.<

12. stuck again! >.<

$0 = 2 + 16t(t^2+1)^-2$

help? how do I get t by itself?

13. I tried! I got down to the lowest part before solving for t, I dont think tis possiable.

I did this

$0=2+16t(t^2+1)^-2$

$-2=\frac {16t}{(t^2+1)^2}$

$-2(t^2+1)^2=16t$
$-(t^2+1)^2=8t$
$-(t^4+2t^2+1)=8t$
$-t^4+2t^2-1=8t$
$-t^4-2t^2+8t+1=0$

well???

14. $v(t) = 2+\frac{16t}{(t^2+1)^2}$

note that $v(t) > 0$ for all values of t in the given interval $0 < t < 2$.

to find the max and min velocity ...

$v'(t) = \frac{16(1-3t^2)}{(t^2+1)^3}$

$v'(t) = 0$ at $t = \sqrt{\frac{1}{3}}$

$v'(t) > 0$ for $t < \sqrt{\frac{1}{3}}$

and

$v'(t) < 0$ for $t > \sqrt{\frac{1}{3}}$

therefore, $v(t)$ is a maximum at $t = \sqrt{\frac{1}{3}}$

$v\left(\sqrt{\frac{1}{3}}\right)$ is the maximum velocity in the given interval.

the minimum velocity would occur at one of the two endpoint times ( $t = 0$ or $t = 2$) if they are included in the interval.

recheck the given interval to see if it is really $0 \le t \le 2$.