# Second derivative help!!

• Mar 2nd 2009, 05:55 PM
rust1477
Second derivative help!!
Hi there, I've been working this problem forever and I cannot come close to the right answer...

A function of the position of a car being driven on a straight road, measured in feet from some fixed point is given by:

$s(t)=1/3(t^2+8)^3/2$

BTW: to avoid any confusion, the $(t^2+8)$ is raised to $3/2$
Sorry, I could not get it to look right.
I am to determine the acceleration of the car when t=1. I know I have to take the second derivative of s(t) to get the right equation to plug t into, but I can't get it for the life of me... Any suggestions? (Worried)
• Mar 2nd 2009, 06:10 PM
arpitagarwal82
$s'(t) = 1/3 * 2t * (t^2 + 8)^(1/2)$

$s''(t) = 2/3 ( (t^2 + 8)^(1/2) + t * 2t * (t^2 + 8)^(-1/2)$

at t = 1, $s''(t) = 2/3 ( \sqrt 9 + 2/\sqrt 9)$
=2/3 (3 + 2/3)
= 2/3 * 11/3
= 22/9
• Mar 2nd 2009, 06:11 PM
arpitagarwal82
Above solution is assuming $(t^2+8)^{3/2}$ is in numerator of s(t)
• Mar 2nd 2009, 06:22 PM
rust1477
Yes, the original s(t) is in the numerator.

However, any way I interpret your answer is incorrect...

The choices I'm given are:
$8/3 ft/sec^2$
$10/3 ft/sec^2$
$3 ft/sec^2$
$4 ft/sec^2$
• Mar 2nd 2009, 06:34 PM
Quote:

Originally Posted by rust1477
Hi there, I've been working this problem forever and I cannot come close to the right answer...

A function of the position of a car being driven on a straight road, measured in feet from some fixed point is given by:

$s(t)=1/3(t^2+8)^3/2$

BTW: to avoid any confusion, the $(t^2+8)$ is raised to $3/2$
Sorry, I could not get it to look right.
I am to determine the acceleration of the car when t=1. I know I have to take the second derivative of s(t) to get the right equation to plug t into, but I can't get it for the life of me... Any suggestions? (Worried)

• Mar 2nd 2009, 06:35 PM
arpitagarwal82
Yeah
I missed some factors
Here i correct one
$s'(t) = 1/3 * 2t * 3/2 * (t^2 + 8)^(1/2)$

$s''(t) = ( (t^2 + 8)^(1/2) + t * 2t *1/2 * (t^2 + 8)^(-1/2)$

at t = 1, $s''(t) = ( \sqrt 9 + 1/\sqrt 9)$
=(3 + 1/3)
=10/3
• Mar 2nd 2009, 06:43 PM
rust1477
Hey thank you so much for showing me correct answer...

Can you elaborate a little bit on the computations you did to arrive at the second derivative?

When I computed the $s'(t)$, I got
$1/3*3/2*(t^2+8)^1/2*2t$

Is this correct for the first derivative, and if so, can you explain how you arrived at the second derivative?
• Mar 3rd 2009, 07:22 AM
arpitagarwal82
Yes yu ave calculated s'(t) correct.
On simplification $s'(t) = t(t^2 + 8)^{1/2}$

To calculate s''(t) use differentiation by parts.
$s'' = (t^2 + 8)^{1/2} dt/dt + t * d ((t^2 + 8)^{1/2})/dt$
$= (t^2 + 8)^{1/2} + t * 2t * 1/2 * (t^2 + 8)^{-1/2}$
$=(t^2 + 8)^{1/2} + t^2(t^2 + 8)^{-1/2}$

Is that clear now?