Acceleration problem

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August 4th 2009, 01:55 PM
radioheadfan

Acceleration problem

The position equation for the movement of a particle is given by $s(t) = \sqrt{t^3 +1}$ where s is measured in feet and t is measured in seconds. Find the acceleration when t = 2 seconds.

I know that I have to get y'', I got ${3t^2/2} -\sqrt{t^3+1}$ , but when trying to get y'' I just made a huge mess and came out to a wrong answer. Someone want to walk me through the rest of this?
August 4th 2009, 02:05 PM
slider142

Remember to keep the symbols straight; there is no y in the actual problem. (Wink) Though I know what you mean. Acceleration is defined to be the second derivative of position with respect to time t, so we want to find s''(t).
First we find s'(t), which can be found by applying the chain rule, where the square root is the outer function and $t^3 + 1$ is the inner function. To keep things clear, I like to use exponents in place of roots, so I would write it out:
$s(t) = (t^3 + 1)^\frac{1}{2}$
Explicitly, we can see that s(t) = f(g(t)) where $g(t) = t^3 + 1$ and $f(t) = t^{\frac{1}{2}}$ . By the chain rule, s'(t) = f'(g(t))g'(t).
$s'(t) = \frac{1}{2}(t^3 + 1)^{-\frac{1}{2}}(3t^2)$
In order to get the second derivative you will have to use the product rule in addition to the chain rule. Can you find s''(t) from here?
August 4th 2009, 02:10 PM
radioheadfan

Quote:

Originally Posted by slider142

Remember to keep the symbols straight; there is no y in the actual problem. (Wink) Though I know what you mean. Acceleration is defined to be the second derivative of position with respect to time t, so we want to find s''(t).
First we find s'(t), which can be found by applying the chain rule, where the square root is the outer function and $t^3 + 1$ is the inner function. To keep things clear, I like to use exponents in place of square roots, so I would write it out:
$s(t) = (t^3 + 1)^\frac{1}{2}$
Explicitly, we can see that s(t) = f(g(t)) where $g(t) = t^3 + 1$ and $f(t) = t^{\frac{1}{2}}$ . By the chain rule, s'(t) = f'(g(t))g'(t).
$s'(t) = \frac{1}{2}(t^3 + 1)^{-\frac{1}{2}}(3t^2)$
Explicitly, we can see that s(t) = f(g(t)) where g(t) = t^3 + 1 and f(t) = t^(1/2)
In order to get the second derivative you will have to use the product rule in addition to the chain rule. Did you find your problem here?

Yes sir, whenever I integrate the power rule and the chain rule together I usually have a difficult time. So can you explain this to me so I can understand the concept better?
August 4th 2009, 02:20 PM
slider142

In the above, we have
$<br /> s'(t) = \frac{3}{2}(t^3 + 1)^{-\frac{1}{2}}(t^2)<br />$
Since we do not know its derivative offhand, we note that it is a product of two functions, each of whose derivative we do know. Specifically
$f(t) = (t^3 + 1)^{-\frac{1}{2}}$
and
$g(t) = t^2$
so that we have
$s'(t) = \frac{3}{2}f(t)g(t)$
By the product rule, we know that
$s''(t) = \frac{3}{2}(f'(t)g(t) + f(t)g'(t))$
The only place you will need that chain rule here is in finding f'(t). Note how similar f(t) is to s(t). What do you propose s''(t) is?
August 4th 2009, 02:34 PM
radioheadfan

Okay I did it the way you asked me to and came out with ${-9t^4/2}+{3(t^3+1)^-3/2/2}+3t^4 +3t$
August 4th 2009, 02:46 PM
slider142

It looks like you're missing some small steps. Don't skip any steps until you get more experience with these rules. The first step is to get the parts of the product rule that you do not have, f'(t) and g'(t). Do not worry about how they interact with the rest of the equation yet. What did you get for f'(t)? What did you get for g'(t)?
August 4th 2009, 02:55 PM
radioheadfan

for f' I got 6t and for g' I got $-1/2(-\sqrt{(t^3+1)}) (3t^2)$
August 4th 2009, 03:12 PM
slider142

Quote:

Originally Posted by radioheadfan

for f' I got 6t and for g' I got $-1/2(-\sqrt{(t^3+1)}) (3t^2)$

Okay, so in your case, you switched them around from my post and you let
$f(t) = 3t^2$
and
$g(t) = (t^3 + 1)^{-\frac{1}{2}} = \frac{1}{\sqrt{t^3 + 1}}$ .
Your f'(t) is correct, but your g'(t) is not. g'(t) should be
$-\frac{1}{2}(t^3 + 1)^{-\frac{1}{2} - 1}(3t^2) = -\frac{1}{2}(t^3 + 1)^{-\frac{3}{2}}(3t^2)$
The outer function gets taken down one unit to the -3/2 power by the power rule of differentiation.
August 4th 2009, 03:25 PM
radioheadfan

Quote:

Originally Posted by slider142

Okay, so in your case, you switched them around from my post and you let
$f(t) = 3t^2$
and
$g(t) = (t^3 + 1)^{-\frac{1}{2}} = \frac{1}{\sqrt{t^3 + 1}}$ .
Your f'(t) is correct, but your g'(t) is not. g'(t) should be
$-\frac{1}{2}(t^3 + 1)^{-\frac{1}{2} - 1}(3t^2) = -\frac{1}{2}(t^3 + 1)^{-\frac{3}{2}}(3t^2)$
The outer function gets taken down one unit to the -3/2 power by the power rule of differentiation.

Just to make sure I'm doing this correctly
right now I have
$<br /> \frac{1}{2}[6t(t^3+1)^{-\frac{1}{2}}+{-\frac{9t^4}{2}}(t^3+1)^{-\frac{3}{2}}]$
August 4th 2009, 03:31 PM
slider142

Quote:

Originally Posted by radioheadfan

Just to make sure I'm doing this correctly
right now I have
$<br /> \frac{1}{2}[6t(t^3+1)^{-\frac{1}{2}}+{-\frac{9t^4}{2}}(t^3+1)^{-\frac{3}{2}}]$

That is exactly correct. (Nod)
August 4th 2009, 05:40 PM
radioheadfan

Quote:

Originally Posted by slider142

That is exactly correct. (Nod)

Awesome, okay so the final answer is?

$3t(t^3+1)^\frac{-1}{2}-\frac{9t^4(t^3+1)^\frac{-3}{2}}{2}$
August 4th 2009, 06:27 PM
slider142

Quote:

Originally Posted by radioheadfan

Awesome, okay so the final answer is?

$3t(t^3+1)^\frac{-1}{2}-\frac{9t^4(t^3+1)^\frac{-3}{2}}{2}$

Almost. The second term should also be multiplied by 1/2.
August 4th 2009, 07:24 PM
radioheadfan

$3t(t^3+1)^\frac{-1}{2}-\frac{9t^4(t^3+1)^\frac{-3}{2}}{4}$
August 5th 2009, 02:14 AM
slider142

Quote:

Originally Posted by radioheadfan

$3t(t^3+1)^\frac{-1}{2}-\frac{9t^4(t^3+1)^\frac{-3}{2}}{4}$

Yep, you've got it.